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Goshia [24]
2 years ago
6

Can someone please evaluate (-12)²​

Mathematics
2 answers:
kow [346]2 years ago
8 0

Answer:

(-12)²=(-12×-12)=144

lorasvet [3.4K]2 years ago
8 0

Answer: 144

Step-by-step explanation:

(-12)^2

= -12(-12)

= 12(12)

= 144

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Because 3*6= 18 and 18/2 equals 9. So therefore they are NOT equal.
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Consider the trinomial 6x2 13x 6. The value of ac is. The value of b is. The two numbers that have a product of ac and a sum of
Verdich [7]

The value of ac is 36 and the value of b is 13. Then the value of ac + b is 49. The two numbers are 4 and 9.

<h3>What is polynomial?</h3>

Polynomial is an algebraic expression that consists of variables and coefficients. Variable are called unknown. We can apply arithmetic operations such as addition, subtraction, etc. But not divisible by variable.

Consider the trinomial 6x² + 13x + 6.

Compare with the standard equation, we have

ax² + bx + c

Then we have

a = 6

b = 13

c = 6

The product of ac will be

ac = 6 × 6 = 36

The sum of ac and b will be

ac + b = 36 + 13 = 49

The two numbers that have a product of ac and a sum of b are 4 and 9.

More about the polynomial link is given below.

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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%.

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In the proof, what is the reason for (3)?
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Answer:

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