WAEC SSCE - General Mathematics - 1992
Tambaya 1 Rahoto
Simplify (3/4 + 1/3) x 41/3 + 31/4
Bayanin Amsa
To simplify this expression, we need to follow the order of operations, which is Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction (PEMDAS). First, we can simplify the expression inside the parentheses: (3/4 + 1/3) x 41/3 + 31/4 = (9/12 + 4/12) x 41/3 + 31/4 = 13/12 x 41/3 + 31/4 Next, we can simplify the exponent 41/3. To do this, we can convert the mixed number to an improper fraction: 41/3 = (4 x 3) + 1/3 = 13/3 Now we can substitute the value of 41/3 into the expression: 13/12 x 41/3 + 31/4 = 13/12 x 13/3 + 13/4 To simplify this, we can first simplify the multiplication of fractions: 13/12 x 13/3 = 169/36 Now we can add the two fractions: 169/36 + 13/4 = 169/36 + 39/36 = 208/36 = 13}{9 Therefore, the simplified expression is 13}{9, which corresponds to option (E).
Tambaya 3 Rahoto
If cos θ = 5/13, what is the value of tan \(\theta\) for 0 < θ < 90° ?
Bayanin Amsa
We can use the following trigonometric identity: \[\tan \theta = \frac{\sin \theta}{\cos \theta}\] We are given that $\cos \theta = 5/13$. To find $\sin \theta$, we can use the Pythagorean identity: \[\sin^2 \theta + \cos^2 \theta = 1\] Substituting $\cos \theta = 5/13$, we get: \[\sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1\] Simplifying, we get: \[\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}\] Taking the square root of both sides, we get: \[\sin \theta = \frac{12}{13}\] Substituting into the formula for $\tan \theta$, we get: \[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \boxed{\frac{12}{5}}\]
Tambaya 4 Rahoto
Write as a single fraction \(\frac{1}{1 - x} + \frac{2}{1 + x}\)
Bayanin Amsa
To add the fractions \(\frac{1}{1-x}\) and \(\frac{2}{1+x}\), we need to have a common denominator. We can obtain this by multiplying the first fraction by \((1+x)/(1+x)\) and the second fraction by \((1-x)/(1-x)\). This gives us: $$ \frac{1}{1-x} \cdot \frac{1+x}{1+x} + \frac{2}{1+x} \cdot \frac{1-x}{1-x} = \frac{1+x}{1-x^2} + \frac{2(1-x)}{1-x^2} $$ Combining the two fractions gives us: $$ \frac{1+x+2(1-x)}{1-x^2} = \frac{3-x}{1-x^2} $$ Therefore, the single fraction is \(\boxed{\frac{3-x}{1-x^2}}\).
Tambaya 5 Rahoto
Simplify: \((\frac{16}{81})^{\frac{1}{4}}\)
Bayanin Amsa
To simplify \((\frac{16}{81})^{\frac{1}{4}}\), we need to take the fourth root of \(\frac{16}{81}\). The fourth root of a number is the number that, when multiplied by itself four times, gives that number. So, to find the fourth root of \(\frac{16}{81}\), we need to find a number that, when multiplied by itself four times, gives \(\frac{16}{81}\). We can rewrite \(\frac{16}{81}\) as \((\frac{2}{3})^4\), since \(2^4=16\) and \(3^4=81\). Therefore, \((\frac{16}{81})^{\frac{1}{4}} = ((\frac{2}{3})^4)^{\frac{1}{4}} = \frac{2}{3}\), which is.
Tambaya 6 Rahoto
The angles marked in the diagram above are measured in degrees. Find x
Bayanin Amsa
In the given diagram, we have a triangle with two sides of equal length, which means that the angles opposite these sides are also equal. Thus, we can conclude that: 4x - 6 = 3x + 12 (using the fact that angles in a triangle add up to 180 degrees) x = 18 Therefore, the value of x is 18 degrees. Answer: (b) 24°.
