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Lady_Fox [76]
3 years ago
9

Mrs. Anderson just received a promotion. She will get a 25% raise to her annual salary of $68,477. What is Mrs. Anderson's new a

nnual salary?
Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0

Answer:

Mrs. Anderson's new annual salary is \$85,596.25

Step-by-step explanation:

Let

x------> Mrs. Anderson's new annual salary

we know that

100\%+25\% =125\% =\frac{125}{100}=1.25

so

The linear equation that represent this situation is

x=1.25*\$68,477

x=\$85,596.25

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each cup cost 1.25 on Tuesday

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In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use
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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)  

p_v =P(Z>7.36)=9.19x10^{-14}  

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 .  

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  

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 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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The linear function should be y=2 i apologize if i am wrong
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