WAEC SSCE - Further Mathematics - 2006
Pregunta 1 Informe
Find the equation of the line passing through (0, -1) and parallel to the y- axis.
Detalles de la respuesta
The equation of a line can be represented in the form y = mx + b, where m is the slope of the line and b is the y-intercept (the point where the line intersects the y-axis). Since the line is parallel to the y-axis, its slope is undefined (division by zero is not defined). Therefore, the equation of the line cannot be represented in the form y = mx + b. However, we know that the line passes through the point (0, -1). This means that the x-coordinate of every point on the line is 0 (since the line is parallel to the y-axis and does not move horizontally). Therefore, the equation of the line is x = 0. This equation means that every point on the line has an x-coordinate of 0, which is the y-axis. So, the line is simply the y-axis itself. Thus, the answer is x = 0.
Pregunta 2 Informe
A particle is projected vertically upwards from a height 45 metres above the ground with a velocity of 40 m/s. How long does it take it to hit the ground? [Take g = \(10 ms^{-2}\)].
Detalles de la respuesta
We can use the kinematic equation: h = ut + 1/2gt^2 where h is the initial height (45m), u is the initial velocity (40 m/s), g is the acceleration due to gravity (10 m/s^2), and t is the time taken for the particle to hit the ground. At the point where the particle hits the ground, h = 0, so we can solve for t: 0 = 40t - 1/2(10)t^2 Simplifying this equation gives: 0 = 5t(8 - t) This equation has two solutions: t = 0 and t = 8. We can ignore the t = 0 solution because it corresponds to the initial moment when the particle is projected upward. So, the particle will hit the ground after 8 seconds. Therefore, the answer is: 9s (since the particle was projected from a height of 45m, it takes 8 seconds to fall and hit the ground. However, the total time from the projection until it hits the ground would be 9 seconds).
Pregunta 3 Informe
Find the coordinates of the point on the curve \(y = x^{2} + 4x - 2\), where the gradient is zero.
Detalles de la respuesta
Pregunta 5 Informe
A committee of 4 is to be selected from a group of 5 men and 3 women. In how many ways can this be done if the chairman of the committee must be a man?
Detalles de la respuesta
To form a committee of 4 with a chairman, we first need to select a man to be the chairman from the 5 men available. This can be done in 5 ways. Then, we need to select 3 people from the remaining 7 people (4 men and 3 women) to complete the committee. This can be done in 7 choose 3 ways, which is equal to 35. Therefore, the total number of ways to form a committee of 4 with a chairman from a group of 5 men and 3 women is: 5 * 35 = 175 So the correct answer is option (D) 175.
Pregunta 6 Informe
| Age(in years) | 1 - 5 | 6 - 10 | 11 - 15 |
| Frequency | 3 | 5 | 2 |
Calculate the standard deviation of the distribution.
Detalles de la respuesta
Pregunta 7 Informe
A particle of mass 2.5 kg is moving at a speed of 12 m/s. If a force of magnitude 10 N acts against it, find how long it takes to come to rest.
Detalles de la respuesta
To find the time it takes for the particle to come to rest, we can use the formula: time = (initial velocity) / (deceleration) where the deceleration is the rate at which the particle's velocity decreases due to the force acting against it. The deceleration can be calculated using Newton's Second Law, which states that force is equal to mass times acceleration: force = mass x acceleration Rearranging this formula, we get: acceleration = force / mass Substituting the given values, we get: acceleration = 10 N / 2.5 kg = 4 m/s^2 Therefore, the time it takes for the particle to come to rest is: time = 12 m/s / 4 m/s^2 = 3.0 s So, the answer is 3.0 s.
Pregunta 8 Informe
A and B are two independent events such that \(P(A) = \frac{2}{5}\) and \(P(A \cap B) = \frac{1}{15}\). Find \(P(B)\).
Detalles de la respuesta
Pregunta 9 Informe
The parallelogram PQRS has vertices P(-2, 3), Q(1, 4), R(2, 6) and S(-1,5). Find the coordinates of the point of intersection of the diagonals.
Detalles de la respuesta
To find the point of intersection of the diagonals of a parallelogram, we need to find the midpoint of the segment that connects the two diagonals. This midpoint is the point of intersection of the diagonals. Let's start by finding the equation of the line containing the diagonal PR. The slope of PR is: $$m_{PR}=\frac{y_R-y_P}{x_R-x_P}=\frac{6-3}{2-(-2)}=\frac{3}{4}$$ Using point-slope form, the equation of line PR is: $$y-3=\frac{3}{4}(x+2)$$ Simplifying, we get: $$y=\frac{3}{4}x+\frac{15}{4}$$ Now let's find the equation of the line containing the diagonal QS. The slope of QS is: $$m_{QS}=\frac{y_S-y_Q}{x_S-x_Q}=\frac{5-4}{-1-1}=\frac{-1}{2}$$ Using point-slope form, the equation of line QS is: $$y-4=\frac{-1}{2}(x-1)$$ Simplifying, we get: $$y=-\frac{1}{2}x+\frac{9}{2}$$ To find the point of intersection of the diagonals, we need to find the midpoint of the segment that connects the two diagonals. Let's call this midpoint M. The midpoint M is the average of the coordinates of the endpoints of the segment, which are the intersection points of the diagonals with their respective sides. To find the intersection point of PR with side QS, we solve the system of equations formed by the equations of lines PR and QS: $$\begin{cases}y=\frac{3}{4}x+\frac{15}{4}\\y=-\frac{1}{2}x+\frac{9}{2}\end{cases}$$ Solving for x and y, we get: $$\begin{cases}x=\frac{6}{5}\\y=\frac{39}{10}\end{cases}$$ So the intersection point of PR with side QS is (\(\frac{6}{5}, \frac{39}{10}\)). To find the intersection point of QS with side PR, we solve the system of equations formed by the equations of lines QS and PR: $$\begin{cases}y=-\frac{1}{2}x+\frac{9}{2}\\y=\frac{3}{4}x+\frac{15}{4}\end{cases}$$ Solving for x and y, we get: $$\begin{cases}x=-\frac{6}{5}\\y=\frac{21}{10}\end{cases}$$ So the intersection point of QS with side PR is (-\(\frac{6}{5}\), \(\frac{21}{10}\)). The midpoint M is the average of these two points, so we have: $$M=(\frac{\frac{6}{5}-\frac{6}{5}}{2}, \frac{\frac{39}{10}+\frac{21}{10}}{2})=(0, \frac{3}{2})$$ Therefore, the coordinates of the point of intersection of the diagonals is (0, \(\frac{3}{2}\)). So the answer is option (C), \((0, 4\frac{1}{2})\).
Pregunta 11 Informe
Given that \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\), find the value of P.
Detalles de la respuesta
We have the equation \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\). We can solve for P by performing matrix multiplication and solving the resulting system of linear equations: \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} -6 - 3P \\ P + 4P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\) This gives us two equations: \(-6 - 3P = 3\) \(5P = -26\) Solving for P, we get: \(-6 - 3P = 3 \Rightarrow -3P = 9 \Rightarrow P = -3\) \(5P = -26 \Rightarrow P = \frac{-26}{5}\) Therefore, the value of P is -3. Note that the system of linear equations is consistent, meaning that there is a unique solution.
Pregunta 13 Informe
A binary operation ♦ is defined on the set R, of real numbers by \(a ♦ b = \frac{ab}{4}\). Find the value of \(\sqrt{2} ♦ \sqrt{6}\).
