WAEC SSCE - Further Mathematics - 2009
Ajụjụ 1 Ripọtì
The roots of a quadratic equation are -3 and 1. Find its equation.
Akọwa Nkọwa
To find the equation of a quadratic equation, we can use the factored form, which is given by: $$(x-r_1)(x-r_2)=0$$ where $r_1$ and $r_2$ are the roots of the equation. In this case, the given roots are -3 and 1, so we have: $$(x-(-3))(x-1)=0$$ $$(x+3)(x-1)=0$$ Expanding this expression gives us: $$x^2+3x-x-3=0$$ Simplifying, we get: $$x^2+2x-3=0$$ So the equation of the quadratic equation is $x^2+2x-3=0$. Therefore, option (C) is the correct answer.
Ajụjụ 2 Ripọtì
In the diagram, a ladder PS leaning against a vertical wall PR makes angle x° with the horizontal floor. The ladder slides down to a point QT such that angle QTR = 30° and SNT = y°. Find an expression for tan y.
Akọwa Nkọwa
Let's consider the right triangle PST. We know that angle PST = 90°, angle PSR = x°, and angle PTR = 30°. Therefore, angle PTS = 60° - x°. We also know that the ladder has length L and that it slides down the wall to a point Q such that TQ = x. Now, consider the right triangle QRT. We know that angle QRT = 90° and angle QTR = 30°. Therefore, angle QRT = 60°. We also know that TQ = x and QR = L - x. Finally, consider the right triangle NST. We know that angle NST = 90°, angle NTS = y°, and angle SNT = y° - (60° - x°) = x° + y° - 60°. We also know that NS = QR = L - x. Now, we can use the tangent function to find an expression for tan y: tan y = NS/NT = (L - x)/(TQ + QR) = (L - x)/(x + (L - x)) = (L - x)/(L) We can simplify this expression by multiplying the numerator and denominator by \(\frac{1}{\sqrt{3}}\): tan y = \(\frac{\frac{1}{\sqrt{3}} (L - x)}{\frac{1}{\sqrt{3}} L}\) = \(\frac{\sqrt{3} \tan x - 1}{\sqrt{3} + \tan x}\) Therefore, the correct answer is.
Ajụjụ 3 Ripọtì
The coordinates of the centre of a circle is (-2, 3). If its area is \(25\pi cm^{2}\), find its equation.
Akọwa Nkọwa
We know that the equation of a circle with center coordinates \((a,b)\) and radius \(r\) is given by \((x-a)^2 + (y-b)^2 = r^2\). From the problem statement, we are given that the center of the circle is at \((-2,3)\), so we can substitute \(a=-2\) and \(b=3\) in the equation of the circle as \((x+2)^2 + (y-3)^2 = r^2\). We are also given that the area of the circle is \(25\pi cm^2\). We know that the area of a circle is given by the formula \(A = \pi r^2\), where \(r\) is the radius of the circle. We can solve for the radius by substituting the area value, so we have \begin{align*} A &= \pi r^2\\ 25\pi &= \pi r^2\\ 25 &= r^2\\ r &= 5 \end{align*} Now we have the coordinates of the center and the radius, so we can substitute them in the equation of the circle to get $$(x+2)^2 + (y-3)^2 = 25$$ Expanding the square terms, we have $$x^2+4x+4+y^2-6y+9=25$$ Simplifying, we get $$x^2+y^2+4x-6y-12=0$$ Therefore, the equation of the circle is \boxed{x^2+y^2+4x-6y-12=0}.
Ajụjụ 4 Ripọtì
Calculate, correct to the nearest degree, the angle between the vectors \(\begin{pmatrix} 13 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\).
Akọwa Nkọwa
To find the angle between two vectors, we can use the dot product formula: \(\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta\), where \(\mathbf{a}\) and \(\mathbf{b}\) are the vectors, \(\|\mathbf{a}\|\) and \(\|\mathbf{b}\|\) are their magnitudes, and \(\theta\) is the angle between them. So, for the given vectors, we have: \(\begin{pmatrix} 13 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \sqrt{13^2 + 1^2} \sqrt{1^2 + 4^2} \cos \theta\) Simplifying, we get: \(53 = \sqrt{170} \sqrt{17} \cos \theta\) \(\cos \theta = \frac{53}{\sqrt{170} \sqrt{17}}\) Taking the inverse cosine of both sides, we get: \(\theta \approx 73^\circ\) So, the answer is closest to 72°.
