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Bogdan [553]
2 years ago
10

Given the functions f(x) = 7x + 13 and g(x) = x + 2, which of the following functions represents f[g(x)] correctly?

Mathematics
2 answers:
oksian1 [2.3K]2 years ago
6 0
If you want to find f[g(x)]  just substitute "x+2"  for  an "x" in equation f(x)=7x+13. You'll get:

f[g(x)]= 7(x+2)+13=7x + 14 + 13 = 7x + 27

Choose second answer
LenKa [72]2 years ago
4 0

Answer:

B. f(g(x)) = 7x + 27

Step-by-step explanation:

We have, f(x) = 7x+13 and g(x) = x+2.

So, the function f(g(x)) is obtained by substituting the function g(x) = x+2 in f(x) = 7x+13,

i.e. f(g(x)) = f(x+2)

i.e. f(g(x)) = 7 × (x+2) + 13

i.e. f(g(x)) = 7x + 14 + 13

i.e. f(g(x)) = 7x + 27

Thus, f(g(x)) = 7x + 27

Hence, option B is correct.

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The potential energy, P, in a spring is represented using the formula P = A equals StartFraction one-half EndFraction b h.kx2. L
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4 0
1 year ago
The probability of a train arriving on time and leaving on time is 0.8. The probability that the train arrives on time and leave
kkurt [141]

Answer:

<u>0.9524</u>

Step-by-step explanation:

<em>Note enough information is given in this problem. I will do a similar problem like this. The problem is:</em>

<em>The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?</em>

<em />

<u>Solution:</u>

This is conditional probability.

Given:

  • Probability train arrive on time and leave on time = 0.8
  • Probability train arrive on time = 0.84
  • Probability train leave on time = 0.86

Now, according to conditional probability formula, we can write:

P(Leave \ on \  time | arrive \  on \ time) = P(arrive ∩ leave) / P(arrive)

Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"

and

P(arrive) means probability arriving on time given as 0.84, so:

0.8/0.84 = <u>0.9524</u>

<u></u>

<u>This is the answer.</u>

5 0
2 years ago
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