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3 years ago

15

(-2,1) and (4,9), (-3,8) and (5,2) Are the lines parallel, perpendicular, or neither?

1 answer:

marissa [1.9K]3 years ago

6 0

Answer:

perpendicular

Step-by-step explanation:

if you find the slope of the points, they are completely opposite 4/3 and -3/4

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In the solid figure below, what is the volume of the right section of the figure?

Elden [556K]

The answer is d hope the helped

3 0

3 years ago

How many solutions does the equation y + 12 = y + 10 + 2 have?

coldgirl [10]

Y + 12 = y + 10 + 2

y + 12 = y + 12

y - y = 12 - 12

0 = 0

True for all y

hope this helps!

8 0

3 years ago

Input<br>output<br>17<br>rule: 3x-1

Marrrta [24]

Answer:

x = 6

Step-by-step explanation:

3x - 1 = 17

add 1 to both sides:

3x = 18

divide by 3:

x = 6

7 0

3 years ago

Find the standard deviation of the data set below. Round to the nearest tenth,<br> 6, 5, 2, 5,8

ASHA 777 [7]

Answer:

The standard deviation = 1.9

Step-by-step explanation:

standard deviation, ρ = sqrt ( $\frac{sum(x_{i} - u)^{2} }{N}$ )

Where u is the mean of the data, $x_{i}$ each given value, and N is the number of data given.

Mean, u = $\frac{6+5+2+5+8}{5}$

= 5.2

So that;

$(6-5.2)^{2}$ = $(0.8)^{2}$ = 0.64

$(5-5.2)^{2}$ = $(-0.2)^{2}$ = 0.04

$(2-5.2)^{2}$ = $(-3.2)^{2}$ = 10.24

$(5-5.2)^{2}$ = $(-0.2)^{2}$ = 0.04

$(8-5.2)^{2}$ = $(2.8)^{2}$ = 7.84

Sum = 18.8

So that;

$\sqrt{\frac{18.8}{5} }$ = $\sqrt{3.76}$

= 1.9391

= 1.9

The standard deviation of the given data is 1.9

6 0

3 years ago

Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be r

vlabodo [156]

Answer:

The sample size required is $n = (\frac{1.96*6.84}{M})^2$ , in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

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.

Use 6.84 days as a planning value for the population standard deviation.

This means that $\sigma = 6.84$ .

What sample size would be required to obtain a margin of error of M days?

This is n for the given value of M. So

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

$M = 1.96\frac{6.84}{\sqrt{n}}$

$M\sqrt{n} = 1.96*6.84$

$\sqrt{n} = \frac{1.96*6.84}{M}$

$(\sqrt{n})^2 = (\frac{1.96*6.84}{M})^2$

$n = (\frac{1.96*6.84}{M})^2$

The sample size required is $n = (\frac{1.96*6.84}{M})^2$ , in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

8 0

3 years ago