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Drupady [299]
2 years ago
15

Emma earns $6 each time she mows the lawn and $8 per hour for babysitting. She is saving up to buy a new pair of jeans that cost

$48. If she mows the lawn x times and babysits for y hours, which graph shows the amount of work she needs to complete to earn at least enough to purchase the new jeans?

Mathematics
2 answers:
dangina [55]2 years ago
6 0

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

x------> the number of times Emma mows the lawn

y------> the number of hours Emma babysits

we know that

6x+8y\geq 48 ------> inequality that represent the situation

The solution is the shade area above the solid line between the values of x and y positive

The equation of the solid line is equal to 6x+8y=48

The slope of the line is negative m=-\frac{3}{4}

The y-intercept of the line is the point (0,6) (value of y when the value of x is equal to zero)

The x-intercept of the line is the point (8,0) (value of x when the value of y is equal to zero)

so

The graph in the attached figure

Ilya [14]2 years ago
4 0

Answer:

Its C on ed2020

Step-by-step explanation:

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a) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

H_0: \mu=32.48\\\\H_a:\mu< 32.48

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565

The degrees of freedom for this sample size are:

df=n-1=65-1=64

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

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<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

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