Tambaya 7 Rahoto
In the diagram above, O is the center of the circle PQRS and ?QPS = 36°. Find ?QOS
Bayanin Amsa
Tambaya 8 Rahoto
In the diagram above PQT is a tangent to the circle QRS at Q. Angle QTR = 48° and ?QRT = 95°. Find ?QRT
Bayanin Amsa
Tambaya 10 Rahoto
A regular polygon has each of its exterior angles as 18o. How many sides has the polygon
Bayanin Amsa
The exterior angles of a polygon add up to 360 degrees. If each exterior angle of a regular polygon is 18 degrees, then the polygon has 360/18 = 20 exterior angles. Since a regular polygon has the same number of sides as exterior angles, the polygon has 20 sides. Therefore, the answer is (d) 20.
Tambaya 11 Rahoto
A bricklayer measured the length of a wall and obtained 4.10m. If the actual length of the wall is 4.25m, find his percentage error.
Bayanin Amsa
To find the percentage error, we need to find the absolute difference between the measured value and actual value, divide it by the actual value, and then multiply by 100 to get a percentage. Absolute difference = |actual value - measured value| = |4.25 - 4.10| = 0.15 Percentage error = (|actual value - measured value| / actual value) x 100% Percentage error = (0.15 / 4.25) x 100% = 3.529% (rounded to 3 decimal places) Therefore, the percentage error is approximately 3.529%, which corresponds to 3 9/17%.
Tambaya 12 Rahoto
Construct a quadratic equation whose roots are \(-\frac{1}{2}\) and 2.
Tambaya 13 Rahoto
A group of students measured a certain angle
(to the nearest degree) and obtained the following
results: 75o, 76o, 72o, 73o, 74o, 79o, 72o, 72o, 77o,
72o, 71o, 70o, 78o, 73o. Find the mode
Bayanin Amsa
To find the mode of the given set of data, we need to identify the value that appears most frequently. From the given set of measurements, we can see that 72o appears 4 times, which is more than any other value. Therefore, the mode of the data set is 72o. So, the correct option is (e) 72o.
Tambaya 14 Rahoto
Evaluate \(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)
Bayanin Amsa
We can use the following property of logarithms: - \(\log_{a}b + \log_{a}c = \log_{a}(bc)\) - \(\log_{a}b - \log_{a}c = \log_{a}\left(\dfrac{b}{c}\right)\) Using this property, we can rewrite the expression as: \begin{align*} \log_{10} 25 + \log_{10} 32 - \log_{10} 8 &= \log_{10}(25\cdot 32) - \log_{10} 8 \\ &= \log_{10}(800) - \log_{10} 8 \\ &= \log_{10}\left(\frac{800}{8}\right) \\ &= \log_{10}(100) \\ &= 2 \end{align*} Therefore, the value of the expression is 2. Thus, the answer is (2).
Tambaya 15 Rahoto
The angle of a sector of a circle of radius 8cm is 240°. This sector is bent to form a cone. Find the radius of the base of the cone.
Bayanin Amsa
The sector of the circle with an angle of 240° forms a cone. To find the radius of the base of the cone, we need to use the formula relating the arc length of the sector and the circumference of the base of the cone. The arc length of the sector is (240/360) * 2π * 8 = (4/3) * π * 8 = (32/3)π cm. The circumference of the base of the cone is equal to the arc length of the sector, so it is (32/3)π cm. We know that the circumference of a circle is equal to 2πr, where r is the radius of the circle. Therefore, we can write: (32/3)π = 2πr Simplifying the equation, we get: r = (16/3) cm So, the radius of the base of the cone is 16/3 cm. Therefore, the correct option is: 16/3cm
Tambaya 16 Rahoto
A cliff on the bank of a river is 300 metres
high. if the angle of depression of a point on the
opposite side of the river is 60° find the width of
the river.
Bayanin Amsa
Tambaya 17 Rahoto
The data below shows the frequency distribution
of marks scored by a group of students in a class
test.
