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Law Incorporation [45]
3 years ago
13

Solve for g: 5=5g+5h Show all your work please.

Mathematics
1 answer:
natka813 [3]3 years ago
5 0
5 = 5 (g + h)
1 = g + h
1 - h = g
g = 1 - h

Answer: g = 1 - h.

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A cone with radius 12ft and height 5 ft what is the volume ?
Mrac [35]

Answer:

the volume is 753.98

Formula:

V=1/3hπr2

4 0
3 years ago
The cost of a new CD is $14.95, and the cost of a video game is $39.99. How much would c CDs and v video games cost?
8090 [49]

Answer:

14.95c+39.99v

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
4(-2d-3)=12, someone please solve this for me asap
wlad13 [49]

4(-3 + -2d) = 12

(-3 * 4 + -2d * 4) = 12

(-12 + -8d) = 12

-12 + -8d = 12

-12 + 12 + -8d = 12 + 12

-12 + 12 = 0

0 + -8d = 12 + 12

-8d = 12 + 12

12 + 12 = 24

-8d = 24

d = -3

5 0
2 years ago
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
[8 + (24 x 3)] divided by 7 <br><br> Use PEMDAS &amp;’ show work .
posledela

Answer:

11.43

Step-by-step explanation:

[8 + (24 x 3)] divided by 7

[8 + (72)] divided by 7

80/7

11.42857142857143

11.43 (I rounded it)

6 0
3 years ago
Read 2 more answers
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