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Digiron [165]
3 years ago
14

one month charlie rented 2 movies 3 video games for a total of $21. The next month he rented 4 movies and 8 video games for a to

tal of $53. Find the rental cost of each movies and each video game.
Mathematics
1 answer:
Aleks [24]3 years ago
4 0
X= cost per movie
y= cost per video game

EQUATION 1:
2x + 3y= $21

EQUATION 2:
4x + 8y= $53

STEP 1:
multiply equation 1 by -2.

-2(2x + 3y)= -2($21)
-4x - 6y= -42


STEP 2:
solve linear equations by elimination. use equation 2 and equation answer from step 1. add the two equations together. the x terms will cancel out. solve for y.

4x + 8y= 53
-4x - 6y= -42

2y= 11
divide both sides by 2
y= 11/2
divide
y= $5.5 per video game


STEP 3:
substitute y answer from step 2 into either original equation to solve for x.
2x + 3y= $21
2x + 3(5.50)= 21
2x + 16.50= 21
subtract 16.50 from both sides
2x= 4.50
divide both sides by 2
x= $2.25 per movie


CHECK:
4x + 8y= $53
4(2.25) + 8(5.50)= 53
9 + 44= 53
53= 53


ANSWER: The cost per movie is $2.25 and the cost per video game is $5.50.

Hope this helps! :)
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A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances
wariber [46]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.

So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.

Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:

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And the Standard deviation S₁= 10.4 feet

X₂: Braking distance of an SUV equipped with tires made with compound two.

Its sample mean is X[bar]₂= 71 feet

And the Standard deviation S₂= 7.6 feet

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The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

The statistic to use to compare these two populations is a pooled Z test

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Z ≈ N(0;1)

Z_{H_0}= \frac{69-71-0}{\sqrt{\frac{108.16}{81} +\frac{57.76}{81} } }= -1.397

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Z_{\alpha  } = Z_{0.05}= -1.648

Decision rule:

If Z_{H_0} > -1.648 , then you do not reject the null hypothesis.

If Z_{H_0} ≤ -1.648 , then you reject the null hypothesis.

Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.

At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.

I hope you have a SUPER day!

4 0
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