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timofeeve [1]
3 years ago
8

-10(x + 21) = -15x + 18 What value of x makes the equation true?

Mathematics
1 answer:
schepotkina [342]3 years ago
7 0
Answer:

x = 45.6

Explanation:

-10(x + 21) = -15x + 18

-10x - 210 = -15x + 18

-210 = -5x + 18

-228 = -5x

x = 45.6

(I’m not sure if this is what you’re looking for, but I hope it helped a bit)
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3 years ago
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Work out the input for this function machine ​
Delicious77 [7]

Answer:

Answer is 12 isnt it

Step-by-step explanation:

12×4=48

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4 0
2 years ago
9-17 = 23<br> q=<br> what does q equal?
UkoKoshka [18]

Answer:

q = - 14 / 17

Step-by-step explanation:

9 - 17q = 23

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3 0
3 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
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mestny [16]
<span>amount rounded to the nearest tenth of 1.248 should be 1.2
Answer is 1.2</span>
6 0
3 years ago
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