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Question 2 Report
If f(x) = 4x\(^3\) + px\(^2\) + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p
Question 3 Report
The table shows the distribution of marks obtained by some students in a test
Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 |
Frequency | 4 | 12 | 16 | 6 | 2 |
Find the modal class mark.
Answer Details
To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).
Question 4 Report
A fair die is tossed 60 times and the results are recorded in the table
Number of die | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 15 | 10 | 14 | 2 | 8 | 11 |
Find the probability of obtaining a prime number.
Answer Details
A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.
Question 6 Report
Given that M = \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) and N = \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\), calculate (3M - 2N)
Answer Details
To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).
Question 7 Report
A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).
Calculate the distance travelled in the third second.
Question 8 Report
In how many ways can 8 persons be seated on a bench if only three seats are available?
Answer Details
The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.
Question 9 Report
Given that f: x --> x\(^2\) - x + 1 is defined on the Set Q = { x : 0 ≤ x < 20, x is a multiple of 5}. find the set of range of F.
Question 10 Report
The first term of an AP is 4 and the sum of the first three terms is 18. Find the product of the first three terms
Answer Details
Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.
Question 11 Report
Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0.
Answer Details
To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.
Question 12 Report
A committee consists of 6 boys and 4 girls. In how many ways can a sub-committee consisting of 3 boys and 2 girls be formed if one particular boy and one particular girl must be on the sub-committee?
Question 13 Report
Given that P = { x: 0 ≤ x ≤ 36, x is a factor of 36 divisible by 3} and Q = { x: 0 ≤ x ≤ 36, x is an even number and a perfect square}, find P n Q.
Question 14 Report
Three forces, F\(_1\) (8N, 030°), F\(_\2) (10N, 150° ) and F\(_\3) ( KN, 240° )are in equilibrium. Find the value of N
Question 15 Report
If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.
Question 16 Report
If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)
Question 17 Report
If sin x = \(\frac{12}{13}\) and sin y = \(\frac{4}{5}\), where x and y are acute angles, find cos (x + y)
Question 18 Report
If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R
Question 20 Report
A binary operation * is defined on the set of real numbers, R, by
P * q = \(\frac{q^2 - p^2}{2pq}\). Find 3 * 2
Answer Details
The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.
Question 21 Report
Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2.
Question 22 Report
A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).
Calculate the maximum height reached.
Question 23 Report
A body of mass 15kg is placed on a smooth plane which is inclined at 60° to the horizontal. If the box is at rest,
calculate the normal reaction to the plane. [ Take g = 10m/s\(^2\) ]
Answer Details
To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.
Question 24 Report
Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)
Answer Details
To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.
Question 25 Report
Simplify ( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)
Question 26 Report
Find the inverse of \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\)
Question 27 Report
If α and β are the roots of 3x\(^2\) - 7x + 6 = 0, find \(\frac{1}{α}\) + \(\frac{1}{β}\)
Question 28 Report
Consider the following statements:
X: Benita is polite
y: Benita is neat
z: Benita is intelligent
Which of the following symbolizes the statement: "Benita is neat if and only if she is neither polite nor intelligent"?
Question 30 Report
Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6
Question 31 Report
A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.
A. 12/91 B. C. D.
Question 32 Report
If 2i +pj and 4i -2j are perpendicular, find the value of p.
Answer Details
To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.
Question 33 Report
In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\).
Question 34 Report
The table shows the distribution of marks obtained by some students in a test
Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 |
Frequency | 4 | 12 | 16 | 6 | 2 |
What is the upper class boundary of the upper quartile class?
Question 35 Report
For what range of values of x is x\(^2\) - 2x - 3 ≤ 0
Answer Details
To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.
Question 39 Report
P and Q are two linear transformations in the X-Y plane defined by
P: (x, y) → (-3x + 6y, 4x + y) and
Q: (x, y) → (2x-3y, -4x - 6y).
(a) Write down the matrices of P and Q. (b) What is the image of (-2,-3) under the transformation Q?