What is the modal score
Bayanin Amsa
We find the modal score by identifying the score that occurs most frequently in the data. Looking at the frequency distribution, we can see that the score 4 appears most frequently, with a frequency of 5. Therefore, the modal score is 4.
Tambaya 18 Rahoto
A fair die is rolled once. What is the probability of
obtaining 4 or 6?
Bayanin Amsa
When you roll a fair die, there are six possible outcomes, each with an equal chance of occurring. These outcomes are the numbers 1, 2, 3, 4, 5, and 6. To find the probability of obtaining 4 or 6, we need to count how many of the outcomes correspond to these numbers. In this case, there are two outcomes that satisfy the condition, namely 4 and 6. Therefore, the probability of obtaining 4 or 6 is 2 out of 6, or 2/6. We can simplify this fraction by dividing both the numerator and denominator by their greatest common factor, which is 2. This gives us 1/3. Therefore, the probability of obtaining 4 or 6 when rolling a fair die is 1/3.
Tambaya 19 Rahoto
If x varies over the set of real numbers, which of the following is illustrated in the diagram above?
Bayanin Amsa
Tambaya 20 Rahoto
| Marks | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No of students | 1 | 3 | 2 | 0 | 1 | 6 | 1 | 0 | 1 | 0 |
The table above shows the scores of 15 students in a Physics test.
How many students scored at least 5?
Tambaya 21 Rahoto
Points X and Y are respectively 12m Nonh and
5m of point Z. Calculate XY
Bayanin Amsa
To find the distance between points X and Y, we can use the Pythagorean theorem, which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we can consider points X, Y, and Z as the vertices of a right triangle, where the right angle is at point Z. Then, XY is the hypotenuse, and we know that XZ = 12m and YZ = 5m. Therefore, we have: XY^2 = XZ^2 + YZ^2 XY^2 = 12^2 + 5^2 XY^2 = 144 + 25 XY^2 = 169 XY = √169 XY = 13m Therefore, the distance between points X and Y is 13m. So the answer is option (C) 13m.
Tambaya 22 Rahoto
Solve the equation \(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)
Bayanin Amsa
First, we simplify both sides of the equation by getting rid of the fractions. To do this, we can multiply both sides by the least common multiple (LCM) of the denominators, which is 12. So we have: \begin{align*} \frac{m}{3} + \frac{1}{2} &= \frac{3}{4} + \frac{m}{4}\\ 4 \cdot \left(\frac{m}{3} + \frac{1}{2}\right) &= 4 \cdot \left(\frac{3}{4} + \frac{m}{4}\right)\\ \frac{4m}{3} + 2 &= \frac{9}{4} + \frac{m}{4} \end{align*} Next, we can simplify further by moving all the terms involving m to one side of the equation, and all the constant terms to the other side: \begin{align*} \frac{4m}{3} - \frac{m}{4} &= \frac{9}{4} - 2\\ \frac{16m}{12} - \frac{3m}{12} &= \frac{9}{4} - \frac{8}{4}\\ \frac{13m}{12} &= \frac{1}{4}\\ 13m &= 3\\ m &= \frac{3}{13} \end{align*} Therefore, the solution to the equation is \(m = \frac{3}{13}\). So the correct answer is 3.
Tambaya 23 Rahoto
The value of tan 315°
Bayanin Amsa
To find the value of tan 315°, we can use the identity: tan (θ + 180°) = tan θ Since tan (315° + 180°) = tan 495°, and 495° is coterminal with 135°, we have: tan 315° = tan (135° + 180°) = tan 135° Using the fact that tan 45° = 1, and tan (θ + 90°) = -cot θ, we have: tan 135° = tan (45° + 90°) = -cot 45° = -1 Therefore, the value of tan 315° is -1.
Tambaya 24 Rahoto
What must be added to the expression x\(^2\) - 18x to make it a perfect square?
