Question 11

HELP ME IM STUCK ANSWER IT QUICK

fiasKO [112]

Answer:

red = 28 apples

green = 13 apples

equations:

r + g = 41

r = g + 15

Step-by-step explanation:

r = number of red

g = number of green

r = g + 15 (the number of red apples is 15 more than the number of green apples)

r + g = 41

Substitute the firs equation into the second and solve for a numerical value of g

(g + 15) + g = 41

2g + 15 = 41

2g = 26

g = 13

Now solve for a numerical value of r

r = g + 15

r = 13 + 15

r = 28

checking the math:

r + g = 41

28 + 13 = 41

Please lmk if you have any questions.

6 0

3 years ago

Write a real world problem where the Unit rate is 6 miles per hr

Galina-37 [17]

This is just an example i came up with randomly:

Emma is going somewhere for vacation. Emma has been traveling for 6 hours and in those 6 hours, she has travelled 36 miles. How many miles did she travel in one hour?

8 0

3 years ago

Which equation, when solved, results in a different value of x than the other three?

Bess [88]

Is there any more information?

4 0

3 years ago

Maria is draining her swimming pool. The depth of the water in the pool changes by −3/4 foot every hour.The depth of the water w

Alex17521 [72]

Answer: 1 1/4feet

Step-by-step explanation:

From the question, we are informed that the depth of the water in the pool changes by −3/4 foot every hour and that the depth of the water was 5 feet when she started draining.

The depth of the water after 5 hours will be:

= 5 - 3/4(5)

= 5 - 15/4

= 5 - 3 3/4

= 1 1/4 feet

7 0

3 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago