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**Question 1**
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The mean and the range of the set of numbers 1.20, 1.00, 0.90, 1.40, 0.80, 1.20, and 1.10 are m and r respectively. Find m + r

**Answer Details**

To find the mean of a set of numbers, we add up all the numbers in the set and divide by the total number of numbers. To find the range, we subtract the smallest number from the largest number in the set. In this case, we have the set of numbers: 1.20, 1.00, 0.90, 1.40, 0.80, 1.20, and 1.10. To find the mean, we add up all the numbers and divide by 7 (the total number of numbers in the set): mean = (1.20 + 1.00 + 0.90 + 1.40 + 0.80 + 1.20 + 1.10) / 7 = 1.14 To find the range, we subtract the smallest number (0.80) from the largest number (1.40): range = 1.40 - 0.80 = 0.60 Therefore, m + r = 1.14 + 0.60 = 1.74. So the correct answer is 1.69.

**Question 2**
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class1−34−67−9Frequency585 $\begin{array}{cccc}class& 1-3& 4-6& 7-9\\ Frequency& 5& 8& 5\end{array}$

Find the standard deviation of the data using the table above

**Answer Details**

class intervalsFre(F)class-marks(x)Fx(x−x)(x−x)2F(x−x)21−351010−39904−6840400007−954040393601890450
$\begin{array}{ccccccc}\text{class intervals}& Fre(F)& \text{class-marks(x)}& Fx& (x-x)& (x-x{)}^{2}& F(x-x{)}^{2}\\ 1-3& 5& 10& 10& -3& 9& 90\\ 4-6& 8& 40& 40& 0& 0& 0\\ 7-9& 5& 40& 40& 3& 9& 360\\ & 18& & 90& & & 450\end{array}$

x = ∑fx∑f
$\frac{\sum fx}{\sum f}$

= 9018
$\frac{90}{18}$

= 5

S.D = ∑f(x−x)2∑f
$\frac{\sum f(x-x{)}^{2}}{\sum f}$

= 45018
$\frac{450}{18}$

= √25
$25$

= 5

**Question 3**
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If x = (1203) $\left(\begin{array}{cc}1& 2\\ 0& 3\end{array}\right)$ and y = (2143) $\left(\begin{array}{cc}2& 1\\ 4& 3\end{array}\right)$. Find xy.

**Answer Details**

To find xy, we need to perform matrix multiplication of x and y. To do this, we take the dot product of each row of x with each column of y, as follows: First row of x: (1 2 0 3) First column of y: ( 2 1 4 3 ) Dot product of the first row of x with the first column of y: 1*2 + 2*1 + 0*4 + 3*3 = 2 + 2 + 9 = 13 This gives us the first element of the product matrix. We repeat this process for the remaining rows and columns to get: xy = ( 1*2 + 2*1 + 0*4 + 3*3 1*1 + 2*4 + 0*3 + 3*2 1*4 + 2*3 + 0*2 + 3*1 1*3 + 2*2 + 0*1 + 3*4 0*2 + 3*1 + 1*4 + 2*3 0*1 + 3*4 + 1*3 + 2*2 0*4 + 3*3 + 1*2 + 2*1 0*3 + 3*2 + 1*1 + 2*4 ) = ( 13 11 10 17 11 19 13 14 ) Hence, xy = (107129). Therefore, the correct option is (A).

**Question 4**
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The variance of the scores 1, 2, 3, 4, 5 is

**Answer Details**

To calculate the variance of a set of numbers, you first need to find the mean (average) of the set. The mean of 1, 2, 3, 4, 5 is: (1 + 2 + 3 + 4 + 5) ÷ 5 = 3 Next, subtract the mean from each number in the set and square the differences. This gives you the squared deviations from the mean: (1 - 3)^2 = 4 (2 - 3)^2 = 1 (3 - 3)^2 = 0 (4 - 3)^2 = 1 (5 - 3)^2 = 4 The sum of the squared deviations is: 4 + 1 + 0 + 1 + 4 = 10 Finally, divide the sum of the squared deviations by the number of items in the set: 10 ÷ 5 = 2 Therefore, the variance of the scores 1, 2, 3, 4, 5 is 2. Therefore, the option that represents the variance of the scores 1, 2, 3, 4, 5 is 2.05.

