WAEC SSCE - Further Mathematics - 2020

Akọwa Nkọwa

Akọwa Nkọwa

The value of p can be found by solving the definite integral for a given value of p and equating it to the given value of 129. The definite integral of the function px^2 + 16 between the limits 0 and 3 can be calculated using antiderivatives. The antiderivative of px^2 is (1/3)px^3 and the antiderivative of 16 is 16x. So, ∫^3_0(px^2 + 16)dx = (1/3)p * 3^3 + 16 * 3 - (1/3)p * 0^3 - 16 * 0 = (1/3)p * 27 + 48 Now, equating this expression to 129 and solving for p, we get: (1/3)p * 27 + 48 = 129 (1/3)p * 27 = 81 p * 27 = 243 p = 243/27 p = 9 So, the value of p is 9.

Akọwa Nkọwa

Akọwa Nkọwa

The function f(x) = 2x - x^2 is increasing on an interval if, as x increases, f(x) also increases. To determine this interval, we can first find the critical points of the function, which are the values of x where the derivative of the function is equal to 0 or undefined. The derivative of f(x) = 2x - x^2 is f'(x) = 2 - 2x. Setting f'(x) = 0, we get 2 - 2x = 0, so x = 1. This is the critical point of the function. To determine whether the function is increasing or decreasing at the critical point, we can use the second derivative test. The second derivative of the function f(x) = 2x - x^2 is f''(x) = -2, which is always negative. So, the first derivative (f'(x)) is decreasing at the critical point. Since the first derivative is decreasing at x = 1, the function f(x) = 2x - x^2 is decreasing to the left of x = 1 and increasing to the right of x = 1. So, the function is increasing on the interval x > 1.

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Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To find the median of a set of numbers, we need to arrange the numbers in order from least to greatest. Arranging the given set of numbers in order, we get: 2, 2, 5, 7, 9, 9, 9, 10, 10, 12, 18 There are 11 numbers in the set. Since 11 is an odd number, the median is the middle number when the set is arranged in order. The middle number is the sixth number, which is 9. Therefore, the median of the set of numbers 9, 7, 5, 2, 12, 9, 9, 2, 10, 10, and 18 is 9.

Akọwa Nkọwa

Let's first recall the formula for the sum of the roots of a quadratic equation in the form ax\(^2\) + bx + c = 0: Sum of roots = -b/a In the given equation, 2x\(^2\) + 5mx + n = 0, the coefficient of x\(^2\) is 2, which means a = 2. Therefore, the sum of the roots can be expressed as: Sum of roots = - (5m) / 2 We are given that the sum of the roots is 5, so we can set up an equation: 5 = - (5m) / 2 Multiplying both sides by -2, we get: -10 = 5m Dividing both sides by 5, we obtain: m = -2 Therefore, the value of m is -2. So the answer is -2.0.

Akọwa Nkọwa

Akọwa Nkọwa

To find the area enclosed by the line 3x - 5y + 4 = 0 and the axes, we need to find the points where the line intersects with the x-axis and y-axis. First, let's find the x-intercept. To do this, we set y = 0 and solve for x: 3x - 5(0) + 4 = 0 3x + 4 = 0 3x = -4 x = -4/3 So the x-intercept is (-4/3, 0). Next, let's find the y-intercept. To do this, we set x = 0 and solve for y: 3(0) - 5y + 4 = 0 -5y + 4 = 0 -5y = -4 y = 4/5 So the y-intercept is (0, 4/5). Now we can draw a triangle with vertices at the origin (0,0), the x-intercept (-4/3, 0), and the y-intercept (0, 4/5). The base of the triangle is the x-intercept, which has a length of 4/3. The height of the triangle is the y-intercept, which has a length of 4/5. Therefore, the area of the triangle is: (1/2) * base * height = (1/2) * (4/3) * (4/5) = 8/15 = 0.53 (rounded to two decimal places) Therefore, the answer is option C: 0.53 square units.

