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2 years ago

Write the polar coordinates (3, π) in rectangular form.

1 answer:

6 0

$\bf (\stackrel{\stackrel{r}{\downarrow }}{3}~~,~~\stackrel{\stackrel{\theta }{\downarrow }}{\pi })~~\begin{cases}x=rcos(\theta )\\y=rsin(\pi )\end{cases}\implies [3cos(\pi )~~,~~3sin(\pi )]\\[2em][3(-1)~~,~~3(0)]\implies \boxed{(-3,0)}$

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2 years ago

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Answer:

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Step-by-step explanation:

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1 year ago

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2 years ago

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2 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:

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

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2 years ago