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Archy [21]
3 years ago
15

I REALLY NEED HELP WORTH 10 POINTS URGENT!!!!!

Mathematics
1 answer:
polet [3.4K]3 years ago
8 0

Answer:

(f\circ g)(4)=31

Step-by-step explanation:

<u>Composite Function</u>

Given f(x) and g(x) real functions, the composite function named fog(x) is defined as:

(f\circ g)(x)=f(g(x))

For practical purposes, it can be found by substituting g into f.

We have:

f(x)=3x+1

g(x)=x^2-6

Computing the composite function:

(f\circ g)(x)=f(g(x))=3(x^2-6)+1

Operating:

(f\circ g)(x)=3x^2-18+1

Operating:

(f\circ g)(x)=3x^2-17

Now evaluate for x=4

(f\circ g)(4)=3(4)^2-17=48-17

\boxed{(f\circ g)(4)=31}

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Express 10500 in terms of its prime factors​
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10- 3/4z = 8<br><br> Somebody, please help me! I don't get this!
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6 0
2 years ago
A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.
Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

a)

The probability that the birth is a male can be written as

p(m) = 0.52 (which corresponds to 52%)

While the probability that the birth is a female can be written as

p(f) = 0.48 (which corresponds to 48%)

Here we want to calculate the probability that over  2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male  does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

p(mm)=p(m)\cdot p(m)

And substituting, we find

p(mm)=0.52\cdot 0.52 = 0.27

So, 27%.

b)

In this case, we want to find the probability that both children are female, so the probability

p(ff)

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

p(ff)=p(f)\cdot p(f)

And substituting

p(f)=0.48

We find:

p(ff)=0.48\cdot 0.48=0.23

Which means 23%.

c)

In this case, we want to find the probability they have exactly one male and exactly one female child. This is given by the sum of two probabilities:

- The probability that 1st child is a male and 2nd child is a female, namely p(mf)

- The probability that 1st child is a female and 2nd child is a male, namely p(fm)

So, this probability is

p(mf Ufm)=p(mf)+p(fm)

We have:

p(mf)=p(m)\cdot p(f)=0.52\cdot 0.48=0.25

p(fm)=p(f)\cdot p(m)=0.48\cdot 0.52=0.25

Therefore, this probability is

p(mfUfm)=0.25+0.25=0.50

So, 50%.

Learn more about probabilities:

brainly.com/question/5751004

brainly.com/question/6649771

brainly.com/question/8799684

brainly.com/question/7888686

#LearnwithBrainly

5 0
3 years ago
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