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Anna71 [15]
3 years ago
15

The Air conditioner unit in the classroom uses 120 volts. it requires 2000 watts of usage on the average, and it stays on for 8

hours every day. Find out the Amps, Watt hrs/Day, Watts hrs/yr, Kilowatt Hrs/Yr and cost/Yr if the cost of elelctricity per kilowatt is $0.25
Mathematics
1 answer:
nalin [4]3 years ago
5 0

Step-by-step explanation:

P = IV

2000 W = I (120 V)

I = 16.7 A

P = E/t

P = (2000 W × 8 hr) / (1 day)

P = 16,000 W·hr/day

P = 16000 W·hr/day × (365 day / yr)

P = 5,840,000 W·hr/yr

P = 5,840,000 W·hr/yr × (1 kW / 1000 W)

P = 5840 kW·hr/yr

C = 5840 kW·hr/yr × (0.25 $/kW/hr)

C = $1460/yr

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Masja [62]

Answer:

D          SD= sqrt(22.5D)   Average Rate of Change            

47                       32.52  

50                       33.5                                            0.34

56                       35.50                                   0.33

60                       36.74                                   0.31

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65                       38.24                                   0.3

In the context of this problem, the average rate of change is the rate at which the car decelerates as it skids. As the distance that the car skids increases, the rate at which the car decelerates decreases at a gradual rate.

Step-by-step explanation:

3 0
3 years ago
4. On the day Alex was born, his father invested $5000 in an account with a 1.2% annual growth rate.
mihalych1998 [28]

Answer and Step-by-step explanation:

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B) When we plug 21 into the equation,  we get 106260.

I hope this helps!

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4 years ago
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A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and
Hitman42 [59]

Answer:

0.13% probability that this selected sample has an average thickness greater than 53

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 50, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

What is the probability that this selected sample has an average thickness greater than 53?

This is 1 subtracted by the pvalue of Z when X = 53. So

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

By the Central Limit Theorem

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

Z = \frac{53 - 50}{1}

Z = 3

Z = 3 has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% probability that this selected sample has an average thickness greater than 53

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