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3 years ago

6

In a linear function, which of the following represents the average rate of change of the function? Slope y-intercept independen

t variable dependent variable

1 answer:

Alenkasestr [34]3 years ago

7 0

Answer:

the slope

Step-by-step explanation:

If the function is linear, the average rate of change of y with respect to x is the slope of the line.

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Write the conversion factor for seconds to minutes. Use the factor to convert 135 seconds to minutes.

OverLord2011 [107]

Answer

Find out the conversion factor for seconds to minutes and convert 135 seconds to minutes.

To prove

1 minute = 60 second

for seconds to minutes.

$1 second = \frac{1}{60}\ minutes$

Therefore the conversion factor for seconds to minutes be

$= \frac{1}{60}\ minutes$

Now convert 135 seconds to minutes.

$= \frac{135\times 1}{60}$

= 2.25 minutes

Hence proved

4 0

3 years ago

Read 2 more answers

HELP PLEASE HELP HELP

kupik [55]

I think it’s 0but not s

5 0

2 years ago

Your bowl contains 8 red balls, and 9 blue balls, and you draw 4 balls. in how many ways can the selection be made so that at le

kumpel [21]

<span>Your bowl contains 8 red balls, and 9 blue balls, and you draw 4 balls. in how many ways can the selection be made so that at least one of each color is drawn.</span>

5 0

3 years ago

As a real estate agent kensho earns 4.5% commission on the houses he sells in the first quarter of the year he sold two houses o

Natalka [10]

175000 / 100 = 1750 this step finds 1% of 175000
1750 x 4.5 = 7875 this step finds 4.5% of 175000
Commission for first house $7875
199000 / 100 = 1990 this step finds 1% of 199000
1990 x 4.5 = 8955 this step finds 4.5% of 199000
Commission for second house $8955
$7875 + $8955 = $16,830

Kensho's commission for this quarter is $16,830

4 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