Bayanin Amsa
To make the expression x\(^2\) - 18x a perfect square, we need to add a constant term. The constant term should be half of the coefficient of x squared, which is (18/2)\(^2\) = 81. Therefore, we need to add 81 to the expression to make it a perfect square. Adding 81 to the expression gives: x\(^2\) - 18x + 81 = (x - 9)\(^2\) Thus, the answer is 81.
Tambaya 25 Rahoto
The data below shows the number of worker?
employed in the various sections of a construction
company in Lagos.
Carpenters 24 Labourers 27
Plumbers 12 Plasterers 15
Painters 9 Messengers 3
Bricklayers 18
If one of the workers is absent on a certain day,
what is the probability that he is a bricklayer?
Bayanin Amsa
The total number of workers in the company is 24+27+12+15+9+3+18 = 108. The probability of an absent worker being a bricklayer is the ratio of the number of bricklayers to the total number of workers, which is 18/108 = 1/6. Therefore, the probability that the absent worker is a bricklayer is 1/6. Answer: 1/6.
Tambaya 26 Rahoto
Three balls are drawn one after the other with replacement, from a bag containing 5 red, 9white and 4 blue identical balls, What is the probability that they are one red, one white and one blue?
Bayanin Amsa
The probability of drawing one red ball out of 18 balls is 5/18. Similarly, the probability of drawing one white ball is 9/18 (or 1/2) and the probability of drawing one blue ball is 4/18 (or 2/9). Since we are drawing three balls, we can find the probability of getting one red, one white, and one blue by multiplying the probabilities of each event happening together. Since we are drawing with replacement, the probabilities do not change from draw to draw. Thus, the probability of getting one red, one white, and one blue is: (5/18) x (1/2) x (2/9) = 5/162 Therefore, the answer is option C: 5/162.
Tambaya 27 Rahoto
The volume of a cone of height 9cm is 1848cm\(^3\). Find its radius. [Take π = 22/7]
Bayanin Amsa
Tambaya 28 Rahoto
What is the probability of throwing a number greater than 2 with a single fair die
Bayanin Amsa
The total number of outcomes when rolling a fair die is 6 since there are six sides on the die. The numbers that are greater than 2 are 3, 4, 5, and 6. So, there are four possible outcomes that are greater than 2. The probability of getting a number greater than 2 with a single fair die is the number of favorable outcomes divided by the total number of possible outcomes, which is 4/6 or 2/3. Therefore, the answer is (d) 2/3.
Tambaya 29 Rahoto
The data below shows the frequency distribution
of marks scored by a group of students in a class
test.
Find the mean mark.
Bayanin Amsa
Tambaya 32 Rahoto
The curved surface area of a cylindrical tin is 704cm\(^2\). Calculate the height when the radius is 8cm. [Take π = 22/7]
Bayanin Amsa
Tambaya 33 Rahoto
The diagram above shows the shaded segment of a circle of radius 7cm. if the area of the triangle OXY is 12\(\frac{1}{4}\)cm\(^2\), calculate the area of the segment
[Take ? = 22/7]
Bayanin Amsa
Tambaya 34 Rahoto
Town P is on bearing 315o from town Q while town R is south of town P and west of town Q. lf town R is 60km away from Q, how far is R from P?
Bayanin Amsa
Tambaya 35 Rahoto
In the diagram above, O is the center of the circle QOR is a diameter and ?PSR is 37°. Find ?PRQ
Bayanin Amsa
Since QOR is a diameter, it divides the circle into two equal halves, which means that ?QOR is a right angle (90°). Thus, we have: ?PQR = 180° - ?QOR (sum of angles in a triangle) ?PQR = 180° - 90° (since ?QOR is a right angle) ?PQR = 90° Now, we have a right triangle PQR with right angle at Q. Since we know that ?PSR is 37°, we can use the fact that the sum of angles in a triangle is 180° to find ?PRQ: ?PQR + ?QPR + ?PRQ = 180° 90° + ?QPR + 37° = 180° ?QPR = 180° - 90° - 37° ?QPR = 53° Therefore, the answer is 53°.
