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3 years ago

PLS HELP DUE TODAY

Mathematics

1 answer:

Gekata [30.6K]3 years ago

6 0

Here we go !!

Since it's a regular pentagon, all it's sides are equal so let's solve for x

$\hookrightarrow \: 3(x - 1) = 5x - 6$

$\hookrightarrow \: 3x - 3 = 5x - 6$

$\hookrightarrow \: 3x - 5x = - 6 + 3$

$\hookrightarrow \: - 2x = - 3$

$\hookrightarrow \: x = \dfrac{ - 3}{ - 2}$

$\boxed{\boxed{x = \dfrac{3}{2} }}$

now let's find the measure of each side, i.e

$\hookrightarrow \: 5x - 6$

$\hookrightarrow \: (5 \times \dfrac{3}{2}) - 6$

$\hookrightarrow \: \dfrac{15}{2} - 6$

$\hookrightarrow \: 7.5 - 6$

$\mathrm{\hookrightarrow \: 1.5 \: units}$

perimeter = 5 × side length ( for a regular pentagon )

$\hookrightarrow \: 1.5 \times 5$

$\boxed{ \boxed{ \: \: \: \: \: \: \: \: 7.5 \: \: units \: \: \: \: \: \: \: }}$

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a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

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The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

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And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

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$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

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