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Yuliya22 [10]
3 years ago
5

Solve for b. Put your answer in simplest radical form. Show Steps plz! A = 4πbc + πb^2

Mathematics
1 answer:
sladkih [1.3K]3 years ago
4 0

Answer:

b = -2c ± [√(4π²c² + πA)]/π

Step-by-step explanation:

A = 4πbc + πb^2

A = 4πbc + πb²

πb² + 4πbc - A = 0

Using the quadratic formula to solve this quadratic equation.

The quadratic formula for the quadratic equation, pb² + qb + r = 0, is given as

b = [-q ± √(q² - 4pr)] ÷ 2p

Comparing

πb² + 4πbc - A = 0 with pb² + qb + r = 0,

p = π

q = 4πc

r = -A

b = [-q ± √(q² - 4pr)] ÷ 2p

b = {-4πc ± √[(4πc)² - 4(π)(-A)]} ÷ 2π

b = {-4πc ± √[16π²c² + 4πA]} ÷ 2π

b = (-4πc/2π) ± {√[16π²c² + 4πA] ÷ 2π}

b = -2c ± [√(4π²c² + πA)]/π

Hope this Helps!!!

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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:

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#LearnwithBrainly

5 0
3 years ago
(2x-7)^2 subtracted from 2
weqwewe [10]

Write out the expression 2 - (2x-7)^2.  That's it.

But if you want to go further and remove the parentheses, first expand (2x-7)^2:   (2x-7)^2 = 4x^2 - 28x + 49,

and then subtract this result from 2:

2 - (4x^2 - 28x + 49) (It's important to use parentheses here)

Now, following the distributive property of multiplication, remove the parentheses:

2 - 4x^2 + 28x - 49

Combining the constants, we get the final answer:  - 4x^2 + 28x - 47

4 0
2 years ago
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