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Pani-rosa [81]
3 years ago
9

Factor the following equation to find its zeros. y = x² - 36

Mathematics
1 answer:
alukav5142 [94]3 years ago
3 0

Answer:

x = 6, x = - 6

Step-by-step explanation:

Given

y = x² - 36

To find the zeros let y = 0, that is

x² - 36 = 0 ← x² - 36 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 36 = 0

x² - 6² = 0

(x - 6)(x + 6) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 6 = 0 ⇒ x = - 6

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8 0
3 years ago
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

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
2 years ago
Hey i need help this assignment is like 3 pages long
slavikrds [6]

Answer:

c=1.96

Step-by-step explanation:

2.1/c=1.5/1.4

2.1*1.4=1.5*c

2.94/1.5=1.5c/1.5

c=1.96

7 0
2 years ago
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