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Explanation: The mode of a data set is the value that appears most frequently.

In the given data set:

⇒ 15 appears 2 times

⇒ 12 appears 3 times

⇒ 18 appears 2 times

⇒ 10 appears 1 time

The value 12 appears more often than any other value. Therefore, the mode is 12.

The formula for the sum to infinity of a G.P. is $S_\infty = \frac{a}{1-r}$ , where $a$ is the first term and $r$ is the common ratio, provided $|r| < 1$ .

Given $a = 16$ and $r = \frac{1}{2}$ . Since $|\frac{1}{2}| < 1$ , the sum to infinity exists.

$S_\infty = \frac{16}{1 - \frac{1}{2}}$

$S_\infty = \frac{16}{\frac{1}{2}}$

$S_\infty = 16 \times 2$

$S_\infty = 32$

The permutation formula is $nP_r = \frac{n\!}{(n-r)\!}$ .

The given ratio can be written as:

$\frac{12P_{k+1}}{11P_{k+1}} = \frac{12}{5}$

Expand the permutation terms:

$\frac{\frac{12!}{(12 - (k+1))!}}{\frac{11!}{(11 - (k+1))!}} = \frac{12}{5}$

Simplify the denominators:

$\frac{\frac{12!}{(11 - k)!}}{\frac{11!}{ (10 - k)!}} = \frac{12}{5}$

Rewrite the division as multiplication by the reciprocal:

$\frac{12!}{(11 - k)!} \times \frac{(10 - k)!}{11!} = \frac{12}{5}$

Use the property $n\! = n \times (n-1)\!$ :

$12\! = 12 \times 11\!$

$(11 - k)\! = (11 - k) \times (10 - k)\!$

Substitute these into the equation:

$\frac{12 \times 11!}{(11 - k) \times (10 - k)!} \times \frac{(10 - k)!}{11!} = \frac{12}{5}$

Cancel out $11\! and (10 - k)\!$ :

$\frac{12}{11 - k} = \frac{12}{5}$

Since the numerators are equal (12 on both sides), the denominators must also be equal:

11 - k = 5

Solve for k:

k = 11 - 5

k = 6

Explanation:

⇒ In a right-angled triangle, the Pythagorean theorem states that the square of the hypotenuse (q) is equal to the sum of the squares of the other two sides (p and r):$q^2 = p^2 + r^2$

⇒ Given:$q = 13 \text{ cm}, r = 5 \text{ cm}$

$13^2 = p^2 + 5^2$

$169 = p^2 + 25$

$p^2 = 169 - 25$

$p^2 = 144$

$p = \sqrt{144} = 12 \text{ cm}$

For a $2 \times 2 matrix A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $ad - bc$ .

For matrix $A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ ,

Determinant of $A = (2)(3) - (1)(4)$

$= 6 - 4$

$= 2$ .

Explanation:

1. Calculate the Loss:

$\text{Loss} = \text{Cost Price} - \text{Selling Price}$

$\text{Loss} = \text{₦}80,000 - \text{₦}64,000 = \text{₦}16,000$

2. Calculate Percentage Loss:

$\text{Percentage Loss} = \frac{\text{Loss}}{\text{Cost Price}} \times 100\%$

$\text{Percentage Loss} = \frac{\text{₦}16,000}{\text{₦}80,000} \times 100\%$

$\text{Percentage Loss} = \frac{16}{80} \times 100\%$

$\text{Percentage Loss} = \frac{1}{5} \times 100\%$

$\text{Percentage Loss} = 20\%$

The relationship is $P = k_1 Q + k_2 Q^2$ .

Using $P=10, Q=1$ :

$10 = k_1(1) + k_2(1)^2 \Rightarrow 10 = k_1 + k_2$ (Equation 1)

Using $P=28, Q=2$ :

$28 = k_1(2) + k_2(2)^2 \Rightarrow 28 = 2k_1 + 4k_2$ (Equation 2)

From Equation 1, $k_1 = 10 - k_2$ . Substitute into Equation 2:

$28 = 2(10 - k_2) + 4k_2$

$28 = 20 - 2k_2 + 4k_2$

$28 - 20 = 2k_2$

$8 = 2k_2 \Rightarrow k_2 = 4$ .

Substitute $k_2=4$ into $k_1 = 10 - k_2$ :

$k_1 = 10 - 4 = 6$ .

So, the relationship is $P = 6Q + 4Q^2$ .

