SACE Stage 2 Mathematical Methods topics 1 to 6: exam-style practice quiz

$5(3x^2 + 1)^4$

$30x(3x^2 + 1)^4$

$6x(3x^2 + 1)^4$

$30x(3x^2 + 1)^5$

$\dfrac{e^x(x - 1)}{x^2}$

$\dfrac{e^x(1 - x)}{x^2}$

$e^x$

$\dfrac{e^x(x + 1)}{x^2}$

increasing and concave up

decreasing and concave down

increasing and concave down

at a local maximum

It has a vertical asymptote at $x = 2$ and passes through $(3, 1)$

It has a vertical asymptote at $x = -2$ and passes through $(-1, 1)$

It has a horizontal asymptote at $y = 1$ and passes through $(3, 1)$

It has a vertical asymptote at $x = 2$ and passes through $(2, 1)$

$x = \frac{7}{2} = 3.5$

$x = \ln\frac{7}{2} \approx 1.25$

$x = \frac{\ln 2}{\ln 7} \approx 0.36$

$x = \frac{\ln 7}{\ln 2} \approx 2.81$

$e - 1$

-1

1

0

9 square units

-9 square units

27 square units

18 square units

mean 0.3 and variance 0.09

mean 0.3 and variance 0.21

mean 0.7 and variance 0.21

mean 0.3 and variance 0.46

Test B is the relatively better result, because 72 is a higher mark than 64

Test B is the relatively better result, with z = 1.2 compared with z = 1.5

The results are equally good, because both marks are above their test means

Test A is the relatively better result, with z = 1.5 compared with z = 1.2

strongly right-skewed, centred on $\mu$, with standard deviation $\sigma$

approximately normal, centred on $\mu$, with standard deviation $\frac{\sigma}{\sqrt{50}}$

approximately normal, centred on $\mu$, with standard deviation $\sigma$

approximately normal, centred on the median, with standard deviation $\frac{\sigma}{50}$

96

10

97

49

About 95 percent of households use between 50.3 kL and 54.3 kL of water each quarter

There is a 95 percent probability that the sample mean lies between 50.3 and 54.3 kL

The population mean is certainly 52.3 kL, the centre of the confidence interval

About 95 percent of intervals built this way from repeated samples would contain the true mean