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2 years ago

An electrician needs 18.25 feet of wire for each outlet she installs. How man can she install with a 50 ft spool of wire?

Mathematics

2 answers:

S_A_V [24]2 years ago

7 0

Step-by-step explanation:

the answer is 2

jolli1 [7]2 years ago

3 0

Answer:

Step-by-step explanation:

Ok, so if we know how much wire is needed for one outlet, we must figure out how many 18.25s fit into 50 ft of wire.

18.25*2=36.5

18.25*3=54.75 which is over 50 so it cannot possibly be more than 2.

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Answer:

Step-by-step explanation:

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2 years ago

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bezimeni [28]

Pete needs one more quart of paint, leaving two quarts in total.

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2 years ago

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3 years ago

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To mix a particular shade of purple paint, read paint and blue paint are mixed in the ratio 5:3. To make 80 gallons of this shad

zubka84 [21]

Answer:

50 gallons of red

30 gallons of blue

Step-by-step explanation:

5 parts of the whole must be red, 3 parts of the whole must be blue, and there are 5+3=8 parts in total.

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4 0

2 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

2 years ago