Explanation:

1. Total possible outcomes: When two dice are tossed, the total number of possible outcomes is 6 \times 6 = 36

2. Possible sums (from 2 to 12):

3. Identify prime numbers among these sums: The prime numbers between 2 and 12 are 2, 3, 5, 7, 11.

4. List outcomes that sum to these prime numbers:

⇒ Sum = 2: (1,1) - 1 outcome

⇒ Sum = 3: (1,2), (2,1) - 2 outcomes

⇒ Sum = 5: (1,4), (4,1), (2,3), (3,2) - 4 outcomes

⇒ Sum = 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) - 6 outcomes

⇒ Sum = 11: (5,6), (6,5) - 2 outcomes

5. Total favorable outcomes: 1 + 2 + 4 + 6 + 2 = 15 outcomes.

6. Calculate the probability:

$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{15}{36}$

7. Simplify the fraction: Divide both numerator and denominator by their greatest common divisor, 3.

$\frac{15 \div 3}{36 \div 3} = \frac{5}{12}$

Using the formula for grouped mode:

$\text{Mode} = L + \left(\frac{f_m - f_1}{2f_m - f_1 - f_2}\right) \times w$

Modal class: $20-29 (f\_m = 28)$

Lower class boundary (L) = 19.5

$f_1 = 15$

$f_2 = 22$

Class width (w) = 10

$\text{Mode} = 19.5 + \left(\frac{28 - 15}{2(28) - 15 - 22}\right) \times 10$

$\text{Mode} = 19.5 + \left(\frac{13}{56 - 37}\right) \times 10$

$\text{Mode} = 19.5 + \left(\frac{13}{19}\right) \times 10$

$\text{Mode} = 19.5 + \frac{130}{19}$

$\text{Mode} = 19.5 + 6.8421...$

$\text{Mode} \approx 26.34$

Rounding to one decimal place: 26.3.

Cost of 200 shares = 200 shares $\times$ ₦50/share = ₦10,000.

VAT amount = 5% of ₦10,000 = $\frac{5}{100} \times 10000 = ₦500$.

Total amount paid = Cost of shares + VAT amount = ₦10,000 + ₦500 = ₦10,500.

Explanation:

⇒ To differentiate $y = 3\cos(5x)$ , we use the chain rule.

⇒ Let u = 5x. Then $\frac{du}{dx} = 5$ .

⇒ The function becomes $y = 3\cos(u)$ .

⇒ Differentiate y with respect to u:

$\frac{dy}{du} = 3(-\sin(u)) = -3\sin(u)$ .

⇒ By the chain rule,

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ .

$\frac{dy}{dx} = (-3\sin(u)) \times 5 = -15\sin(u)$

⇒ Substitute u = 5x back: $\frac{dy}{dx} = -15\sin(5x)$ .

First, convert the fractions to decimals or common fractions:

$\frac{2}{0.05} = \frac{2}{\frac{5}{100}} = \frac{200}{5} = 40$

==> $\frac{1}{0.015} = \frac{1}{\frac{15}{1000}} = \frac{1000}{15} = \frac{200}{3}$

Now substitute these values into the expression:

$\left|40 - \frac{200}{3}\right| = \left|\frac{120}{3} - \frac{200}{3}\right| = \left|-\frac{80}{3}\right|$

The absolute value is:

$\left|-\frac{80}{3}\right| = \frac{80}{3}$

Convert to decimal and round to 2 decimal places:

$\frac{80}{3} \approx 26.666... \approx 26.67$

Total hours worked = 40 hours.

Overtime hours = h hours.

Standard work hours = Total hours - Overtime hours = 40 - h hours.

Earning from standard work = Standard hours \\times standard rate

$\text{Standard Earning} = (40 - h) \times p = (40 - h)p$

Earning from overtime = Overtime hours x overtime rate (1.5p)

==> $\text{Overtime Earning} = h \times (1.5p) = 1.5hp$

Total weekly earning = Standard Earning + Overtime Earning

==> $\text{Total Earning} = (40 - h)p + 1.5hp$

Expand the expression:

$\text{Total Earning} = 40p - hp + 1.5hp$

Combine like terms:

$\text{Total Earning} = 40p + 0.5hp$

Factor out p:

$\text{Total Earning} = p(40 + 0.5h)$

Explanation:

1. Median: The data set has 8 values (an even number). The median is the average of the two middle values, which are the 4th and 5th terms (x and 40).

⇒ $\text{Median} = \frac{x + 40}{2}$

⇒ Given, Median = 35.

$35 = \frac{x + 40}{2}$

$70 = x + 40$

$x = 70 - 40 = 30$

2. Mean: The mean is the sum of all values divided by the count.

⇒ $\text{Sum} = 20 + 25 + 30 + x + 40 + y + 55 + 60$

⇒ Substitute x=30:

$\text{Sum} = 20 + 25 + 30 + 30 + 40 + y + 55 + 60 = 260 + y$

⇒ Given, Mean = 40.

