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timurjin [86]
3 years ago
6

A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle's largest angle?

Mathematics
2 answers:
Rama09 [41]3 years ago
6 0

1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

     i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

      ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

       a^{2}=b ^{2}+c ^{2}-2bc(cosA)

2.

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA)   150=-234(cosA)

cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

Taya2010 [7]3 years ago
4 0

Answer:

Measurement of this triangle's largest angle = 129.87°

Step-by-step explanation:

Let a = 9 in, b = 13 in and c = 20 in

We have cosine rule cosA=\frac{b^2+c^2-a^2}{2bc}

    cosA=\frac{13^2+20^2-9^2}{2\times 13\times 20}=0.938\\\\A=20.21^0\\\\cosB=\frac{9^2+20^2-13^2}{2\times 9\times 20}=0.867\\\\B=29.93^0\\\\cosC=\frac{9^2+13^2-20^2}{2\times 9\times 13}=-0.641\\\\C=129.87^0\\\\

Measurement of this triangle's largest angle = 129.87°

   

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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} $

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$ 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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shutvik [7]

Answer:

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Step-by-step explanation:

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__

4. The explicit formula for an arithmetic sequence with first term a1 and common difference d is ...

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Answer:

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Answer:

Number of sodas sold: 78

Number of hot dogs sold: 24

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