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Yuri [45]
3 years ago
9

Hellpppppp due today will give brainliest ​

Mathematics
1 answer:
Dovator [93]3 years ago
8 0

Answer:

The last option. <3

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Complete the table to summarize what you know about each player's throw: (2 points: 1 point for each row of the chart) tre and h
Thepotemich [5.8K]

Answer:

Tre: Tre’s position was on the pitcher’s mound and he threw the ball to 3rd base.

Hector: Hector’s position was in right field and he threw the ball to 2nd base.

5 0
3 years ago
I NEED AN ANSWER PLEASE :(((
Alekssandra [29.7K]

Answer:

80 lemons

Step-by-step explanation:

Let x represent the amount of lemons that were on the tree.

We can use this to set up an equation:

40\%x=32

Note that percentages can also be written as that number over 100.

\frac{40}{100}x=32

Simplify the fraction.

\frac{2}{5}x=32

Multiply both sides by 5

2x=160

Divide both sides by 2

x=80

There were 80 lemons on the tree before he picked the lemons.

6 0
2 years ago
Complete the solution of the equation. Find the value of y when x equals 4. -7x + y = -22
xz_007 [3.2K]

-7x + y = -22

-7*4 + y = -22

y = -22 + 28

y = 6

4 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
Scott earned 20 points more on his most recent math test than he earned on the previous test. If he earned at least 80 points on
jeyben [28]

Answer:

60

Step-by-step explanation:

80-20=60

8 0
3 years ago
Read 2 more answers
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