WAEC SSCE - Further Mathematics - 2021

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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.

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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.

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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.

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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).

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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.

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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.

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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.

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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.

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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.

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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) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

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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.

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Reading the expression as \(\left(p+\tfrac12\sqrt3\right)(1-\sqrt3)^{2}=3-\sqrt3\).

First expand the square:

\[(1-\sqrt3)^{2}=1-2\sqrt3+3=4-2\sqrt3\]

So:

\[\left(p+\tfrac{\sqrt3}{2}\right)(4-2\sqrt3)=3-\sqrt3\]

Note \(4-2\sqrt3=2(2-\sqrt3)\). Divide both sides:

\[p+\frac{\sqrt3}{2}=\frac{3-\sqrt3}{4-2\sqrt3}\]

Rationalise the right side by \(\times\dfrac{4+2\sqrt3}{4+2\sqrt3}\): denominator \(16-12=4\); numerator \((3-\sqrt3)(4+2\sqrt3)=12+2\sqrt3-6=6+2\sqrt3\). So the right side \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).

\[p=\frac{3+\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{3}{2}\]

\[\boxed{p=\tfrac32}\]

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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 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 716 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 615 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 416 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 315

= 112 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}{}$ + 740 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ +  120 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

=  10+21+6120 $\frac{<br>\mathrm{<br>}}{}$ =  37120 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

 (b) All different colors = 1-p (AIl the same colour) =

1 - 37120 $\frac{37}{120}$ = 83120 $\frac{83}{120}$

(c) p(At least one red)  = p(RR) + p(RB) + p(RG)

516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 715 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}{}$ + 516 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ * 415 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

= 112 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ + 748 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ +  112 $\frac{\mathrm{<br>}}{}$

8+748 $\frac{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}{\mathrm{<br>}}$ = 1548 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

= 516

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(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

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Write the position vectors as coordinates: \(P(4,-5)\), \(Q(1,3)\), \(R(-5,2)\).

(a) Coordinates of M

In the parallelogram \(PQRM\) the vertices are taken in order \(P\to Q\to R\to M\), so the diagonals \(PR\) and \(QM\) bisect each other. Equating their midpoints:

\[\text{mid}(PR)=\left(\tfrac{4+(-5)}{2},\tfrac{-5+2}{2}\right)=\left(-\tfrac{1}{2},-\tfrac{3}{2}\right).\]

\[\text{mid}(QM)=\left(\tfrac{1+m_1}{2},\tfrac{3+m_2}{2}\right).\]

So \(1+m_1=-1\Rightarrow m_1=-2\) and \(3+m_2=-3\Rightarrow m_2=-6\).

\[\boxed{M(-2,-6)}.\]

(b) Acute angle between \(\overline{PM}\) and \(\overline{PQ}\)

\[\overline{PM}=M-P=(-2-4,\,-6-(-5))=(-6,-1),\]

\[\overline{PQ}=Q-P=(1-4,\,3-(-5))=(-3,8).\]

Using the scalar (dot) product,

\[\overline{PM}\cdot\overline{PQ}=(-6)(-3)+(-1)(8)=18-8=10,\]

\[|\overline{PM}|=\sqrt{(-6)^2+(-1)^2}=\sqrt{37},\qquad |\overline{PQ}|=\sqrt{(-3)^2+8^2}=\sqrt{73}.\]

\[\cos\theta=\frac{10}{\sqrt{37}\,\sqrt{73}}=\frac{10}{\sqrt{2701}}\approx 0.1924.\]

\[\theta=\cos^{-1}(0.1924)\approx 78.9^{\circ}\approx 79^{\circ}.\]

Since this value is already acute, the acute angle is \(79^{\circ}\).

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Reading the polynomial as \(f(x)=2x^{3}+px^{2}+qx-5\) (the two linear terms in the printed stem are a typographical slip for a quadratic and a linear term).

Since \((x-1)\) is a factor, \(f(1)=0\):

\[2+p+q-5=0\Rightarrow p+q=3\quad(\text{i})\]

The remainder is \(27\) when divided by \((x+2)\), so \(f(-2)=27\):

\[2(-8)+p(4)+q(-2)-5=27\Rightarrow -16+4p-2q-5=27\]

\[4p-2q=48\Rightarrow 2p-q=24\quad(\text{ii})\]

Add (i) and (ii): \(3p=27\Rightarrow p=9\). Then \(q=3-9=-6\).

\[\boxed{p=9,\ q=-6}\]

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(a) Horizontal projectile from a cliff

Horizontally the stone travels at a constant \(30\,\text{m/s}\); vertically it starts with zero vertical velocity and falls under gravity. First find the time of flight from the vertical motion, using \(h=\tfrac{1}{2}gt^{2}\):

\[50=\tfrac{1}{2}(10)t^{2}\;\Rightarrow\; 50=5t^{2}\;\Rightarrow\; t^{2}=10\;\Rightarrow\; t=\sqrt{10}\approx 3.16\,\text{s}.\]

The horizontal distance (range) is

\[R=\text{(horizontal speed)}\times t=30\sqrt{10}\approx 94.9\,\text{m}.\]

The stone lands about \(94.9\,\text{m}\) from the foot of the cliff.

