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yaroslaw [1]
2 years ago
9

Enter a fraction equivalent to 0.4. Use only whole numbers for numerators and denominators.

Mathematics
1 answer:
77julia77 [94]2 years ago
6 0

Answer:

2/5

Step-by-step explanation

!s the answer.

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Simplify the expression 3+2^2 / 2+3
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Ans

8 option b.

Step-by-step explanation:

3+4/2+3

=3+2+3

=5+3

=8

3 0
2 years ago
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Answer:

A ✓5 is the irrational number.

Step-by-step explanation:

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Increase 14187.13 by 14.5
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Answer:

14201.63

Step-by-step explanation:

14187.63

+   14.50

=14201.63

7 0
3 years ago
HELLPPPP!! GIVING ALL MY POINTS FOR THIS QUESTION!<br> Identify m∠AMB.
Arlecino [84]

Answer:

I think it is the 3 one

Step-by-step explanation:

8 0
3 years ago
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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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