Ajụjụ 5 Ripọtì
A body is acted upon by forces \(F_{1} = (10 N, 090°)\) and \(F_{2} = (6 N, 180°)\). Find the magnitude of the resultant force.
Akọwa Nkọwa
To find the magnitude of the resultant force, we first need to find the horizontal and vertical components of the two given forces. For the first force, we have: - Horizontal component: $$F_{1x} = F_{1} \cos 90° = 0$$ - Vertical component: $$F_{1y} = F_{1} \sin 90° = 10 N$$ For the second force, we have: - Horizontal component: $$F_{2x} = F_{2} \cos 180° = -6 N$$ - Vertical component: $$F_{2y} = F_{2} \sin 180° = 0$$ The horizontal and vertical components of the resultant force can be found by adding the corresponding components of the two given forces: - Horizontal component: $$F_{x} = F_{1x} + F_{2x} = 0 - 6 N = -6 N$$ - Vertical component: $$F_{y} = F_{1y} + F_{2y} = 10 N + 0 = 10 N$$ The magnitude of the resultant force can be found using the Pythagorean theorem: $$|F| = \sqrt{F_{x}^{2} + F_{y}^{2}} = \sqrt{(-6)^{2} + (10)^{2}} \approx 11.66 N$$ Therefore, the magnitude of the resultant force is approximately 11.7 N. The answer is (B) 11.7 N.
Ajụjụ 6 Ripọtì
The gradient of point P on the curve \(y = 3x^{2} - x + 3\) is 5. Find the coordinates of P.
Akọwa Nkọwa
To find the coordinates of point P, we need to differentiate the given equation with respect to x to get the derivative of the function. The derivative of y with respect to x gives us the gradient or slope of the tangent at any point on the curve. Taking the derivative of y with respect to x, we get: \[\frac{dy}{dx} = 6x - 1\] We are given that the gradient at point P is 5. So, we can equate the derivative of y with 5 and solve for x. \[6x - 1 = 5\] \[6x = 6\] \[x = 1\] Now that we know x = 1, we can find the corresponding y-coordinate by substituting x=1 into the original equation for y. \[y = 3x^{2} - x + 3\] \[y = 3(1)^{2} - 1 + 3\] \[y = 5\] Therefore, the coordinates of point P are (1, 5). To summarize, we found the x-coordinate of point P by equating the derivative of the equation with the given gradient of 5 and solved for x. Then, we found the corresponding y-coordinate by substituting the value of x into the original equation.
Ajụjụ 7 Ripọtì
In the diagram, a ladder PS leaning against a vertical wall PR makes angle x° with the horizontal floor. The ladder slides down to a point QT such that angle QTR = 30° and SNT = y°. Find the relation between x and y.
Akọwa Nkọwa
Ajụjụ 8 Ripọtì
Given that \(2^{x} = 0.125\), find the value of x.
Akọwa Nkọwa
We can write 0.125 as a fraction: 0.125 = 1/8 So, we have: 2^x = 1/8 To find the value of x, we need to determine what power of 2 gives us 1/8. We can rewrite 1/8 as a power of 2 by using the fact that: 1/8 = 2^(-3) Therefore, we have: 2^x = 2^(-3) For the above equation to be true, x must be equal to -3. Therefore, the answer is option (D) -3.
Ajụjụ 10 Ripọtì
A stone is projected vertically with a speed of 10 m/s from a point 8 metres above the ground. Find the maximum height reached. \([g = 10 ms^{-2}]\).
Akọwa Nkọwa
We can solve this problem using the equations of motion for a particle moving vertically under gravity. The key idea is that at the maximum height, the vertical velocity of the stone will be zero. We can use the following equation to find the time taken by the stone to reach the maximum height: \[-u/g = -10/10 = -1\] where u is the initial velocity and g is the acceleration due to gravity. The time taken to reach the maximum height is 1 second. We can now use the following equation to find the maximum height reached: \[h = u t - \frac{1}{2} g t^{2} = 10\times 1 - \frac{1}{2}\times 10\times 1^{2} = 5\] where h is the maximum height reached. Therefore, the maximum height reached by the stone is 5 metres above the initial height of 8 metres, which gives us a total height of 13 metres. Hence, the answer is option (A) 13 metres.
Ajụjụ 11 Ripọtì
p and q are statements such that \(p \implies q\). Which of the following is a valid conclusion from the implication?
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
Solve \(x^{2} - 2x - 8 > 0\).
