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jenyasd209 [6]
3 years ago
6

Help Me!!

Mathematics
1 answer:
iogann1982 [59]3 years ago
6 0

check the picture below.


so we know Bethany's payrate, her slope is 11.75, namely $11.75/hr, for for every hour, she makes 11.75 bucks.


now, let's use those red-circled points, to get Grant's slope, and therefore his payrate,


\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{38.70})\qquad 
(\stackrel{x_2}{7}~,~\stackrel{y_2}{90.30})
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{90.30-38.70}{7-3}\implies \cfrac{51.6}{4}\implies 12.9


low and behold, Grant makes 12.90 or 12.9 bucks every hour.


well, Bethany does 11.75 hmmm what's their difference?


12.90 - 11.75

1.15


so 12.90 is greater than 11.75 by 1.15, meaning, Grant makes 1.15 bucks more every hour than Bethany.

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At a family reunion, a family friend brings 20 pounds of ice and 30 cups. Which expression shows the greatest common factor, usi
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The expression for the greatest common factor of 20 and 30 using the distributive property is 10(2 + 3)

Pounds of ice = 20

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The relation for the distributive property :

a × (b + c)

Expand ;

a × (b + c) = (a × b) + (a × c) = ab + ac

Finding the greatest common factor of 20 and 30 ;

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According to the distributive property

(2 × 10) + (3 × 10) = 10(2 + 3)

Therefore, the expression for the greatest common factor is 10(2 + 3)

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6 0
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Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance
lana66690 [7]

Answer:

Null Hypothesis, H_0 : p = 0.20  

Alternate Hypothesis, H_a : p > 0.20  

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20​% of users develop nausea.

<em>Let p = population proportion of users who develop nausea</em>

So, <u>Null Hypothesis,</u> H_0 : p = 0.20  

<u>Alternate Hypothesis</u>, H_a : p > 0.20  

Here, <u><em>null hypothesis</em></u> states that 20​% of users develop nausea.

And <u><em>alternate hypothesis</em></u> states that more than 20​% of users develop nausea.

The test statistics that would be used here is <u>One-sample z proportion</u> test statistics.

                     T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }   ~ N(0,1)

where,  \hat p = proportion of users who develop nausea in a sample of 241 subjects =  \frac{54}{241}  

             n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

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Answer:

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Step-by-step explanation:

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