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MatroZZZ [7]
3 years ago
11

A video game store was getting rid of old games,selling them 7 for 100.66. .If they sold 2 games ,how much money would they have

made?
Mathematics
1 answer:
Pani-rosa [81]3 years ago
4 0
Answer: They would make $28.76.

Explanation: If 7 video games are worth $100.66, when you divide 100.66 by 7(100.66/7) you get the amount of money that 1 video game would be worth which is $14.38. Multiply 14.38 by 2(the number of video games they sold) and you get the answer of 28.76(the total money they made).
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Take the regular price, multiply it by .2, subtract that off of the regular price. What it means is 20% of the regular price is taken away from the regular price.
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Find the area of each sector. Show your work.
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Solve the formula for the indicated variable. EASY 40 POINTS!!!!!!!!!!!!!!!!
vlada-n [284]

Answer:

d = Nc + x

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N = \frac{d-x}{c}

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Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t
Amanda [17]

Answer:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

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t_{\alpha/2}= -2.08

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Step-by-step explanation:

Data given

\bar X=33.9 represent the sample mean

s=4.1 represent the sample standard deviation

n=22 sample size  

\mu_o =26 represent the value that we want to test

\alpha=0.025 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:  

Null hypothesis:\mu \geq 36  

Alternative hypothesis:\mu < 36  

The statistic is given by:

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

And replacing we got:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

Now we can calculate the critical value but first we need to find the degreed of freedom:

df = n-1= 22-1=21

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

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