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Kruka [31]
1 year ago
8

Anthony's math teacher finds that there's roughly a linear relationship between the amount of time students spend on their homew

ork and their weekly quiz scores. This relationship can be represented by the equation y=57+7.4x, where y represents the expected quiz score and x represents hours spent on homework that week. What is the meaning of the x-value when y=90
Mathematics
1 answer:
Helen [10]1 year ago
3 0

The meaning of the x-value when y=90 means that Anthony will use 3.1 hours.

<h3>What is an equation?</h3>

An equation is the statement that illustrates that the variables given. In this case, two or more components are taken into consideration to describe the scenario.

The relationship that can be represented by the equation y=57+7.4x,

where

y represents the expected quiz score

x represents hours spent on homework that week.

When y = 90, the value of x will be:

y=57+7.4x

57 + 7.4x = 90

Collect like terms

7.4x = 90 - 57

7.4x = 23

Divide

x = 23/7.4

x = 3.1

He'll use 3.1 hours.

Learn more about equations on:

brainly.com/question/2972832

#SPJ1

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a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}&#10;And in order to find the sample mean we just need to use this formula:&#10;[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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