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CTJan27 Online Year 8 Practice Test 18122025 - Test 02

45:00

Student Name

Student Email

Complete All The Questions

$9x^{10}y^{10}$ $\frac{x^{10}}{9y^{10}}$ $\frac{9y^{10}}{x^{10}}$ $\frac{x^{5}}{3y^{5}}$

$x = 4$ $x = -\frac{18}{5}$ $x = \frac{24}{7}$ $x = 6$

$-1 \le x < 5$ $3 < x \le 10$ $\frac{4}{3} < x \le 4$ $1 < x \le 5$

The solution does not include $x = -4$. The solution is $x < -4$. The solution includes the interval $(-\infty, -4)$. The solution includes $x = -5$.

$(-\infty, 3] \cup [7, \infty)$ $x \ne 7$ $[3, 7) \cup (7, \infty)$ $x \ge 3$

$(-\infty, 5]$ $(-\infty, \infty)$ $[-1, 5]$ $[5, \infty)$

$(0, -5)$ $(3, 0)$ $(0, 5)$ $(0, -3)$

$y = \frac{5}{2}x + 11$ $y = -\frac{5}{2}x - 9$ $y = \frac{2}{5}x + \frac{13}{5}$ $y = \frac{5}{2}x + 9$

$k = -4$ $k = 2$ $k = 8$ $k = 1$

$(-\frac{1}{2}, \frac{5}{8})$ $(2, -2)$ $(1, -1)$ $( -1, 1)$

$(3, -1)$ $(4, 2)$ $(\frac{51}{17}, -\frac{1}{17})$ $(2, -4)$

$7.0 \times 10^{-7}$ $7.0 \times 10^{-8}$ $0.7 \times 10^{-7}$ $7.0 \times 10^{-1}$

$\frac{a^6}{b^{12}}$ $\frac{b^{12}}{a^6}$ $a^6 b^{12}$ $\frac{1}{a^6 b^{12}}$

$3x^2 - 8xy + 8y^2$ $3x^2 - 2xy + 6y^2$ $7x^2 - 8xy + 6y^2$ $3x^2 + 2xy + 8y^2$

$2x^3 + 11x^2 - 16x + 3$ $2x^3 - x^2 - 16x + 3$ $2x^3 - 11x^2 + 14x + 3$ $2x^3 + 2x^2 - 16x + 3$

$(6x + 5)(x + 1)$ $(6x - 1)(x - 5)$ $(3x - 5)(2x - 1)$ $(3x - 1)(2x - 5)$

$4x(x^2 - 9)$ $4x(x - 3)(x + 3)$ $(2x - 6)(2x + 6)$ $4(x^3 - 9x)$

The boundary line is solid, shaded above the line. The boundary line is dashed, shaded above the line. The boundary line is dashed, shaded below the line. The boundary line is solid, shaded below the line.

$r = \frac{A}{P + t}$ $r = \frac{A}{Pt} - P$ $r = A - P - Pt$ $r = \frac{A - P}{Pt}$

$$3x + 5 = 16; x = 3.67$$ $$3(x - 5) = 16; x = 6.33$$ $$3x - 5 = 16; x = 7$$ $$5 - 3x = 16; x = -3.67$$

$20$ meters $14$ meters $12$ meters $10$ meters

$65$ km/h $80$ km/h $75$ km/h $70$ km/h

$6$ L $4$ L $10$ L $3$ L

$16$ years old $22$ years old $14$ years old $17$ years old

$$\$2200$$ $$\$1800$$ $$\$1600$$ $$\$2000$$

$x \ge 95$ $x \ge 97$ $x > 90$ $x \ge 98$

$2 \text{ L}$ $10 \text{ L}$ $5 \text{ L}$ $4 \text{ L}$

$20 \text{ km/h}$ $15 \text{ km/h}$ $25 \text{ km/h}$ $30 \text{ km/h}$

$2500 \text{ m}^2$ $1000 \text{ m}^2$ $625 \text{ m}^2$ $1250 \text{ m}^2$

$12 \text{ hours}$ $24 \text{ hours}$ $4 \text{ hours}$ $6 \text{ hours}$

$6 \text{ cm}$ $10 \text{ cm}$ $12 \text{ cm}$ $9 \text{ cm}$

$21 \text{ years}$ $12 \text{ years}$ $15 \text{ years}$ $18 \text{ years}$

$72 \text{ cm}^2$ $50 \text{ cm}^2$ $56 \text{ cm}^2$ $60 \text{ cm}^2$

$\$300$ $\$150$ $\$250$ $\$200$

$4 \text{ hours}$ $6 \text{ hours}$ $5 \text{ hours}$ $3 \text{ hours}$

The Interquartile Range (IQR) must be close to zero, suggesting very low overall variability. The data is approximately symmetric, with high variability in the lower 50\%. The data is positively skewed (right-skewed), indicating that the majority of values are concentrated in the lower 50\%. The data is negatively skewed (left-skewed), indicating that the bulk of the values are high.

$0.1$ $0.4$ $0.2$ $0.8$

$32,000$ $60,000$ $50,000$ $40,000$

$285$ $190$ $150$ $165$

Cannot be determined without the minimum data value. $100$ $85$ $70$

$30\%$ $20\%$ $50\%$ $10\%$

$3780$ $34,650$ $6300$ $1890$

Yes, because $P(A) = 0.6$ and $P(B) = 0.7$. No, because $P(A \cap B) \neq P(A)P(B)$. Yes, because $P(A \cup B) = 0.9$. No, because $P(A') + P(B') \neq 1$.

$\frac{1}{5}$ $\frac{1}{6}$ $\frac{1}{4}$ $\frac{3}{10}$