Tambaya 36 Rahoto
In the diagram above, TR = TS and ?TRS = 60o. Find the value of x
Bayanin Amsa
Since TR = TS and ?TRS = 60o, then ?TSR = ?TRS = 60o. Therefore, the angles of triangle TSP add up to 180o: ?TPS + ?TSP + ?TSR = 180o. Substituting the known values: x + 60 + 60 = 180. Simplifying: x = 60. Hence, the value of x is 60.
Tambaya 37 Rahoto
The nth term of a sequence is given by 3.2\(^{n-2}\). Write down the first three terms of the sequence.
Tambaya 38 Rahoto
A ladder leans against a vertical wall at an angle
60° to the wall, if the foot of the ladder is 5 metres
away from the wall, calculate the length of the
ladder.
Bayanin Amsa
Tambaya 39 Rahoto
The data above shows the frequency distribution
of marks scored by a group of students in a class
test.
How many students took the test?
Bayanin Amsa
Tambaya 40 Rahoto
Find the value of t in the diagram above
Bayanin Amsa
Using the Pythagorean theorem, we can find that the length of the hypotenuse of the right triangle on the left is: $$\sqrt{5^2+12^2} = 13$$ Since the triangle on the right is similar to the triangle on the left, we can set up the following proportion: $$\frac{5}{12} = \frac{t}{13}$$ Cross-multiplying, we get: $$12t = 65$$ Therefore: $$t = \frac{65}{12} \approx 5.4$$ So the answer is closest to 5.6cm. Therefore the correct option is (A).
Tambaya 41 Rahoto
Find the median of the following set of scores:
65, 72, 55, 48, 78
Bayanin Amsa
To find the median of a set of scores, we need to arrange the scores in order of magnitude from smallest to largest. The scores in this set are: 48, 55, 65, 72, 78 There are five scores in this set, so the median will be the middle score. Since the scores are already in order, we can simply find the middle score, which is 65. Therefore, the median of this set of scores is 65.
Tambaya 42 Rahoto
A number is chosen at random from the set (1 ,2,3
....,9, 10). What is the probability that the number
is greater than or equal to 7?
Bayanin Amsa
There are 10 numbers in the set (1, 2, 3, ..., 9, 10). Out of these, the numbers that are greater than or equal to 7 are 7, 8, 9, and 10. So, there are 4 numbers that satisfy the given condition. Therefore, the probability of choosing a number that is greater than or equal to 7 is the number of numbers that satisfy the condition divided by the total number of numbers in the set, which is: 4/10 = 2/5 Hence, the probability that the number chosen at random from the set (1, 2, 3, ..., 9, 10) is greater than or equal to 7 is 2/5.
Tambaya 43 Rahoto
Sin 60° has the same value as I. Sin 120° II. cos 240° III. -sin 150° IV. cos 210° V. sin 240°
Bayanin Amsa
Tambaya 44 Rahoto
Let U = {1, 2, 3, 4}, P = {2, 3} and Q = {2, 4}. What is (P∩Q)'?
Bayanin Amsa
In set theory, the symbol ∩ denotes the intersection of two sets, meaning the set of elements that are in both sets. Therefore, P∩Q denotes the set of elements that are in both P and Q, which is {2}. The complement of a set A, denoted by A', is the set of elements that are not in A but are in the universal set U. Therefore, (P∩Q)' denotes the set of elements that are in U but are not in P∩Q, which is {1, 3, 4}. Thus, the correct answer is: (1, 3, 4).
Tambaya 46 Rahoto
The data below shows the number of worker?
employed in the various sections of a construction
company in Lagos.
Carpenters 24 Labourers 27
Plumbers 12 Plasterers 15
Painters 9 Messengers 3
Bricklayers 18
If a worker is retrenched, what is the probability that he is a plumber or plasterer.
