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2 years ago

Find the following f(g(x)) when f(x)=2x+4 and g(x)=2x

Mathematics

2 answers:

klasskru [66]2 years ago

7 0

Answer:

4 + 4x = f(g(x))

Step-by-step explanation:

According to the compositional function, we plug the g(x) function into the f(x) function in replacement for every x we see:

$2(2x) + 4 = 4x + 4 \\f(g(x)) = 4x + 4$

The above answer is written in reverse, which is the exact same result.

I am joyous to assist you anytime.

KonstantinChe [14]2 years ago

6 0

Answer:

f(2x)= 2(2x)+4= 4x + 4 is the solution

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Answer:

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Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

Radda [10]

Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

Step-by-step explanation:

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.

In this question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

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

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

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2 years ago