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zhannawk [14.2K]
2 years ago
13

The graph of a linear relationship has a slope of -5 and passes through the point (2, -12). Write the equation of the line in sl

ope-intercept form.
Mathematics
1 answer:
DerKrebs [107]2 years ago
5 0
<h3>Answer:</h3>

y=-5x-2

<h3>Solution:</h3>
  • We are given the line's slope and a point that it passes through.
  • Using the given information, we can write the equation of the line in Point-Slope Form:
  • y-y_1=m(x-x_1)
  • In this formula, y₁ stands for the y-coordinate of the point, m stands for the slope of the line, and x₁ stands for the x-coordinate of the point.
  • In this case, y₁ is equal to -12, m is equal to -5, and x₁ is equal to 2.
  • Plug in the values:
  • y-(-12)=-5(x-2)
  • y+12=-5(x-2)
  • Now, convert into Slope-Intercept Form, y=mx+b:
  • y+12=-5x+10
  • y=-5x+10-12
  • y=-5x-2

Hope it helps.

Do comment if you have any query.

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A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he
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(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

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The margin of error, <em>MOE</em> = 0.04.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

The critical value of <em>z</em> for 95% confidence interval is: z_{\alpha/2}=1.96

Compute the minimum sample size required as follows:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, <em>MOE</em> = 0.02.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

The critical value of <em>z</em> for 95% confidence interval is: z_{\alpha/2}=1.96

Compute the minimum sample size required as follows:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

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3 years ago
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