Which statement is true regarding the graphed functions?

Mathematics

1 answer:

anyanavicka [17]2 years ago

3 0

Answer:

the last option: $f(-2)=g(-2)$

Step-by-step explanation:

Make sure you have the numerical answer for each of the functional expressions that are shown among the possible solution choices:

f(4) = -14 (what the blue function reads [its y-value] when x is 4)

g(4) = 10 (what the red function reads [its y-value] for x=4)

g(-2) = 4 (y-value of the red function when x is -2)

f(2) = -8 (y-value of the blue function for x = 2)

f(-2) = 4 (y-value of the blue function for x =-2)

use them to compare the options they give you, and the only one that matches is the last option.

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A circular hedge surrounds a sculpture on a square base. The radius of the hedge is 6x. The side length of the square sculpture

andrey2020 [161]

Hi, thank you for posting your question here at Brainly.

Since the hedge is surrounding the square base, the area of the base can be determined by subtracting the area of the base from the hedge.

A = pi*r^2 - (4x)^2
A = pi*(6x)^2 - (4x)^2
A = 2x^2*(3pi - 2)

I hope I was able to answer your question. Have a good day.

6 0

3 years ago

Suppose you have an experiment where you flip a coin three times. You then count the number of heads. a.)State the random variab

WITCHER [35]

Answer:

a) X=number of heads observed when flipped the coin 3 times

b) the probability distribution is

P(X=x) = 3/4 * (1/[(3-x)!*x!)])

P(X)=1/8 for x=0 and x=3 and P(X)=3/8 for x=1 and x=2

Step-by-step explanation:

the random variable will be X=number of heads observed when flipped the coin 3 times . Since the result from each flip is independent of the others , then X has a binomial probability distribution , such that

P(X=x)= n!/[(n-x)!*x!)*p^x * (1-p)^(n-x)

where

n= number of times the coin is flipped = 3

p= probability of getting heads in a flip of the coin = 1/2 (assuming that the coin is fair)

therefore

P(X=x)= 3!/[(3-x)!*x!)*(1/2)^(3-x) * (1/2)^x = 3!/[(3-x)!*x!) * (1/2)³ = 3/4 * (1/[(3-x)!*x!)])

P(X=x)= 3/4 * (1/[(3-x)!*x!)]) , for x=[0,1,2,3]

for x=0 and x=3 → P(X)=3/4 * (1/[3!*0!)]) = 1/8

for x=1 and x=2 → P(X)=3/4 * (1/[2!*1!)]) = 3/8

we can verify that is correct since the sum of all the probabilities from x=0 to x=3 is 1/8 + 3/8+ 3/8+ 1/8 = 1

8 0

2 years ago