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tia_tia [17]
3 years ago
10

Jesse owns a sporting goods store that sells skis and snowboards. The store earns a profit of $52 for each ski sold and a profit

of $64 for each snowboard sold. If Jesse's store sells a total of 83 pairs of skis and snowboards and earns a profit of $4892 in November, how many pairs of skis and snowboards did the store sell that month
Mathematics
1 answer:
vladimir2022 [97]3 years ago
6 0
Ok so 4892=83(52)+64x
4892=4316+64x
576=64x
9=x
So he should 9 snowboards.
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<u>ANSWER:</u>

Kari bought 3 boxes of cookies to share. The algebraic expression is \frac{36}{x}

<u>Solution:</u>

Given, Kari bought 3 boxes of cookies to share with a book club.  

Each box contains 12 cookies.  

So, in total we have 3 x 12 cookies = 36 cookies.

Now, we have to find how many cookies can each person p will get.

Let, the total number of persons be x.

Then, after equally sharing the cookies,

\text { Number of cookies with each person }=\frac{\text {number } of \text { cookies available.}}{\text {total number of persons}}

=\frac{36}{x} \text { cookies. }

Hence, the algebraic expression is \frac{36}{x}

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3 years ago
Read 2 more answers
The number of events is 29​, the number of trials is 298​, the claimed population proportion is​ 0.10, and the significance leve
Nina [5.8K]

Answer:

z=\frac{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155  

p_v =2*P(Z  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

Step-by-step explanation:

1) Data given and notation

n=298 represent the random sample taken

X=29 represent the events claimed

\hat p=\frac{29}{298}=0.0973 estimated proportion

p_o=0.1 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 proportion is 0.1 or no.:  

Null hypothesis:p=0.1  

Alternative hypothesis:p \neq 0.1  

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{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155  

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 bilateral test the p value would be:  

p_v =2*P(Z  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

We can do the test also in R with the following code:

> prop.test(29,298,p=0.1,alternative = c("two.sided"),conf.level = 1-0.05,correct = FALSE)

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Answer:

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Step-by-step explanation:

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