(b) Body in equilibrium on two strings

The weight of the body is \(W=mg=2\times 10=20\,\text{N}\), acting vertically downward. String \(AP\) (tension \(T_1\)) makes \(56^{\circ}\) with the upward vertical, and string \(AR\) (tension \(T_2\)) is horizontal. Resolve the forces at \(A\).

Vertical equilibrium: only \(T_1\) has a vertical component, and it supports the weight:

\[T_1\cos 56^{\circ}=20\;\Rightarrow\; T_1=\frac{20}{\cos 56^{\circ}}=\frac{20}{0.5592}\approx 35.8\,\text{N}.\]

Horizontal equilibrium: the horizontal component of \(T_1\) is balanced by \(T_2\):

\[T_2=T_1\sin 56^{\circ}=20\tan 56^{\circ}=20(1.4826)\approx 29.7\,\text{N}.\]

Hence \(T_1\approx 35.8\,\text{N}\) and \(T_2\approx 29.7\,\text{N}\).

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(a) Uniform retardation, so use \(v^{2}=u^{2}-2as\) with \(u=45,\ v=5,\ a=10\):

\[5^{2}=45^{2}-2(10)s\Rightarrow 25=2025-20s\]

\[20s=2000\Rightarrow s=100\text{ m}\]

(b) The bucket is pulled upward, so applying Newton's second law upward with \(m=16\text{kg},\ T=240\text{N},\ g=10\text{m/s}^{2}\):

\[T-mg=ma\Rightarrow 240-16(10)=16a\]

\[80=16a\Rightarrow a=5\text{ m/s}^{2}\ (\text{upward})\]

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(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

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Because selection is with replacement, each draw is independent with a constant probability of a bad mango:

\[p=\frac{6}{24}=\frac14,\qquad q=1-p=\frac34\]

Let \(X\) be the number of bad mangoes in \(n=6\) draws; \(X\sim\text{Bin}(6,\tfrac14)\). We need \(P(X\le3)\), which is easiest via the complement \(P(X\le3)=1-P(X\ge4)\).

\[P(4)=\binom{6}{4}\left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}=\frac{15\times9}{4096}=\frac{135}{4096}\]

\[P(5)=\binom{6}{5}\left(\tfrac14\right)^{5}\left(\tfrac34\right)=\frac{6\times3}{4096}=\frac{18}{4096}\]

\[P(6)=\left(\tfrac14\right)^{6}=\frac{1}{4096}\]

\[P(X\ge4)=\frac{135+18+1}{4096}=\frac{154}{4096}=\frac{77}{2048}\]

\[P(X\le3)=1-\frac{77}{2048}=\frac{1971}{2048}\approx0.9624\]

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We can use the formula for combinations to solve for y:

^5yC₂ = ^(5y)! / _(2!(5y-2)!) = 190

Expanding the factorials:

^(5y)(5y-1) / 2 = 190

Solving for y:

(5y)(5y-1) = 380

25y^2 - 5y = 380

25y^2 - 5y - 380 = 0

Using the quadratic formula:

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-(−5) ± √((-5)^2 - 4(25)(-380)))/(2(25))

y = (5 ± √(25 + 90000))/(50)

y = (5 ± √(90025))/(50)

Taking the positive root:

y = (5 + 300)/(50)

y = 305/50

y = 6.1

So the value of y is approximately 6.1.

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9∫1 ${}^{}{<br>}_{}$ x(2x−3)√x $\frac{<br>}{}$ dx = 2x2−3x)x1/2 $\frac{{\mathrm{<br>}}^{}<br>}{{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"><br></span>}}^{}}$

= x −1/2 ${}^{ \; \; \; }$ (2x2 ${}^2$ - 3x)

= 2x3/2 ${}^{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}$ - 3x1/2 ${}^{<br>\; <span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}$

= 2x3/23/2+1 $\frac{{\mathrm{<br>}}^{\mathrm{<br>}}}{}$ - 3x1/21/2+1 $\frac{{\mathrm{<br>}}^{}}{\mathrm{<br>}}$

= 2x3/25/2 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$ - 3x1/23/2 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$

= 9∫1 ${}^{}{<br>}_{}$ [4x3/25 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$ - 6x1/23 $\frac{{\mathrm{<br>}}^{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br></span>}}}{}$] 

[ 4(9)3/25 $\frac{<br>}{}$- 6(9)1/23 $\frac{{<br>}^{}}{}$] - [ 4(9)1/25 $\frac{<br>}{}$ - 6(1)1/23 $\frac{<br>}{}$]

= [ 4(27)5 $\frac{<br>}{}$- 6(3)3 $\frac{<br>}{}$- [ 45 - 2 ]

=  [ 1085- 2 ] -  [ 65 ]

= 785 + 65= 845 $\frac{\mathrm{<span\; style="word-spacing:\; normal;\; font-weight:\; var(--bs-body-font-weight);"><br><br></span>}}{}$

164/5

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(a) (i) Cumulative frequency table

Height class (cm)	Frequency	Upper class boundary (cm)	Cumulative frequency
100–109	27	109.5	27
110–119	58	119.5	85
120–129	130	129.5	215
130–139	105	139.5	320
140–149	50	149.5	370
150–159	25	159.5	395
160–169	5	169.5	400

The total number of pupils is \(N=400\). Include the starting point \((99.5,0)\).

