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bogdanovich [222]
3 years ago
13

Please help and Solve for x.

Mathematics
1 answer:
Serhud [2]3 years ago
7 0

Answer:

tan70 = \frac{opp}{adj}  \\  \:  \:  =  \frac{15}{x }  \\  \\  x =  \frac{15 }{tan70 } \\   \\  =  \frac{15}{2.747}  \\  \\  = 5.46

since 5.46 is equal to 5.5...

I think the answer is the first one

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Eric has 12 trophies in his room 9 of them are for football what percentage of the total number of sports trophies are his footb
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The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they
Airida [17]

Answer:

a) Null hypothesis:\mu \geq 14  

Alternative hypothesis:\mu < 14

H0: μ -greater than or equal to 14

Ha: μ - less than 14

b) p_v =P(t_{(49)}  

c) For this case since the p value is higher than the significance level given of 0.05 the answer would be NO

We can't conclude that the actual mean waiting time is significantly less than the claim of 14 minutes made by the taxpayer advocate

Step-by-step explanation:

Data given

\bar X=12 represent the sample mean

\sigma=10 represent the population standard deviation

n=50 sample size  

\mu_o =14 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

Part a: System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 14 minute, the system of hypothesis would be:  

Null hypothesis:\mu \geq 14  

Alternative hypothesis:\mu < 14

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Calculate the statistic

If we replace we got:

t=\frac{12-14}{\frac{10}{\sqrt{50}}}=-1.414    

Part b: P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=50-1=49  

Since is a one sided lower test the p value would be:  

p_v =P(t_{(49)}  

Part c

For this case since the p value is higher than the significance level given of 0.05 the answer would be NO

We can't conclude that the actual mean waiting time is significantly less than the claim of 14 minutes made by the taxpayer advocate

8 0
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