Detalles de la respuesta
The binary operation ♦ is defined as \(a ♦ b = \frac{ab}{4}\). So, to find the value of \(\sqrt{2} ♦ \sqrt{6}\), we just need to substitute \(\sqrt{2}\) for \(a\) and \(\sqrt{6}\) for \(b\), and then simplify the expression. \[\sqrt{2} ♦ \sqrt{6} = \frac{\sqrt{2}\cdot\sqrt{6}}{4} = \frac{\sqrt{12}}{4} = \frac{\sqrt{4}\cdot\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}\] Therefore, the value of \(\sqrt{2} ♦ \sqrt{6}\) is \(\frac{\sqrt{3}}{2}\).
Pregunta 14 Informe
A lift moving upwards with a uniform acceleration of 5\(ms^{-2}\) carries a body of mass p kg. If the reaction on the floor is 480 N, find the value of p. [Take g = \(10 ms^{-2}\)].
Detalles de la respuesta
Pregunta 15 Informe
Simplify \(\frac{^{n}P_{4}}{^{n}C_{4}}\)
Detalles de la respuesta
The expression \(\frac{^{n}P_{4}}{^{n}C_{4}}\) represents the number of ways to choose and arrange 4 objects out of a total of n objects, divided by the number of ways to choose 4 objects out of n objects without arranging them. The number of ways to choose and arrange 4 objects out of n objects is given by the formula for permutations, which is \(^{n}P_{4} = \frac{n!}{(n-4)!}\). The number of ways to choose 4 objects out of n objects without arranging them is given by the formula for combinations, which is \(^{n}C_{4} = \frac{n!}{4!(n-4)!}\). So, \(\frac{^{n}P_{4}}{^{n}C_{4}} = \frac{\frac{n!}{(n-4)!}}{\frac{n!}{4!(n-4)!}} = \frac{4!}{1!} = 24\). Therefore, the answer is 24.
Pregunta 16 Informe
The equation of the line of best fit for variables x and y is \(y = 19.33 + 0.42x\), where x is the independent variable. Estimate the value of y when x = 15.
Detalles de la respuesta
The equation given is in the form y = mx + b, where m is the slope and b is the y-intercept of the line of best fit. In this case, the slope is 0.42 and the y-intercept is 19.33. To estimate the value of y when x = 15, we simply substitute x = 15 into the equation and solve for y: y = 19.33 + 0.42(15) y = 19.33 + 6.3 y = 25.63 Therefore, the estimated value of y when x = 15 is 25.63.
Pregunta 18 Informe
A force of 32 N is applied to an object of mass m kg which is at rest on a smooth horizontal surface. If the acceleration produced is 8\(ms^{-2}\), find the value of m.
Detalles de la respuesta
We can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. The formula can be written as F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the force acting on the object is 32 N, and the acceleration produced is 8 \(ms^{-2}\). Since the object is at rest, its initial velocity is zero. So, we can use the formula F = ma to find the value of m. Rearranging the formula, we get m = F/a. Substituting the given values, we get m = 32 N / 8 \(ms^{-2}\) = 4 kg. Therefore, the value of m is 4 kg.
Pregunta 19 Informe
Find the sum of the exponential series \(96 + 24 + 6 +...\)
Detalles de la respuesta
The given series is 96 + 24 + 6 + ... , and we need to find its sum. First, we can observe that each term in the series is obtained by dividing the previous term by 4. So, the common ratio between consecutive terms is 1/4. Let S be the sum of the series. Then we have: S = 96 + 24 + 6 + ... Dividing both sides by 4, we get: S/4 = 24 + 6 + 1.5 + ... Now, if we subtract the second equation from the first, we get: S - S/4 = 96 Simplifying, we get: 3S/4 = 96 Multiplying both sides by 4/3, we get: S = 128 Therefore, the sum of the given series is 128. So, the correct option is (B) 128.
Pregunta 20 Informe
Solve the inequality \(2x^{2} + 5x - 3 \geq 0\).
Detalles de la respuesta
To solve this inequality, we can start by finding the roots of the quadratic equation \(2x^2+5x-3=0\). We can do this by factoring the quadratic or by using the quadratic formula. Factoring the quadratic, we get: \begin{align*} 2x^2+5x-3&=0\\ (2x-1)(x+3)&=0 \end{align*} So the roots are \(x=\frac{1}{2}\) and \(x=-3\). These are the values of \(x\) that make the left-hand side of the inequality equal to zero. Now, we can use these roots to determine the sign of the expression \(2x^2+5x-3\) for different values of \(x\). We can do this by testing values of \(x\) in the intervals between the roots and outside the roots. For example, if we pick a value of \(x\) less than \(-3\), say \(x=-4\), then we have: \begin{align*} 2x^2+5x-3&=2(-4)^2+5(-4)-3\\ &=32-20-3\\ &=9 \end{align*} Since \(9\) is positive, this means that any value of \(x\) less than \(-3\) makes the expression \(2x^2+5x-3\) positive. Similarly, if we pick a value of \(x\) between \(-3\) and \(\frac{1}{2}\), say \(x=0\), then we have: \begin{align*} 2x^2+5x-3&=2(0)^2+5(0)-3\\ &=-3 \end{align*} Since \(-3\) is negative, this means that any value of \(x\) between \(-3\) and \(\frac{1}{2}\) makes the expression \(2x^2+5x-3\) negative. Finally, if we pick a value of \(x\) greater than \(\frac{1}{2}\), say \(x=1\), then we have: \begin{align*} 2x^2+5x-3&=2(1)^2+5(1)-3\\ &=4+5-3\\ &=6 \end{align*} Since \(6\) is positive, this means that any value of \(x\) greater than \(\frac{1}{2}\) makes the expression \(2x^2+5x-3\) positive. Putting all of this together, we can see that the inequality \(2x^2+5x-3\geq 0\) is true when \(x\leq -3\) or \(x\geq \frac{1}{2}\). Therefore, the correct option is: - \(x \leq -3\) or \(x \geq \frac{1}{2}\)
Pregunta 21 Informe
Evaluate \(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
Detalles de la respuesta
To evaluate the limit \(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\), we can start by substituting \(x = 1\) directly into the expression. However, this results in an expression of the form \(\frac{0}{0}\), which is undefined. To evaluate such a limit, we can use algebraic techniques to simplify the expression before taking the limit. In this case, we can factor the denominator to obtain: \[\frac{1 - x}{x^{2} - 3x + 2} = \frac{1 - x}{(x - 1)(x - 2)} = -\frac{1}{x - 2}\] Now, we can take the limit as \(x\) approaches 1 by substituting \(x = 1\) directly into the simplified expression. This gives: \[\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} -\frac{1}{x - 2} = -\frac{1}{-1} = 1\] Therefore, the limit of the expression as \(x\) approaches 1 is equal to 1. So, the correct option is 1.
Pregunta 22 Informe
If \(f(x) = \frac{1}{2 - x}, x \neq 2\), find \(f^{-1}(-\frac{1}{2})\).