Akọwa Nkọwa

Akọwa Nkọwa

When a force is acting at an angle to the horizontal, we can break it down into two components: the horizontal component and the vertical component. The horizontal component of the force is the part of the force that is acting in the horizontal direction, while the vertical component is the part of the force that is acting in the vertical direction. To find the horizontal component of the force of 230N acting at an angle of 65\(^o\), we use the formula: Horizontal component = Force × cos(angle) where angle is the angle between the force and the horizontal direction. Substituting the values given in the problem, we have: Horizontal component = 230N × cos(65\(^o\)) Using a calculator, we can find that cos(65\(^o\)) is approximately 0.4226. Therefore, the horizontal component of the force is: Horizontal component = 230N × 0.4226 ≈ 97.2N Therefore, the answer is approximately 97.2N.

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The given equation represents the distance covered by a particle in motion at any time t seconds. We are asked to find the distance covered by the particle before it comes to rest, which means the particle stops moving. To find the distance covered by the particle before coming to rest, we need to find the time when the particle stops moving. The particle stops moving when its velocity becomes zero. We can find the velocity of the particle at any time t seconds by differentiating the given equation with respect to time: v = dS/dt = 120 - 32t The particle stops moving when v = 0, so we can set the velocity equation to zero and solve for t: 120 - 32t = 0 t = 3.75 seconds Now we know that the particle stops moving after 3.75 seconds. To find the distance covered by the particle before it stops moving, we can substitute this value of t in the original equation: S = 120t - 16t^2 S = 120(3.75) - 16(3.75)^2 S = 225 meters Therefore, the distance covered by the particle before coming to rest is 225 meters. Option (D) is the correct answer.

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To find the angle between the vectors i + 5j and 5i - j, we can use the dot product formula: a · b = |a||b| cos θ where a and b are the two vectors, |a| and |b| are their magnitudes, and θ is the angle between them. First, we need to find the dot product of the two vectors: (i + 5j) · (5i - j) = (1)(5) + (5)(-1) = 0 Next, we need to find the magnitudes of the two vectors: | i + 5j | = √(1² + 5²) = √26 | 5i - j | = √(5² + (-1)²) = √26 Now we can substitute the dot product and magnitudes into the dot product formula to solve for θ: 0 = (√26)(√26) cos θ cos θ = 0 θ = 90° Therefore, the angle between i + 5j and 5i - j is 90 degrees. Option (D) is the correct answer.

Akọwa Nkọwa

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To solve for p and q, we can equate the corresponding elements of both matrices: p + q = 2 1 = 1 p - q = 8 From the first and third equations, we can obtain: p = 5 q = -3 Therefore, the correct answer is p = 5, q = -3.

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Akọwa Nkọwa

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To calculate the probability that the product of two numbers selected at random with replacement from the set {-5,-2,4,8} is positive, we need to find the total number of ways to select two numbers from the set and the number of ways to select two numbers whose product is positive. There are four numbers in the set, so there are 4 x 4 = 16 possible ways to select two numbers with replacement. For example, we could select -5 and -2, or we could select 4 and 8. To find the number of ways to select two numbers whose product is positive, we need to consider the following cases: 1. Both numbers are positive: There are two positive numbers in the set, so there are 2 x 2 = 4 ways to select two positive numbers. 2. Both numbers are negative: There are two negative numbers in the set, so there are 2 x 2 = 4 ways to select two negative numbers. Therefore, there are a total of 4 + 4 = 8 ways to select two numbers whose product is positive. The probability of selecting two numbers whose product is positive is the number of ways to select two numbers whose product is positive divided by the total number of ways to select two numbers, which is 8/16 or 1/2. Therefore, the answer is, which is \(\frac{1}{2}\).

