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beks73 [17]
4 years ago
7

In triangle ABC, the measure of ∠B is 90°, BC=16, and AC=20. Triangle DEF is similar to triangle ABC, where vertices D, E, and F

correspond to vertices A, B, and C, respectively, and each side of triangle DEF is 1/3 the length of the corresponding side of triangle ABC. What is the value of sinF?
SAT
1 answer:
patriot [66]4 years ago
5 0

Answer:

sinF =  3 /5  or 0.6.

Explanation:

Triangle ABC is a right triangle with its right angle at B. Therefore, AC is the hypotenuse of right triangle ABC, and AB and BC are the legs of right triangle ABC. According to the Pythagorean theorem,

AB=√202−162=√400−256=√144=12

Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of angle∠F equals the measure of angle∠C. Therefore, sinF=sinC. From the side lengths of triangle ABC,

sinF =  oppositeside /hypotenuse  =  AB /AC  =  12/ 20  =  3 /5

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company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul
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Sampling distribution involves the proportions of a data element in a given sample.

  • <em>The proportion of Good TV set is 0.67</em>
  • <em>The number of ways of selecting 5 from 6 TV sets is 6</em>
  • <em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

n = 6

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

Good = 4

So, the proportion (p) that are good are:

p = \frac{Good}{n}

p = \frac{4}{6}

p = 0.67

<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

Where

r = 5

So, we have:

^6C_5 = \frac{6!}{(6 - 5)!5!}

^6C_5 = \frac{6!}{1!5!}

^6C_5 = \frac{6 \times 5!}{1 \times 5!}

^6C_5 = \frac{6}{1}

^6C_5 = 6

Hence, there are 6 ways

<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

Where

r = 4

So, we have:

^6C_4 = \frac{6!}{(6 - 4)!4!}

^6C_4 = \frac{6!}{2!4!}

^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}

^6C_4 = \frac{30}{2}

^6C_4 = 15

So, the table is as follows:

\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

p = \frac{Good}{n}

<u>(d) The sampling distribution</u>

In (a), we have:

p = 0.67 --- proportion of Good TV

The sampling error is calculated as follows:

SE_n = |p - p_n|

So, we have:

\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]

Read more about sampling distributions at:

brainly.com/question/10554762

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