(c) Obtain a single transformation representing the transformation Q followed by P.
(d) Find the image of (1,4) when transformed by Q followed by P.
(e) Find the image P\(^1\) of the point (-√2,2√2) under an anticlockwise rotation of 225° about the origin.
Answer Details
(a) The matrix representation of a linear transformation can be obtained by writing the image of the standard basis vectors (1,0) and (0,1). The matrix of P is given by: P = [[-3, 6], [4, 1]] Similarly, the matrix of Q is given by: Q = [[2, -3], [-4, -6]] (b) To find the image of a point (-2,-3) under the transformation Q, we first write the point as a column vector: (-2,-3) = [-2, -3]^T Next, we multiply the matrix Q with the vector: Q * [-2, -3]^T = [2, -3] * [-2, -3]^T = [-12, 18]^T So the image of (-2,-3) under the transformation Q is (-12, 18). (c) To find the single transformation representing the transformation Q followed by P, we multiply the matrices Q and P: Q * P = [[2, -3], [-4, -6]] * [[-3, 6], [4, 1]] = [[36, -18], [-24, -30]] So the single transformation representing the transformation Q followed by P is given by the matrix [[36, -18], [-24, -30]]. (d) To find the image of (1,4) when transformed by Q followed by P, we first write the point as a column vector: (1, 4) = [1, 4]^T Next, we multiply the matrix representing Q followed by P with the vector: [[36, -18], [-24, -30]] * [1, 4]^T = [36, -18] * 1 + [-24, -30] * 4 = [36, -18] + [-96, -120] = [-60, -138]^T So the image of (1,4) when transformed by Q followed by P is (-60, -138). (e) To find the image P^1 of the point (-√2,√2) under an anticlockwise rotation of 225° about the origin, we can first rotate the point by 225° in the counterclockwise direction and then apply the transformation P. The counterclockwise rotation of 225° can be represented by the matrix: R = [[cos(225), -sin(225)], [sin(225), cos(225)]] = [[-√2/2, √2/2], [-√2/2, -√2/2]] Next, we multiply the matrix R with the vector representing the point (-√2,√2): R * [-√2, √2]^T = [[-√2/2, √2/2], [-√2/2, -√2/2]] * [-√2, √2]^T = [-√2/2 * -√2 + √2/2 * √2, -√2/2 * -√2 - √2/2 * √2]^T = [√2, -√2]^T So the image of the point (-√2,√2) under the counterclockwise rotation of 225° is (√2, -√2). Finally, to find
Question 40 Report
The polynomial f(x) =2x\(^3\) + px+ qx - 5 has (x-1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.
To find the values of p and q, we can use the Remainder Theorem and synthetic division. The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is equal to f(a).
So, in this case, when we divide f(x) by (x - 1), the remainder is 27, meaning that f(1) = 27. We can use this information to find one of the constants, p.
Next, we can use synthetic division to divide f(x) by (x + 2) and find the remainder. The process of synthetic division involves dividing each term of the polynomial by the leading coefficient of the divisor, and then using the coefficients to find the remainder.
In this case, the polynomial f(x) is divided by (x + 2), giving us:
2x^3 + px + qx - 5 ÷ x + 2
2 1 p/2 (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)
Since the remainder is 27, we have:
-2x^2 + (p - 4)x + (q - 5) = 27
Solving for p and q, we have:
p - 4 = 0 p = 4
q - 5 = 27 q = 32
So, the values of p and q are p = 4 and q = 32.
Answer Details
To find the values of p and q, we can use the Remainder Theorem and synthetic division. The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is equal to f(a).
So, in this case, when we divide f(x) by (x - 1), the remainder is 27, meaning that f(1) = 27. We can use this information to find one of the constants, p.
Next, we can use synthetic division to divide f(x) by (x + 2) and find the remainder. The process of synthetic division involves dividing each term of the polynomial by the leading coefficient of the divisor, and then using the coefficients to find the remainder.
