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DENIUS [597]
3 years ago
5

Question 5 is attached! 10+5 pts ~!

Mathematics
1 answer:
Deffense [45]3 years ago
7 0
17k - 6k- 5k - 2k - k = 15

17k - 6k is 11k.

so, 11k - 5k - 2k - k =15
11k-5k is 6k

so 6k - 2k - k
6k - 2k is 4k

so 4k - k
4k - k is 3k

so 3k = 15
Take 3 to the other side --   k= 15/3
final answer is k = 5
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ANSWERR ASAPP!!! I GIVE BRAINLIESTT!!<br><br>Answer b and correct c and a if wrong ​
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Answer:

b is (4,55). c is correct I believe.

Step-by-step explanation:

You need to find where the lines cross on the graph. You see that at point (4,55)

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What Difference between hundreds and hundredths?
nordsb [41]

hundred means 100 or 200. Like I have 100 cats.

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hundred=100

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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
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xxTIMURxx [149]

Answer:

5/9 & -10/3

Step-by-step explanation:

Set each to 0 and solve:

9x-5 = 0

9x=5

x=5/9

6x+20=0

6x=-20

x=-20/6

x= -10/3   or -3 1/3

7 0
3 years ago
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