Akọwa Nkọwa

The values of x between 0° and 360° that satisfy cos x = -0.7133 are 135.5° and 224.5°. The cosine function has a range of values between -1 and 1, and for a given value of cosine, there are two angles in the interval [0°, 360°] that satisfy the equation. These angles are found by using the inverse cosine function, also known as arccos. The arccos of -0.7133 is equal to 135.5° and 224.5°, which are the two values of x that satisfy the equation cos x = -0.7133 in the interval [0°, 360°].

Akọwa Nkọwa

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To find the inverse of a matrix, we need to use the formula: \[A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\] where \(\det(A)\) is the determinant of matrix \(A\) and \(\text{adj}(A)\) is the adjugate of matrix \(A\). In this case, we have: \[A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\] The determinant of \(A\) is: \[\det(A) = \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} = (3\times2) - (5\times1) = 1\] The adjugate of \(A\) is: \[\text{adj}(A) = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\] Therefore, the inverse of \(A\) is: \[A^{-1} = \frac{1}{\det(A)}\text{adj}(A) = \frac{1}{1}\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\] So the answer is (B) \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\).

Akọwa Nkọwa

A linear sequence is a sequence in which each term differs from the previous term by a constant amount. In other words, if we subtract any term from the term that comes after it, we get the same value. To find the nth term of the linear sequence (5y + 1), (2y + 1), (1 - y), ..., we need to first find the common difference between consecutive terms. We can do this by subtracting the second term from the first term, and then subtracting the third term from the second term: (2y + 1) - (5y + 1) = -3y (1 - y) - (2y + 1) = -3y - 1 Since both subtractions give us the same value of -3y, we know that the common difference between consecutive terms is -3y. Now, we need to find the formula for the nth term of the sequence. We can use the general formula for an arithmetic progression: nth term = a\(_1\) + (n - 1)d where a\(_1\) is the first term, d is the common difference, and n is the term we want to find. In our case, the first term is (5y + 1), the common difference is -3y, and we want to find the nth term. Therefore, we have: nth term = (5y + 1) + (n - 1)(-3y) Simplifying this expression, we get: nth term = 5y + 1 - 3ny + 3y nth term = (8 - 3n)y + 1 Therefore, the nth term of the linear sequence (5y + 1), (2y + 1), (1 - y), ... is (8 - 3n)y + 1.

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The given equation is V = plog\(_x\), (M + N). We need to express N in terms of X, P, M, and V. We can start by isolating the term containing N: V = plog\(_x\), (M + N) V = p(log\(_x\) M + log\(_x\) N) V - plog\(_x\) M = plog\(_x\) N log\(_x\) N = (V - plog\(_x\) M) / p Now we can solve for N by exponentiating both sides with base x: N = x\(^{\frac{V - plog_x M}{p}}\) Therefore, the expression for N in terms of X, P, M, and V is N = x\(^{\frac{V - plog_x M}{p}}\). Option (A) is the correct answer.

Akọwa Nkọwa

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The question presents a table showing how 10 candidates' essays were ranked by three examiners (I, II, and III). The question requires finding the Spearman's rank correlation coefficient for each pair of examiners to determine which examiners agree most.

(a) The Spearman's rank correlation coefficient measures the strength and direction of the relationship between two variables. It is given by the formula:

r = 1 - (6Σd²)/(n(n²-1))

where d is the difference between the ranks of the two examiners for each candidate, and n is the number of candidates.

(i) To find the correlation coefficient between examiners I and II, we calculate the differences between their ranks for each candidate and square them, then sum the squared differences. Using the formula above, we get r = 0.23, which indicates a weak positive correlation.

(ii) To find the correlation coefficient between examiners I and III, we repeat the same process as above and obtain r = -0.02, which indicates no correlation or a very weak negative correlation.

(iii) To find the correlation coefficient between examiners II and III, we repeat the same process as above and obtain r = 0.23, which indicates a weak positive correlation.

(b) Comparing the correlation coefficients, we can conclude that examiners I and II agree most since they have the highest correlation coefficient of 0.23. However, this correlation is still considered weak.

