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Diano4ka-milaya [45]
3 years ago
11

Simplify the expression. Write answer using only positive exponents. (1/3a^3)^-4

Mathematics
2 answers:
Fantom [35]3 years ago
8 0

Answer:

81/a^16

Step-by-step explanation:

(1/3a^3)^-4

1/3^-4a^-16

3^4 x 1/a^16

81/a^16

Artemon [7]3 years ago
3 0

Answer:

81/a^12

Step-by-step explanation:

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Answer:x=3

Step-by-step explanation:

the 3 angles of a triangle equal 180

you add 105 and 46, which equals 151

you subtract 151 from 180 to see what the missing angle is (29)

you would make 6x+11 equal to 29

-subtract 11 from 29= 18

6x/6, 18/6

the answer is 3

hope this helps! :)

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3 years ago
Math Write the number 63 in four different ways.
Rina8888 [55]

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Our new training room can seat 150 employees. If 6 tables consisting of 6 chairs are now taken, what percentage of our new facil
lina2011 [118]

Answer: 24%

Step-by-step explanation:

Since there are 6 tables with six chairs at each, multiply 6 x 6, which equals 36.

Only a fraction of the room is full, so you need to create a fraction for this problem. In this case, it would be 36/150 (36 chairs taken, 150 total chairs).

To convert this fraction into a percent, divide the numerator by the denominator, then multiply by 100.

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6 0
3 years ago
Determine the probability of selecting a two-child family with one boy and one girl assuming boys and girls are equally likely
Fittoniya [83]

Using the binomial distribution, it is found that there is a 0.5 = 50% probability of selecting a two-child family with one boy and one girl.

For each child, there are only two possible outcomes, either it is a boy, or it is a girl. The probability of a child being a boy or being a girl is independent of any other child, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • Two children, hence n = 2.
  • Equally as likely to be a boy or a girl, hence p = 0.5.

The probability of one of each is P(X = 1), hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{2,1}.(0.5)^{1}.(0.5)^{1} = 0.5

0.5 = 50% probability of selecting a two-child family with one boy and one girl.

A similar problem is given at brainly.com/question/24863377

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795 divided by 42 does it have a remainder and what’s the answer?
salantis [7]

Answer:

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