Detalles de la respuesta
To find \(f^{-1}(-\frac{1}{2})\), we need to find the value of x such that \(f(x) = -\frac{1}{2}\). We know that \(f(x) = \frac{1}{2 - x}\), so we set \(\frac{1}{2 - x} = -\frac{1}{2}\) and solve for x: \[\frac{1}{2 - x} = -\frac{1}{2}\] \[2 - x = -2\] \[x = 4\] Therefore, \(f^{-1}(-\frac{1}{2}) = 4\). Note that we were able to solve for \(f^{-1}(y)\) by first setting \(y = f(x)\), then solving for x. This is because the inverse function of f, denoted by \(f^{-1}\), is defined such that \(f^{-1}(f(x)) = x\) for all x in the domain of f.
Pregunta 23 Informe
The mean and median of integers x, y, z and t are 5 and z respectively. If x < y < z < t and y = 4, find (x + t).
Detalles de la respuesta
Pregunta 24 Informe
Find the least value of the function \(f(x) = 3x^{2} + 18x + 32\).
Detalles de la respuesta
Pregunta 25 Informe
Find the coefficient of \(x^{4}\) in the binomial expansion of \((1 - 2x)^{6}\).
Detalles de la respuesta
We can use the binomial theorem to find the coefficient of \(x^{4}\) in the expansion of \((1 - 2x)^{6}\). The binomial theorem states that the \(n\)th power of a binomial \((a + b)^{n}\) can be expressed as the sum of the terms of the form \(a^{n-r}b^{r}\), where \(r\) ranges from 0 to \(n\). In this case, the binomial is \((1 - 2x)\), so we can expand it using the binomial theorem as follows: \[(1 - 2x)^{6} = \sum_{r=0}^{6} {6 \choose r} (1)^{6-r} (-2x)^{r}\] where \({6 \choose r}\) is the binomial coefficient. To find the coefficient of \(x^{4}\), we need to find the term where \(r\) is 2. So, we can substitute \(r = 2\) into the formula above to get: \[{6 \choose 2} (1)^{4} (-2x)^{2} = 15 \times 4x^{2} = 60x^{2}\] However, we are looking for the coefficient of \(x^{4}\), not \(x^{2}\). Since the exponent of \(x\) is 2 in the term we just calculated, we need to multiply it by another power of \(x^{2}\) to get \(x^{4}\). This means we need to find the term where \(r\) is 4. Substituting \(r = 4\) into the formula above, we get: \[{6 \choose 4} (1)^{2} (-2x)^{4} = 15 \times 16x^{4} = 240x^{4}\] Therefore, the coefficient of \(x^{4}\) in the binomial expansion of \((1 - 2x)^{6}\) is 240. Hence, the answer is 240.
Pregunta 26 Informe
\(P = {x : 1 \leq x \leq 6}\) and \(Q = {x : 2 < x < 9}\) where \(x \in R\), find \(P \cap Q\).
Detalles de la respuesta
\(P = {x : 1 \leq x \leq 6}\) means that P is the set of all real numbers between 1 and 6 (including 1 and 6). Similarly, \(Q = {x : 2 < x < 9}\) means that Q is the set of all real numbers between 2 and 9 (excluding 2 and 9). To find \(P \cap Q\), we need to find the set of all elements that are common to both P and Q. From the definitions of P and Q, we can see that the range of values that satisfy both P and Q is from 2 to 6 (including 2 and 6). Therefore, \(P \cap Q\) is the set of all real numbers between 2 and 6 (including 2 and 6). Hence, the correct option is (d) \({x : 2 < x \leq 6}\).
Pregunta 27 Informe
The mean age of n men in a club is 50 years. Two men aged 55 and 63 years left the club, and the mean age reduced by 1 year. Find the value of n.
Detalles de la respuesta
Let's denote the sum of ages of the n men in the club by S. Then we know that: Mean age of n men = S/n = 50 Multiplying both sides by n, we get: S = 50n After two men aged 55 and 63 years leave the club, the sum of ages of the remaining men becomes: S' = S - 55 - 63 = 50n - 118 The new mean age is reduced by 1 year, so we have: S'/(n-2) = 50 - 1 Simplifying this equation and substituting S', we get: (50n - 118)/(n-2) = 49 Multiplying both sides by n-2, we get: 50n - 118 = 49(n-2) Simplifying this equation, we get: n = 20 Therefore, there were originally 20 men in the club.
Pregunta 28 Informe
Find the coordinates of the centre of the circle \(4x^{2} + 4y^{2} - 5x + 3y - 2 = 0\).
Detalles de la respuesta
To find the center of the given circle, we need to write the equation in standard form, which is: \[(x - h)^2 + (y - k)^2 = r^2\] Where (h, k) is the center of the circle and r is the radius. We start by completing the square for both x and y terms, by moving the constant term to the right side and grouping the x and y terms together: \[4x^2 - 5x + 4y^2 + 3y = 2\] To complete the square for the x terms, we need to add and subtract \(\frac{5}{4}\) times the square of the coefficient of x (which is 2), inside the first bracket: \[4\left(x^2 - \frac{5}{4}x + \left(\frac{5}{4}\right)^2\right) + 4y^2 + 3y = 2 + 4\left(\frac{5}{4}\right)^2\] We do the same for the y terms, by adding and subtracting \(\frac{3}{8}\) times the square of the coefficient of y (which is \(\frac{3}{2}\)), inside the second bracket: \[4\left(x^2 - \frac{5}{4}x + \left(\frac{5}{4}\right)^2\right) + 4\left(y^2 + \frac{3}{8}y + \left(\frac{3}{16}\right)^2\right) = 2 + 4\left(\frac{5}{4}\right)^2 + 4\left(\frac{3}{16}\right)^2\] Now we have the equation in standard form, and we can read off the center and radius: Center = \(\left(\frac{5}{4}, -\frac{3}{8}\right)\) Radius = \(\sqrt{\frac{129}{16}} = \frac{3\sqrt{43}}{4}\) Therefore, the correct answer is option (C) \((\frac{5}{8}, -\frac{3}{8})\).
Pregunta 29 Informe
Two forces, each of magnitude 16 N, are inclined to each other at an angle of 60°. Calculate the magnitude of their resultant.
Detalles de la respuesta
Pregunta 30 Informe
The area of a sector of a circle is 3\(cm^{2}\). If the sector subtends an angle of 1.5 radians at the centre, calculate the radius of the circle.
Detalles de la respuesta
To solve this problem, we need to use the formula for the area of a sector of a circle, which is: Area of sector = (θ/2) × r² where θ is the central angle subtended by the sector, r is the radius of the circle, and the angle is measured in radians. In this problem, we are given the area of the sector and the central angle, so we can plug in these values and solve for the radius: 3 = (1.5/2) × r² Multiplying both sides by 2/1.5, we get: 4 = r² Taking the square root of both sides, we get: r = 2 Therefore, the radius of the circle is 2 cm. We can check our answer by plugging it back into the formula for the area of the sector: Area of sector = (1.5/2) × 2² = 1.5 × 2 = 3 This confirms that our answer is correct.
Pregunta 31 Informe
Two functions f and g are defined by \(f : x \to 3x - 1\) and \(g : x \to 2x^{3}\), evaluate \(fg(-2)\).
Detalles de la respuesta
Pregunta 32 Informe
Given that \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = p + q\sqrt{6}\), find q.
Detalles de la respuesta
Pregunta 33 Informe
Calculate, correct to one decimal place, the angle between 5i + 12j and -2i + 3j.
