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Pepsi [2]
3 years ago
5

The function is used to convert temperature from Celsius (C) to Fahrenheit (F). The function C(K) = K − 273.15 is used to conver

t temperature from Kelvin (K) to Celsius. Which is the correct function for converting temperature from Kelvin to Fahrenheit?
Mathematics
1 answer:
ANTONII [103]3 years ago
8 0

Answer:

F(k)= \frac{9}{5}(k - 273.15) + 32

 

Step-by-step explanation:

Given

F(c)= \frac{9}{5}c + 32

C(k) = k - 273.15

Required

Function from K to F

This implies that, we calculate F(k)

Given that:

F(c)= \frac{9}{5}c + 32

Substitute c(k) for c

F(c(k))= \frac{9}{5}c(k) + 32

Substitute: C(k) = k - 273.15

F(c(k))= \frac{9}{5}(k - 273.15) + 32

Hence:

F(k)= \frac{9}{5}(k - 273.15) + 32

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Answer:

a) 0.1587 = 15.87% probability that a battery will break down during the warranty period.

b) The expected profit of the auto store on a battery is of $15.239.

c) The expected monthly profit on batteries if the auto store sells an average of 500 batteries a month is of $7,619.50.

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.

The life of batteries is known to be normally distributed with a mean and a standard deviation of 40 and 16 months, respectively.

This means that \mu = 40, \sigma = 16

a. What is the probability that a battery will break down during the warranty period?

Two year warranty, that is, 24 months. This probability is the pvalue of Z when X = 24. So

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

Z = \frac{24 - 40}{16}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.1587 = 15.87% probability that a battery will break down during the warranty period.

b. What is the expected profit of the auto store on a battery?

1 - 0.1587 = 0.8413 probability of a profit of $20.

0.1587 probability of a loss of $10. So

E = 0.8413*20 - 0.1587*10 = 15.239

The expected profit of the auto store on a battery is of $15.239.

c. What is the expected monthly profit on batteries if the auto store sells an average of 500 batteries a month?

Multiplying the average for a battery by 500. So

15.239*500 = $7,619.50

The expected monthly profit on batteries if the auto store sells an average of 500 batteries a month is of $7,619.50.

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