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2 years ago

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Which equation represents a line that passes through (–2, 4) and has a slope of 2/5? y – 4 = 2/5 (x 2) y 4 = 2/5 (x – 2) y 2 = 1

/3 (x – 4) y – 2 = 1/3 (x 4)

2 answers:

Natali5045456 [20]2 years ago

7 0

Equation of a line is given by y - y1 = m(x - x1); where m is the slope.
y - 4 = 2/5 (x - (-2))
y - 4 = 2/5 (x + 2)

Alla [95]2 years ago

4 0

Answer:

The required equation of line is:

$y-4=\frac{2}{5}(x+2)$

Therefore, option 1 is correct.

Step-by-step explanation:

We have been given a slope and a point

And we need to form the equation of line

We will use the formula for equation of line which is:

$y-y_1=m(x-x_1)$

Where m is the slope of the line.

here, $y_1=4,x_1=-2$

On substituting the values we get:

$y-4=\frac{2}{5}(x-(-2))$

$y-4=\frac{2}{5}(x+2)$

Therefore, the required equation of line is:

$y-4=\frac{2}{5}(x+2)$

Therefore, option 1 is correct.

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Answer:

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The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

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For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

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We can calculate the sample mean and deviation with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

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$s=2.41$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =200$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :

Null hypothesis: $\mu = 200$

Alternative hypothesis: $\mu = 200$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =2*P(t_{(9)}>2.32)=0.0455$

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

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