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andriy [413]
3 years ago
7

What is the quotient of 15p^-4q^-6/-20p^-12q^-3 in simplified form?

Mathematics
2 answers:
musickatia [10]3 years ago
3 0
\cfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}} =\cfrac{3p^{-4+12}}{-4q^{-3+6}} =- \cfrac{3p^{8}}{4q^{3}}
Novay_Z [31]3 years ago
3 0

Answer:

\dfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}\Rightarrow \dfrac{3p^{8}}{-4q^{3}}

Step-by-step explanation:

We are given rational expression.

\text{Expression: }\dfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}

We need to simplify and write as quotient.

Using exponent law simplify the expression

a^m\div a^n=a^{m-n}

a^m\times a^n=a^{m+n}

\Rightarrow \dfrac{3p^{-4+12}q^{-6+3}}{-4}

\Rightarrow \dfrac{3p^{8}q^{-3}}{-4}

\Rightarrow \dfrac{3p^{8}}{-4q^{3}}

Now we write quotient of simplest form.

Thus, simplified form is \Rightarrow \dfrac{3p^{8}}{-4q^{3}}


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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.

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3 years ago
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valkas [14]
First, turn the fraction into a decimal to get a better look at the problem. To do that, divide the numerator (top half) by the denominator (bottom half). 1 divided by 3 = 0.33
0.33 is larger than 0.3. If it helps, add a 0 to 0.3 to get 0.30 and you'll see that 33 is in fact a larger number than 30. (Adding the zero does NOT change the number, it holds no place value it's just there for looks.)
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