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Licemer1 [7]
3 years ago
11

Use elimination to solve each system of equations. 1/3x-y=-1 and 1/5x-2/5y=-1

Mathematics
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

x=−9 and y=−2

Step-by-step explanation:

Multiply the first equation by 2,and multiply the second equation by -5.

2(1/3x−y=−1)

−5(1/5x−2/5y=−1)

Becomes: 2/3x−2y=−2

−x+2y=5

Add these equations to eliminate y:

−1/3  x=3

Then solve −1/3x=3 for x:

−1/3x        3

−1/3    = −1/3

(Divide both sides by (-1)/3)

x=−9

Now that we've found x let's plug it back in to solve for y.

Write down an original equation:

1/3x−y=−1

Substitute−9 for x in 1/3x−y=−1:

1/3(−9)−y=−1

−y−3=−1(Simplify both sides of the equation)

−y−3+3=−1+3(Add 3 to both sides)

−y=2

−y

−1=

2

−1

(Divide both sides by -1)

y=−2

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Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is t
svet-max [94.6K]

Answer:

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

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 = 450, \sigma = 100

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

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

Z = \frac{550 - 450}{100}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 350

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

Z = \frac{350 - 450}{100}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

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

1.88 = \frac{X - 450}{100}

X - 450 = 1.88*100

X = 638

A score of 638(or higher).

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X when Z has a pvalue of 0.60. So it is X when Z = 0.253

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

0.253 = \frac{X - 450}{100}

X - 450 = 0.253*100

X = 475.3

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

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

-0.385 = \frac{X - 450}{100}

X - 450 = -0.385*100

X = 411.5

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

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

0.385 = \frac{X - 450}{100}

X - 450 = 0.385*100

X = 488.5

The middle 30% of the test scores is between 411.5 and 488.5.

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The ratio of the number of boys to the number of girls in a school is 5:7.If there are 600 students in the school,how many girls
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So,

The secret to solving problems with ratios is to find the value of one unit.

5:7 = 12 units total

To find one unit, divide the total number of students by the total number of units.
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50 = a

The value of each unit is 50.

Now, multiply the units by the numbers in the ratio.
50(5) = b
250 = boys

50(7) = x
350 = x

There are 350 girls.
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