Detalles de la respuesta
To find the angle between two vectors, we can use the dot product formula: cos θ = (A dot B) / (|A| |B|) where A and B are the two vectors, A dot B is their dot product, and |A| and |B| are their magnitudes. Let's apply this formula to the given vectors: A = 5i + 12j B = -2i + 3j A dot B = (5*-2) + (12*3) = -10 + 36 = 26 |A| = sqrt(5^2 + 12^2) = 13 |B| = sqrt((-2)^2 + 3^2) = sqrt(13) cos θ = (26) / (13 * sqrt(13)) Using a calculator, we can find that cos θ ≈ 0.801 To find the angle θ itself, we can take the inverse cosine (cos^-1) of this value: θ ≈ cos^-1(0.801) ≈ 37.8° However, this is only half of the actual angle between the two vectors. To get the full angle, we need to take into account the direction of the vectors. If we draw the two vectors on a coordinate plane, we can see that they are in different quadrants: A is in the second quadrant and B is in the fourth quadrant. The angle between them will be the difference between their angles relative to the positive x-axis. Using the inverse tangent (tan^-1) function, we can find the angles of each vector: Angle of A = tan^-1(12/5) ≈ 67.4° Angle of B = tan^-1(-3/2) ≈ -56.3° (Note that we use a negative angle for B because it is in the fourth quadrant, where angles are negative.) The angle between the vectors will be the difference between these two angles: θ = |67.4° - (-56.3°)| = 123.7° However, this is the full angle between the vectors, and the question only asks for the acute angle (the smaller of the two possible angles). To find the acute angle, we subtract the full angle from 180°: Acute angle = 180° - 123.7° ≈ 56.3° Therefore, the answer is option B: 56.3°.
Pregunta 34 Informe
Evaluate \(\log_{0.25} 8\)
Detalles de la respuesta
We can use the change of base formula to evaluate \(\log_{0.25} 8\). The change of base formula states that for any positive numbers \(a\), \(b\), and \(c\) with \(a \neq 1\), we have: \(\log_{a} b = \frac{\log_{c} b}{\log_{c} a}\) In this case, we can choose any base we like, but it is convenient to use base 2 because 8 and 0.25 are powers of 2. So, we can rewrite the expression as: \(\log_{0.25} 8 = \frac{\log_{2} 8}{\log_{2} 0.25}\) Now, we can evaluate the logarithms on the right-hand side using the rules of logarithms. We know that \(2^{3} = 8\) and \(2^{-2} = 0.25\), so we have: \(\log_{2} 8 = 3\) and \(\log_{2} 0.25 = -2\) Substituting these values into the equation above, we get: \(\log_{0.25} 8 = \frac{3}{-2} = -\frac{3}{2}\) Therefore, the value of \(\log_{0.25} 8\) is \(-\frac{3}{2}\). Hence, the answer is \(-\frac{3}{2}\).
Pregunta 35 Informe
If \((x - 3)\) is a factor of \(2x^{3} + 3x^{2} - 17x - 30\), find the remaining factors.
Detalles de la respuesta
If \((x - 3)\) is a factor of \(2x^{3} + 3x^{2} - 17x - 30\), then we can use long division or synthetic division to divide the polynomial by \((x - 3)\) and find the remaining factors.
Using long division, we have:
2x^2 + 9x + 10
------------------------
x - 3 | 2x^3 + 3x^2 - 17x - 30
- (2x^3 - 6x^2)
--------------
9x^2 - 17x
- (9x^2 - 27x)
-----------
10x - 30
- (10x - 30)
---------
0
The result of the division is \(2x^{2} + 9x + 10\), which is a quadratic polynomial. Therefore, the remaining factors are given by factoring this quadratic polynomial. We can factor it as follows:
\[2x^{2} + 9x + 10 = (2x + 5)(x + 2)\]Therefore, the complete factorization of \(2x^{3} + 3x^{2} - 17x - 30\) is:
\[2x^{3} + 3x^{2} - 17x - 30 = (x - 3)(2x + 5)(x + 2)\]So, the correct option is (2x + 5)(x + 2).
Pregunta 36 Informe
If \(a = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\) and \(b = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\), find a vector c such that \(4a + 3c = b\).
Detalles de la respuesta
To solve this problem, we want to find a vector c that satisfies the equation 4a + 3c = b. We can rearrange this equation to get: 3c = b - 4a Now we just need to solve for c. We can do this by dividing both sides by 3: c = (b - 4a) / 3 Substituting the values of a and b given in the problem, we get: c = (\begin{pmatrix} -3 \\ 5 \end{pmatrix} - 4\begin{pmatrix} 3 \\ 2 \end{pmatrix}) / 3 c = (\begin{pmatrix} -3 \\ 5 \end{pmatrix} - \begin{pmatrix} 12 \\ 8 \end{pmatrix}) / 3 c = \begin{pmatrix} -15/3 \\ -3/3 \end{pmatrix} c = \begin{pmatrix} -5 \\ -1 \end{pmatrix} Therefore, the answer is option B, \(\begin{pmatrix} -5 \\ -1 \end{pmatrix}\).
Pregunta 37 Informe
The roots of the equation \(2x^{2} + kx + 5 = 0\) are \(\alpha\) and \(\beta\), where k is a constant. If \(\alpha^{2} + \beta^{2} = -1\), find the values of k.
Detalles de la respuesta
The sum of the roots of the quadratic equation \(ax^{2} + bx + c = 0\) is given by the formula \(-\frac{b}{a}\), and the product of the roots is given by the formula \(\frac{c}{a}\). Therefore, for the quadratic equation \(2x^{2} + kx + 5 = 0\), we have: \[\alpha + \beta = -\frac{k}{2}\] \[\alpha \beta = \frac{5}{2}\] Squaring the first equation, we get: \[(\alpha + \beta)^{2} = \alpha^{2} + 2\alpha\beta + \beta^{2} = \frac{k^{2}}{4}\] Substituting the given value of \(\alpha^{2} + \beta^{2} = -1\), we have: \[-1 + 2\alpha\beta = \frac{k^{2}}{4}\] Substituting the value of \(\alpha \beta = \frac{5}{2}\), we have: \[-1 + 2\left(\frac{5}{2}\right) = \frac{k^{2}}{4}\] Simplifying the left-hand side, we get: \[4 = \frac{k^{2}}{4}\] Multiplying both sides by 4, we get: \[16 = k^{2}\] Taking the square root of both sides, we get: \[k = \pm 4\] Therefore, the values of k are \(\pm 4\).
Pregunta 38 Informe
Simplify \(8^{n} \times 2^{2n} \div 4^{3n}\)
Detalles de la respuesta
We can simplify the expression by first rewriting each of the bases as powers of 2: $$8^{n} \times 2^{2n} \div 4^{3n} = (2^{3})^{n} \times 2^{2n} \div (2^{2})^{3n}.$$ Using the laws of exponents, we can simplify this to: $$(2^{3n}) \times 2^{2n} \div 2^{6n} = 2^{3n+2n-6n} = 2^{-n}.$$ Therefore, the simplified expression is \(2^{-n}\).
Pregunta 39 Informe
ABCD is a square. Forces of magnitude 14N, 4N, 2N and \(2\sqrt{2} N\) act along the sides AB, BC, CD and DA respectively. Find in Newtons, the magnitude of the resultant of the forces.
Detalles de la respuesta
Pregunta 40 Informe
In a firing contest, the probabilities that Kojo and Kwame hit the target are \(\frac{2}{5}\) and \(\frac{1}{3}\) respectively. What is the probability that none of them hit the target?
