Pls help with question 4

The stopping distance d of an automobile is directly proportional to the square of its speed s. On one road, a car requires 75 f

Anton [14]

Answer:

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

Step-by-step explanation:

Direct proportion means that two quantities increase or decrease in the same ratio

If y is directly proportional to x (y ∝ x) , then $\frac{y_{1}}{y_{2}}=\frac{x_{1}}{x_{2}}$ <em>OR</em> y = k x, where k is the constant of proportionality

∵ The stopping distance d of an automobile is directly

proportional to the square of its speed s

- That means d ∝ s²

∴ $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

∵ A car requires 75 feet to stop from a speed of 30 miles per hour

∴ d = 75 feet

∴ s = 30 miles/hour

- Change the mile to feet

∵ 1 mile = 5280 feet

∴ 30 miles/hour = 30 × 5280 = 158400 feet/hour

∵ The car require to stop from a speed of 48 miles per hour

on the same road

- Change the mile to feet

∴ 48 miles/hour = 48 × 5280 = 253440 feet/hour

∵ $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

- Substitute the values of $d_{1}$ by 75 feet, $s_{1}$ by 158400 feet/hour

and $s_{2}$ by 253440 feet/hour

∴ $\frac{75}{d_{2}}=\frac{(158400)^{2}}{(253440)^{2}}$

∴ $\frac{75}{d_{2}}=\frac{25}{64}$

- By using cross multiplication

∴ 25 × $d_{2}$ = 75 × 64

- Divide both sides by 25

∴ $d_{2}$ = 192 feet

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

4 0

2 years ago

A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the d

maria [59]

Answer:

(0.6613, 8.3387)

Step-by-step explanation:

Let the heights of men be the first population and the heights of women be the second population. Then

We have $n_{1} = 15$ , $\bar{x}_{1} = 69.4$ , $s_{1} = 3.09$ and

$n_{2} = 7$ , $\bar{x}_{2} = 64.9$ , $s_{2} = 2.58$ . The pooled

estimate is given by

$s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(3.09)^{2}+(7-1)(2.58)^{2}}{15+7-2} = 8.68$

The 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{7}}$ , i.e.,