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tensa zangetsu [6.8K]
2 years ago
11

What is the y-intercept of the line perpendicular to (1,1) and (-3,-1) and passes through (-3,-2)

Mathematics
1 answer:
marin [14]2 years ago
5 0

Answer:

-8

Step-by-step explanation:

Let's first establish the reference line (the one that the second line will be perpendicular to).  We are told that this line passes through two points:

(1,1) and (-3,-1).

We'll find a line equation using the point slope form format to start:  

(y - y1) = m * (x - x1), where m is the slope and the x and y are from two points.

(x,y) = (1,1)

(x1,y1) = (-3,-1)

Rearrange the equation:

(y - y1) = m * (x - x1)

m = (y - y1)/ (x - x1)

m = (1-(-1))/(1-(-3))

m = 2/4, or 1/2:  The slope is 1/2.    [<u>This is "m."]</u>

We can use the slope-intercept form for this line (y=mx + b) and then calculate b, the y-intercept:

y = (1/2)x + b

Use either of the two given points.  I'll use (1,1) since I have memorized the "1" math tables.

y = (1/2)x + b  

1 = (1/2)(1) + b for (1,1)

b = 1/2

This makes the reference line:  y = (1/2)x+(1/2)

===

The line perpendicular must have a slope that is the negative inverse of (1/2).  This would be -(2/1), or -2.

We can then write y = -2x + b

To find be, enter the one given point for this line:  (-3,-2)

y = -2x + b

-2 = -2(-3) + b

-2 = 6 + b

b = -8

The perpendicular line is thus:

y = -2x - 8

It has a y-intercept of -8

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Community college students conduct a survey at their college. They ask "Do you plan to transfer to a university to pursue a bacc
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Answer:

0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697

0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

For this case the conditions we have the conditions to assume a normal distribution for the population proportion since:

np=100*0.8=80 >10 , n(1-p) = 10*(1-0.8) =20 >10

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The estimated proportion is given by \hat p =\frac{80}{100} =0.8. If we replace the values obtained we got:

0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697

0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

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