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dolphi86 [110]
3 years ago
5

Which of the numbers below is greater than 313.587?

Mathematics
1 answer:
Marrrta [24]3 years ago
5 0

Answer:

314

Step-by-step explanation:

Because 314 is bigger than 313.587

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Please help with #12
ExtremeBDS [4]

Answer:

a. 1 1/8 b. 8/9

Step-by-step explanation:

You can set this up as a proportion to solve.  For part a. we know that 2/3 of the road is 3/4 mile long.  2/3 + 1/3 = the whole road, so we need how many miles of the road is 1/3 its length.  Set up the proportion like this:

\frac{\frac{2}{3} }{\frac{3}{4} } =\frac{\frac{1}{3} }{x}

Cross multiplying gives you:

\frac{2}{3}x=\frac{1}{3}*\frac{3}{4}

The 3's on the right cancel out nicely, leaving you with

\frac{2}{3}x=\frac{1}{4}

To solve for x, multiply both sides by 3/2:

\frac{3}{2}*\frac{2}{3}x=\frac{1}{4}*\frac{3}{2} gives you

x=\frac{3}{8}

That means that the road is still missing 3/8 of a mile til it's finished.  The length of the road is found by adding the 3/4 to the 3/8:

\frac{3}{4}+\frac{3}{8}=\frac{6}{8}+\frac{3}{8}=\frac{9}{8}

So the road is a total of 1 1/8 miles long.

For b. we need to find out how much of 1 1/8 is 1 mile:

1 mile = x * 9/8 and

x = 8/9.  When 1 mile of the road is completed, that is 8/9 of the total length of the road completed.

8 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
Write 1 1/4 as a decimal and percent? decimal_____________ percent_____________
jarptica [38.1K]
Decimal 1.25
percent 125%
7 0
3 years ago
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its breaking in or crash


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