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lianna [129]
3 years ago
15

PLEASE HELP! WILL MARK BRAINLIEST

Mathematics
1 answer:
omeli [17]3 years ago
4 0

1)

v - 6 ≥ 4

<u>   +6  </u> <u>+6 </u>

v      ≥ 10

Graph: 10 -----------------→  <em>the dot at 10 is filled in because of the "equal to" symbol</em>

********************************************************************************************************

2)

-5x < 15

\frac{-5x}{-5} > \frac{15}{-5}  <em>divided by a negative so the symbol flipped </em>

x > -3

Graph: -3 o------------------→ <em>the dot at -3 is NOT filled in because it's NOT "equal to" </em>

********************************************************************************************************

3)

3k > 5k + 12

<u>-5k </u>  <u>-5k       </u>

-2k >          12

\frac{-2k}{-2} < \frac{12}{-2}   <em>divided by a negative so the symbol flipped </em>

k < -6

Graph: ←------------o -6   <em>the dot at -6 is NOT filled in because it's NOT "equal to" </em>

********************************************************************************************************

5)

2t ≤ 4      or    7t ≥ 49

\frac{2t}{2} \leq\frac{4}{2}      or    \frac{7t}{7} \geq \frac{49}{7}

 t ≤ 2      or      t ≥ 7

Graph: ←------- 2        7 ---------→   <em>the dots at 2 and 7 are filled in</em>

********************************************************************************************************

6)

| n + 2 | = 4

n + 2 = 4      or     n + 2 = -4

<u>     -2</u>   <u>-2 </u>             <u>     -2 </u>   <u>-2 </u>

n      = 2        or     n       = -6

Answer: n = {2, -6}

********************************************************************************************************

7)

| 2x - 7 | > 1

2x - 7 > 1       or     2x - 7 < -1

<u>      +7</u>  <u>+7 </u>             <u>     +7 </u>   <u>+7 </u>

2x      > 8       or     2x     <  6

    \frac{2x}{2} > \frac{8}{2}             or       \frac{2x}{2} < \frac{6}{2}

    x > 4          or         x < 3

Graph: ←----------o 3      4 o------------→

********************************************************************************************************

8)

A = {1, 2, 3, 4, 5, 6, 7, 8, 9}    B = {2, 4, 6, 8}

A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9}    <em>union combines both sets</em>

A ∩ B =  {2, 4, 6, 8}   <em>intersection includes only those that are in BOTH sets</em>

********************************************************************************************************

9)

P = {1, 5, 7, 9, 13}   R = { 1, 2, 3, 4, 5, 6, 7}    Q = {1, 3, 5}

P ∩ R ∩ Q = {1, 5}

P ∩ R = {7}  <em>disregard 1 & 5 since they are already in </em>P∩Q∩R

R ∩ Q = {3}  <em>disregard 1 & 5 since they are already in </em>P∩Q∩R

P ∩ Q = { }  <em>disregard 1 & 5 since they are already in </em>P∩Q∩R

P no intersection = {9, 13)

R no intersection = {2, 4, 6}

Q no intersection = {  }

If you can't figure out how to draw it based on the information I provided, please see the attached diagram. <em>Note: Usially, you would only identify P, Q, R.  I identified the intersections so you would understand why those specific numbers are placed in the designated sections.</em>





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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).

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Step-by-step explanation:

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

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

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

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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.

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