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vekshin1
3 years ago
12

What is the answer to (12^2)^3 times 12^-4 and 12^7 times 12^-5 over 12^2

Mathematics
2 answers:
raketka [301]3 years ago
7 0

Answer:  12² = 144

12² ÷ 12² = 1

Step-by-step explanation:

I'll go through it backwards: 12^7 times 12^-5  You subtract the -5 from 7 in the exponents to get 12^2.  12^2 divided by 12^2  is 1.

(12^2)^3 works out to 12^6 because you multiply the exponents when its a power raised by a power. Then you subtract the 4 from the 6 (or add -4 +6) to get 12^2.

That means 12 squared, same as 12 times 12

That's 144.

If you have to multiply the two parts together, It's 144 times 1.

If "and" means "add them together"-- YOU KNOW!

shtirl [24]3 years ago
6 0

Answer:

1

Step-by-step explanation:

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andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

6 0
3 years ago
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Step-by-step explanation:

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