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ser-zykov [4K]
3 years ago
12

Combine Like Terms and Evaluate

Mathematics
2 answers:
rosijanka [135]3 years ago
6 0

Answer:

29

Step-by-step explanation:

-2x+4+3y-5x-2y

Combine like terms

-7x+1y+4

evaluate

-7(-3) 1(4) 4

Solve

21+4+4

29

Nady [450]3 years ago
5 0
<h3 /><h2>- 2x + 4 + 3y - 5x - 2y</h2>

if x = - 3 and y = 4

<h3><u>simplification</u> </h3>

<h3>- 2x - 5x + 3y - 2y + 4</h3>

<h3>- 7x + y + 4</h3>

<h3><u>evaluation</u><u> </u></h3>

<h3>- 7x + y + 4</h3>

<h3>- 7 ( - 3 ) + 4 + 4</h3>

<h3>21 + 4 + 4</h3>

<h2>= 29</h2>

hope that helps !

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Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2) , and X(1, −2) . Trapezoid WKLX ​ is translated 4 units right and 3 uni
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There are 7 7/8 cups of skittles in a Jar. If 2 1/4 cups are used for a party, how many skittes are left in the jar?
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2.A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation
Vesna [10]

Answer:

Step-by-step explanation:

Given that:

population mean = 10

standard deviation = 0.1

sample mean = 9.8 < x > 10.2

The z score can be computed as:

z = \dfrac{\bar x - \mu}{\sigma}

if x > 10.2

z = \dfrac{10.2- 10}{0.1}

z = \dfrac{0.2}{0.1}

z = 2

If x < 9.8

z = \dfrac{9.8- 10}{0.1}

z = \dfrac{-0.2}{0.1}

z = -2

The p-value = P (z ≤ 2) + P (z ≥ 2)

The p-value = P (z ≤ 2) + ( 1 -  P (z ≥ 2)

p-value = 0.022750 +(1 -   0.97725)

p-value = 0.022750 +  0.022750

p-value = 0.0455

Therefore; the probability of defectives  = 4.55%

the probability of acceptable = 1 - the probability of defectives

the probability of acceptable = 1 - 0.0455

the probability of acceptable = 0.9545

the probability of acceptable = 95.45%

4.55% are defective or 95.45% is acceptable.

sampling distribution of proportions:

sample size n=1000

p = 0.0455

The z - score for this distribution at most 5% of the items is;

z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}

z = 0.6828

The p-value = P(z ≤ 0.6828)

From the z tables

p-value = 0.7526

Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526

The z - score for this distribution for at least 85% of the items is;

z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = −15.86

p-value = P(z ≥  -15.86)

p-value = 1 - P(z <  -15.86)

p-value = 1 - 0

p-value = 1

Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1

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