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tamaranim1 [39]

1 year ago

5

The perimeter of a rectangle A is 540 inches. The ratio of the lengths ofa large and small rectangle is 7:3. Find the perimeter

of rectangle B, thesmaller rectangle. Round to the tenths.

1 answer:

gulaghasi [49]1 year ago

7 0

The ratio between the lengths of 2 rectangles is the same between their perimeters

$l_1\colon l_2=P_1\colon P_2$

Since the ratio between the length of rectangle A and the length of rectangle B is 7: 3, then

The ratio between the perimeter of rectangle A to the perimeter of rectangle B is 7: 3 too

$\begin{gathered} l_A\colon l_B=7\colon3 \\ P_A\colon P_B=7\colon3 \end{gathered}$

Since the perimeter of rectangle A is 540, then

$540\colon P_B=7\colon3$

We will write them as a fraction

$\frac{540}{P_B}=\frac{7}{3}$

By using the cross multiplication

$\begin{gathered} P_B\times7=540\times3 \\ 7P_B=1620 \end{gathered}$

Divide both sides by 7

$\begin{gathered} \frac{7P_B}{7}=\frac{1620}{7} \\ P_B=231.4285714 \end{gathered}$

Round it to the nearest tenth

$P_B=231.4\text{ inches}$

The perimeter of the smaller rectangle B is 231.4 inches

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Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

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Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em> $\mu_1$ <em> = true mean number of sales at location A.</em>

<em /> $\mu_2$ = <em>true mean number of sales at location B</em>

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, <u><em>test statistics</em></u> = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

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