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11Alexandr11 [23.1K]
2 years ago
14

1-5 questions algebra

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
5 0

Answer:

wdljgfwdjbcwshcnqsknwswscnkgqjc w jlwx

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Over the past ten years, the town's population doubled in size. The population is currently 12,000. What was the population ten
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6,000 right? I think
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When using base-10 blocks to divide 452 by 2, how many hundreds flats will be in each equal group? hundreds flats Divide 452 by
daser333 [38]
<h3>Given</h3>

4 hundreds flats; 5 tens rods; 2 ones cubes

<h3>Find</h3>

The number of hundreds flats in each of 2 equal piles

<h3>Solution</h3>

When 4 flats are divided into two equal groups, each group will have ...

... 2 flats

_____

You can imagine doing this the way a card dealer might: first put 1 flat in each of 2 piles, then do the same for the remaining 2 flats. Each pile will end up with 2 flats.

— — — — —

You will have a problem if you continue with the tens rods. There is an odd number of those, so one of them will have to be exchanged for 10 ones cubes.

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3 years ago
Solve for ttt.<br> 2(t+1) = 102(t+1)=10
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Answer:

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distribuite 2t+2=10

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2 years ago
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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

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.

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2 years ago
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8 compliments/ 17 reviews
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