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4 years ago

Find the qoutient of 1170 and 45

Mathematics

2 answers:

statuscvo [17]4 years ago

7 0

Answer:

Step-by-step explanation:

MissTica4 years ago

5 0

Quotient is another word for divide
1170/45
the answer is 26

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16 increased by twice a number is -24

wlad13 [49]

X-the number

16 + 2x = -24 |subtract 16 from both sides

2x = -40 |divide both sides by 2

<u>x = -20</u>

Answer: <em>The number is -24</em>

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3 years ago

When bivariate data from a survey is graphed what is graph called ?

Vinvika [58]

Answer:

Step-by-step explanation:

Scatter plot (Use excel with any bivariate data you can find)

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3 years ago

Factor. 2xy+5x−12y−30

worty [1.4K]

2xy + 5x -12y -30
x(2y + 5) - 6( 2y + 5)
(2y+5) (x-6)

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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.

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3 years ago

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klio [65]

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3 years ago