In this case, the polynomial f(x) is divided by (x + 2), giving us:
2x^3 + px + qx - 5 ÷ x + 2
2 1 p/2 (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)
Since the remainder is 27, we have:
-2x^2 + (p - 4)x + (q - 5) = 27
Solving for p and q, we have:
p - 4 = 0 p = 4
q - 5 = 27 q = 32
So, the values of p and q are p = 4 and q = 32.
Question 41 Report
Given that x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors
x = (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$
Changing x and y to the form xi+yj
we have:
x= -4i + 3j and y = -9i - 15j
using:
cosØ = xy|x|ly| $\frac{\mathit{x}\mathit{y}}{|\mathit{x}|\mathit{l}\mathit{y}|}$
where Ø is the angle between x and y
xy = (-4i+3j) (-9i - 15j)
-36 + 60 x 0 - 27 x 0- 45 = -81
|x| = √(42 ${}^{2}$ +32 ${}^{2}$) =√(16 +9) = √25 = 5
|y| = √(-9)2 ${}^{2}$ + (-15)2 ${}^{2}$ = √(81 +225) = √306
cosØ =−815√306 $\frac{-81}{5\surd 306}$
= −9√306170 $\frac{-9\surd 306}{170}$
17.49∗9170 $\frac{17.49\ast 9}{170}$
Ø = cos−1 ${}^{-1}$(-0.1029);
Ø = 95.91°
Ø = 96°
Answer Details
x = (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$
Changing x and y to the form xi+yj
we have:
x= -4i + 3j and y = -9i - 15j
using:
cosØ = xy|x|ly| $\frac{\mathit{x}\mathit{y}}{|\mathit{x}|\mathit{l}\mathit{y}|}$
where Ø is the angle between x and y
xy = (-4i+3j) (-9i - 15j)
-36 + 60 x 0 - 27 x 0- 45 = -81
|x| = √(42 ${}^{2}$ +32 ${}^{2}$) =√(16 +9) = √25 = 5
|y| = √(-9)2 ${}^{2}$ + (-15)2 ${}^{2}$ = √(81 +225) = √306
cosØ =−815√306 $\frac{-81}{5\surd 306}$
= −9√306170 $\frac{-9\surd 306}{170}$
17.49∗9170 $\frac{17.49\ast 9}{170}$
Ø = cos−1 ${}^{-1}$(-0.1029);
Ø = 95.91°
Ø = 96°
Question 42 Report
Evaluate: \(^9{?}_1\) \(\frac{x(2x-3)}{?x}\) dx
9∫1 ${}^{}{}_{}$ x(2x−3)√x $\frac{}{}$ dx = 2x2−3x)x1/2 $\frac{{}^{}}{}$
= x −1/2 ${}^{}$ (2x2 ${}^{2}$ - 3x)
= 2x3/2 ${}^{}$ - 3x1/2 ${}^{}$
= 2x3/23/2+1 $\frac{{\mathrm{}}^{}}{}$ - 3x1/21/2+1 $\frac{{}^{}\mathrm{}}{}$
= 2x3/25/2 $\frac{{\mathrm{}}^{}}{}$ - 3x1/23/2 $\frac{{\mathrm{}}^{}}{}$
= 9∫1 ${}^{}{}_{}$ [4x3/25 $\frac{{\mathrm{}}^{}}{}$ - 6x1/23 $\frac{{\mathrm{}}^{}}{}$]
[ 4(9)3/25 $\frac{}{}$- 6(9)1/23 $\frac{{}^{}}{}$] - [ 4(9)1/25 $\frac{}{}$ - 6(1)1/23 $\frac{}{}$]
= [ 4(27)5 $\frac{}{}$- 6(3)3 $\frac{}{}$- [ 45 - 2 ]
= [ 1085- 2 ] - [ 65 ]
= 785 + 65= 845 $\frac{\mathrm{}}{}$
164/5
Answer Details
9∫1 ${}^{}{}_{}$ x(2x−3)√x $\frac{}{}$ dx = 2x2−3x)x1/2 $\frac{{}^{}}{}$
= x −1/2 ${}^{}$ (2x2 ${}^{2}$ - 3x)
= 2x3/2 ${}^{}$ - 3x1/2 ${}^{}$
= 2x3/23/2+1 $\frac{{\mathrm{}}^{}}{}$ - 3x1/21/2+1 $\frac{{}^{}\mathrm{}}{}$
= 2x3/25/2 $\frac{{\mathrm{}}^{}}{}$ - 3x1/23/2 $\frac{{\mathrm{}}^{}}{}$
= 9∫1 ${}^{}{}_{}$ [4x3/25 $\frac{{\mathrm{}}^{}}{}$ - 6x1/23 $\frac{{\mathrm{}}^{}}{}$]
[ 4(9)3/25 $\frac{}{}$- 6(9)1/23 $\frac{{}^{}}{}$] - [ 4(9)1/25 $\frac{}{}$ - 6(1)1/23 $\frac{}{}$]
= [ 4(27)5 $\frac{}{}$- 6(3)3 $\frac{}{}$- [ 45 - 2 ]
= [ 1085- 2 ] - [ 65 ]
= 785 + 65= 845 $\frac{\mathrm{}}{}$
164/5
Question 43 Report
The table shows the frequency distribution of heights (in cm) of pupils in a certain school.
Heights |
