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.

Daily Math Problems — June 6, 2026: Complex Numbers, Descriptive Statistics & Sampling, Calculus

dailymath

Problems & worked solutions

RO M1

Problem 1 — Complex Numbers

The value of $(1 + i)^{2}$ is:

Show answer & worked solution

A. $1$
B. $1 + 2 i$
C. $2 i$ ✓ correct
D. $1 - 2 i$

$(1 + i)^{2} = 1 + 2 i + i^{2} = 1 + 2 i - 1 = 2 i$ .

RO M1

Problem 2 — Descriptive Statistics & Sampling

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

A. $\overset{ˉ}{x}$
B. $\overset{ˉ}{x} + 5$ ✓ correct
C. $5 \overset{ˉ}{x}$
D. $\overset{ˉ}{x} + \frac{5}{n}$

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

RO M1

Problem 3 — Continuity

The equation $\cos x = x$ has at least one solution in:

Show answer & worked solution

A. $(- 1, 0)$
B. $(0, π / 2)$ ✓ correct
C. $(π, 2 π)$
D. $nowhere$

$g (0) = 1 > 0$ , $g (π / 2) = 0 - π / 2 < 0$ . Since $g$ is continuous, by the IVT there is $c \in (0, π / 2)$ with $g (c) = 0$ , i.e. $\cos c = c$ .

Problems & worked solutions

RO M1

Problem 1 — Complex Numbers

The value of $(1 + i)^{2}$ is:

Show answer & worked solution

A. $1$
B. $1 + 2 i$
C. $2 i$ ✓ correct
D. $1 - 2 i$

$(1 + i)^{2} = 1 + 2 i + i^{2} = 1 + 2 i - 1 = 2 i$ .

RO M1

Problem 2 — Descriptive Statistics & Sampling

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

A. $\overset{ˉ}{x}$
B. $\overset{ˉ}{x} + 5$ ✓ correct
C. $5 \overset{ˉ}{x}$
D. $\overset{ˉ}{x} + \frac{5}{n}$

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

RO M1

Problem 3 — Continuity

The equation $\cos x = x$ has at least one solution in:

Show answer & worked solution

A. $(- 1, 0)$
B. $(0, π / 2)$ ✓ correct
C. $(π, 2 π)$
D. $nowhere$

$g (0) = 1 > 0$ , $g (π / 2) = 0 - π / 2 < 0$ . Since $g$ is continuous, by the IVT there is $c \in (0, π / 2)$ with $g (c) = 0$ , i.e. $\cos c = c$ .