The value of log_2 6 + log_2 (8)/(3) is:

log_2 6 + log_2 (8)/(3) = log_2 (6 · (8)/(3)) = log_2 16 = 4.

If the mean of {x_1, x_2, …, x_n} is x, the mean of {x_1 + 5, x_2 + 5, …, x_n + 5} is:

(1)/(n)Σ (x_i + 5) = (1)/(n)Σ x_i + 5 = x + 5.

Problems & worked solutions

RO M1

The value of $\log_{2} 6 + \log_{2} \frac{8}{3}$ is:

Show answer & worked solution

$\log_{2} 6 + \log_{2} \frac{8}{3} = \log_{2} ⁣ (6 \cdot \frac{8}{3}) = \log_{2} 16 = 4$ .

RO M1

The coefficient of $x^{3}$ in $(x + 2)^{5}$ is:

Show answer & worked solution

$5 - k = 3 \Rightarrow k = 2$ . Coefficient: $(\binom{5}{2}) \cdot 2^{2} = 10 \cdot 4 = 40$ .

RO M1

If the mean of ${x_{1}, x_{2}, \dots, x_{n}}$ is $\overset{ˉ}{x}$ , the mean of ${x_{1} + 5, x_{2} + 5, \dots, x_{n} + 5}$ is:

Show answer & worked solution

$\frac{1}{n} \sum (x_{i} + 5) = \frac{1}{n} \sum x_{i} + 5 = \overset{ˉ}{x} + 5$ .

Problems & worked solutions

RO M1

The value of $\log_{2} 6 + \log_{2} \frac{8}{3}$ is:

Show answer & worked solution

$\log_{2} 6 + \log_{2} \frac{8}{3} = \log_{2} ⁣ (6 \cdot \frac{8}{3}) = \log_{2} 16 = 4$ .

RO M1

The coefficient of $x^{3}$ in $(x + 2)^{5}$ is:

Show answer & worked solution

$5 - k = 3 \Rightarrow k = 2$ . Coefficient: $(\binom{5}{2}) \cdot 2^{2} = 10 \cdot 4 = 40$ .

RO M1

If the mean of ${x_{1}, x_{2}, \dots, x_{n}}$ is $\overset{ˉ}{x}$ , the mean of ${x_{1} + 5, x_{2} + 5, \dots, x_{n} + 5}$ is:

Show answer & worked solution

$\frac{1}{n} \sum (x_{i} + 5) = \frac{1}{n} \sum x_{i} + 5 = \overset{ˉ}{x} + 5$ .