Detalles de la respuesta
The probability that Kojo hits the target is \(\frac{2}{5}\), which means that the probability that he misses the target is \(1-\frac{2}{5}=\frac{3}{5}\). Similarly, the probability that Kwame misses the target is \(1-\frac{1}{3}=\frac{2}{3}\). Now, we want to find the probability that none of them hit the target. Since they are shooting independently, we can multiply their probabilities of missing the target: Probability that both miss the target = Probability that Kojo misses the target AND Probability that Kwame misses the target = \(\frac{3}{5} \times \frac{2}{3} = \frac{6}{15} = \frac{2}{5}\) Therefore, the probability that none of them hit the target is \(\frac{2}{5}\).
Pregunta 41 Informe
(a) A man P has 5 red, 3 blue and 2 white buses. Another man Q has 3 red, 2 blue and 4 white buses. A bus owned by P is involved in an accident with a bus belonging to Q. Calculate the probability that the two buses are not of the same color.
(b) A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?
Detalles de la respuesta
None
Pregunta 42 Informe
(a) A ball P moving with velocity \(2u ms^{-1}\), collides with a similar ball Q, of different mass, which is at rest. After collision, Q moves with \(u ms^{-1}\) and P with velocity \(\frac{1}{2} u ms^{-2}\), in the opposite direction. Find the ratio of the masses of P and Q.
(b) Two forces of magnitudes 3 N and 7 N have a resultant of magnitude 5 N. Calculate, correct to one decimal place, the angle between the two forces.
(c) \(AB = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(CB = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) are two vectors in the XY- plane. If V is the midpoint of AB, find CV.
(a) Let masses be \(m_P\) and \(m_Q\). Taking the original direction of P as positive, after collision P reverses to \(-\tfrac12 u\) and Q moves \(+u\). Conservation of momentum:
\[m_P(2u)+m_Q(0)=m_P\!\left(-\tfrac12 u\right)+m_Q(u).\] \[2m_P u+\tfrac12 m_P u=m_Q u\Rightarrow \tfrac52 m_P=m_Q\Rightarrow \frac{m_P}{m_Q}=\frac{2}{5}.\]Ratio \(m_P:m_Q=\textbf{2:5}\).
(b) For forces 3 N and 7 N with resultant 5 N at angle \(\theta\) between them:
\[R^2=P^2+Q^2+2PQ\cos\theta\Rightarrow 25=9+49+2(3)(7)\cos\theta.\] \[42\cos\theta=25-58=-33\Rightarrow \cos\theta=-0.7857\Rightarrow \theta\approx 141.8^\circ.\](c) \(\vec{AB}=\begin{pmatrix}-4\\6\end{pmatrix},\ \vec{CB}=\begin{pmatrix}2\\-3\end{pmatrix}\). V is the midpoint of AB, so \(\vec{BV}=-\tfrac12\vec{AB}=\begin{pmatrix}2\\-3\end{pmatrix}\). Then
\[\vec{CV}=\vec{CB}+\vec{BV}=\begin{pmatrix}2\\-3\end{pmatrix}+\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}4\\-6\end{pmatrix}.\]Detalles de la respuesta
(a) Let masses be \(m_P\) and \(m_Q\). Taking the original direction of P as positive, after collision P reverses to \(-\tfrac12 u\) and Q moves \(+u\). Conservation of momentum:
\[m_P(2u)+m_Q(0)=m_P\!\left(-\tfrac12 u\right)+m_Q(u).\] \[2m_P u+\tfrac12 m_P u=m_Q u\Rightarrow \tfrac52 m_P=m_Q\Rightarrow \frac{m_P}{m_Q}=\frac{2}{5}.\]Ratio \(m_P:m_Q=\textbf{2:5}\).
(b) For forces 3 N and 7 N with resultant 5 N at angle \(\theta\) between them:
\[R^2=P^2+Q^2+2PQ\cos\theta\Rightarrow 25=9+49+2(3)(7)\cos\theta.\] \[42\cos\theta=25-58=-33\Rightarrow \cos\theta=-0.7857\Rightarrow \theta\approx 141.8^\circ.\](c) \(\vec{AB}=\begin{pmatrix}-4\\6\end{pmatrix},\ \vec{CB}=\begin{pmatrix}2\\-3\end{pmatrix}\). V is the midpoint of AB, so \(\vec{BV}=-\tfrac12\vec{AB}=\begin{pmatrix}2\\-3\end{pmatrix}\). Then
\[\vec{CV}=\vec{CB}+\vec{BV}=\begin{pmatrix}2\\-3\end{pmatrix}+\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}4\\-6\end{pmatrix}.\]Pregunta 43 Informe
If \(2^{2x - 3y} = 32\) and \(\log_{y} x = 2\), find the values of x and y.
Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)
First equation. Since \(32=2^5,\) equate indices:
\[2x-3y=5.\quad(1)\]
Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)
Substitute (2) into (1):
\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]
Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)
A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence
\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]
Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.
Detalles de la respuesta
Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)
First equation. Since \(32=2^5,\) equate indices:
\[2x-3y=5.\quad(1)\]
Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)
Substitute (2) into (1):
\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]
Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)
A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence
\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]
Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.
Pregunta 44 Informe
An object is projected vertically upwards with a velocity of 80 m/s. Find the :
(a) Maximum height reached
(b) Time taken to return to the point of projection. [Take g = \(10 ms^{-2}\)].
Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).
(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):
\[H = \frac{u^2}{2g} = \frac{80^2}{2\times 10} = \frac{6400}{20} = 320\ \text{m}.\]The maximum height reached is 320 m.
(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).
\[T = 2t = 2\times 8 = 16\ \text{s}.\]The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)
Detalles de la respuesta
Take upward as positive, initial velocity \(u = 80\ \text{m/s}\), \(g = 10\ \text{m/s}^2\).
(a) Maximum height. At the highest point the velocity is zero. Using \(v^2 = u^2 - 2gH\) with \(v = 0\):
\[H = \frac{u^2}{2g} = \frac{80^2}{2\times 10} = \frac{6400}{20} = 320\ \text{m}.\]The maximum height reached is 320 m.
(b) Time to return to the point of projection. By symmetry the total time is twice the time to reach the top. Time up: \(v = u - gt \Rightarrow 0 = 80 - 10t \Rightarrow t = 8\ \text{s}\).
\[T = 2t = 2\times 8 = 16\ \text{s}.\]The object returns after 16 s. (Equivalently \(T = \dfrac{2u}{g} = \dfrac{160}{10} = 16\ \text{s}\).)
Pregunta 45 Informe
(a) Using the substitution \(u = 5 - x^{2}\), evaluate \(\int_{1}^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\).
(b) If \(y = px^{2} + qx; \frac{\mathrm d y}{\mathrm d x} = 6x + 7\) and \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6\), find the values of p and q.
(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).
\[\int_{1}^{2}\frac{x}{\sqrt{5-x^2}}\,dx=\int_{4}^{1}\frac{-\tfrac12\,du}{\sqrt{u}}=-\frac12\int_{4}^{1}u^{-1/2}\,du=-\frac12\Big[2\sqrt u\Big]_{4}^{1}=-\Big[\sqrt u\Big]_{4}^{1}.\] \[=-\big(\sqrt1-\sqrt4\big)=-(1-2)=1.\]The value of the integral is 1.
(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).
Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.
So \(p=3,\ q=7\).
