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4 years ago

Perform the indicated operation.

Mathematics

2 answers:

Butoxors [25]4 years ago

6 0

Answer:

The answer to your question is:

$\frac{a^{4} + 6a^{3}- 54a - 81 }{5(a - 1)}$

Step-by-step explanation:

$\frac{(a + 3)^{2} }{a - 1} / \frac{5}{(a^{2} - 9}$

$\frac{(a + 3)^{2}(a^{2} - 9) }{5(a - 1)}$

$\frac{(a^{2}+ 6a + 9)(a^{2} - 9) }{5(a - 1)}$

$\frac{a^{4} + 6a^{3}- 54a - 81 }{5(a - 1)}$

murzikaleks [220]4 years ago

5 0

$\frac{(a+3)^{2} }{a-3}$ ÷ $\frac{5}{a^{2}-9}$ =

$\frac{(a+3)^{2} }{a-3}$ ÷ $\frac{a^{2}-9}{5}$ =

$\frac{(a+3)^{2} }{a-3}$ ÷ $\frac{(a+3)(a-3)}{5}$ =

$\frac{(a+3)^{2} }{(1)}$ ÷ $\frac{(a+3)(1)}{5}$ =

$\frac{(a+3)^{3} }{5}$

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Answer:

See below.

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Unit rate = 78 / 3/4

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3 years ago

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4 years ago

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3 years ago

Suppose that 75% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou

RSB [31]

Answer:

b) 0.2961

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Step-by-step explanation:

We are given the following information:

We treat trucks undergoing a brake inspection passin as a success.

P( trucks undergoing a brake inspection passes the test) = 75% = 0.75

a) Conditions for binomial probability distribution

There are n independent trial.
Each trial have a success probability p
The probability of success is same for all trials.

Then the number of trucks undergoing a brake inspection follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15

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We have to evaluate:

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$P(x = 3)\\= \binom{15}{3}(0.25)^3(1-0.25)^{12}\\=0.2251$

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6 0

3 years ago

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gladu [14]

Find midpoint of segment.
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\\y-y_1=m(x-x_1)
\\
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\\
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\\
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3 years ago