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Ugo [173]
2 years ago
6

What is the value of y? PLEASE EXPLAIN 8.5 = y/3 <------ Fraction

Mathematics
1 answer:
NISA [10]2 years ago
5 0

Answer:

y=25.5

Step-by-step explanation:

To find y, you need to get it by itself. Ask yourself, what is being done to y, then to undo that, do the opposite. For example, if 3 was added to y, you would need to subtract 3 to cancel out the 3. In this case, what is being done to y? It's being divided by 3. So, you multiply by 3 to cancel that out. Whatever you do to one side of an equation, you also have to do to the other. So,

8.5=\frac{y}{3}   Multiply both sides by 3

3(8.5) = 3(\frac{y}{3})  multiply on the left side, and on the right side the 3s cancel out

25.5=y

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The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
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a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
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