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Question 3 Report
A box contains 2 white and 3 blue identical marbles. If two marbles are picked at random, one after the other without replacement, what is the probability of picking two marbles of different colors?
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Question 5 Report
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Question 6 Report
Given that sin \(\theta\) = -0.9063, where O \(\leq\) \(\theta\) \(\leq\) 270°, find \(\theta\).
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Question 7 Report
In the diagram above, O is the center of the circle, |SQ| = |QR| and ?PQR = 68°. Calculate ?PRS
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Question 8 Report
Solve: 6(x - 4) + 3(x + 7) = 3
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We can solve the given equation by using the distributive property of multiplication and combining like terms:
6(x - 4) + 3(x + 7) = 3 6x - 24 + 3x + 21 = 3 9x - 3 = 3 9x = 6 x = 6/9
Simplifying the fraction 6/9, we get:
x = 2/3
Therefore, the solution to the given equation is x = 2/3. So, the correct answer is:
2/3
Question 11 Report
P = {2, 1,3, 9, 1/2}; Q = {1,21/2,3, 7} and R = {5, 4, 21/2}. Find P∩Q∩R
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Question 12 Report
Question 13 Report
In the diagram above , |AD| = 10cm, |DC| = 8cm and |CF| = 15cm. Which of the following is correct?
Question 14 Report
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Question 15 Report
Find the mean of the distribution
Question 16 Report
The population of a village is 5846. Express this number to three significant figures
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Question 17 Report
Which of the following is not a measure of dispersion?
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Question 18 Report
The angles of a pentagon are x°, 2x°, (x + 60)°, (x + 10)°, (x -10)°. Find the value of x.
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In a pentagon, the sum of the angles is equal to (5 - 2) times 180 degrees, which is 540 degrees. So we can set up an equation to solve for x:
x + 2x + (x + 60) + (x + 10) + (x - 10) = 540
Simplifying this equation, we get:
6x + 60 = 540
Subtracting 60 from both sides, we get:
6x = 480
Dividing both sides by 6, we get:
x = 80
Therefore, the value of x is 80. So, the correct answer is:
80
Question 19 Report
P={2, 1,3, 9, 1/2}; Q = {1,21/2,3, 7} and R = {5, 4, 21/2}. Find P∪Q∪R
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Question 21 Report
Find the median of the distribution
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Question 22 Report
If events X and Y are mutually exclusive, . P(X) = 1/3 and P(Y) = 2/5, P(X∩Y) is
Question 24 Report
Mrs. Jones is expecting a baby. The probability that it will be a boy is 1/2 and probability that the baby will have blue eyes is 1/4. What is the probability that she will have a blue-eyed boy?
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Question 25 Report
Find the area of the enclosed region, PXROY correct to the nearest whole number
Question 26 Report
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Question 27 Report
If events X and Y are mutually exclusive, P(X) = 1/3 and P(Y) = 2/5, P(X∪Y) is
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Question 28 Report
In the diagram above , |AD| = 10cm, |DC| = 8cm and |CF| = 15cmIf the area of triangle DCF = 24cm2, find the area of the quadrilateral ABCD.
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Question 29 Report
Question 32 Report
The shaded portion shows the outer boundary
of the half plane defined by the inequality
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To understand this problem, let's first define a half-plane. A half-plane is a part of the plane that lies on one side of a straight line and extends infinitely far in that direction. In this problem, we have an inequality that defines a half-plane.
The inequality is: 4x + 3y ≥ 6
To graph this inequality, we can first plot the line 4x + 3y = 6. To plot this line, we can find two points on the line by setting x = 0 and y = 0 and solving for the other variable.
When x = 0, we get: 3y = 6, y = 2
When y = 0, we get: 4x = 6, x = 3/2
So the two points on the line are (0, 2) and (3/2, 0). We can plot these points and draw a straight line passing through them.
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Now we need to determine which side of the line represents the half-plane defined by the inequality 4x + 3y ≥ 6.
To do this, we can choose a test point not on the line, such as the origin (0,0), and substitute its coordinates into the inequality:
4(0) + 3(0) ≥ 6
0 ≥ 6
Since this is false, the point (0,0) is not in the half-plane defined by the inequality. Therefore, we shade the half-plane that does not include the origin:
|xxxxxxx
|xxxxxxx
|xxxxxxx
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xxxxx|
xxxxx|
xxxxx|
The shaded portion shows the outer boundary of the half-plane defined by the inequality 4x + 3y ≥ 6.
Answer: 4x + 3y ≥ 6.
Question 33 Report
In the diagram above, ATR is a tangent at the point T to the circle center O, if ?TOB = 145°, find ?TAO
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Question 34 Report
Find the number whose logarithm to base 10 is 2.6025
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Question 35 Report
For what value of y is the expression \(\frac{y + 2}{y^{2} - 3y - 10}\) undefined?
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Question 36 Report
Factorize 3a\(^2\) - 11a + 6
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Question 37 Report
In the diagram above, PQ and XY are two concentric arc; center O, the ratio of the length of the two arc is 1:3, find the ratio of the areas of the two sectors OPQ and OXY
Question 38 Report
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Question 39 Report
The mean of 30 observations recorded in an experiment is 5. lf the observed largest value of 34 is deleted, find the mean of the remaining observations
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Question 40 Report
In the diagram above, ?PTQ = ?URP = 25° and XPU = 4URP. Calculate ?USQ.
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Question 42 Report
Question 43 Report
Question 44 Report
Simplify: log6 + log2 - log12
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Question 45 Report
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Question 46 Report
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Question 47 Report
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Question 48 Report
Solve the equation: 3a + 10 = a\(^2\)
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Question 49 Report
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Question 50 Report
In a class of 80 students, every student had to study Economics or Geography or both Economics and Geography. lf 65 students studied Economics and 50 studied Geography, how many studied both subjects?
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We are given that every student had to study Economics or Geography or both. Therefore, the total number of students in the class is 80.
Let's denote the number of students who studied only Economics by 'E', the number of students who studied only Geography by 'G', and the number of students who studied both subjects by 'B'. Then we can use a Venn diagram to represent the information given in the problem:
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| Economics | Geography |
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B G
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| Only Economics | Only Geography |
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We are given that 65 students studied Economics, which includes both those who studied only Economics and those who studied both subjects. So we can write:
E + B = 65
Similarly, we are given that 50 students studied Geography, which includes both those who studied only Geography and those who studied both subjects. So we can write:
G + B = 50
We want to find the value of B, the number of students who studied both subjects. To do this, we can add the two equations above:
E + B = 65
G + B = 50
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E + G + 2B = 115
We know that the total number of students in the class is 80, so we can write:
E + G + B = 80
Substituting the expression for E + G + 2B into this equation, we get:
115 - B = 80
Solving for B, we get:
B = 35
Therefore, 35 students studied both Economics and Geography.
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