**Question 5**
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The graph of f(x) = x2 - 5x + 6 crosses the x-axis at the points

**Answer Details**

To find where the graph of the function crosses the x-axis, we need to find the values of x when f(x) = 0. So we can set the function equal to 0 and solve for x: x^2 - 5x + 6 = 0 We can factor the quadratic equation as: (x - 2)(x - 3) = 0 So the solutions are x = 2 and x = 3. Therefore, the graph of the function crosses the x-axis at the points (2, 0) and (3, 0). Therefore, the correct option is: - (2, 0), (3, 0)

**Question 6**
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For what value of x is the tangent to the curve y = x2 $2$ - 4x + 3 parallel to the x-axis?

**Answer Details**

To find the value of x for which the tangent to the curve y = x^2 - 4x + 3 is parallel to the x-axis, we need to find the point on the curve where the slope of the tangent line is zero. The slope of the tangent to a curve y = f(x) at a point (a, f(a)) is given by f'(a), the derivative of f(x) evaluated at x = a. So, we need to find the derivative of y = x^2 - 4x + 3, which is y' = 2x - 4. Setting y' equal to zero, we get: 2x - 4 = 0 Solving for x, we get: x = 2 Therefore, the value of x for which the tangent to the curve y = x^2 - 4x + 3 is parallel to the x-axis is 2.

**Question 7**
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Factorize completely the expression abx2 + 6y - 3ax - 2byx

**Question 9**
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A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank.

**Answer Details**

To solve this problem, we can use trigonometry. Let h be the height of the water tank, and d be the horizontal distance between the foot of the water tank tower and the position of the boy. Then, we have: tan(60°) = h / d We know that d = 30m, and tan(60°) = √3, so we can solve for h: h = d * tan(60°) h = 30m * √3 h = 30√3 m Therefore, the height of the water tank is 30√3 meters.

**Question 11**
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∣∣ ∣∣−21121k13−1∣∣ ∣∣ $\left|\begin{array}{ccc}-2& 1& 1\\ 2& 1& k\\ 1& 3& -1\end{array}\right|$ = 23

**Answer Details**

To solve the absolute value equation ∣∣ ∣∣−21121k13−1∣∣ ∣∣ = 23, we need to consider two cases: when the expression inside the absolute value bars is positive and when it is negative. Case 1: When the expression inside the absolute value bars is positive: −21121k13−1 = 23 Simplifying the equation, we get: −21121k13 = 24 Multiplying both sides by -1, we have: 21121k13 = -24 Multiplying both sides by 13, we have: 21121k = -312 Therefore, k = -312/21121 Case 2: When the expression inside the absolute value bars is negative: −(−21121k13−1) = 23 Simplifying the equation, we get: 21121k13−1 = 23 Adding 1 to both sides, we have: 21121k13 = 24 This is the same equation we obtained in Case 1. Therefore, the value of k is the same in both cases: k = -312/21121 Therefore, the answer is 2.

**Question 12**
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Find the sum to infinity of the following sequence 1. 910 $\frac{9}{10}$, 2. (910 $\frac{9}{10}$), 3. (910 $\frac{9}{10}$)

**Question 13**
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Find the area bounded by the curve y = 3x2 $2$ - 2x + 1, the coordinates x = 1 and x = 3 and the x-axis

**Question 14**
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Solve the inequality (x - 3)(x - 4) ≤ $\le $ 0

**Answer Details**

To solve this inequality, we need to find the values of x that make the expression (x - 3)(x - 4) less than or equal to 0. To do this, we can use a method called the "sign chart" or "interval notation". First, we can find the critical values of x by setting the expression equal to 0: (x - 3)(x - 4) = 0. This gives us x = 3 and x = 4. Next, we can create a sign chart by dividing the number line into intervals using the critical values of x. We then choose a test point from each interval and evaluate the expression (x - 3)(x - 4) using that test point. If the result is positive, the expression is positive in that interval. If the result is negative, the expression is negative in that interval. Intervals: (-∞, 3), (3, 4), (4, ∞) Test point: -1, 3.5, 5 Expression: (-)(-)=+ (+)(-)= - (+)(+)=+ From the sign chart, we can see that the expression is less than or equal to 0 when x is between 3 and 4, inclusive. Therefore, the solution is: 3 ≤ x ≤ 4 So the correct option is: - 3 ≤ x ≤ 4

**Question 17**
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The pie chart above shows the distribution of students in a secondary school class. If 30 students offered French, how many offered CRK?