Akọwa Nkọwa

To find the values of Q and R, we need to first simplify the right-hand side of the equation and then equate the numerators of both sides.

First, we need to find the common denominator of the fractions on the right-hand side:

P + \(\frac{Q}{x - 1}\) + \(\frac{R}{x + 1}\) = \(\frac{Px^2 - P + Q(x + 1) + R(x - 1)}{(x - 1)(x + 1)}\)

Next, we can equate the numerators of both sides:

3x^2 + 3x - 2 = Px^2 - P + Q(x + 1) + R(x - 1)

Now, we need to solve for Q and R by choosing appropriate values of x.

To find Q, we can choose x = 1:

3(1)^2 + 3(1) - 2 = P(1)^2 - P + Q(1 + 1) + R(1 - 1)

4 = P + 2Q

To find R, we can choose x = -1:

3(-1)^2 + 3(-1) - 2 = P(-1)^2 - P + Q(-1 + 1) + R(-1 - 1)

-2 = P - 2R

We now have two equations with two variables (P and Q), and we can solve for Q by substituting the expression we found for P in the previous step:

4 = (3 + 3 - 2P) + 2Q

2Q = 2P + 1

Q = P + \(\frac{1}{2}\)

To find R, we can substitute the expression we found for P and the expression we found for Q in the equation we found for R:

-2 = (3 + 3 - 2P) - 2(P + \(\frac{1}{2}\))

-2 = 6 - 4P

P = \(\frac{4}{3}\)

Substituting this value of P in the expression we found for Q, we get:

Q = \(\frac{4}{3}\) + \(\frac{1}{2}\) = \(\frac{11}{6}\)

Finally, substituting the values of P and Q in the expression we found for R, we get:

-2 = 6 - 4(\(\frac{4}{3}\)) - 2(\(\frac{11}{6}\))

R = -\(\frac{5}{6}\)

Therefore, the value of Q is \(\frac{11}{6}\) and the value of R is -\(\frac{5}{6}\).

Akọwa Nkọwa

(a) \(y = x^2(1 + x)^{3/2}\). Apply the product rule:

Factor out \((1 + x)^{1/2}\):

(b) Let the centre be \((h, k)\) with \(2k - h = 3\), so \(h = 2k - 3\). The centre is equidistant from \(P(2,3)\) and \(Q(6,7)\):

Expanding and simplifying gives \(8h + 8k = 72\), i.e. \(h + k = 9\). With \(h = 2k - 3\):

Centre \((5, 4)\); radius squared \(= (5 - 2)^2 + (4 - 3)^2 = 10\).

Akọwa Nkọwa

(a) Probability of drawing one red and one green ball:

To find the probability that one ball is red and the other is green, we can use the formula:

P(red and green) = P(red) * P(green)

where P(red) is the probability of drawing a red ball on the first draw, and P(green) is the probability of drawing a green ball on the second draw, given that the first ball was red.

The probability of drawing a red ball on the first draw is 10/18, since there are 10 red balls out of a total of 18 balls in the bag. The probability of drawing a green ball on the second draw, given that the first ball was red, is 8/17, since there are now 8 green balls and 17 balls left in the bag.

Therefore, the probability of drawing one red and one green ball is:

P(red and green) = (10/18) * (8/17) = 0.2353 or approximately 23.53%

(b) Probability of between two and five defective bulbs:

To find the probability that between two and five bulbs are defective, we can use the binomial probability formula:

P(k successes) = (n choose k) * p^k * (1-p)^(n-k)

where n is the number of trials (bulbs selected), k is the number of successes (defective bulbs), p is the probability of success (defective bulb), and (n choose k) is the number of ways to choose k defective bulbs out of n total bulbs.

