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Hoochie [10]
3 years ago
9

Find xA) 12B) 13 C) 11D) 10

Mathematics
1 answer:
Soloha48 [4]3 years ago
5 0

We know that the sum of arc ND and arc ML divided by 2 is equivalent to 6x + 23.

Thus, we can form an equation:

6x + 23 = (11x + 8 + 3x + 14)/2

Simplify:

6x + 23 = (14x + 22)/2

Simply again:

6x + 23 = 7x + 11

Subtract 6x from both sides:

23 = x + 11

Subtract 11 from both sides

x = 12

Option A, or x = 12

-T.B.

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To find the value of x we first need to find the missing angle in the triangle.

To do that we are gonna add the angles we already have.

64 + 45 = 109

Angles in a triangle measures 180 so to get the missing angle we are gonna subtract 109 from 180

180 - 109 = 71 deg

The missing angle in the triangle is 71 deg


Angles on a straight line adds up to 10 so to find x we are gonna subtract 71 from 180

180 - 71 = 109 degrees

Therefore x is 109 degrees.
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How do you convert fractions to decimals and percents
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√(5-22(√5)+121)<br> Simplify and show your steps
JulijaS [17]
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Which choice shows 1/4 of 8
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5 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
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