Bayanin Amsa
To calculate the probability that a retrenched worker is a plumber or plasterer, we need to first find the total number of workers in the company, and then find the number of workers who are plumbers or plasterers. The total number of workers in the company is: 24 (Carpenters) + 27 (Labourers) + 12 (Plumbers) + 15 (Plasterers) + 9 (Painters) + 3 (Messengers) + 18 (Bricklayers) = 108 The number of workers who are plumbers or plasterers is: 12 (Plumbers) + 15 (Plasterers) = 27 Therefore, the probability that a retrenched worker is a plumber or plasterer is: 27 (number of plumbers and plasterers) / 108 (total number of workers) = 1/4 So the correct answer is option D: 1/4.
Tambaya 48 Rahoto
The area of a parallelogram is 513cm\(^2\) and the height is 19cm. Calculate the base.
Bayanin Amsa
The area of a parallelogram is given by the formula: Area = base × height In this problem, the area and height of the parallelogram are given as 513cm\(^2\) and 19cm respectively. We can rearrange the formula to solve for the base: base = Area ÷ height Substituting the given values, we get: base = 513cm\(^2\) ÷ 19cm = 27cm Therefore, the base of the parallelogram is 27cm. The answer is, 27cm.
Tambaya 49 Rahoto
(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.
(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.
| Marks | 4 | 5 | 6 | 7 | 8 | 9 |
| No of candidates | 5 | 8 | 5 | 6 | 4 | 2 |
Find the mean score to the nearest whole number.
Bayanin Amsa
None
Tambaya 50 Rahoto
(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector. [Take \(\pi = \frac{22}{7}\)]
(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.
Tambaya 51 Rahoto
(a) Using a ruler and a pair of compasses only, construct: (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120°; (ii) the locus \(l_{1}\) of points equidistant from A and B; (iii) the locus \(l_{2}\) of points equidistant from AB and AC which passes through triangle ABC .
(b) Label the point P where \(l_{1}\) and \(l_{2}\) intersect.
(c) Measure |CP|.
Tambaya 52 Rahoto
Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the :
(a) distance between Q and R ;
(b) baering of R from Q.
Bayanin Amsa
None
Tambaya 53 Rahoto
(a) If \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\), find x.
(b) Without using Mathematical Tables, evaluate: \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
Tambaya 54 Rahoto
(a) Copy and complete the following table for the relation \(y = \frac{5}{2} + x - 4x^{2}\)
| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | 1 | 2.5 |
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \(-2.0 \leq x \leq 2.0\).
(c) What is the maximum value of y?
(d) From your graph, obtain the roots of the equation \(8x^{2} - 2x - 5 = 0\)
Tambaya 55 Rahoto
(a) If \(17x = 375^{2} - 356^{2}\), find the exact value of x.
(b) If \(4^{x} = 2^{\frac{1}{2}} \times 8\), find x.
(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) The sides PQ and PR of \(\Delta\) PQR are produced to T and S respectively, such that TQR = 131° and < QRS = 98°. Find < QPR.
(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre. [Take \(\pi = \frac{22}{7}\)]
None
Bayanin Amsa
None
Tambaya 57 Rahoto
(a) Solve the following pair of simultaneous equations: \(2x + 5y = 6\frac{1}{2} ; 5x - 2y = 9\)
(b) If \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\), find x.
Bayanin Amsa
None
Tambaya 58 Rahoto
(a) 
Calculate the area of the shaded segment of the circle shown in the diagram [Take \(\pi = \frac{22}{7}\)]
(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.
[Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
None
Tambaya 59 Rahoto
(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?
(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:
(i) a red, then a black marble ; (ii) two black marbles.
None
Bayanin Amsa
None
Tambaya 60 Rahoto
The table below shows the weekly profit in naira from a mini-market.
| Weekly profit (N) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |
| Freq | 6 | 6 | 12 | 11 | 10 | 5 |
(a) Draw the cumulative frequency curve of the data;
(b) From your graph, estimate the : (i) median ; (ii) 80th percentile
(c) What is the modal weekly profit?
None
Bayanin Amsa
None
