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2 years ago

Fine the Slope, please explain!!

Mathematics

2 answers:

Rina8888 [55]2 years ago

5 0

Answer:

-3

Step-by-step explanation:

Because slope is rise over run.

Rise = -3

Run = 1

Rise/Run = -3/1 = -3

julia-pushkina [17]2 years ago

4 0

Answer:

Slope(m)= -3

Step-by-step explanation:

From the given graph, let's take two points,

(0,0) and (1,-3)

$Slope=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right):\left(0,\:0\right)$

$\left(x_2,\:y_2\right):\left(1,\:-3\right)$

$m=\frac{-3-0}{1-0}$

$m=-3$

Hope this helps!

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Solve the inequality x + 2 > 6.

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Answer:

x > 4

Step-by-step explanation:

x + 2 > 6

-2 -2

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Choose the equation that represents a line that passes through points (−1, 2) and (3, 1).

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You did not provide us with equations to select.

Find the slope m.

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m = -1/(3 + 1)

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Use the slope and one of the points and plug into the point-slope formula.

y - 1 = (-1/4)(x - 3)

Isolate y.

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Did you follow?

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3 years ago

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A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before t

aleksklad [387]

Answer:

We cannot say that the mean wake time are different before and after the treatment, with 98% certainty. So the zopiclone doesn't appear to be effective.

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 $t_{0.99,15} = 2.602$ :

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

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

Note that $102.8 \in [71.384 ; 126.416]$ so we cannot say, with 98% confidence, that the mean wake time before treatment is different than the mean wake time after treatment. So the zopiclone doesn't appear to be effective.