Please help me with my work

In Janie's homeroom 54% of the students' families own a dog and 32% own a dog and a cat. What is the probability that a student'

pogonyaev

Answer:

51,111%

Step-by-step explanation:

We know that 45% have a dog but no cat. While 23% have both a dog and a cat.

Then Pr(X has cat|X has dog) = Pr(X has dog and cat)/Pr(X has dog)

= Pr(X has dog and cat)/Pr(X has dog) = 0.23/0.45 = 0.51 ≈ 51,111%

The probability that a student’s family owns a cat if the family owns a dog is 51,111%

6 0

2 years ago

11% as a decimal= 200% as a decimal= 20.07% as a decimal 198.4% as a decimal

Hitman42 [59]

To convert from a percentage to a decimal, move the decimal twice to the left

11% = 0.11
200% = 2.00
20.07% = 0.2007
198.4% = 1.984

Hope this helps!

8 0

3 years ago

Read 2 more answers

I need this math problem done as soon as possible!

Vesna [10]

Answer:

See below

Step-by-step explanation:

Parent function:

y = 3ˣ

Transformed function:

y = 4(3)⁻²ˣ⁺⁸ + 6, (note. I see this as 8, sorry if different but it doesn't make any change to transformation method)

Transformations to be applied:

f(x) → f(-x) reflection over y-axis
f(-x) → f(-2x) stretch horizontally by a factor of 2
f(-2x) → f(-2x + 8) translate 8 units right
f(-2x + 8) → 4f(-2x + 8) stretch vertically by a factor of 4
4f(-2x + 8) → 4f(-2x + 8) + 6 translate 6 units up

7 0

2 years ago

Adrian is training for a 5K race. She

Mars2501 [29]

To get this we divide the total miles she ran for the weeks by the no of weeks she ran.

Total miles =5.5 +7.25+10= 22.25 miles altogether.

no of weeks =3.

Average miles = 22.75 /3=7.583333≅7.58(nearest hundredth)

8 0

2 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

2 years ago