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Serggg [28]
2 years ago
10

Ronnie wants to lower his utility bills to save money on his monthly expenses. He discovers that if he replaces his water heater

with a solar water heater his monthly utility bills can decrease around 50% to 85%. At the end of the year Ronnie wants to calculate the percentage his new solar water heater saved him in the utility bills compared to last year. Ronnie's total utility expenses for this year are $862.40 and last year the utility expenses were $1935.67. What percentage did Ronnie save on his utlility bills by switching to a solar water heater?
Mathematics
2 answers:
Soloha48 [4]2 years ago
5 0

Alright, lets get started.

Last year, Ronnie's utility expenses were = 1935.67 $

After switching to electric water heater to solar water heater, his yearly utility bills get reduced.

This year, Ronnie's utility expenses are = 862.40 $

So the difference on utility bills Ronnie saved = 1935.67 - 862.40

so, difference Ronnie saved = 1073.27 $

So, the percentage of amount Ronnie saved = \frac{1073.27}{1935.67} *100

Percentage of amount Ronnie saved = 55.45 %    

Hence, 55.45 % Ronnie save on his utlility bills by switching to a solar water heater.   :    Answer

Hope it will help :)


marusya05 [52]2 years ago
3 0

Answer:

The answer is C.

Step-by-step explanation:

This is right on ednuity

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Answer:

₹ 37500

Step-by-step explanation:

Let the price before VAT be ₹x.

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7 0
2 years ago
For a certain type of copper wire, it is knownthat, on the average, 1.5 flaws occur per millimeter.Assuming that the number of f
mario62 [17]

Answer:

The probability that no flaws occur in a certain portion of wire of length 5 millimeters =  1.1156 occur / millimeters

Step-by-step explanation:

<u>Step 1</u>:-

Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.

by  Poisson random variable given that λ = 1.5 flaws/millimeter

Poisson distribution P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}

<u>Step 2:</u>-

The probability that no flaws occur in a certain portion of wire

P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}

On simplification we get

P(x=0) = 0.223 flaws occur / millimeters

<u>Conclusion</u>:-

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters

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cestrela7 [59]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

<u>Graph of the function:</u>

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