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hram777 [196]
3 years ago
12

Nine fifths-three fifths

Mathematics
2 answers:
m_a_m_a [10]3 years ago
7 0
Can you please explain what question you have?
Feliz [49]3 years ago
3 0
6:5 it's actually a ritio
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A hospital recorded the weight of babies born at the hospital during one week on a line plot. How many more pounds was the baby
lbvjy [14]

greatest weight 20 lowest weight 5

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2 years ago
How do you subtract nega
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Well think of it like this. lets say your teacher made you do this problem -48 - 10. I would just act like 48 is not an negative and add 10. After that I would put the negative back. So the answer would be -58. :P
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3 years ago
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The total bill for breakfast was $81.30. The group of friends
mr_godi [17]

Answer:

$93.50

Step-by-step explanation:

81.30 x 1.15 = 93.50

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3 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
What fractions are equivalent to 2/3
just olya [345]
4/6
10/15
6/9
8/12
Hope these help you!!! =')
4 0
3 years ago
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