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ioda
3 years ago
14

Please help. i need to pass.

Mathematics
1 answer:
FrozenT [24]3 years ago
5 0
Plug 8 for y

8 = 2x + 4

Subtract 4

4 = 2x

Divide by 2

x = 2

Plug 16 for y

16 = 2x + 4

Subtract by 4

12 = 2x

Divide by 2

x = 6

Substitute 20 for y

20 = 2x + 4

16 = 2x

x = 8

Substitute 22 for y

22 = 2x + 4

18 = 2x

9 = x

So the values are 2, 6, 8, 9
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Write the sum as a product of the GCF of 96 + 18?
worty [1.4K]
96: 1,2,3,4,6,8,12,16,24,32,48,96

18: 1,2,3,6,9,18

GCF: 6

But, if you're asking what 96 + 18 it's 114
7 0
2 years ago
Now dis one dedicated to the girls dem
nirvana33 [79]

Answer:

and dat is onithhh purrrrr.. only of thee most respectfulyy sheesithhh

Step-by-step explanation:

5 0
3 years ago
Markers are sold in boxes packs or as single markers each box has 10 packs each pack has 10 markers
KatRina [158]
100 markers per box

10x10

10- one pack

10- packs in a box
3 0
3 years ago
Read 2 more answers
What is the area of a regular hexagon with a side length of 12 cm<br> With process please
ivanzaharov [21]
Look at the picture.

The area of the hexagon is equal six times the area of the <span>equilateral triangle.

The formula of the </span><span>equilateral triangle with the leg a:

A_\Delta=\dfrac{a^2\sqrt3}{4}

Therefore the formula of the area of the hexagon with a side lenght a is:

A=6\cdot\dfrac{a^2\sqrt3}{4}=\dfrac{3a^2\sqrt3}{2}

We have a = 12cm. Substitute:

A=\dfrac{3\cdot12^2\sqrt3}{2}=\dfrac{3\cdot144\sqrt3}{2}=3\cdot72\sqrt3=216\sqrt3\ cm^2
</span>

4 0
3 years ago
The time for a professor to grade an exam is normally distributed with a mean of 16.3 minutes and a standard deviation of 4.2 mi
dangina [55]

Answer:

A.0.4477

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 16.3, \sigma = 4.2

What is the probability that a randomly selected exam will require between 14 and 19 minutes to​ grade?

This probability is the pvalue of Z when X = 19 subtracted by the pvalue of Z when X = 14. So

X = 19

Z = \frac{X - \mu}{\sigma}

Z = \frac{19 - 16.3}{4.2}

Z = 0.64

Z = 0.64 has a pvalue of 0.7389.

X = 14

Z = \frac{X - \mu}{\sigma}

Z = \frac{14 - 16.3}{4.2}

Z = -0.55

Z = -0.55 has a pvalue of 0.2912

0.7389 - 0.2912 = 0.4477

So the correct answer is:

A.0.4477

7 0
2 years ago
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