Detalles de la respuesta
(a) Let \(u=5-x^2\Rightarrow \dfrac{du}{dx}=-2x\), so \(x\,dx=-\tfrac12\,du\). Limits: \(x=1\Rightarrow u=4\); \(x=2\Rightarrow u=1\).
\[\int_{1}^{2}\frac{x}{\sqrt{5-x^2}}\,dx=\int_{4}^{1}\frac{-\tfrac12\,du}{\sqrt{u}}=-\frac12\int_{4}^{1}u^{-1/2}\,du=-\frac12\Big[2\sqrt u\Big]_{4}^{1}=-\Big[\sqrt u\Big]_{4}^{1}.\] \[=-\big(\sqrt1-\sqrt4\big)=-(1-2)=1.\]The value of the integral is 1.
(b) With \(y=px^2+qx\), \(\dfrac{dy}{dx}=2px+q\) and \(\dfrac{d^2y}{dx^2}=2p\).
Comparing \(\dfrac{dy}{dx}=6x+7\): \(2p=6\Rightarrow p=3\) and \(q=7\). The second derivative \(2p=6\) confirms this.
So \(p=3,\ q=7\).
Pregunta 46 Informe
A body of mass 5 kg resting on a smooth horizontal plane, is acted upon by force 6i + 2j, 5i + 4j and 4i - j. Calculate the:
(a) velocity of the body
(b) Magnitude of its velocity after 4s.
A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.
Resultant force.
\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]
Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)
\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]
(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)
\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]
(b) Magnitude of the velocity after 4 s.
\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]
(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))
Detalles de la respuesta
A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.
Resultant force.
\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]
Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)
\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]
(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)
\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]
(b) Magnitude of the velocity after 4 s.
\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]
(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))
Pregunta 47 Informe
(a) A body of mass 5 kg is placed on a smooth plane inclined at an angle 30° to the horizontal. Find the magnitude of the force: (i) acting parallel to the plane (ii) required to keep the body in equilibrium. [Take \(g = 10 ms^{-2}\)].
(b) A uniform plank PQ of length 10m and mass m kg rests on two supports A and B, where \(|PA| = |BQ| = 1m\). A load of mass 8 kg is placed on the plank at point C such that \(|AC| = 3.5 m\). If the reaction at B is 100 N, calculate the (i) value of m (ii) reaction at A. [Take \(g = 10 ms^{-2}\)].
(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).
(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).
(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).
(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).
Take moments about A (clockwise = anticlockwise):
\[R_B(8)=mg(4)+80(3.5)\Rightarrow 100(8)=40m+280.\] \[800-280=40m\Rightarrow m=\frac{520}{40}=\textbf{13 kg}.\](ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).
\[R_A=210-100=\textbf{110 N}.\]Detalles de la respuesta
(a) Mass 5 kg, weight \(W=50\ \text{N}\), smooth plane at \(30^\circ\).
(i) Component of weight parallel to (down) the plane: \(W\sin 30^\circ=50\times 0.5=\textbf{25 N}\).
(ii) To keep the body in equilibrium a force equal and opposite to this must act up the plane: \(\textbf{25 N}\).
(b) Plank \(PQ=10\ \text{m}\), mass \(m\), \(|PA|=|BQ|=1\ \text{m}\), so \(|AB|=8\ \text{m}\). The plank's weight \(mg\) acts at its centre, 5 m from P i.e. 4 m from A. Load 8 kg (\(=80\ \text{N}\)) at C with \(|AC|=3.5\ \text{m}\). Given reaction at B, \(R_B=100\ \text{N}\).
Take moments about A (clockwise = anticlockwise):
\[R_B(8)=mg(4)+80(3.5)\Rightarrow 100(8)=40m+280.\] \[800-280=40m\Rightarrow m=\frac{520}{40}=\textbf{13 kg}.\](ii) Vertical equilibrium: \(R_A+R_B=mg+80=130+80=210\).
\[R_A=210-100=\textbf{110 N}.\]Pregunta 48 Informe
(a) There are 6 points in a plane. How many triangles can be formed with the points?
(b) A family of 6 is to be seated in a row . In how many ways can this be done if the father and mother are not to be seated together?
(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:
\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]
So 20 triangles can be formed.
(b) Seating 6 people in a row with father and mother NOT together.
Total arrangements of 6 people in a row: \(6!=720.\)
Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:
\[5!\times2!=120\times2=240.\]
Therefore arrangements with them NOT together:
\[720-240=480.\]
Detalles de la respuesta
(a) Triangles from 6 points in a plane. A triangle needs any 3 of the points (assuming no three are collinear), and order does not matter, so use combinations:
\[\binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=20.\]
So 20 triangles can be formed.
(b) Seating 6 people in a row with father and mother NOT together.
Total arrangements of 6 people in a row: \(6!=720.\)
Arrangements with father and mother together: treat the pair as a single block, giving \(5!\) arrangements, and the two of them can swap within the block in \(2!\) ways:
\[5!\times2!=120\times2=240.\]
Therefore arrangements with them NOT together:
\[720-240=480.\]
Pregunta 49 Informe
(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings \(\overrightarrow{XP}\) and \(\overrightarrow{YP}\). The strings are inclined at 60° and 40° respectively to the downward vertical. Find, correct to two decimal places, the tensionsin the strings. [Take g = \(10 ms^{-2}\)].
(b) The height h metres, of a ball thrown into the air is \(2 + 20t + kt^{2}\), after t seconds. If it takes 2 seconds for the ball to reach its highest point, find
(i) the value of k (ii) its highest point from the point of throw.
(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:
\[\frac{T_{XP}}{\sin 140^\circ}=\frac{T_{YP}}{\sin 120^\circ}=\frac{150}{\sin 100^\circ}.\] \[T_{XP}=\frac{150\sin 140^\circ}{\sin 100^\circ}=\frac{150(0.6428)}{0.9848}=97.91\ \text{N}.\] \[T_{YP}=\frac{150\sin 120^\circ}{\sin 100^\circ}=\frac{150(0.8660)}{0.9848}=131.91\ \text{N}.\]Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).
(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):
\[20+2k(2)=0\Rightarrow 20+4k=0\Rightarrow k=-5.\](ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.
Detalles de la respuesta
(a) Weight \(W=mg=15\times 10=150\ \text{N}\). String \(XP\) makes \(60^\circ\) and \(YP\) makes \(40^\circ\) with the downward vertical. The angles between the three forces at P are: between the two tensions \(60^\circ+40^\circ=100^\circ\); between \(T_{XP}\) and the weight \(180^\circ-60^\circ=120^\circ\); between \(T_{YP}\) and the weight \(180^\circ-40^\circ=140^\circ\). By Lami's theorem:
\[\frac{T_{XP}}{\sin 140^\circ}=\frac{T_{YP}}{\sin 120^\circ}=\frac{150}{\sin 100^\circ}.\] \[T_{XP}=\frac{150\sin 140^\circ}{\sin 100^\circ}=\frac{150(0.6428)}{0.9848}=97.91\ \text{N}.\] \[T_{YP}=\frac{150\sin 120^\circ}{\sin 100^\circ}=\frac{150(0.8660)}{0.9848}=131.91\ \text{N}.\]Tension in \(XP\approx\textbf{97.91 N}\), tension in \(YP\approx\textbf{131.91 N}\).
