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Andreyy89
4 years ago
9

Create a real-world problem that uses a multi-step equation.

Mathematics
1 answer:
jarptica [38.1K]4 years ago
3 0

You have 6 bags of candy and they contain 24 pieces of candy in each bag. You divide your candy evenly between your 2 friends and yourself. How many pieces for candy does everyone get?

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Write an exponential growth function to model the situation. A population of 422000 increases by 12% each year
castortr0y [4]
<span> Write an exponential growth function to model the situation. A population of 
422,000 increases by 12% each year. 
2. Describe the transformations (from the parent function) of this exponential 
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3. Find the inverse function of g(x) 
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3 0
4 years ago
A recent national survey found that high school students watched an average (mean) of 7.8 movies per month with a population sta
notka56 [123]

Answer:

Step-by-step explanation:

Given that :

Mean = 7.8

Standard deviation = 0.5

sample size = 30

Sample mean = 7.3 5.4772

The null and the alternative hypothesis is as follows;

\mathbf{ H_o: \mu  \geq  7.8}

\mathbf{ H_1: \mu  <  7.8}

The test statistics can be computed as :

z = \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}

z = \dfrac{7.3- 7.8}{\dfrac{0.5}{\sqrt{30}}}

z = \dfrac{-0.5}{\dfrac{0.5}{5.4772}}

z = - 5.4772

The p-value at  0.05 significance level is:

p-value = 1- P( Z < -5.4772)

p value = 0.00001

Decision Rule:

The decision rule is to reject the null hypothesis if  p value is less than 0.05

Conclusion:

At  the 0.05 significance level,  there is sufficient information to reject the null hypothesis. Therefore ,we  conclude that college students watch fewer movies a month than high school students.

8 0
4 years ago
Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

⇒ Plugging the values.

⇒ z_1=\frac{5.4-5.5}{0.4}

⇒ z_1=-0.25  

For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

4 0
3 years ago
Find the inverse of function f. f(x) = 1/2x +7 / <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B2%7Dx%20%2B%207"
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The answer is C.) f^-1 (x) = 2x- 14 /  \:f^{-1\:}\left(x\right)=2x\:-\:14

5 0
3 years ago
To borrow​ money, you pawn your guitar. Based on the value of the​ guitar, the pawnbroker loans you ​$1080. One month​ later, yo
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The annual rate interest: r = 640.2%

4 0
3 years ago
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