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MakcuM [25]
3 years ago
14

Temperature in degrees Fahrenheit is equal to 32 more than 9/5 times the temperature in degrees Celsius . One day the high tempe

rature in Edison. Nj was 95 degrees Fahrenheit. What was the high temperature in degrees Celsius
Mathematics
1 answer:
bazaltina [42]3 years ago
5 0

The high temperature is 35 degree celsius

<em><u>Solution:</u></em>

Given that, Temperature in degrees Fahrenheit is equal to 32 more than 9/5 times the temperature in degrees Celsius

Let "f" be the temperature in degree Fahrenheit

Let "c" be the temperature in degree celsius

Therefore,

Temperature in degrees Fahrenheit = 32 + \frac{9}{5} times the temperature in degrees Celsius

f = 32 + \frac{9}{5} \times c

One day the high temperature in Edison. Nj was 95 degrees Fahrenheit

f = 95 ; c = ?

Substitute f = 95 in above equation

95 = 32 + \frac{9c}{5}\\\\95=\frac{32 \times 5 +9c}{5}\\\\95 \times 5 = 160+9c\\\\475 = 160 + 9c\\\\9c = 475 - 160\\\\9c = 315\\\\c = 35

Thus high temperature in degrees Celsius is 35 degree celsius

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Answer:

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

b) The minimum score for an A is 84.68.

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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:

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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.

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

X - 61 = 2.96*8

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The minimum score for an A is 84.68.

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3 years ago
Simplify the expression 2.75j + 2.25 - 1.5j + 3
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