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3 years ago

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Solve 18n - 7p-15n = 5p for n. A n = 1 B. n=2p C. n = 1/2 D. n=4p

2 answers:

andrey2020 [161]3 years ago

5 0

Answer:

<h3>The answer is option D</h3>

Step-by-step explanation:

18n - 7p-15n = 5p

To solve for n first make n the subject

Group like terms

That's

18n - 15n = 5p + 7p

Simplify

3n = 12p

Divide both sides by 3 to make n stand alone

That's

3n/3 = 12p/3

We have the final answer as

<h2>n = 4p</h2>

Hope this helps you

azamat3 years ago

3 0

Answer:

D) n = 4p

Step-by-step explanation:

18n - 7p - 15n = 5p

18n - 15n = 5p + 7p

Add like terms

3n = 12p

Divide both sides by 3

3n/3 = 12p/3

n = 4p

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The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 199 times. Here are the observed

Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28, 29, 47, 40, 22, 33.

Probability of an Event

$=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}$

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

$(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12$

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

$(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911$

Result is not significant.

3.

$(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378$

Result is not significant.

4.

$(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349$

Result is not significant.

5.

$(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773$

Result is not significant.

6.

$(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0$

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair die.

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They are not equivalent expression because 3^2 is 9 since 3*3 is 9 but 3*2 is 6, so the expressions are not equivalent. <span />

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3 years ago

What is the surface area of a box that is 6 by 4 inches

san4es73 [151]

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2 years ago

Read 2 more answers

The sum of two numbers is 79. Three times the first number added to 5 times the second number is 283. What are the two numbers?

Karolina [17]

Answer:

<u><em>The two numbers are 56 and 23.</em></u>

Step-by-step explanation:

Let the two numbers be A and B.

We are told that:

A+B = 79, and

2) 3A + 5B = 283

Rearrange the first expression to isolate A:

A = 79-B

Now use this value of B in the second expression:

3A + 5B = 283

3(79-B) + 5B = 283

237-3B + 5B = 283

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B = 23

If B=23, we can find A from (1):

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A+23 = 79

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<u>CHECK:</u>

Does A+B = 79?

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Step-by-step explanation:

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