All categories

3 years ago

350% of 880 is 60% of what number?

Mathematics

1 answer:

Westkost [7]3 years ago

6 0

Answer:

5133 1/3

Step-by-step explanation:

$3.5 \cdot 880 = 0.6x\\3080=0.6x\\x=5133 1/3$

Hope this helps!

You might be interested in

Round 7905.68466806 to the nearest thousand.

mixas84 [53]

Answer:

8000 to the nearest thousand

7905.685 to the thousandth

Step-by-step explanation:

8 0

3 years ago

Calculate the rectangular prism with a length of 5.6cm, a with of 2.1 cm, and a height of 6.6cm.

Inessa05 [86]

5.6 x 2.1 x 6.6 = 77.616 2 significant figures would be 78 cm³ 2 decimal places would be 77.62 cm³

8 0

3 years ago

Find the rate in the following problem. Round the answer to the nearest percent.

Anit [1.1K]

Answer: 75%

Step-by-step explanation:

150/200=.75

200 x .75=150

Now to turn .75 into a percent by moving the . over to the right two times.

75. or 75%

6 0

3 years ago

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

Please help! <br> Simplify. Your answer should contain only positive exponents.

maxonik [38]

Answer:

$\frac{1}{4k^3}$

Step-by-step explanation:

It's already in a form that only ha positive exponents. Just to state it though you could rewrite it as $k*(4k^{4})^{-1}$ or $k*4^{-1}*k^{-4}$ The question asks for positive exponents though so yu don't need to do that.

You also need to know $\frac{x^a}{x^b}=x^{a-b}$ so in this case $\frac{k}{k^4}=\frac{k^1}{k^4} =k^{1-4}=k^{-3}=\frac{1}{k^3}$

so $\frac{k}{4k^4}=\frac{1}{4}*\frac{k}{k^4}=\frac{1}{4}*\frac{1}{k^3}=\frac{1}{4k^3}$

3 0

3 years ago