⇒ $40 = \frac{260 + y}{8}$

⇒ $320 = 260 + y$

⇒ $y = 320 - 260 = 60$

Therefore, x = 30 and y = 60.

Rearrange the equation to isolate one square root:

$\sqrt{y+1} = 1 + \sqrt{y-2}$

Square both sides of the equation:

$(\sqrt{y+1})^2 = (1 + \sqrt{y-2})^2$

Expand the right side (using $(a+b)^2 = a^2 + 2ab + b^2$ ):

$y+1 = 1^2 + 2(1)\sqrt{y-2} + (\sqrt{y-2})^2$

$y+1 = 1 + 2\sqrt{y-2} + y-2$

Simplify the right side:

$y+1 = y - 1 + 2\sqrt{y-2}$

Subtract y from both sides:

$1 = -1 + 2\sqrt{y-2}$

Add 1 to both sides:

$2 = 2\sqrt{y-2}$

Divide by 2:

$1 = \sqrt{y-2}$

Square both sides again to solve for y:

$1^2 = (\sqrt{y-2})^2$

$1 = y-2$

$y = 1 + 2$

$y = 3$

Check the solution: Substitute $y = 3$ back into the original equation.

$\sqrt{3+1} - \sqrt{3-2} = 1$

$\sqrt{4} - \sqrt{1} = 1$

$2 - 1 = 1$

$1 = 1$

Explanation:

⇒ Using the product rule of logarithms: $\log_b X + \log_b Y = \log_b (X \times Y)$

$\log_a 4 + \log_a 16 = \log_a (4 \times 16) = \log_a 64$

⇒ So the equation becomes:

$\log_a 64 = 3$

⇒ By definition of logarithm, if $\log_a X = Y$ , then $a^Y = X$ .

⇒ So, $a^3 = 64.$

⇒ To find a, take the cube root of 64:

⇒ $a = \sqrt[3]{64}$

⇒ Since $4 \times 4 \times 4 = 64$ ,

⇒ a = 4

We can use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$ .

$(\frac{3}{5})^2 + \cos^2 \theta = 1$

$\frac{9}{25} + \cos^2 \theta = 1$

$\cos^2 \theta = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$

$\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$ (since $\theta$ is acute, $\cos \theta$ is positive).

Now, $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$.

1. Find the union of B and C $(B \cup C)$ :

$B = \{b, c, d, e\}$

$C = \{a, c, e, g\}$

$B \cup C = \{a, b, c, d, e, g\}$

2. Find the intersection of $A and (B \cup C) (A \cap (B \cup C))$ :

$A = \{a, b, c\}$

$B \cup C = \{a, b, c, d, e, g\}$

$A \cap (B \cup C) = \{a, b, c\}$

Explanation:

1. Evaluate the terms with exponents:

$3^{-1} = \frac{1}{3}$

$2^2 = 4$

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

2. Substitute these values into the expression:

$\frac{\frac{1}{3} + 4}{\frac{1}{4}}$

3. Simplify the numerator:

$\frac{1}{3} + 4 = \frac{1}{3} + \frac{12}{3} = \frac{13}{3}$

4. Perform the division:

$\frac{\frac{13}{3}}{\frac{1}{4}} = \frac{13}{3} \times \frac{4}{1} = \frac{52}{3}$

1. Find the reference angle:

First, find the acute angle (reference angle, let's call it $\beta$ ) for which $\cos \beta = \frac{\sqrt{3}}{2}$ .

$\cos \beta = \frac{\sqrt{3}}{2} \implies \beta = 30^\circ$

2. Determine the quadrants:

Since $\cos \alpha$ is negative $(-\frac{\sqrt{3}}{2})$ , $\alpha$ must lie in the quadrants where cosine is negative. These are the second (Q2) and third (Q3) quadrants.

3. Calculate $\alpha$ in Q2:

In the second quadrant, $\alpha$ $= 180^\circ - \beta$ .

$\alpha_1 = 180^\circ - 30^\circ = 150^\circ$

4. Calculate $\alpha$ in Q3:**

In the third quadrant, $\alpha = 180^\circ + \beta$ .

$\alpha_2 = 180^\circ + 30^\circ = 210^\circ$

Therefore, the values of $\alpha$ for which $\cos \alpha = -\frac{\sqrt{3}}{2}$ in the given range are $150^\circ$ and $210^\circ.$