In this case, n = 12, p = 0.2 (since 20% of the bulbs are defective), and we want to find the probability of between two and five successes (defective bulbs). We can calculate this probability by adding the probabilities of getting exactly 2, 3, 4, or 5 defective bulbs:

P(2-5 defects) = P(2 defects) + P(3 defects) + P(4 defects) + P(5 defects)

P(2 defects) = (12 choose 2) * 0.2^2 * 0.8^10 = 0.301

P(3 defects) = (12 choose 3) * 0.2^3 * 0.8^9 = 0.318

P(4 defects) = (12 choose 4) * 0.2^4 * 0.8^8 = 0.185

P(5 defects) = (12 choose 5) * 0.2^5 * 0.8^7 = 0.058

Therefore, the probability of between two and five bulbs being defective is:

P(2-5 defects) = 0.301 + 0.318 + 0.185 + 0.058 = 0.862 or approximately 86.2%

Akọwa Nkọwa

(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):

(b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).

(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).

(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).

The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).

Akọwa Nkọwa

To find the magnitude of the resultant force, we need to first calculate the x and y components of each force.

For F1:

• The magnitude is 10 N
• The angle is 90 degrees
• The x component is 10 cos(90) = 0 N
• The y component is 10 sin(90) = 10 N

For F2:

• The magnitude is 20 N
• The angle is 210 degrees
• The x component is 20 cos(210) = -17.3 N (negative because it points to the left)
• The y component is 20 sin(210) = -10 N (negative because it points downwards)

For F3:

• The magnitude is 4 N
• The angle is 330 degrees
• The x component is 4 cos(330) = 3.5 N (positive because it points to the right)
• The y component is 4 sin(330) = -2 N (negative because it points downwards)

To find the total x and y components, we can add up the individual x and y components:

• The total x component is 0 - 17.3 + 3.5 = -13.8 N
• The total y component is 10 - 10 - 2 = -2 N

To find the magnitude of the resultant force, we can use the Pythagorean theorem:

resultant force = sqrt((-13.8)^2 + (-2)^2) = 13.9 N

Therefore, the magnitude of the resultant force is 13.9 N.

Akọwa Nkọwa

To find the vector "z" in the direction of the sum of vectors "w", "x", and "-2y", we can first find the sum of "w", "x", and "-2y", then find the unit vector in that direction and multiply it by the magnitude of "z".

First, let's find the sum of "w", "x", and "-2y":

w = 8i + 3j
x = 6i - 5j
y = 2i + 3j

-2y = -4i - 6j

w + x - 2y = (8i + 3j) + (6i - 5j) - (-4i - 6j) = 18i - 8j

Next, let's find the unit vector in the direction of the sum:

The unit vector in the direction of a vector can be found by dividing the vector by its magnitude.

The magnitude of the sum is:
√(18^2 + (-8)^2) = 20

So the unit vector in the direction of the sum is:
(18i - 8j) / 20 = 9/10i - 4/10j

Finally, let's multiply the unit vector by the magnitude of "z" to find "z":

z = 41 * (9/10i - 4/10j) = 36.9i - 16.4j

So the vector "z" in the direction of the sum of "w", "x", and "-2y" is approximately 36.9i - 16.4j.

Akọwa Nkọwa

(a) To find the angle between the string and the horizontal, we can use the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C

where c is the length of the string (4.2m), a is the distance from Y to the body (2.4m), and b is the distance from X to Y (3m).

We can rearrange the equation to solve for cos C:

cos C = (a^2 + b^2 - c^2) / 2ab

cos C = (2.4^2 + 3^2 - 4.2^2) / 2 * 2.4 * 3 = 0.8

Finally, we can find the angle using the inverse cosine function:

C = arccos 0.8 = 37.38°

So the angle between the string and the horizontal is 37.38°.

(b) To find the magnitude of the force P, we can use Newton's second law (F = ma), where F is the net force, m is the mass of the body (800g = 0.8kg), and a is the acceleration due to gravity (9.8m/s^2).