100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
Frequency |
27 | 58 | 130 | 105 | 50 | 25 | 5 |
(a) (i) Construct a cumulative frequency table. (ii) Use the table to draw a cumulative frequency curve.
(b) Using the curve, estimate the: (i)median height; (ii) inter quartile range (iii) percentage of students whose heights are most 130cm.
(a) (i) To construct a cumulative frequency table, we need to add up the frequencies for each height interval and all the intervals before it. The cumulative frequency for each interval is shown in the table below:
Heights |
100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
Frequency |
27 | 58 | 130 | 105 | 50 | 25 | 5 |
Cumulative Frequency |
27 | 85 | 215 | 320 | 370 | 395 | 400 |
(a) (ii) To draw a cumulative frequency curve, we plot the cumulative frequency for each interval against the upper limit of the interval. We then join the points with straight lines to form the curve, as shown below:
(b) (i) To estimate the median height, we locate the value on the cumulative frequency curve that corresponds to the 50th percentile. This is the point where the curve intersects the line at y = 0.5. We draw a line down to the x-axis to find the corresponding height. From the graph, we can see that the median height is around 124-125cm.
(b) (ii) To estimate the interquartile range (IQR), we need to find the heights corresponding to the 25th and 75th percentiles on the cumulative frequency curve. We draw lines down from these points to the x-axis to find the corresponding heights. The IQR is then the difference between these heights. From the graph, we can estimate the 25th percentile at around 116-117cm and the 75th percentile at around 134-135cm. Therefore, the IQR is around 18-19cm.
(b) (iii) To estimate the percentage of students whose heights are most 130cm, we locate the point on the cumulative frequency curve that corresponds to 130cm. We then draw a line across to the y-axis to find the corresponding percentile. From the graph, we can see that the percentage of students whose heights are most 130cm is around 60-65%.
Answer Details
(a) (i) To construct a cumulative frequency table, we need to add up the frequencies for each height interval and all the intervals before it. The cumulative frequency for each interval is shown in the table below:
Heights |
100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
Frequency |
27 | 58 | 130 | 105 | 50 | 25 | 5 |
Cumulative Frequency |
27 | 85 | 215 | 320 | 370 | 395 | 400 |
(a) (ii) To draw a cumulative frequency curve, we plot the cumulative frequency for each interval against the upper limit of the interval. We then join the points with straight lines to form the curve, as shown below:
(b) (i) To estimate the median height, we locate the value on the cumulative frequency curve that corresponds to the 50th percentile. This is the point where the curve intersects the line at y = 0.5. We draw a line down to the x-axis to find the corresponding height. From the graph, we can see that the median height is around 124-125cm.
