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jeka57 [31]
2 years ago
11

PLEASE HELP WORTH 20 POINTS Find the unknown side of the triangle below.

Mathematics
1 answer:
Nata [24]2 years ago
5 0
The unknown side of the triangle is 37.0 m
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A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit
Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}  

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • p is the probability of a success on a single trial.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

E = \frac{q(1-p)}{p}

The variance is:

V = \frac{q(1-p)}{p^2}

In this problem, 0.2 probability of a finding a person who attended the last football game, thus p = 0.2.

Item a:

  • None of the first three attended, which is P(X = 0) when n = 3.
  • Fourth attended, with 0.2 probability.

Thus:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512

0.2(0.512) = 0.1024

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

  • One person, thus q = 1.

The expected value is:

E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4

The variance is:

V = \frac{0.8}{0.04} = 20

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0
1 year ago
How so you round 7,895,164 to the nearest 1,000
lana [24]

Answer: 7,895,164 rounded to the nearest 1,000 is 7,895,000

5 0
3 years ago
What is the median of the data?
spin [16.1K]
Our data looks like:

3,3,4,4,5,5,5,6,7,8

The median, true to its name is the middle number of the set of data.
The median in this case in in between 5 and 5. Average it out (5+5)/2 = 5.
Your answer should be 5. Hope this helps!
3 0
2 years ago
Read 2 more answers
A rectangular lawn is 2 m longer than it is wide.
PtichkaEL [24]

Let's breadth be x

  • Length be x+2

ATQ

\\ \sf\longmapsto x(x+2)=21

\\ \sf\longmapsto x^2+2x=21

  • Use completing the square method

\\ \sf\longmapsto 4(x^2+2x)=4(21)

\\ \sf\longmapsto 4x^2+8x=84

\\ \sf\longmapsto (2x)^2+2(2x)(2)=84

\\ \sf\longmapsto (2x)^2+2(2x)(2)+2^2=2^2+84

\\ \sf\longmapsto (2x+2)^2=88

\\ \sf\longmapsto 2x+2=\pm\sqrt{88}

  • As it's dimensions we will take positive

\\ \sf\longmapsto 2x+2\approx 9.7

\\ \sf\longmapsto 2x=9.7-2=7.7

\\ \sf\longmapsto x=3.8

  • x+2=3.8+2=5.8

Now

\\ \sf\longmapsto Perimeter=2(L+B)

\\ \sf\longmapsto perimeter=2(3.8+5.8)

\\ \sf\longmapsto perimeter=2(9.6)

\\ \sf\longmapsto Perimeter=19.2m

  • Length of 1 strip=13m

Total strips

\\ \sf\longmapsto \dfrac{19.2}{13}

\\ \sf\longmapsto 1.47strips

  • She needs 2 strips
3 0
2 years ago
Read 2 more answers
HELP PLEASE please omg
Nikolay [14]
Drop the questions it’s blank
6 0
2 years ago
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