Ajụjụ 13 Ripọtì
Simplify \((216)^{-\frac{2}{3}} \times (0.16)^{-\frac{3}{2}}\)
Akọwa Nkọwa
We can simplify this expression using the rules of exponents. First, let's simplify \((216)^{-\frac{2}{3}}\). We know that \((216)^{\frac{2}{3}}\) is the cube root of 216 squared. So, \((216)^{-\frac{2}{3}}\) is simply the reciprocal of that value. \[(216)^{-\frac{2}{3}} = \frac{1}{(216)^{\frac{2}{3}}} = \frac{1}{(6^3)^{\frac{2}{3}}} = \frac{1}{6^2}\] Next, let's simplify \((0.16)^{-\frac{3}{2}}\). We can rewrite \(0.16\) as \(\frac{16}{100}\), and then apply the exponent to both the numerator and denominator: \[(0.16)^{-\frac{3}{2}} = \left(\frac{16}{100}\right)^{-\frac{3}{2}} = \frac{(100)^{\frac{3}{2}}}{(16)^{\frac{3}{2}}}\] Now, we can multiply these two simplified expressions: \[\frac{1}{6^2} \times \frac{(100)^{\frac{3}{2}}}{(16)^{\frac{3}{2}}} = \frac{1}{36} \times \frac{1000}{64} = \frac{125}{288}\] Therefore, the answer is \(\frac{125}{288}\).
Ajụjụ 14 Ripọtì
The roots of the quadratic equation \(2x^{2} - 5x + m = 0\) are \(\alpha\) and \(\beta\), where m is a constant. Find \(\alpha^{2} + \beta^{2}\) in terms of m.
Akọwa Nkọwa
To find \(\alpha^{2} + \beta^{2}\), we need to use the following identities: \[\alpha + \beta = \frac{-b}{a}\quad\text{and}\quad \alpha\beta = \frac{c}{a}\] where a, b, and c are the coefficients of the quadratic equation \(ax^{2} + bx + c = 0\). In this case, the quadratic equation is \(2x^{2} - 5x + m = 0\), so a = 2, b = -5, and c = m. Using the identity \(\alpha + \beta = \frac{-b}{a}\), we have: \[\alpha + \beta = \frac{-(-5)}{2} = \frac{5}{2}\] Using the identity \(\alpha\beta = \frac{c}{a}\), we have: \[\alpha\beta = \frac{m}{2}\] Now, we can use the identity: \[\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta\] Substituting the values we found earlier, we get: \[\alpha^{2} + \beta^{2} = \left(\frac{5}{2}\right)^{2} - 2\left(\frac{m}{2}\right)\] Simplifying, we get: \[\alpha^{2} + \beta^{2} = \frac{25}{4} - m\] Therefore, the answer is \(\frac{25}{4} - m\), which is.
Ajụjụ 15 Ripọtì
A force F acts on a body of mass 12kg increases its speed from 5 m/s to 35 m/s in 5 seconds. Find the value of F.
Akọwa Nkọwa
We can use the formula: $$F = ma$$ where F is the force applied, m is the mass of the body, and a is the acceleration produced. To find the acceleration, we can use the formula: $$a = \frac{v_f - v_i}{t}$$ where v_f is the final velocity, v_i is the initial velocity, and t is the time taken. Substituting the given values, we get: $$a = \frac{35 - 5}{5} = 6 \text{ m/s}^2$$ Substituting this value of acceleration and the given mass into the formula for force, we get: $$F = ma = 12 \times 6 = 72 \text{ N}$$ Therefore, the value of F is 72 N. So, the answer is 72 N.
Ajụjụ 16 Ripọtì
The derivative of a function f with respect to x is given by \(f'(x) = 3x^{2} - \frac{4}{x^{5}}\). If \(f(1) = 4\), find f(x).
Akọwa Nkọwa
To find the function f(x) from its derivative, we need to integrate the derivative with respect to x. \[f(x) = \int f'(x) dx = \int (3x^{2} - \frac{4}{x^{5}}) dx = x^{3} + \frac{1}{x^{4}} + C\] where C is the constant of integration. We can use the given information that f(1) = 4 to find the value of C: \[f(1) = 1 + 1 + C = 4\] \[C = 2\] Substituting this value of C, we get: \[f(x) = x^{3} + \frac{1}{x^{4}} + 2\] Therefore, the answer is option (B) \(f(x) = x^{3} + \frac{1}{x^{4}} + 2\).
