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Write each percent as a fraction for 88%

GREYUIT [131]

88 = 88/100 or 22/25

6 0

3 years ago

A circle has a circumference of 7.850 units. What is its radius?

Roman55 [17]

$\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=7.850 \end{cases}\implies 7.850=2\pi r\implies \cfrac{7.85}{2\pi }=r\implies 1.25\approx r$

8 0

3 years ago

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Raul has $460 in his checking account. Each month, $45 is automatically deducted from his account to pay for his cell phone plan

nexus9112 [7]

So if he wants to have more then 100 in his bank then what i would do is take 100 from 460 and you get 360 then divide that by 45 and you get 8 but as said before he wants more then 100 so he would be able to pay 7 months in a row and have over 100 in his bank account still

4 0

3 years ago

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Use the Statistical Applet: P ‑Value for a Test of One Proportion to answer the question.You have taken a sample of ????=1000 in

umka21 [38]

Answer:

$z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08$

$p_v =P(z>5.08)=1.88x10^{-7}$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is highr than 0.67 or 67% .

Step-by-step explanation:

1) Data given and notation

n=20 represent the random sample taken

X=15 represent the people who eat breakfast

$\hat p=\frac{15}{20}=0.75$ estimated proportion of people who eat breakfast

$p_o=0.67$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 67%:

Null hypothesis: $p\leq 0.67$

Alternative hypothesis: $p > 0.67$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>5.08)=1.88x10^{-7}$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is highr than 0.67 or 67% .

6 0

3 years ago

Please please help me out ASAP

likoan [24]

The answer are c. x=-1 and e.x=1

4 0

4 years ago