(b) \(h=2+20t+kt^2\), so \(\dfrac{dh}{dt}=20+2kt\). At the highest point the velocity is zero at \(t=2\):
\[20+2k(2)=0\Rightarrow 20+4k=0\Rightarrow k=-5.\](ii) Highest point: \(h(2)=2+20(2)+(-5)(2)^2=2+40-20=\textbf{22 m}\) above the point of throw.
Pregunta 50 Informe
(a) Two pupils are chosen at random from a group of 4 boys and 5 girls. Find the probability that the two pupils chosen would be boys.
(b) Twenty percent of the total production of transistors produced by a machine are below standard. If a random sample of 6 transistors produced by the machine is taken, what is the probability of getting (i) exactly 2 standard transistors (ii) exactly 1 standard transistor (iii) at least 2 standard transistors (iv) at most 2 standard transistors?
(a) Choosing 2 from 4 boys and 5 girls (9 pupils):
\[P(\text{both boys})=\frac{\binom{4}{2}}{\binom{9}{2}}=\frac{6}{36}=\frac16.\](b) Let \(p=P(\text{standard})=0.8\) and \(q=P(\text{below standard})=0.2\), with \(n=6\). This is binomial \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\).
(i) exactly 2 standard: \(\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.0154\).
(ii) exactly 1 standard: \(\binom{6}{1}(0.8)(0.2)^5=6(0.8)(0.00032)=0.0015\).
(iii) at least 2 standard: \(1-P(0)-P(1)\), where \(P(0)=(0.2)^6=0.000064\).
\[=1-0.000064-0.001536=0.9984.\](iv) at most 2 standard: \(P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.0170\).
Detalles de la respuesta
(a) Choosing 2 from 4 boys and 5 girls (9 pupils):
\[P(\text{both boys})=\frac{\binom{4}{2}}{\binom{9}{2}}=\frac{6}{36}=\frac16.\](b) Let \(p=P(\text{standard})=0.8\) and \(q=P(\text{below standard})=0.2\), with \(n=6\). This is binomial \(P(X=r)=\binom{6}{r}(0.8)^r(0.2)^{6-r}\).
(i) exactly 2 standard: \(\binom{6}{2}(0.8)^2(0.2)^4=15(0.64)(0.0016)=0.0154\).
(ii) exactly 1 standard: \(\binom{6}{1}(0.8)(0.2)^5=6(0.8)(0.00032)=0.0015\).
(iii) at least 2 standard: \(1-P(0)-P(1)\), where \(P(0)=(0.2)^6=0.000064\).
\[=1-0.000064-0.001536=0.9984.\](iv) at most 2 standard: \(P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.0170\).
Pregunta 51 Informe
(a) The gradient of the tangent to the curve \(y = 4x^{3}\) at points P and Q is 108. Find the coordinates of P and Q.
(b) Given that \(A = 45°, B = 30°, \sin (A + B) = \sin A \cos B + \sin B \cos A\) and \(\cos (A + B) = \cos A \cos B - \sin A \sin B\)
(i) Show that \(\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}\) and \(\cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4}\)
(ii) hence find \(\tan 15°\).
(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:
\[12x^2=108\Rightarrow x^2=9\Rightarrow x=\pm 3.\]At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).
So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).
(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):
\[\sin 75^\circ=\sin A\cos B+\sin B\cos A=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\] \[\cos 75^\circ=\cos A\cos B-\sin A\sin B=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}.\]Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.
(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:
\[\tan 15^\circ=\frac{(\sqrt6-\sqrt2)^2}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{6-2\sqrt{12}+2}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3.\]So \(\tan 15^\circ=2-\sqrt3\).
Detalles de la respuesta
(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:
\[12x^2=108\Rightarrow x^2=9\Rightarrow x=\pm 3.\]At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).
So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).
(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):
\[\sin 75^\circ=\sin A\cos B+\sin B\cos A=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\] \[\cos 75^\circ=\cos A\cos B-\sin A\sin B=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}.\]Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.
(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:
\[\tan 15^\circ=\frac{(\sqrt6-\sqrt2)^2}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{6-2\sqrt{12}+2}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3.\]So \(\tan 15^\circ=2-\sqrt3\).
Pregunta 52 Informe
If (x + 2) and (x - 1) are factors of \(f(x) = 6x^{4} + mx^{3} - 13x^{2} + nx + 14\), find the
(a) values of m and n.
(b) remainder when f(x) is divided be (x + 1).
Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)
(a) Values of m and n.
\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)
So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)
\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)
Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)
\[m=12,\qquad n=-19.\]
(b) Remainder when f(x) is divided by \((x+1).\)
By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)
\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]
Remainder \(=14.\)
Detalles de la respuesta
Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)
(a) Values of m and n.
\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)
So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)
\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)
Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)
\[m=12,\qquad n=-19.\]
(b) Remainder when f(x) is divided by \((x+1).\)
By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)
\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]
Remainder \(=14.\)
Pregunta 53 Informe
The table shows the distribution of the ages of a group of people in a village.
| Ages (in years) | 15 - 18 | 19 - 22 | 23 - 26 | 27 - 30 | 31 - 34 | 35 - 38 |
| Frequency | 40 | 33 | 25 | 10 | 8 | 4 |
Using an assumed mean of 24.5, calculate the mean of the distribution.
The assumed-mean (working-mean) method for grouped data uses \(\bar{x}=A+\dfrac{\sum fd}{\sum f}\), where \(d=x-A\), \(x\) is the class mid-value and \(A=24.5\) is the assumed mean.
Each class has width 4, so the mid-value is the average of the class limits, e.g. \(\tfrac{15+18}{2}=16.5\).
| Ages | Frequency \(f\) | Mid-value \(x\) | \(d=x-24.5\) | \(fd\) |
|---|---|---|---|---|
| 15 - 18 | 40 | 16.5 | -8 | -320 |
| 19 - 22 | 33 | 20.5 | -4 | -132 |
| 23 - 26 | 25 | 24.5 | 0 | 0 |
| 27 - 30 | 10 | 28.5 | 4 | 40 |
| 31 - 34 | 8 | 32.5 | 8 | 64 |
| 35 - 38 | 4 | 36.5 | 12 | 48 |
| Total | 120 | -300 |
\[\bar{x}=A+\frac{\sum fd}{\sum f}=24.5+\frac{-300}{120}=24.5-2.5=22.0\]
The mean age of the distribution is 22 years.
Detalles de la respuesta
The assumed-mean (working-mean) method for grouped data uses \(\bar{x}=A+\dfrac{\sum fd}{\sum f}\), where \(d=x-A\), \(x\) is the class mid-value and \(A=24.5\) is the assumed mean.
Each class has width 4, so the mid-value is the average of the class limits, e.g. \(\tfrac{15+18}{2}=16.5\).
| Ages | Frequency \(f\) | Mid-value \(x\) | \(d=x-24.5\) | \(fd\) |
|---|---|---|---|---|
| 15 - 18 | 40 | 16.5 | -8 | -320 |
| 19 - 22 | 33 | 20.5 | -4 | -132 |
| 23 - 26 | 25 | 24.5 | 0 | 0 |
| 27 - 30 | 10 | 28.5 | 4 | 40 |
| 31 - 34 | 8 | 32.5 | 8 | 64 |
| 35 - 38 | 4 | 36.5 | 12 | 48 |
| Total | 120 | -300 |
\[\bar{x}=A+\frac{\sum fd}{\sum f}=24.5+\frac{-300}{120}=24.5-2.5=22.0\]
The mean age of the distribution is 22 years.
