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3 years ago

Are these lines parallel perpendicular or neither

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Answer:

perpendicular

Step-by-step explanation:

start but putting each line into the standard line format

2x-y=-5

-y=-2x-5

y=2x+5

x+2y=-6

2y=-x-6

y=-(1/2)x-3

if lines are parallel the slopes would be the same. 2 is not equal to -1/2

if lines are perpendicular the slope of one would be the opposite inverse of the other.

negative inverse of 2 = - (1/2)

so the lines are perpendicular

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If you subtract 12<br> from my number and<br> multiply the difference by -3,<br> the result is -54.

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Answer:

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Step-by-step explanation:

Let the number be x

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-3x + 36 = -54

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Answer:

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(b) Margin of error ( E) = $1,000 , n = 216

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Step-by-step explanation:

Given -

Standard deviation $\sigma$ = $7,500

$\alpha$ = 1 - confidence interval = 1 - .95 = .05

$Z_{\frac{\alpha}{2}}$ = $Z_{\frac{.05}{2}}$ = 1.96

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(a) Margin of error ( E) = $2,000

Margin of error ( E) = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

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Squaring both side

$E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}$

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n = 54 ( approximately)

(b) Margin of error ( E) = $1,000

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

1000 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}$

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500 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}$

n = 864

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