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MAVERICK [17]
3 years ago
8

What is the y-intercept of the exponential function? f(x)=−32(2)x−3+3 Enter your answer in the box.

Mathematics
2 answers:
Bogdan [553]3 years ago
6 0

y-intercept for x = 0.

Substitute x = 0 to the equation of the function:

f(x)=-32(2)^{x-3}+3\\\\f(0)=-32(2)^{0-3}+3=-32(2)^{-3}+3=-32\cdot\dfrac{1}{2^3}+3=-32\cdot\dfrac{1}{8}+3\\\\=-4+3=-1\\\\Answer:\ \boxed{y-intercept=-1\to(0,\ -1)}

Rudik [331]3 years ago
5 0

Answer:

-1

Step-by-step explanation:

We are given that a function

f(x)=-32(2)^{x-3}+3

We have to find the y- intercept of the exponential function.

To find the y- intercept of given exponential we will substitute x=0

Substitute x=0 then, we get

f(0)=-32(2)^{0-3}+3

f(0)=-32(2)^{-3}+3

f(0)=-\frac{32}{(2)^3}+3

By using property a^{-x}=\frac{1}{a^x}

f(0)=-\frac{32}{8}+3

f(0)=-4+3=1

Hence, the y- intercept of given exponential =-1

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Answer:

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Step-by-step explanation:

GCF: 3

Find all the prime factors between 3 and 15

3: 1,3 1*3

15: 1,3,5,15 (1*15 3*5)

3 is the GCF between the 2 numbers.

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A school has a population of 2500 students. The number of students is increasing at continuous growth rate of 4.5% each year. Wh
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2 years ago
Find the area of each figure below
Ugo [173]

Answer:

25. 27a^{3}b^{2}

26. 3\pi x^{3}y

Step-by-step explanation:

For 25:

A = \frac{a+b}{2} h  (area of a trapezoid)

A=\frac{11a^{2}b+7a^{2}b}{2} *3ab  (substitute terms)

= \frac{18a^{2}b}{2}*3ab  (collect like terms)

= 9a^{2}b*3ab  (reduce the fraction by crossing out 2)

= 27a^{3}b^{2}  (calculate)

For 26:

A=\pi r^{2}  (equation of area of a circle)

A=\pi (3x^{3} y)^{2}  (enter the radius)

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3 years ago
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You can use a calculator online.
Here's the link but I already did it for you

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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.

4 0
3 years ago
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