Pregunta 54 Informe
(a) Simplify \(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\), leaving your answer in the form \(a + b\sqrt{c}\); where a, b and c are rational numbers.
(b) The points (7, 3), (2, 8) and (-3, 3) lie on a circle. Find the (i) equation and (ii) radius of the circle.
(a) Simplify \(\dfrac{\sqrt{75}-3}{\sqrt{3}+1}\).
Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):
\[\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{15-5\sqrt{3}-3\sqrt{3}+3}{3-1}=\frac{18-8\sqrt{3}}{2}=9-4\sqrt{3}.\]So \(a=9,\ b=-4,\ c=3\).
(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).
\((7,3):\ 58+14g+6f+c=0\) (i)
\((-3,3):\ 18-6g+6f+c=0\) (ii)
\((2,8):\ 68+4g+16f+c=0\) (iii)
(i) - (ii): \(40+20g=0\Rightarrow g=-2\).
(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\) (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\) (v).
(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).
(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).
(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).
Detalles de la respuesta
(a) Simplify \(\dfrac{\sqrt{75}-3}{\sqrt{3}+1}\).
Since \(\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}\), the expression is \(\dfrac{5\sqrt{3}-3}{\sqrt{3}+1}\). Rationalise by multiplying by \(\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\):
\[\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{15-5\sqrt{3}-3\sqrt{3}+3}{3-1}=\frac{18-8\sqrt{3}}{2}=9-4\sqrt{3}.\]So \(a=9,\ b=-4,\ c=3\).
(b) Let the circle be \(x^2+y^2+2gx+2fy+c=0\).
\((7,3):\ 58+14g+6f+c=0\) (i)
\((-3,3):\ 18-6g+6f+c=0\) (ii)
\((2,8):\ 68+4g+16f+c=0\) (iii)
(i) - (ii): \(40+20g=0\Rightarrow g=-2\).
(ii): \(18-6(-2)+6f+c=0\Rightarrow 6f+c=-30\) (iv).
(iii): \(68+4(-2)+16f+c=0\Rightarrow 16f+c=-60\) (v).
(v) - (iv): \(10f=-30\Rightarrow f=-3\), then \(c=-30-6(-3)=-12\).
(i) Equation: \(x^2+y^2-4x-6y-12=0\), i.e. \((x-2)^2+(y-3)^2=25\).
(ii) Radius: centre \((-g,-f)=(2,3)\), \(r=\sqrt{g^2+f^2-c}=\sqrt{4+9+12}=\sqrt{25}=\textbf{5}\).
Pregunta 55 Informe
The sum of the 2nd and 5th terms of an arithmetic progression (AP) is 42. If the difference between the 6th and 3rd term is 12, find the
(i) Common difference
(ii) first term
(iii) 20th term.
Let the first term be \(a\) and common difference \(d.\) The \(n\)th term is \(T_n=a+(n-1)d.\)
Set up the equations. Sum of 2nd and 5th terms is 42:
\[(a+d)+(a+4d)=42\Rightarrow 2a+5d=42.\]
Difference between 6th and 3rd terms is 12:
\[(a+5d)-(a+2d)=12\Rightarrow 3d=12.\]
(i) Common difference. \(3d=12\Rightarrow d=4.\)
(ii) First term. Substitute into \(2a+5d=42:\)
\[2a+5(4)=42\Rightarrow 2a=22\Rightarrow a=11.\]
(iii) 20th term.
\[T_{20}=a+19d=11+19(4)=11+76=87.\]
Detalles de la respuesta
Let the first term be \(a\) and common difference \(d.\) The \(n\)th term is \(T_n=a+(n-1)d.\)
Set up the equations. Sum of 2nd and 5th terms is 42:
\[(a+d)+(a+4d)=42\Rightarrow 2a+5d=42.\]
Difference between 6th and 3rd terms is 12:
\[(a+5d)-(a+2d)=12\Rightarrow 3d=12.\]
(i) Common difference. \(3d=12\Rightarrow d=4.\)
(ii) First term. Substitute into \(2a+5d=42:\)
\[2a+5(4)=42\Rightarrow 2a=22\Rightarrow a=11.\]
(iii) 20th term.
\[T_{20}=a+19d=11+19(4)=11+76=87.\]
Pregunta 56 Informe
Find the equation of the tangent to the curve \(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\) at the point (1, 0).
Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)
Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)
\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]
Gradient at the point. At \(x=1:\)
\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]
Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)
\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]
Detalles de la respuesta
Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)
Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)
\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]
Gradient at the point. At \(x=1:\)
\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]
Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)
\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]
Pregunta 57 Informe
The table shows the distribution of hours spent at work by the employees of a factory in a week.
| Time (in hours) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 |
| No of persons | 8 | 11 | 23 | 25 | 8 | 5 |
(a) Draw an ogive for the distribution.
(b) Using your graph, estimate the (i) lower quartile (ii) median (iii) 40th percentile (iv) number of employees that spent at least 50 hours 30 minutes at work.
(a) Cumulative-frequency table
| Time (hours) | Frequency | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 20–29 | 8 | 19.5–29.5 | 8 |
| 30–39 | 11 | 29.5–39.5 | 19 |
| 40–49 | 23 | 39.5–49.5 | 42 |
| 50–59 | 25 | 49.5–59.5 | 67 |
| 60–69 | 8 | 59.5–69.5 | 75 |
| 70–79 | 5 | 69.5–79.5 | 80 |
Plot cumulative frequency against the upper class boundary, beginning at .5, 0. The resulting less-than ogive is:
(b) Total number of employees, N=80.
- The lower quartile corresponds to cumulative frequency \(\frac{N}{4}=20\). From the ogive, \(Q_1\approx 40\) hours.
- The median corresponds to cumulative frequency \(\frac{N}{2}=40\). From the ogive, median \(\approx 49\) hours.
- The 40th percentile corresponds to cumulative frequency \(0.40\times80=32\). From the ogive, \(P_{40}\approx45.5\) hours.
- At 50 hours 30 minutes, that is, 50.5 hours, the cumulative frequency read from the ogive is approximately 45. Hence, the number who spent at least 50.5 hours at work is \[80-45=\boxed{35\text{ employees}}.\]
Detalles de la respuesta
(a) Cumulative-frequency table
| Time (hours) | Frequency | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 20–29 | 8 | 19.5–29.5 | 8 |
| 30–39 | 11 | 29.5–39.5 | 19 |
| 40–49 | 23 | 39.5–49.5 | 42 |
| 50–59 | 25 | 49.5–59.5 | 67 |
| 60–69 | 8 | 59.5–69.5 | 75 |
| 70–79 | 5 | 69.5–79.5 | 80 |
Plot cumulative frequency against the upper class boundary, beginning at .5, 0. The resulting less-than ogive is:
(b) Total number of employees, N=80.
- The lower quartile corresponds to cumulative frequency \(\frac{N}{4}=20\). From the ogive, \(Q_1\approx 40\) hours.
- The median corresponds to cumulative frequency \(\frac{N}{2}=40\). From the ogive, median \(\approx 49\) hours.
- The 40th percentile corresponds to cumulative frequency \(0.40\times80=32\). From the ogive, \(P_{40}\approx45.5\) hours.
- At 50 hours 30 minutes, that is, 50.5 hours, the cumulative frequency read from the ogive is approximately 45. Hence, the number who spent at least 50.5 hours at work is \[80-45=\boxed{35\text{ employees}}.\]