(b) (ii) To estimate the interquartile range (IQR), we need to find the heights corresponding to the 25th and 75th percentiles on the cumulative frequency curve. We draw lines down from these points to the x-axis to find the corresponding heights. The IQR is then the difference between these heights. From the graph, we can estimate the 25th percentile at around 116-117cm and the 75th percentile at around 134-135cm. Therefore, the IQR is around 18-19cm.
(b) (iii) To estimate the percentage of students whose heights are most 130cm, we locate the point on the cumulative frequency curve that corresponds to 130cm. We then draw a line across to the y-axis to find the corresponding percentile. From the graph, we can see that the percentage of students whose heights are most 130cm is around 60-65%.
Question 44 Report
(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres.
(i) How many days will it take the jogger to reach a distance of 15km in training?
(ii) Calculate the total distance he would have run in the training.
(b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.
(a)
(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.
Let's call the number of days it takes to reach this distance "n."
On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.
In general, on the nth day, he runs 500 + 250(n-1) meters.
So we can set up an equation:
500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000
Simplifying, we get:
250n^2 + 250n - 15000 = 0
Dividing both sides by 250:
n^2 + n - 60 = 0
This equation can be factored as:
(n + 6)(n - 10) = 0
Since we're looking for a positive value for n, the answer is n = 10.
So it will take the jogger 10 days to reach a distance of 15km in training.
(ii) We can use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.
We also know that the sum to infinity is 25/2, so S = 25/2.
Plugging in these values and solving for r:
25/2 = (-1500)(1 - r^10)/(1 - r)
r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)
So the common ratio is 1/2.
The total distance the jogger would have run in the training is the sum of the terms of the geometric series:
S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.
(b)
We know that the second term of the GP is -3, so we can write the first two terms as:
a, ar
where ar = -3.
We also know that the sum to infinity is 25/2, so we can use the formula:
S = a/(1 - r)
25/2 = a/(1 - r)
a = 25/2 - 25r/2
Substituting this value of a into the equation ar = -3:
(25/2 - 25r/2)r = -3
Simplifying:
25r^2 - 50r + 6 = 0
We can solve for r using the quadratic formula:
r = (50
Answer Details
(a)
(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.
Let's call the number of days it takes to reach this distance "n."
On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.
In general, on the nth day, he runs 500 + 250(n-1) meters.
So we can set up an equation:
500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000
Simplifying, we get:
250n^2 + 250n - 15000 = 0
Dividing both sides by 250:
n^2 + n - 60 = 0
This equation can be factored as:
(n + 6)(n - 10) = 0
Since we're looking for a positive value for n, the answer is n = 10.
So it will take the jogger 10 days to reach a distance of 15km in training.
(ii) We can use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.
We also know that the sum to infinity is 25/2, so S = 25/2.
Plugging in these values and solving for r:
25/2 = (-1500)(1 - r^10)/(1 - r)
r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)
So the common ratio is 1/2.
The total distance the jogger would have run in the training is the sum of the terms of the geometric series:
S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.
(b)
We know that the second term of the GP is -3, so we can write the first two terms as:
a, ar
where ar = -3.
We also know that the sum to infinity is 25/2, so we can use the formula:
S = a/(1 - r)
25/2 = a/(1 - r)
a = 25/2 - 25r/2
Substituting this value of a into the equation ar = -3:
(25/2 - 25r/2)r = -3
Simplifying:
25r^2 - 50r + 6 = 0
We can solve for r using the quadratic formula:
r = (50
Question 45 Report
A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.
Calculate the probability of picking:
(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.
Total bulbs:
n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs)
= p(RR) Or p(BB) or p(GG)
516
$\frac{\mathrm{\mathrm{}}}{}$ * 415
$\frac{\mathrm{\mathrm{}}}{}$ + 716
$\frac{\mathrm{\mathrm{}}}{}$ * 615
$\frac{\mathrm{\mathrm{}}}{}$ + 416
$\frac{\mathrm{\mathrm{}}}{}$ * 315
= 112 $\frac{\mathrm{}}{}$ + 740