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dezoksy [38]
3 years ago
15

Nancy is checking to determine if the expressions x + 4 + x and 6 + 2 x minus 2 are equivalent. When x = 3, she correctly finds

that both expressions have a value of 10. When x = 5, she correctly evaluates the first expression to find that x + 4 + x =14.
What is the value of the second expression when x = 5, and are the two expressions equivalent?
The value of the second expression is 8, so the expressions are not equivalent.
The value of the second expression is 14, so the expressions are equivalent.
The value of the second expression is 16, so the expressions are equivalent.
The value of the second expression is 18, so the expressions are not equivalent.
Mathematics
2 answers:
nataly862011 [7]3 years ago
7 0

Answer:

The value of the second expression is 14, so the expressions are equivalent.

Step-by-step explanation:

x=5

6+2(5)-2=

6+10-2=

16-2=

14

The other expression also came out to be 14, so they are equivalent.

adell [148]3 years ago
5 0

Answer:

The value of the second expression is 14, so the expressions are equivalent.

Step-by-step explanation:

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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

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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.

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3 years ago
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4 years ago
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On the exam, Rebecca successfully created 20 out of 25 vases.
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\text{Hello there! :]}

\large\boxed{125 \text { vases}}

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Step-by-step explanation:

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