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Leya [2.2K]
2 years ago
10

Round your answer to the nearest hundredth.

Mathematics
1 answer:
Natali5045456 [20]2 years ago
5 0

Answer:

uh, I think it's 4.13

Step-by-step explanation:

5*sin(65)=4.13 now this seems a bit off but this was the only one that actually worked

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Experimental probability =(number of occurrence of event)/(total number of trials made)
=9/30
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thus the number of seeds that will sprout given that 20 seeds we planted will be:
3/10×20
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2 years ago
The mean of Alice's last 5 math test scores is 88. The first 4 of these 5 math test scores were 87, 75, 93, and 85. What score d
stiks02 [169]

Answer:

Step-by-step explanation:

x is the score she got on the fifth test.

Since her average score is 88, the sum of all five tests is 5×88.

x + 87 + 75 + 93 + 85 = 5×88 = 440

x = 440 -  87 - 75 - 93 - 85

x = 100

She got a 100 on her fifth test.

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2 years ago
Simple math. <br> Question 11.
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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.

8 0
2 years ago
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