**Question 18**
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A die has four of it's faces coloured white and the remaining two coloured black. What is the probability that when the die is thrown two consecutive time, the top face will be white in both cases?

**Answer Details**

The probability of getting a white face on a single roll of the die is 4/6 or 2/3, since four of the six faces are white. Similarly, the probability of getting a white face on a second roll of the die is also 2/3. To find the probability of both events occurring (getting a white face on both rolls), we multiply the probabilities of each event together. So, the probability of getting a white face on the first roll AND getting a white face on the second roll is (2/3) x (2/3) = 4/9. Therefore, the correct answer is not provided in the options given.

**Question 19**
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Find T in terms of K, Q and S if S = 2r(π $\pi $QT + K)

**Answer Details**

s24r2
$\frac{{s}^{2}}{4{r}^{2}}$ = QTπ
$\pi $ + KT

s24r2
$\frac{{s}^{2}}{4{r}^{2}}$ - kπ
$\pi $ = QTπ
$\pi $

T = s24Qπr2 $\frac{{s}^{2}}{4Q\pi {r}^{2}}$ - k

**Question 20**
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In a triangle XYZ, < YXZ = 44° and < XYZ = 112°. Calculate the acute angle between the internal bisectors of < XYZ and < XZY.

**Question 21**
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Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR

**Question 22**
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Class Interval1.5?1.92.0?2.42.5?2.93.0?2.93.5?3.94.0?4.44.5?4.9Frequency2214151053Class boundaries1.45?1.951.95?2.452.45?2.952.95?3.453.45?3.953.95?4.454.45?4.95ClassMid?point1.72.22.73.23.74.24.7

Find the mode of the distribution above to find the mode of the distribution.

**Answer Details**

Mode = a + (b - a)(fm - Fb)

2Fm - Fa - Fb

= 3.0 + (3.4?3)(15?4)2(15)?4?10
$\frac{(3.4?3)(15?4)}{2(15)?4?10}$

= 3 + (6.4)(11)30?14
$\frac{(6.4)(11)}{30?14}$

= 3 + 4.416
$\frac{4.4}{16}$

= 3 + 0.275

= 3.275

= 3.3cm

**Question 23**
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Class IntervalFrequencyClass boundariesClass Mid-point1.5−1.921.45−1.951.72.0−2.4211.95−2.452.22.5−2.942.45−2.952.73.0−2.9152.95−3.453.23.5−3.9103.45−3.953.74.0−4.453.95−4.454.24.5−4.934.45−4.954.7 $\begin{array}{cccc}\text{Class Interval}& Frequency& \text{Class boundaries}& \text{Class Mid-point}\\ 1.5-1.9& 2& 1.45-1.95& 1.7\\ 2.0-2.4& 21& 1.95-2.45& 2.2\\ 2.5-2.9& 4& 2.45-2.95& 2.7\\ 3.0-2.9& 15& 2.95-3.45& 3.2\\ 3.5-3.9& 10& 3.45-3.95& 3.7\\ 4.0-4.4& 5& 3.95-4.45& 4.2\\ 4.5-4.9& 3& 4.45-4.95& 4.7\end{array}$

The median of the distribution above is

**Answer Details**

Median = a+bfm
$\frac{a+b}{fm}$ (12∑f−CFb
$\frac{1}{2}\sum f-C{F}_{b}$)

= 2.95 + 0.515
$\frac{0.5}{15}$(2.-7)

= 2.95 + 0.515
$\frac{0.5}{15}$ x 13

= 2.95 + 0. 43

= 3.38

= 3.4

**Question 24**
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Convert 3.1415926 to 5 decimal places

**Answer Details**

To convert a number to a specific number of decimal places, we need to look at the digit in the next decimal place and round the original number accordingly. In this case, the digit in the sixth decimal place is 2, which is less than 5. Therefore, we do not need to round up the fifth decimal place, which is 9. Thus, the correct answer is 3.14159, which is the same as the original number but rounded to 5 decimal places.