Ajụjụ 17 Ripọtì
The ages, in years, of 5 boys are 5, 6, 6, 8 and 10. Calculate, correct to one decimal place, the standard deviation of their ages.
Akọwa Nkọwa
To calculate the standard deviation, we need to follow these steps: 1. Find the mean (average) of the ages. 2. Find the difference between each age and the mean. 3. Square each of these differences. 4. Find the average of these squared differences. 5. Take the square root of this average. Let's first find the mean of the ages: mean = (5+6+6+8+10)/5 = 7 years Now we can find the difference between each age and the mean: |5 - 7| = 2 |6 - 7| = 1 |6 - 7| = 1 |8 - 7| = 1 |10 - 7| = 3 Next, we square each of these differences: 2^2 = 4 1^2 = 1 1^2 = 1 1^2 = 1 3^2 = 9 Now we find the average of these squared differences: average = (4+1+1+1+9)/5 = 2.4 Finally, we take the square root of this average: standard deviation = sqrt(2.4) ≈ 1.5 Therefore, the correct answer is not among the given options. The closest option is (d) 1.8 years, but this is not within one decimal place of the calculated standard deviation.
Ajụjụ 18 Ripọtì
Simplify \(2\log_{3} 8 - 3\log_{3} 2\)
Ajụjụ 20 Ripọtì
Eight football clubs are to play in a league on home and away basis. How many matches are possible?
Ajụjụ 21 Ripọtì
Given that \(\log_{3}(x - y) = 1\) and \(\log_{3}(2x + y) = 2\), find the value of x.
Akọwa Nkọwa
We can use the properties of logarithms to simplify the given expressions and then solve for x. Firstly, from the first equation, we have: \[\log_{3}(x - y) = 1\] \[x - y = 3\] Similarly, from the second equation, we have: \[\log_{3}(2x + y) = 2\] \[2x + y = 9\] Now, we have a system of two equations with two variables, x and y. We can solve for y by subtracting the first equation from the second: \[(2x + y) - (x - y) = 9 - 3\] \[x + 2y = 6\] \[y = 3 - \frac{1}{2}x\] Substituting this value of y into the first equation: \[x - (3 - \frac{1}{2}x) = 3\] \[\frac{3}{2}x = 6\] \[x = 4\] Therefore, the value of x is 4, which is.
Ajụjụ 22 Ripọtì
The position vectors of A and B are (2i + j) and (-i + 4j) respectively; find |AB|.
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
The velocity \(v ms^{-1}\) of a particle moving in a straight line is given by \(v = 3t^{2} - 2t + 1\) at time t secs. Find the acceleration of the particle after 3 seconds.
Akọwa Nkọwa
To find the acceleration of the particle, we need to take the derivative of the velocity function with respect to time (t). The derivative of velocity with respect to time gives us the acceleration. Taking the derivative of v with respect to t, we get: \[\frac{dv}{dt} = 6t - 2\] Now, we need to find the acceleration of the particle after 3 seconds. We can do this by substituting t = 3 into the expression we just found for the derivative of v: \[\frac{dv}{dt} = 6t - 2\] \[\frac{dv}{dt}\bigg|_{t=3} = 6(3) - 2\] \[\frac{dv}{dt}\bigg|_{t=3} = 16\] Therefore, the acceleration of the particle after 3 seconds is 16 \(ms^{-2}\). To summarize, we found the derivative of the velocity function with respect to time to get the acceleration function. Then, we substituted t = 3 into the acceleration function to find the acceleration of the particle after 3 seconds.
Ajụjụ 24 Ripọtì
If \(P = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\), find \((P^{2} + P)\).
Akọwa Nkọwa
To find \((P^{2} + P)\), we first need to find the value of \(P^{2}\). \(P^{2} = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix}\) Now, we can find \((P^{2} + P)\) by adding \(P^{2}\) and \(P\). \((P^{2} + P) = \begin{vmatrix} 3 & 2 \\ 4 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\) Therefore, the answer is \(\begin{vmatrix} 4 & 3 \\ 6 & 4 \end{vmatrix}\).
Ajụjụ 25 Ripọtì
Find the coefficient of \(x^{4}\) in the binomial expansion of \((2 + x)^{6}\).
