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frozen [14]
3 years ago
8

A stylist at a salon must pay $35 per day to rent a station at the salon. The stylist makes $22.50 on each haircut. If h represe

nts the number of haircuts, which equation can be used to find how many haircuts she must give per day to earn $122.50 after paying rent?
A - 22.50h+35=122.50

B - 35h−22.50=122.50

C - 35h+22.50h=122.50

D - 22.50h−35=122.50

2. How many haircuts must she give per day to earn $122.50 after paying rent?

A - 4

B -5

C - 7

D - 14
Mathematics
2 answers:
Alex17521 [72]3 years ago
8 0
So one is C and for the second one is D
Anna71 [15]3 years ago
3 0
The first is second is d

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Answer:

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

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

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Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

\hat p=\frac{492}{778}=0.632 estimated proportion of people that chose a window seat.

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

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 majority of adults prefer window seats when they fly:  

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statisitc, 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{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36  

4) Statistical decision  

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The significance level provided \alpha=0.01. 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>7.36)=9.19x10^{-14}  

5) Conclusion

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

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