(a) (ii) Less-than cumulative frequency curve (ogive)

Plot the upper class boundaries against their cumulative frequencies and join successive points with a smooth increasing curve.

(b) Estimates from the ogive

(i) Median height

\[\frac{N}{2}=\frac{400}{2}=200\]

At cumulative frequency 200, the corresponding height is approximately \(128.3\text{ cm}\).

Median height \(\approx 128.3\text{ cm}\).

(ii) Interquartile range

\[Q_1=\frac{N}{4}=100,\qquad Q_3=\frac{3N}{4}=300\]

From the curve, \(Q_1\approx120.7\text{ cm}\) and \(Q_3\approx137.6\text{ cm}\).

\[\text{Interquartile range}=Q_3-Q_1=137.6-120.7\approx16.9\text{ cm}.\]

(iii) Percentage of pupils whose heights are at most \(130\text{ cm}\)

At \(130\text{ cm}\), the cumulative frequency is approximately \(220\).

\[\text{Percentage}=\frac{220}{400}\times100\%\approx55\%.\]

Therefore, approximately \(55\%\) of the pupils have heights at most \(130\text{ cm}\).

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(a) Histogram

Since the class intervals have unequal widths, the heights of the rectangles are the frequency densities:

\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}.\]

Monthly income (₦'000)	Class boundaries (₦'000)	Frequency	Class width	Frequency density
135–139	134.5–139.5	20	5	4.0
140–149	139.5–149.5	42	10	4.2
150–154	149.5–154.5	28	5	5.6
155–164	154.5–164.5	38	10	3.8
165–169	164.5–169.5	22	5	4.4

Plot the class boundaries on the horizontal axis and frequency density on the vertical axis. The resulting histogram is:

(b) Estimated mode

The modal class is \(149.5\text{–}154.5\), since it has the greatest frequency density, \(5.6\).

Using the standard intersecting-diagonals construction on the modal rectangle gives

\[\begin{aligned} \text{Mode} &=149.5+\frac{5.6-4.2}{(5.6-4.2)+(5.6-3.8)}\times 5\\ &=149.5+\frac{1.4}{3.2}\times5\\ &=151.6875\approx151.7. \end{aligned}\]

Therefore, the estimated modal monthly income is \(151.7\) thousand naira, that is, approximately ₦151,700.

Antwoorddetails

(a) To find the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis, we first need to find the x-coordinate of the point where the curve intersects the X-axis. This is where y = 0, so we can solve the equation (x² - x + 1)(x - 2) = 0 to find the roots of the equation. The roots are x = 1 ± i√3 and x = 2, but since we want the point where the curve intersects the X-axis, we take x = 2.

Next, we need to find the gradient of the curve at this point. We can do this by differentiating the equation y = (x² - x + 1)(x - 2) with respect to x, giving:

dy/dx = 3x² - 8x + 1

Substituting x = 2, we get:

dy/dx = 3(2)² - 8(2) + 1 = -7

Therefore, the gradient of the curve at the point where it intersects the X-axis is -7.

Since the normal to the curve is perpendicular to the tangent at the point of intersection, we know that the gradient of the normal is the negative reciprocal of the gradient of the tangent. So the gradient of the normal is 1/7.

Finally, we can find the equation of the normal using the point-slope form of the equation of a straight line:

y - 0 = (1/7)(x - 2)

Simplifying this equation gives:

y = (1/7)x - 2/7

So the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis is y = (1/7)x - 2/7.

(b) To find the equation of the line joining Q and the midpoint of PR, we first need to find the coordinates of the midpoint of PR. The midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). So the midpoint of PR is ((-1 + 3)/2, (2 - 4)/2), which simplifies to (1, -1).

Next, we need to find the gradient of the line joining Q and the midpoint of PR. We can use the gradient formula, which gives:

m = (y₂ - y₁)/(x₂ - x₁)

Substituting the coordinates of Q and the midpoint of PR gives:

m = (1 - (-1))/(5 - 1) = 1/2

Antwoorddetails

Use \(\cos\theta=\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|\,|\mathbf{y}|}\) with \(\mathbf{x}=\begin{pmatrix}-4\\3\end{pmatrix},\ \mathbf{y}=\begin{pmatrix}-9\\15\end{pmatrix}\).

Dot product: \(\mathbf{x}\cdot\mathbf{y}=(-4)(-9)+(3)(15)=36+45=81\).

Magnitudes: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).

\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]

\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]