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Masteriza [31]
3 years ago
9

The distance from Mark’s apartments to his job is 27 miles. Mark works 5 days per week. How many miles does Mark drive to and fr

om work each week?
Mathematics
1 answer:
svlad2 [7]3 years ago
8 0
27 (miles) × 2 (times per day) = 54 (miles per day) × 5 (days each week) =

270 miles per week
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Javier used the expression below to represent his score in a game of mini-golf
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Sorry but u should write ur expression so i can solve it....
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Natasha_Volkova [10]

Answer:

x = 2, y = -1.

Step-by-step explanation:

2x + y = 3

x  -   y  = 3

Adding the 2 equations eliminates y:

3x = 6

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Checks OK.

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company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26
podryga [215]

Answer:

t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

p_v =P(t_{(49)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation  

\bar X=25.02 represent the sample mean

s=4.83 represent the sample standard deviation

n=50 sample size  

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

\alpha 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is at least 26 mpg, the system of hypothesis would be:  

Null hypothesis:\mu \geq 26  

Alternative hypothesis:\mu < 26  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

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 test the p value would be:  

p_v =P(t_{(49)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

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3 years ago
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