<em>The word usage in last sentence of the question is not well presented. However, I can understand that you intend to solve for the width of the park</em>

Answer:

$Width = 2\ km$

Step-by-step explanation:

Given

$Length = \frac{3}{8}km$

$Area = \frac{3}{4}km^2$

Required

Determine the Width

For a rectangle:

$Area = Length * Width$

Substitute for Area and Length

$\frac{3}{4} = \frac{3}{8} * Width$

Multiply both sides by $\frac{8}{3}$

$\frac{8}{3} * \frac{3}{4} = \frac{3}{8} * \frac{8}{3}* Width$

$\frac{8}{3} * \frac{3}{4} = Width$

$\frac{24}{12} = Width$

$2 = Width$

$Width = 2\ km$

6 0

3 years ago

.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and

Leni [432]

Answer:

a) Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

b) $t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean for the amount spent each shopper

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

5 0

3 years ago

|-4b — 81 + 1 -1 - 2 | + 263​

|-4b — 81 + 1 -1 - 2 | + 263