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Vladimir [108]
3 years ago
6

A dog walker charges a flat rate of $6 per walk plus an hourly rate of $30. How much does the dog walker charge for a 45 minute

walk? Write an equation in function notation for the situation and then use it to solve the problem. Determine if the given statement is true or flase.
The dependent variable is the number of hours true or false

The function for the walker fee is f(h)= 30h+6 true or false

The dog walker charges $22.5 for a 45min walk true or flase
Mathematics
1 answer:
zvonat [6]3 years ago
5 0

Step 1

<u>Find the equation in function notation</u>

Let

h-------> the number of hours

y-------> the function for the walker fee in dollars

we know that

the hourly rate is 30\frac{\$}{hour}

y=6+30h

in this linear equation

the independent variable is the variable h

the dependent variable is the variable y

<u>Convert to function notation</u>

Let

f(h)=y

f(h)=6+30h

Step 2

Find how much does the dog walker charge for a 45 minute walk

Convert the time in hours

1\ hour=60\ minutes

45\ minutes=45/60=0.75\ hours

substitute in the equation

For h=0.75\ hours

f(0.75)=6+30*(0.75)=\$28.5

therefore

<u>the answer is</u>

the dog walker charge for a 45 minute walk \$28.5

<u>Statements</u>

Step 3

<u>The dependent variable is the number of hours true or false</u>

The statement is false,

because the independent variable is the number of hours and the dependent variable is the walker fee in dollars

Step 4

<u>The function for the walker fee is f(h)= 30h+6 true or false</u>

The statement is True --------> see the Step 1

Step 5

<u>The dog walker charges $22.5 for a 45 min walk true or false</u>

The statement is False

Because the dog walker charges for a 45 minute walk \$28.5

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Step-by-step explanation:

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Step-by-step explanation:

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30/82.50 = 1/x
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6 0
3 years ago
The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls
Elanso [62]

Answer:

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=22 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =12.9^2=166.41 represent the sample variance obtained

\sigma^2_0 =16.1^2 =259.21 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: \sigma^2 \geq 259.21

Alternative hypothesis: \sigma^2

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{22-1}{259.21} 166.41 =13.482

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

p_v =P(\chi^2

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

3 0
3 years ago
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