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3 years ago

Can someone please help me!! And can you show your work

Mathematics

1 answer:

Debora [2.8K]3 years ago

4 0

Answer:

1) 8.073 (multiply 0.621 by 13km)

2)1,320,000,000(5280ft times 250,000)

3)37.850(10,000 divided by 264.2)

4)4hrs

5)157680000

(60 seconds in one minute and 60 min in one hr so 60 times 60 =3600 in one hr. then multiply 3600 by 24 for the entire day which is 86400sec. Then multiply that day by 365 which is 31536000 times 5 years which gets us our total of 157680000 seconds.

6) 758000 L

1 gal = 3.79 L so we have to multiply 3.79 times 200,000 which gets us 758000

Step-by-step explanation:

1) To get this answer we need to find out how many miles are in one kilometer. 0.621 mile is 1 km so we need to put 13 for km which will get 8.073 miles. SO the answer is 8.073

2) we need to find how many feet are in a mile. 1 mile= 5,280. so 5280 times 250,000 = 1320000000

3) 264.2 gal= 1 cubic meter so 10,000/264.2= 37.850

4) 1 hr= 60 min = 4hrs and 58 minutes roughly

5) 60 sec is 1 minute= so 60 times 60(the minutes in

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Answer:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

Step-by-step explanation:

1) Data given and notation

$\bar X=23500$ represent the sample mean

$s=3900$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =2000$ 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 determine if the true mean is higher than 20000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2000$

Alternative hypothesis: $\mu > 2000$

We don't know the population deviation, so for this case 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:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=100-1=99$

Since is a one-side upper test the p value would be:

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

Conclusion

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Let R be the greater radius and r be the smaller radius

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=(11^2-9^2)

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