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arlik [135]
3 years ago
7

Can someone please help me on this?

Mathematics
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

x = 7.5

Step-by-step explanation:

4.5 (1/3) = 1.5

5 (1.5) = 7.5

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(10v^3 + 71v^2 + 11v+23) / (v+ 7)
vlabodo [156]

tell us the question so we can help.

7 0
3 years ago
Henlo please help!?! Thanks!!<br> ASAP? :D
erik [133]

Answer:

6 feet

Step-by-step explanation:

Volume = w × h × l

120 = 4 × 5 × l

l = 120 / 20

l = 6 ft

4 0
3 years ago
Read 2 more answers
Please helpppp thank y’all
julsineya [31]

Answer:

105

Step-by-step explanation:

since its  a straight line it would add up to 180 degrees so 52 + 23 + 75 and you need 180 so 180-75 which gives 105 which is the missing angle

6 0
3 years ago
Read 2 more answers
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 is the length of the longest side of a triangle that has the vertices (2, 6), (-4, 6), and (-4, 4)? a. 2/10 units b. 40 uni
rewona [7]
Length (2, 6) to (-4, 6) is sqrt((x2 - x1))^2 + (y2 - y1)^2) = sqrt((-4 -2)^2 + (6 - 6)^2) = sqrt((-6)^2 + 0) = 6

Length (2, 6) to (-4, 4) is sqrt((-4 - 2)^2 + (4 - 6)^2) = sqrt((-6)^2 + (-2)^2) = sqrt(36 + 4) = sqrt(40)  = 2sqrt(10) units

Length (-4, 6) to (-4, 4) is sqrt((-4 - (-4))^2 + (4 - 6)^2) = sqrt(0^2 + (-2)^2) = 2

Therefore, the length of the longest side is 2sqrt(10) units
4 0
3 years ago
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