The net force is equal to the mass of the body multiplied by the acceleration due to gravity, so:

F = ma = 0.8 * 9.8 = 7.84N

So the magnitude of the force P is 7.84N.

(c) To find the tension T in the string, we can use the fact that the tension must be equal in both parts of the string to keep the system in equilibrium.

Let's call the tension in the part of the string from Y to O "T1" and the tension in the part of the string from O to X "T2".

T1 * cos 37.38° = T2 * cos (180° - 37.38°) = T2 * cos 142.62°

T1 = T2 * cos 142.62° / cos 37.38°

Since T1 = T2, we can substitute T2 for T1:

T2 = T2 * cos 142.62° / cos 37.38°

So, T2 / T2 = cos 142.62° / cos 37.38°

Since T2 is not equal to 0, we can divide both sides by T2:

1 = cos 142.62° / cos 37.38°

To find the tension T in the string, we can use either T1 or T2, since they are equal. Let's use T1:

T1 = T2 = 7.84 / (cos 142.62° / cos 37.38°) = 7.84 / 0.19 = 41.16 N

So the tension in the string is 41.16 N.

Akọwa Nkọwa

(a) If (x + 2) is a factor of g(x), it means that (x + 2) is a root of the polynomial g(x). To find the zeros of g(x), we need to factorize the polynomial by using long division or synthetic division.

Using synthetic division, we get:

-2 | 2 11 -1 -30
   |-----------
   | 2 -6 11 | 0

Therefore, g(x) can be factorized as g(x) = 2x^2 - 6x + 11. To find the zeros of g(x), we need to solve the equation 2x^2 - 6x + 11 = 0.

Using the quadratic formula, we get:

x = \(\frac{6 \pm \sqrt{6^2 - 4(2)(11)}}{4}\)

x = \(\frac{3}{2} \pm \frac{\sqrt{23}}{2}\)

Therefore, the zeros of g(x) are \(\frac{3}{2} + \frac{\sqrt{23}}{2}\), and \(\frac{3}{2} - \frac{\sqrt{23}}{2}\).

(b) To solve the simultaneous equations 3(2^x) + 3^(y - 2) = 25 and 2x - 3^(y + 1) = -19, we need to isolate one variable in terms of the other and substitute the expression into the other equation.

From the second equation, we have:

2x = 3^(y + 1) - 19

Substituting this expression into the first equation, we get:

3(2^x) + 3^(y - 2) = 25

3(2^x) + 3^(y - 2) = 3^2

Dividing both sides by 3^2, we get:

2^x + 3^(y - 2 - 2) = 1

2^x + 3^(y - 4) = 1

Substituting the expression for 2x from the second equation, we get:

2^x + 3^(y - 4) = 1 + 3^(y + 1) - 19

2^x + 3^(y - 4) = 3^(y + 1) - 18

Rearranging this equation, we get:

2^x = 3^(y + 1) - 3^(y - 4) - 18

Using logarithms, we can solve for x:

x = log₂(3^(y + 1) - 3^(y - 4) - 18)

Substituting this expression for x into the second equation, we get:

2log₂(3^(y + 1) - 3^(y - 4) - 18) - 3^(y + 1) = -19

Simplifying this

Akọwa Nkọwa

(a) Total members \(= 6 + 8 = 14\). Ways to choose 4 \(= \binom{14}{4} = 1001\).

Use the complement of "fewer than 2 farmers":

(b) With \(P(X) = \tfrac{1}{5},\ P(Y) = \tfrac{1}{12},\ P(Z) = \tfrac{1}{6}\):

(i) None:

(ii) All:

(iii) At least one:

Akọwa Nkọwa

(a) To find the inverse of -1 under *, we need to find a real number x such that -1 * x = x * (-1) = 0, where 0 is the identity element under *.

Substituting -1 and x into the given formula, we get:

-1 * x = -1 + x - \(\frac{(-1)x}{2}\) = 0

Simplifying this equation, we get:

x - \(\frac{x}{2}\) = 1

\(\frac{x}{2}\) = 1

x = 2

Therefore, the inverse of -1 under * is 2.

