Use the order of operations to find the value of the expression.

On a coordinate plane, a parabola opens up. It goes through (negative 2, 4), has a vertex at (0.25, negative 6), and goes throug

Oksana_A [137]

Answer:

The function has two x-intercepts.

The vertex of the function is (one-quarter, negative 6 and one-eighth).

Step-by-step explanation:

Given that the vertex is located at (0.25, -6) and the parabola opens up, the function has two x-intercepts.

f(x) = 2x² - x - 6 has the following coefficients:

a = 2

b = -1

c = -6

x-coordinate of the vertex is:

x = -b/(2a)

x = 1/(2*2) = 1/4

y-coordinate of the vertex is:

f(1/4) = 2(1/4)² - 1/4 - 6 = -6 1/8

3 0

3 years ago

Read 2 more answers

The profits for the week was $1200. the amount spend was $975 . find the revenue for the week

Kryger [21]

You get $1200-$975=225

6 0

4 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$