Akọwa Nkọwa
To find the coefficient of \(x^{4}\) in the binomial expansion of \((2 + x)^{6}\), we can use the formula for the binomial expansion: \[\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}\] where in this case, \(n = 6\), \(a = 2\), and \(b = x\). So, we have \[(2+x)^6 = \sum_{k=0}^{6}\binom{6}{k}(2)^{6-k}x^{k}.\] To find the coefficient of \(x^{4}\), we need to look at the term where \(k = 4\), which is \[\binom{6}{4}(2)^{6-4}x^{4} = 15\cdot 4x^{4} = 60x^{4}.\] Therefore, the coefficient of \(x^{4}\) in the binomial expansion of \((2 + x)^{6}\) is 60. So the answer is option C, 60.
Ajụjụ 26 Ripọtì
Three defective bulbs got mixed up with seven good ones. If two bulbs are selected at random, what is the probability that both are good?
Akọwa Nkọwa
We can solve this problem using the concept of probability. Let's begin by finding the total number of ways to select two bulbs out of ten. We can do this using the combination formula: $${10 \choose 2} = \frac{10!}{2!(10-2)!} = 45$$ So there are 45 ways to select two bulbs out of ten. Now let's find the number of ways to select two good bulbs out of the seven good ones. We can use the combination formula again: $${7 \choose 2} = \frac{7!}{2!(7-2)!} = 21$$ So there are 21 ways to select two good bulbs out of seven. Finally, we can find the probability of selecting two good bulbs by dividing the number of ways to select two good bulbs by the total number of ways to select two bulbs: $$P(\text{both bulbs are good}) = \frac{21}{45} = \frac{7}{15}$$ Therefore, the answer is, the probability that both bulbs are good is $\frac{7}{15}$.
Ajụjụ 27 Ripọtì
Given \(\sin \theta = \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd form.
Akọwa Nkọwa
We know that:
\(\sin \theta = \frac{{\text{{opposite}}}}{{\text{{hypotenuse}}}}\)
We can draw a right triangle with an angle of \(\theta\), opposite side of \(\sqrt{3}\), and hypotenuse of 2. Using the Pythagorean theorem, we can find the adjacent side to be 1. Therefore, the triangle looks like:
/|
/ |
1 / |sqrt(3)
/ |
/____|
2
Using the double-angle formula for tangent, we have:
\(\tan 2\theta = \frac{{2 \tan \theta}}{{1 - \tan^{2} \theta}}\)
Using the formula for tangent:
\(\tan \theta = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}} = \sqrt{3}\)
Substituting in the formula for double-angle tangent, we have:
\(\tan 2\theta = \frac{{2 \sqrt{3}}}{{1 - (\sqrt{3})^{2}}} = \frac{{2 \sqrt{3}}}{{1 - 3}}\)
Simplifying, we get:
\(\tan 2\theta = \frac{{2 \sqrt{3}}}{{-2}} = -\sqrt{3}\)
Therefore, the correct answer is.
Ajụjụ 28 Ripọtì
Three men, P, Q and R aim at a target, the probabilities that P, Q and R hit the target are \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{3}{4}\) respectively. Find the probability that exactly 2 of them hit the target.
Akọwa Nkọwa
To find the probability that exactly 2 of them hit the target, we need to consider all possible combinations of two men hitting the target while the other misses. There are three such combinations: PQ, QR, and PR. For the combination PQ, the probability that P hits the target and Q misses is \(\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\). The probability that P misses and Q hits is \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). Therefore, the probability that exactly PQ hit the target is \(\frac{1}{3} + \frac{1}{6} = \frac{1}{2}\). Similarly, for QR, the probability that Q hits and R misses is \(\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}\). The probability that Q misses and R hits is \(\frac{2}{3} \times \frac{3}{4} = \frac{1}{2}\). Therefore, the probability that exactly QR hit the target is \(\frac{1}{12} + \frac{1}{2} = \frac{7}{12}\). Finally, for PR, the probability that P hits and R misses is \(\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\). The probability that P misses and R hits is \(\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}\). Therefore, the probability that exactly PR hit the target is \(\frac{1}{8} + \frac{3}{8} = \frac{1}{2}\). The total probability that exactly 2 of them hit the target is the sum of the probabilities for each combination, which is \(\frac{1}{2} + \frac{7}{12} + \frac{1}{2} = \frac{5}{6}\). Therefore, the answer is (C) \(\frac{5}{12}\).
Ajụjụ 29 Ripọtì
Evaluate \(\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix}\).
