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Bezzdna [24]
2 years ago
15

Graph the line with slope 2. passing through the point (-3, 3). 3 1 ? 13

Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
3 0

The graph of the line with the given slope that passes through (-3, 3) is shown in the image attached below.

<h3>What is the Equation of a Line in Slope-intercept Form?</h3>

The equation in slope-intercept form of a line, is y = mx + b, where:

b = y-intercept

m = slope = change in y/change in x.

Given a line with slope = 2 and passes through (-3, 3), plug in m = 2 and (a, b) = (-3, 3) into y - b = m(x - a) to write the equation of the line

y - 3 = 2(x - (-3))

Rewrite in slope-intercept form

y - 3 = 2x + 6

y = 2x + 6 - 3

y = 2x + 3

Therefore, the graph of the line with the given slope that passes through (-3, 3) is shown in the image attached below.

Learn more about slope-intercept form on:

brainly.com/question/1884491

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Z is inversely proportional to X is the same as
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3 years ago
4(10b-4) <br><br> Explain please
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<h3>Answer:   40b - 16</h3>

Work Shown:

4(10b-4)

4*10b - 4*4

40b - 16

I multiplied the outer term 4 by each term inside. Refer to the distribution property (aka distributive property).

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A highway engineer specifies that a certain section of roadway covering a horizontal distance of 2km should have a downgrade of
Zigmanuir [339]

Answer:

0.16km

Step-by-step explanation:

A highway engineer wants to compute  the change in elevation of a section of road. The horizontal distance of this section of road is 2km and downgrade is 8%

The slope formula is given by

m=\frac{rise}{run}=\frac{x}{y}

m = 8% = 8/100

run = y = 2km

m =\frac{x}{y}

\frac{8}{100} =\frac{x}{2}

\frac{2*8}{100} =x

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Verification:

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Answer:

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Step-by-step explanation:

The goal of this analysis is to determine if the mean wake time before the treatment is statistically significant. The question informed us the mean wake time before and after the treatment, the number of subjects and the standard deviation of the sample after treatment. So using the formula, we can calculate the confidence interval as following:

IC[\mu ; 98\%] = \overline{y} \pm t_{0.99,n-1}\sqrt{\frac{Var(y)}{n}}

Knowing that t_{0.99,15} = 2.602:

IC[\mu ; 98\%] = 98.9 \pm 2.602\frac{42.3}{4} \Rightarrow 98.9 \pm 27.516

IC[\mu ; 98\%] = [71.387 ; 126,416]

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