**Question 25**
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The derivative of cosec x is

**Answer Details**

The derivative of cosec x is -cot x cosec x. Cosec x is the reciprocal of sin x, so we can rewrite it as (1/sin x). Using the quotient rule of differentiation, we can find the derivative of cosec x as follows: d/dx (cosec x) = d/dx (1/sin x) = (-1/sin^2x) * d/dx (sin x) Now, using the chain rule, we can find the derivative of sin x: d/dx (sin x) = cos x Substituting this back into our original expression, we get: d/dx (cosec x) = (-1/sin^2x) * cos x Simplifying this, we get: d/dx (cosec x) = -cot x cosec x

**Question 26**
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In the diagram, find PQ if the area of triangle PQR is 35cm2

**Answer Details**

To find the length of PQ, we need to use the formula for the area of a triangle, which is: Area = 1/2 * base * height In this case, the area is given as 35cm^2, and we know that the height is PQ. We also know that the base is QR, since the height is perpendicular to the base. So, we can write: 35 = 1/2 * QR * PQ Multiplying both sides by 2 and dividing by PQ, we get: 70/PQ = QR We don't have the value of QR, but we can use the Pythagorean theorem to find it. In triangle PQR, we have: PR^2 = PQ^2 + QR^2 We don't know the value of PR, but we can express it in terms of PQ and QR using the fact that PQR is a right-angled triangle. We have: PR = sqrt(PQ^2 + QR^2) Substituting this expression into the Pythagorean theorem, we get: PQ^2 + QR^2 = (sqrt(PQ^2 + QR^2))^2 Simplifying, we get: PQ^2 + QR^2 = PQ^2 + QR^2 This is an identity, which means it is true for all values of PQ and QR. But we can use it to eliminate QR from the equation we got earlier: 70/PQ = QR Multiplying both sides by QR, we get: 70 = PQ^2 + QR^2 Substituting QR^2 = 70 - PQ^2, we get: 70/PQ = sqrt(70 - PQ^2) Squaring both sides, we get: 4900/PQ^2 = 70 - PQ^2 Rearranging, we get: PQ^4 - 70PQ^2 + 4900 = 0 This is a quadratic equation in PQ^2, which we can solve using the quadratic formula: PQ^2 = (70 ± sqrt(70^2 - 4*1*4900))/2 PQ^2 = (70 ± 20sqrt(3))/2 We can discard the negative solution, since PQ is a length and can't be negative. So, we have: PQ^2 = 35 + 10sqrt(3) Taking the square root, we get: PQ ≈ 7.98 So, the closest option is 8cm, which corresponds to the first option, "7cm".

**Question 27**
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Simplify ?12??3?12+?3

**Question 28**
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PT is a tangent to the circle TYZX. YT = YX and < PTX = 50o. Calculate < TZY

**Answer Details**

XTY = 50o = TXY = 130o2
$\frac{{130}^{o}}{2}$

= 65o

= 65o

**Question 29**
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Find the distance between two towns p(45o N, 30oW) and Q(15oS, 30oW) if the radius of the earth is 7000km. [π=227 $\pi =\frac{22}{7}$]

**Question 30**
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If x–1 $x\u20131$ and x+1 $x+1$ are both factors of the equation x3+px2+qx+6=0 ${x}^{3}+p{x}^{2}+qx+6=0$, evaluate p $p$ and q

**Question 31**
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Express the product of 0.0014 and 0.011 in standard form

**Answer Details**

To express the product of 0.0014 and 0.011 in standard form, we need to multiply these two numbers together and then express the result in the form of a x 10^n, where 1 ≤ a < 10 and n is an integer. 0.0014 multiplied by 0.011 gives us 0.0000154. We can then express this in standard form by moving the decimal point to the right until there is only one non-zero digit to the left of the decimal point. This gives us 1.54 x 10^-5, which is option D. Therefore, the answer is: - 1.54 x 10-5

**Question 32**
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Let p be a probability function on set S, where S = (a1, a2, a3, a4). Find P(a1) if P(a2) = 13 $\frac{1}{3}$, p(a3) = 16 $\frac{1}{6}$ and p(a4) = 15