Akọwa Nkọwa
To evaluate the given expression, we need to perform matrix multiplication. We multiply the first row of the matrix on the left with the column matrix on the right as shown below: \[\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} (2 \times 2) + (3 \times 3) \\ (4 \times 2) + (1 \times 3) \end{pmatrix} = \begin{pmatrix} 13 \\ 11 \end{pmatrix}\] Therefore, the answer is \(\begin{pmatrix} 13 \\ 11 \end{pmatrix}\).
Ajụjụ 30 Ripọtì
Express the force F = (8 N, 150°) in the form (a i + b j) where a and b are constants.
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
The line \(y = mx - 3\) is a tangent to the curve \(y = 1 - 3x + 2x^{3}\) at (1, 0). Find the value of the constant m.
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Which of the following is the semi- interquartile range of a distribution?
Akọwa Nkọwa
The semi-interquartile range of a distribution is the measure of dispersion or spread of data around the median. It is half of the difference between the upper quartile and the lower quartile. Therefore, the correct answer is \(\frac{1}{2}(\text{Upper quartile - Lower quartile})\). To find the interquartile range (IQR), we subtract the lower quartile (Q1) from the upper quartile (Q3), i.e., IQR = Q3 - Q1. Then, the semi-interquartile range (SIQR) is half of the IQR, i.e., SIQR = (Q3 - Q1) / 2. The other options listed in the question are not measures of dispersion or spread around the median, so they are not correct answers to the question.
Ajụjụ 33 Ripọtì
Two fair dices, each numbered 1, 2, ..., 6, are tossed together. Find the probability that they both show even numbers.
Akọwa Nkọwa
When two dice are tossed, there are 6 × 6 = 36 possible outcomes, since each die has 6 possible outcomes. Out of these 36 possible outcomes, there are 3 even numbers on each die (2, 4, 6), so the probability of rolling an even number on a single die is 3/6 = 1/2. To find the probability that both dice show even numbers, we need to multiply the probability of rolling an even number on the first die (1/2) by the probability of rolling an even number on the second die (also 1/2), since the two events are independent. So, the probability that both dice show even numbers is: 1/2 × 1/2 = 1/4 Therefore, the answer is option (B) \(\frac{1}{4}\).
Ajụjụ 34 Ripọtì
Two balls are drawn, from a bag containing 3 red, 4 white and 5 black identical balls. Find the probability that they are all of the same colour.
Akọwa Nkọwa
There are different ways to approach this problem, but one possible method is to use combinations. First, we need to find the total number of ways to draw two balls from the bag, without replacement. This can be calculated using the formula for combinations: \(\text{Total number of ways} = \binom{12}{2} = \frac{12!}{2!10!} = 66\) Next, we need to count the number of ways to draw two balls of the same color. There are three possible cases: both red, both white, or both black. For two red balls, we can choose 2 balls from the 3 red balls in the bag, giving us: \(\text{Number of ways for two red balls} = \binom{3}{2} = 3\) Similarly, we can count the number of ways for two white balls and two black balls: \(\text{Number of ways for two white balls} = \binom{4}{2} = 6\) \(\text{Number of ways for two black balls} = \binom{5}{2} = 10\) Therefore, the total number of ways to draw two balls of the same color is: \(\text{Number of ways for two same color balls} = 3 + 6 + 10 = 19\) Finally, we can calculate the probability of drawing two balls of the same color by dividing the number of ways for two same color balls by the total number of ways: \(\text{Probability of two same color balls} = \frac{19}{66} \approx 0.288\) Therefore, the answer is, which is \(\frac{19}{66}\).
Ajụjụ 35 Ripọtì
If \(\frac{^{8}P_{x}}{^{8}C_{x}} = 6\), find the value of x.
Akọwa Nkọwa
The formula for permutations is: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) The formula for combinations is: \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) We are given that: \(\frac{^{8}P_{x}}{^{8}C_{x}} = 6\) Substituting the formulas for permutations and combinations, we get: \(\frac{\frac{8!}{(8-x)!}}{\frac{8!}{x!(8-x)!}} = 6\) Simplifying, we get: \(\frac{x!}{(8-x)!} = \frac{1}{6}\) Multiplying both sides by \(\frac{(8-x)!}{x!}\), we get: \(\frac{8!}{x!(8-x)!} = 6\) Dividing both sides by 8!, we get: \(\frac{1}{^{8}C_{x}} = \frac{1}{720}\) Multiplying both sides by 720, we get: \(^{8}C_{x} = 720\) We can use the formula for combinations to find the value of x: \(^{8}C_{x} = \frac{8!}{x!(8-x)!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{x!(8-x)!}\) Since \(^{8}C_{x} = 720\), we have: \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{x!(8-x)!} = 720\) Dividing both sides by 720, we get: \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{720} = x!(8-x)!\) Simplifying, we get: \(x!(8-x)! = 40320\) The only value of x that satisfies this equation is x = 3. Therefore, the correct answer is.
