Can someone help me? I’ll give brainliest to whoever gets it right first

Consider the following function. f(x)=x^2+2x+1.

Gekata [30.6K]

Answer:

A) x² + 2x + 6

B) x² + 2x - 7

C) ¼(x²+2x+1))

D) 6x²+12x+6

E) -x²-2x-1

Step-by-step explanation:

A) f(x) + 5 =x²+2x+1 + 5

= x² + 2x + 6

B) f(x)-8,=x^2+2x+1-8

= x² + 2x - 7

C) ¼f(x) = ¼(x²+2x+1)

D) 6f(x) = 6(x²+2x+1) = 6x²+12x+6

E) -f(x) = -(x²+2x+1) = -x²-2x-1

4 0

3 years ago

Can someone help me pls i put rhe question as a picture i would help alot

GuDViN [60]

The answer would be D

8 0

3 years ago

Read 2 more answers

The answer I neeed it now

slava [35]

Anser is B). Subtract smallest number

5 0

3 years ago

joe drove 408 miles using 16 gallons of gas at this ratr how many mokrs would he drive using 11 gallons of gas

stiv31 [10]

I was on the right track, lol. 280.5 miles.

7 0

3 years ago

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At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

2 years ago