(b) To find the truth set of m * 7 = m * 5, we need to find all real numbers m that satisfy the equation.

Substituting m and 7 into the given formula, we get:

m * 7 = m + 7 - \(\frac{7m}{2}\)

Substituting m and 5 into the given formula, we get:

m * 5 = m + 5 - \(\frac{5m}{2}\)

Setting these two expressions equal to each other, we get:

m + 7 - \(\frac{7m}{2}\) = m + 5 - \(\frac{5m}{2}\)

Simplifying this equation, we get:

\(\frac{m}{2}\) = 2

m = 4

Therefore, the truth set of m * 7 = m * 5 is {4}, which means that the only real number that satisfies the equation is 4.

Akọwa Nkọwa

(a) Evaluate each logarithm to base 2:

(b) Let the A.P. have first term \(a\) and common difference \(d\). The 1st, 3rd and 7th terms are \(a,\ a + 2d,\ a + 6d\). These are consecutive G.P. terms, so:

Since \(d \neq 0\), \(a = 2d\). The sum of the first two A.P. terms is 6:

Substituting \(a = 2d\): \(4d + d = 6 \Rightarrow d = \tfrac{6}{5} = 1.2\), and \(a = 2d = \tfrac{12}{5} = 2.4\).

(i) First term \(= 2.4\). (ii) Common difference \(= 1.2\).

Akọwa Nkọwa

The classes have equal width (10). Convert to continuous boundaries by subtracting/adding 0.5.

(a) Histogram. Draw contiguous rectangles of heights \(2,2,2,8,13,11,12,10,4\) over the boundaries above (frequency on the vertical axis). Total \(=64\).

(b) Modal score. The modal class is \(50-59\) (highest frequency 13, boundaries \(49.5-59.5\)), with \(d_1=13-8=5\) and \(d_2=13-11=2\):\[\text{Mode}=L+\frac{d_1}{d_1+d_2}c=49.5+\frac{5}{7}\times 10=49.5+7.14=56.6\]On the histogram this is found by joining the top corners of the modal bar to those of the adjacent bars and reading down to the axis.

The estimated modal score is approximately 56.6.

Akọwa Nkọwa

Integrate \( (3x - 2)^5 \) by reversing the chain rule (raise the power, divide by the new power and by the derivative of the bracket, \(3\)):

\[ \int (3x - 2)^5\,dx = \frac{(3x - 2)^6}{6 \times 3} = \frac{(3x - 2)^6}{18}. \]

Evaluate between \(1\) and \(3\):

\[ \left[\frac{(3x - 2)^6}{18}\right]_1^3 = \frac{(7)^6}{18} - \frac{(1)^6}{18} = \frac{117649 - 1}{18} = \frac{117648}{18} = 6536. \]

The value of the integral is \( 6536 \).

Akọwa Nkọwa

(a) To find QOP: we need to apply the functions in the correct order. QOP means we apply Q first, then O, then P.

Q: y → y - 2, so QOP: y → 2(y - 2) + 3 = 2y - 1. Therefore, QOP: y → 2y - 1.

(b) To form a four-digit odd number greater than 4000: the leftmost digit must be 7. We can choose any of the remaining digits for the thousands place, giving us 4 choices. For the hundreds, tens, and ones places, we can choose any of the five digits (1, 3, 7, 8, 2) with repetition allowed, giving us 5 choices for each of those places.

Therefore, the total number of four-digit odd numbers greater than 4000 that can be formed with these digits is:

4 (choices for the thousands place) × 5 (choices for the hundreds place) × 5 (choices for the tens place) × 5 (choices for the ones place) = 500.

So, there are 500 four-digit odd numbers greater than 4000 that can be formed from 1, 7, 3, 8, 2 if repetition is allowed.