Which equation matches the graph below?

1 answer:

6 0

Answer:Given that the graph shows tha the functión at x = 0 is below the y-axis, the constant term of the function has to be negative. This leaves us two possibilities:

y = 8x^2 + 2x - 5 and y = 2x^2 + 8x - 5

To try to discard one of them, let us use the vertex, which is at x = -2.

With y = 8x^2 + 2x - 5, you get y = 8(-2)^2 + 2(-2) - 5 = 32 - 4 - 5 = 23 , which is not the y-coordinate of the vertex of the curve of the graph.

Test the other equation, y = 2x^2 + 8x - 5 = 2(-2)^2 + 8(-2) - 5 = 8 - 16 - 5 = -13, which is exactly the y-coordinate of the function graphed.

Step-by-step explanation:

You might be interested in

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:

$P(X > x) = 0.08$