The mean of five numbers is \(\frac{\text{sum of numbers}}{5}\). Solving, \(12 \times 5 = 60\). The sum of the four known numbers is \(9 + 10 + 15 + 14 = 48\). The fifth number is \(60 – 48 = 12\).

The probability of picking two red balls is \(\frac{\binom{5}{2}}{\binom{12}{2}}\). Calculating, \(\binom{5}{2} = 10\) and \(\binom{12}{2} = 66\). Thus, the probability is \(\frac{10}{66} = \frac{5}{33}\).

To get a sum of \(9\), possible pairs are [(3,6), (4,5), (5,4), (6,3)], totaling \(4\) favorable outcomes. The total outcomes when rolling two dice are \(36\). The probability is \(\frac{4}{36} = \frac{1}{9}\).

Students who study only math are those who study math but not both subjects. The total studying math is \(18\), and those studying both are \(10\). Thus, students studying only math is \(18 – 10 = 8\).

Let the cost price of each item be \(C_1\) and \(C_2\). For the profit item, \(S_1 = 1.2C_1\), so \(C_1 = \frac{240}{1.2} = 200\). For the loss item, \(S_2 = 0.8C_2\), so \(C_2 = \frac{240}{0.8} = 300\). Total cost is \(200 + 300 = 500\). Total revenue is \(480\). Loss is \(\frac{500 – 480}{500} = 4\%\).

Let the width be \(x\). Then the length is \(x + 4\). The area is \(x(x + 4) = 96\). Solving, \(x^2 + 4x – 96 = 0\). Factoring, \((x – 8)(x + 12) = 0\), so \(x = 8\) (positive root). The dimensions are \(8\) cm by \(12\) cm.

The combined rate is \(\frac{1}{3} + \frac{1}{5} = \frac{5}{15} + \frac{3}{15} = \frac{8}{15}\) of the tank per hour. To fill \(75\%\) or \(\frac{3}{4}\) of the tank, time required is \(\frac{\frac{3}{4}}{\frac{8}{15}} = \frac{3 \times 15}{4 \times 8} = \frac{45}{32} \approx 1.4\) hours.

The original speed is \(\frac{120}{2} = 60\) km/h. After a \(20\%\) decrease, the new speed is \(60 \times 0.8 = 48\) km/h. The time to travel \(150\) km is \(\frac{150}{48} \approx 3.125\) hours, which rounds to \(3.25\) hours or \(3.75\) hours.

The surface area of a sphere is \(4\pi r^2\). Setting \(4\pi r^2 = 144\pi\), divide by \(\pi\): \(4r^2 = 144\). Simplify: \(r^2 = 36 \implies r = 6\) cm.

Using the Pythagorean theorem, \(x^2 + 6^2 = 10^2\). Simplify: \(x^2 + 36 = 100 \implies x^2 = 64 \implies x = 8\) cm.

The triangle is a right triangle because \(8^2 + 15^2 = 17^2\). The area is \(\frac{1}{2} \cdot 8 \cdot 15 = 60\) cm\(^2\).

The volume of a cone is given by \(V = \frac{1}{3}\pi r^2 h\). Substituting the given values, \(V = \frac{1}{3}\pi (4)^2 (6) = \frac{1}{3}\pi (16)(6) = 32\pi\).

Find two numbers that multiply to \(6 \cdot 6 = 36\) and add to \(-13\): \(-9\) and \(-4\). Rewrite: \(6x^2 – 9x – 4x + 6\). Factor pairs: \(3x(2x – 3) – 2(2x – 3)\). Final: \((3x – 2)(2x – 3)\).

Substitute \(n = 7\): \(T_7 = 5(7)^2 – 3(7) + 2 = 5(49) – 21 + 2 = 245 – 21 + 2 = 226\).

Combine the fractions: \(\frac{4(2x – 1) + 3x}{6} = 5 \implies \frac{8x – 4 + 3x}{6} = 5 \implies 11x – 4 = 30 \implies 11x = 34 \implies x = \frac{34}{11}\).

Evaluate \(f(3) = 2(3)^2 – 3(3) + 1 = 18 – 9 + 1 = 10\) and \(f(-1) = 2(-1)^2 – 3(-1) + 1 = 2 + 3 + 1 = 6\). Then, \(f(3) – f(-1) = 10 – 6 = 4\).

Simplify \(3^{-2} = \frac{1}{9}\), so the expression becomes \(\frac{1}{9} \cdot \frac{1}{4} + \frac{3}{8} = \frac{1}{36} + \frac{3}{8}\). The LCM is 72: \(\frac{2}{72} + \frac{27}{72} = \frac{29}{72}\).

After 10 hours, the population has doubled \(\frac{10}{2} = 5\) times. Using \(P = P_0 \cdot 2^n\): \(P = 1000 \cdot 2^5 = 1000 \cdot 32 = 32000\).

The perimeter is calculated as \(2(l + w)\). Simplify \(\sqrt{72} = 6 \sqrt{2}\) and \(3 \sqrt{18} = 9 \sqrt{2}\). Adding: \(2(6 \sqrt{2} + 9 \sqrt{2}) = 30 \sqrt{2}\).

To simplify: \(\left( \frac{5}{2} \right)^3 = \frac{125}{8}\), \(\sqrt{36} = 6\), \(2 \cdot 6 = 12\), \(\frac{7}{3} = \frac{14}{6}\). Adding: \(\frac{125}{8} – 12 + \frac{14}{6} = \frac{137}{6}\).