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ss7ja [257]
3 years ago
15

Tony solves the linear equation 3(x - 3) + 2x + 9 = 2x + 2(x - 1) using the step shown below.

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

D.)

Step-by-step explanation:

because when you have numbers inside the parentheses, you have to multiply both of those numbers by the number to the right of the parentheses, not just the X.

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Dylan's water bottle has a greater capacity. There are 1,000 Milliliters in a Liter. So Dylan's water bottle holds 400ml more. 
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Solve the given inequality. Describe the solution set using the set-builder or interval notation. Then, graph the
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4m + 24.25 < 34.25

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Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of
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Answer:

t = 31.29

Step-by-step explanation:

Given

\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

\bar x =\frac{\sum x}{n}

n, in both datasets in 6

For A

\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}

\bar x_A =\frac{157}{6}

\bar x_A =26.17

For B

\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}

\bar x_B =\frac{101.4}{6}

\bar x_B =16.9

Next, calculate the sample standard deviation

This is calculated using:

s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

For A

s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}

s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}

s_A = \sqrt{\frac{1.7134}{5}}

s_A = \sqrt{0.34268}

s_A = 0.5854  

For B

s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}

s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}

s_B = \sqrt{\frac{0.92}{5}}

s_B = \sqrt{0.184}

s_B = 0.4290

Calculate the pooled variance

S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}

S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}

S_p^2 = \frac{2.6336708}{10}

S_p^2 = 0.2634

Lastly, calculate the test statistic using:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

We set

\mu_A = \mu_B

So, we have:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}

The equation becomes

t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}

t = \frac{9.27}{\sqrt{0.0878}}

t = \frac{9.27}{0.2963}

t = 31.29

<em>The test statistic is 31.29</em>

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Answer:

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0.8 = 80%

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