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dimaraw [331]
2 years ago
8

Rewrite this equation so that it demonstrates the division property of equality: 12m=42

Mathematics
2 answers:
elixir [45]2 years ago
8 0
Answer:
12m ÷ 12 = 42 ÷ 12
alexandr402 [8]2 years ago
8 0

Answer:

m=30

Step-by-step explanation:

subtract or divide 12 from 12 and 42, then you're left with m and 32. m=32

12m=42

-12. -12.

m. 30

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Richard has enrolled in a 401(k) savings plan. He intends to deposit $250 each month; his employer does not contribute to his ac
astraxan [27]

Answer:

Richard will have $60,000 in his account in 20 years.

Step-by-step explanation:

(1) Multiply $250 x 12

(2) Multiply the answer of $250 x 12 which is 3000 by 20

(3) Final answer would be $60,000

7 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
What is the definition of symbolism?
notka56 [123]

Answer:

The use of symbols to represent ideas or qualities.

5 0
3 years ago
The expression p/x^2 - 5x +6 simplifies to x+4/x-2. Which expression doees p represent?
gladu [14]
We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d

we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor

p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)

therefor
p=d(x+4) and
x^2-5x+6=d(x-2)

we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub


p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
p= \frac {(x-6)(x+1)(x+4)}{(x-2)}

7 0
3 years ago
A square root is categorized as what kind of number?<br><br> Choices:<br> Irrational <br> Rational
anyanavicka [17]
<span>Rational i believe is the answer i ain't for sure though</span>
6 0
3 years ago
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