Ajụjụ 36 Ripọtì
An arc of length 10.8 cm subtends an angle of 1.2 radians at the centre of a circle. Calculate the radius of the circle.
Akọwa Nkọwa
In a circle, the length of an arc is given by:
l = rθ
where r is the radius of the circle, and θ is the angle subtended by the arc at the centre of the circle in radians.
In this question, we are given the length of the arc (l = 10.8 cm) and the angle subtended by the arc (θ = 1.2 radians). We can rearrange the formula above to solve for the radius r:
r = l/θ
Substituting the values given, we get:
r = 10.8/1.2 = 9
Therefore, the radius of the circle is 9 cm.
Ajụjụ 37 Ripọtì
Express \(\frac{2}{3 - \sqrt{7}} \text{ in the form} a + \sqrt{b}\), where a and b are integers.
Akọwa Nkọwa
We can begin by multiplying both the numerator and denominator by the conjugate of the denominator, which is \(3 + \sqrt{7}\). This is done to eliminate the square root in the denominator. \[\frac{2}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}} = \frac{2(3 + \sqrt{7})}{9 - 7} = \frac{2(3 + \sqrt{7})}{2} = 3 + \sqrt{7}\] Therefore, \(\frac{2}{3 - \sqrt{7}} = 3 + \sqrt{7}\), and the answer is (B) \(3 + \sqrt{7}\).
Ajụjụ 38 Ripọtì
If the mean of -1, 0, 9, 3, k, 5 is 2, where k is a constant, find the median of the set of numbers.
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
If (x + 3) is a factor of the polynomial \(x^{3} + 3x^{2} + nx - 12\), where n is a constant, find the value of n.
Akọwa Nkọwa
If (x + 3) is a factor of the polynomial, then we can write the polynomial as: \[(x+3)(ax^2+bx+c)\] Expanding the above equation, we get: \[ax^3 + (b+3a)x^2 + (c+3b)x + 3c = x^3 + 3x^2 + nx - 12\] Equating the coefficients of corresponding powers of x, we get: \[a = 1, b + 3a = 3, c + 3b = n, 3c = -12\] From the last equation, we have: \[c = -4\] Using this value of c in the third equation, we have: \[n = c + 3b = -4 + 3b\] From the second equation, we have: \[b + 3a = 3\] \[b + 3(1) = 3\] \[b = 0\] Hence, we have: \[n = -4 + 3b = -4\] Therefore, the value of n is -4.
Ajụjụ 40 Ripọtì
The first term of a geometric progression is 350. If the sum to infinity is 250, find the common ratio.
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
The coordinates of the vertices of triangle ABC are A(-2, 1), B(4, -2) and C(1, 8) respectively. If D(x, y) is the foot perpendicular from A to BC, find
(a) an equation connecting x and y ;
(b) the unit vector in the direction of BC.
Ajụjụ 42 Ripọtì
The table below shows the corresponding values of two variables X and Y.
| X | 33 | 31 | 28 | 25 | 23 | 22 | 19 | 17 | 16 | 14 |
| Y | 4 | 6 | 4 | 10 | 12 | 10 | 14 | 15 | 18 | 22 |
(a) Plot a scatter diagram to represent the data.
(b) Calculate \(\bar{x}\), the mean of X and \(\bar{y}\), the mean of Y.
(c) Draw the line of best fit to pass through \((\bar{x}, \bar{y})\).
(d) From your graph in (c), determine the (i) relationship between X and Y ; (ii) value of Y when X is 24.
Ajụjụ 43 Ripọtì
(a) The position vectors of points L and M are (5i + 6j) and (13i + 4j) respectively. If point K lies on LM such that LK : KM is 2 : 3, find the position vector of K.
(b) Three poles are situated at points A, B and C on the same horizontal plane such that \(AB = (8km, 060°)\) and \(BC = (12km, 130°)\). Calculate,
(i) |AC|, correct to three significant figures ; (ii) the bearing of C from A, correct to the nearest degree.
Ajụjụ 44 Ripọtì
(a) Simplify \(^{n + 1}C_{3} - ^{n - 1}C_{3}\)
(b) A fair die is thrown five times. Calculate, correct to three decimal places, the probability of obtaining (i) at most two sixes ; (ii) exactly three sixes.
