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Andre45 [30]
3 years ago
9

75,300,207 in word form

Mathematics
1 answer:
sineoko [7]3 years ago
8 0
Seventy-five million three hundred thousand two hundred seven
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CAN SOMEONE PLEASE JUST ANSWER THIS ASAP FOR BRAINLIEST!!!
Julli [10]

Maybe the answer is -4 ....? I just can't find any more logic to this problem.

5 0
3 years ago
What are the next three terms of each number pattern: 2,8,32,128
Sloan [31]
Each number is 4 times the previous number"
2*4 = 8
8*4 = 32
32*4 = 128

so next 3 are:
128 * 4 = 512
512*4 = 2048
2048 * 4 = 8192

512 , 2048,  8192
8 0
3 years ago
Read 2 more answers
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
Write the definition of the vertex of an angle as a biconditional statement.
san4es73 [151]
The corner of a shape where two side meet
6 0
3 years ago
What property says that -12+12+0
natali 33 [55]

Answer:

Looks like <u>inverse property of addition </u>to me

4 0
3 years ago
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