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3 years ago

(need done quickly)

1 answer:

3 0

The error is that he probably forgot the decimal point thus causing him to say '"hundred" instead of "and" for the decimal point.

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2 years ago

Will give Brainlyest. ASAP

crimeas [40]

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3 years ago

A quarter throws a football from a height of 6.5 feet with an initial velocity of 64 feet per second

ohaa [14]

Answer:

0.10s

Step-by-step explanation:

Given data

DIstance= 6.5ft

velocity = 64 feet per second

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3 years ago

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.

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

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