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Likurg_2 [28]
3 years ago
7

Solve for x. Show each step of the solution. 4(8-x)-32=98-5(3x+24)

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer: x=−2

Step-by-step explanation:

Let's solve your equation step-by-step.

4(8−x)−32=98−5(3x+24)

Step 1: Simplify both sides of the equation.

4(8−x)−32=98−5(3x+24)

(4)(8)+(4)(−x)+−32=98+(−5)(3x)+(−5)(24)(Distribute)

32+−4x+−32=98+−15x+−120

(−4x)+(32+−32)=(−15x)+(98+−120)(Combine Like Terms)

−4x=−15x+−22

−4x=−15x−22

Step 2: Add 15x to both sides.

−4x+15x=−15x−22+15x

11x=−22

Step 3: Divide both sides by 11.

11x /11 = −22 /11

x=−2

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Answer:

If Q=2.1(R)+5, then it would be 2.1(5)+5=15.5 ?

Step-by-step explanation:

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Which of the following statements must be true about this diagram?
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Answer:

A, C, D

Step-by-step explanation:

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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
2 years ago
Rita has $b. Sharron has $11.50 less than Rita.
taurus [48]

Answer:

h = b - 11.50

Step-by-step explanation:

Sharron(h) has 11.50 less than Rita(b)

h = b - 11.50

3 0
3 years ago
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