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2 years ago

Write the equation of the line in fully simplified slope-intercept form.

Mathematics

2 answers:

Alexxandr [17]2 years ago

5 0

Use 2 points on the line to find the slope.

Slope = change in y / change in x

The two points used: (-2,5) and ( 1,-10)

slope = (-10 -5) /(1- -2) = -15/3 = -5

the y intercept is where the line is at in the y direction when it crosses the x axis at 0.

the y intercept is -5

The equation is y = -5x -5

Oksanka [162]2 years ago

5 0

Answer:

y=-9/2x-4

Step-by-step explanation:

see explanation in photo below

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The price of a technology stock was $9.63 yesterday. Today, the price rose to $9.74. Find the percentage increase. Round your an

aev [14]

0.10 is the answer pretty sure

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2 years ago

The sum of two numbers is 24. The second number minus the first number equals 2.

Sonbull [250]

Answer:

(11, 13)

Step-by-step explanation:

First Equation: x+y=24

Second Equation: y=x+2

x + (x + 2) = 24

2x + 2 - 2 = 24 - 2

2x/2 = 22/2

x = 11

y = (11) + 2

y = 13

Therefore, the solution to the system is (11, 13)

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3 years ago

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and

Naddika [18.5K]

Answer:

Using a 90% confidence level

A. A sample size of 68 should be used.

B. A sample size of 98 should be used.

Step-by-step explanation:

I think there was a small typing mistake and the confidence level was left out. I will use a 90% confidence level.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 72 seconds, what sample size should be used?

We have the standard deviation in minutes, so the margin of error should be in minutes.

72 seconds is 72/60 = 1.2 minutes.

So we need a sample size of n, and n is found when M = 1.2. We have that $\sigma = 6$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$1.2 = 1.645*\frac{6}{\sqrt{n}}$

$1.2\sqrt{n} = 6*1.645$

$\sqrt{n} = \frac{6*1.645}{1.2}$

$(\sqrt{n})^{2} = (\frac{6*1.645}{1.2})^{2}$

$n = 67.65$

Rounding up.

A sample size of 68 should be used.

B. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used?

Same logic as above, just use M = 1.

$M = z*\frac{\sigma}{\sqrt{n}}$

$1 = 1.645*\frac{6}{\sqrt{n}}$