Akọwa Nkọwa
None
Ajụjụ 45 Ripọtì
(a) The distribution of the lives (in days) of 40 transitor batteries is shown in the table:
| Battery life (in days) |
26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
| Frequency | 4 | 7 | 13 | 8 | 6 | 2 |
(a) Draw a histogram for the distribution.
(b) Use your graph in (a) to determine the mode for the distribution.
(c) Using an assumed mean of 43 days, calculate the mean of the distribution.
(d) What percentage number of batteries will live for less than 31 days or more than 45 days?
Ajụjụ 46 Ripọtì
If the quadratic equation \((x + 1)(x + 2) = k(3x + 7)\) has equal roots, find the possible values of the constant k.
Ajụjụ 47 Ripọtì
Coplanar force 4N, 8N, 6N, 4N and 5N act at a point as shown in the diagram. If the 6N force acts in the direction 090°, calculate the :
(a) magnitude of the resultant force;
(b) direction of the resultant force.
Ajụjụ 48 Ripọtì
(a) Use the trapezium rule with five ordinates to evaluate \(\int_{0} ^{1} \frac{3}{1 + x^{2}} \mathrm {d} x\), correct to four significant figures.
(b) If \(A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\), find the image of the point (1, 2) under the linear transformation \(A^{2} + A + 2I\), where I is the \(2 \times 2\) unit matrix.
Akọwa Nkọwa
None
Ajụjụ 49 Ripọtì
A student representative council consists of 8 girls and 6 boys. If an editorial board consisting of 5 persons is to be formed, what is the probability that the board consists of
(a) 3 girls and 2 boys ;
(b) either all girls or all boys.
Ajụjụ 50 Ripọtì
Solve the simultaneous equations : \(\log_{2} x - \log_{2} y = 2 ; \log_{2} (x - 2y) = 3\)
Ajụjụ 51 Ripọtì
(a) The position vectors of points A, B and C are \(i + 5j , 3i + 9j\) and \(-i + j\) respectively. (i) Show that points A, B and C are collinear; (ii) Determine the ratio \(|AB| : |BC|\).
(b) A uniform beam XY of mass 10 kg and length 24m is hunged horizontally from a cross bar by teo vertical inextensible strings, one attached to X and the other at a point M, 4m away from Y. A mass of 50kg is suspended at a point N which is 8m from X. If the system remains in equilibrium, calculate the tensions in the strings.
Ajụjụ 52 Ripọtì
(a) The 3rd and 6th terms of a Geometric Progression (G.P) are 2 and 54 respectively. Find the : (i) common ratio ; (ii) first term ; (iii) sum of the first 10 terms, correct to the nearest whole number.
(b) The ratio of the coefficient of \(x^{4}\) to that of \(x^{3}\) in the binomial expansion of \((1 + 2x)^{n}\) is \(3 : 1\). Find the value of n.
Ajụjụ 53 Ripọtì
The probabilities of Rotey obtaining the highest mark in Mathematics, Physics and Biology tests are 0.9, 0.75 and 0.8 respectively. Calculate the probability of getting the highest marks in at least two of the subjects.
Ajụjụ 54 Ripọtì
(a) Using the same axes, sketch the curves \(y = 6 - x - x^{2}\) and \(y = 3x^{2} - 2x + 3\).
(b) Find the x- coordinates of the points of intersection of the two curves in (a).
(c) Calculatethe area of the finite region bounded by the two curves in (a).
Ajụjụ 55 Ripọtì
The position vector of a body, with respect to the origin, is given by \(r = 4ti + (12 - 3t)j\) at any time t seconds.
(a) Find the velocity of the body ;
(b) Calculate the magnitude of the displacement between t = 0 and t = 5.
Ajụjụ 56 Ripọtì
Given that \(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\), evaluate \(\tan 15°\), leaving your answer in surd form.
Ajụjụ 57 Ripọtì
(a) Evaluate \(\int_{1} ^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\)
(b)(i) Evaluate: \(\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}\)
(ii) Using your answer in b(i), solve the simultaneous equations :
\(2x - 3y + z = 10\)
\(y - 2z = -7\)
\(x + 2y - 3z = -9\)
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) Solve : \(2x^{2} + x - 6 < 0\)
(b) Express \(\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}}\) in the form \(m\sqrt{2} + n\sqrt{5}\) where m and n are rational numbers.
