I need help brosssssssssssssss

Maggie needs to spend at least six hours each week practicing the piano. She has already practiced three and one fourth hours th

vfiekz [6]

Answer:

three and one fourth + 2x ≥ 6

Step-by-step explanation:

Let

x -----> the minimum number of hours he needs to practice on each of the two days

we know that

needs to spend at least seven hours each week practicing the drums

so

$3\frac{1}{4}+2x\geq 6\ hours$

Convert mixed number to an improper fraction

$3\frac{1}{4}\ hours=\frac{3*4+1}{4}=\frac{13}{4}\ hours$

substitute

$\frac{13}{4}+2x\geq 6\ hours$

Subtract 13/4 both sides

$2x\geq 6-\frac{13}{4}$

$2x\geq \frac{11}{4}$

Divide by 2 both sides

$x\geq \frac{11}{8}$

therefore

The minimum number of hours he needs to practice on each of the two days is $\frac{11}{8}\ hours$

5 0

3 years ago

Mike wants to fence in part of his backyard. He wants the length of the fenced-in area to be at least 20 feet long, l ≥ 20. He h

quester [9]

W = 10 ft; l = 50 ft is Correct

w = 90 ft; l = 30 ft is Correct

w = 50 ft; l = 40 ft is Correct

5 0

3 years ago

Pleae answer this fast asap

BARSIC [14]

Answer:

I think its (2, 1) .

Step-by-step explanation:

the 2 represents the x axis

the 1 represents the y axis

7 0

3 years ago

Help pleaseee I need help filling the blank spaces with either true/false

vesna_86 [32]

Answer:

1. true 2. true 3. false 4. true 5. false 6. false 7. false

Step-by-step explanation:

im not a 100% sure this is right but ik some that are right im just not sure about 3. and 5. so ye UWU

8 0

2 years ago

Read 2 more answers

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