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gladu [14]
2 years ago
15

Margaret works at a clothing store. Today she has sold only shirt and pants.The ratio of the number of shirts sold to the number

of the items sold to 2/5. Which model shows this to ratio
Mathematics
1 answer:
poizon [28]2 years ago
6 0

Answer:

Shirt : Pant = 2 : 3 ,  & Pant : Shirt = 3 : 2 , Linear model can be used.

Step-by-step explanation:

Ratio of ' Number of shirts sold : Number of items sold' = 2 : 5

Let the common ratio be = x

Then, number of shirts sold = 2x. And, number of total items sold = 5x.

So, number of pants sold = Number of items sold - number of shirts sold = 5x - 2x = 3x

So, shirt to pant ratio = 2x / 3x = 2 : 3 .And,  pant to shirt ratio = 3 : 2

This model can be shown in linear equation, as all are linear powered.

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8n + 12 - 5n = 21<br> n=?
insens350 [35]

Answer:

n=3

Step-by-step explanation:

Simplifying

8n + 12 + -5n = 21

Reorder the terms:

12 + 8n + -5n = 21

Combine like terms: 8n + -5n = 3n

12 + 3n = 21

Solving

12 + 3n = 21

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '-12' to each side of the equation.

12 + -12 + 3n = 21 + -12

Combine like terms: 12 + -12 = 0

0 + 3n = 21 + -12

3n = 21 + -12

Combine like terms: 21 + -12 = 9

3n = 9

Divide each side by '3'.

n = 3

Simplifying

n = 3

5 0
3 years ago
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mafiozo [28]

Answer:

the answer is 24 because if you 6 x 4 it's 24

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Step-by-step explanation:

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2 years ago
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Write an equation for the quotient of a number and -12
ivanzaharov [21]
Quotient Rule. Objectives: In this tutorial, we derive the formula for finding the derivative of a quotient of two functions and apply this formula to several examples. After working through these materials, the student should be able to derive the quotient rule and apply it.
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3 years ago
If it's 7:30 what time will it gets in an hour and a quarter?
slava [35]

Answer:

8:55

Step-by-step explanation:

7:30 plus 1 hour is 8:30. Quarter is 25 or 1/4. Then you add 8:30 plus 25  which gives you 8:55.

7 0
2 years ago
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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.

   

8 0
3 years ago
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