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jekas [21]
3 years ago
10

What is the solution to this system of equations? Remember to write your solution as a a point (x, y).

Mathematics
1 answer:
Butoxors [25]3 years ago
6 0

Answer:

(1,-3) 1 is x and 3 is y because the two arrows intersept in that coordinate

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Step-by-step explanation:

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A math professor notices that scores from a recent exam are normally distributed with a mean of 61 and a standard deviation of 8
Alexeev081 [22]

Answer:

a) 25% of the students exam scores fall below 55.6.

b) The minimum score for an A is 84.68.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 61 and a standard deviation of 8.

This means that \mu = 61, \sigma = 8

(a) What score do 25% of the students exam scores fall below?

Below the 25th percentile, which is X when Z has a p-value of 0.25, that is, X when Z = -0.675.

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

-0.675 = \frac{X - 61}{8}

X - 61 = -0.675*8

X = 55.6

25% of the students exam scores fall below 55.6.

(b) Suppose the professor decides to grade on a curve. If the professor wants 0.15% of the students to get an A, what is the minimum score for an A?

This is the 100 - 0.15 = 99.85th percentile, which is X when Z has a p-value of 0.9985. So X when Z = 2.96.

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

2.96 = \frac{X - 61}{8}

X - 61 = 2.96*8

X = 84.68

The minimum score for an A is 84.68.

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3 years ago
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Step-by-step explanation:

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Is this right please help me Is this right please help me
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