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3 years ago

Which number line shows the solution to the inequality? y - 2 < -5

Mathematics

2 answers:

xxMikexx [17]3 years ago

6 0

Answer: is the first option .

Step-by-step explanation:

Tatiana [17]3 years ago

5 0

For this case we have the following inequality:
$y - 2 \ \textless \ -5
$
We will solve the inequality algebraically.
We have then:
We add 2 to both sides of the inequality:
$y - 2 + 2 \ \textless \ -5 + 2
$
Rewriting we have:
$y \ \textless \ - 3
$
Therefore the solution is:
(-inf, -3)
Answer:
(-inf, -3)
See attached image.

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(8 x 3) + (8 x 2) = 8 x (blank + blank)

kakasveta [241]

Step-by-step explanation:

(8 x 3) + (8 x 2) = 8 x (3 + 2)

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8 0

3 years ago

Simplify

Alika [10]

Simplify 6 – 4x – 2 + x

Combine like terms:

6 - 2 = 4

-4x + x = -3x

-3x + 4, or (A) is your answer

----------------------------------------------------------------------------------------------------------------

Simplify 5 – 2x – 3 + x

Combine like terms:

5 - 3 = 2

-2x + x = -x

-x + 2, or (B) is your answer

----------------------------------------------------------------------------------------------------------------

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4 0

3 years ago

Read 2 more answers

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr

aniked [119]

Answer:

(a) $E(X) = 0.383$

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) $P(x = 4) = 0.000169$

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$E(X) = \frac{n \times r}{N}$

$E(X) = \frac{10 \times 8}{209}$

$E(X) = \frac{80}{209}$

$E(X) = 0.383$

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}}$

$P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}}$

$$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}$

$P(x = 4) = 0.000169$

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0

3 years ago