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Softa [21]
2 years ago
14

Look at the figure below: an image of a right triangle is shown with an angle labeled y degrees If sin y° = s divided by 8 and t

an y° = s divided by t, what is the value of cos y°?
cos y° = 8s
cos y° = 8t
cos y°= t / 8
cos y°=8 / t
Mathematics
1 answer:
Elena L [17]2 years ago
5 0

Answer:

Cos y = t / 8

Step-by-step explanation:

Using the hints given in the question, the omitted tribagke will look like the triangle attached on the picture ;

From trigonometry :

Sin y = opposite / hypotenus

Sin y = s / 8

Opposite side = s ; hypotenus = 8

Tan y = opposite / Adjacent

Tan y = s / t

Adjacent side = t

Then ;

Cos y = Adjacent / hypotenus

Hence,

Cos y = t / 8

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The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

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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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