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2 years ago

Divide 3983 /4 give quotient and remainder

Mathematics

1 answer:

alukav5142 [94]2 years ago

4 0

Answer:

quotient = 995

remainder = 3

Step-by-step explanation:

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If the sale price is $40 and the discount is 25%, what is the original price?

sveta [45]

Use this equation to solve the problem:

x - (x * .25) = 40

This equation is saying the original price times itself by 25% is 40.

x - (x * .25) = 40
Simplify.
x - .25x = 40
.75x = 40
Divide both sides by .75.
x = 53.33

The original price was $53.33

3 0

3 years ago

Can someone help me ?

V125BC [204]

10/3 x 5 + 10/3 x 1/4

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3 years ago

1/2 - 20 [ 14- 3 4/15] / 6/5 ^2

lidiya [134]

Step-by-step explanation:

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6 0

3 years ago

What is the term of a 7-term geometric series if the first term is -11 the last term is -45,056 and the common ratio is -4

shutvik [7]

I assume you mean to ask how to find the sum of the seven terms?

You have

$a_2=ra_1$
$a_3=ra_2=r^2a_1$
$a_4=ra_3=r^3a_1$
...
$a_7=ra_6=r^6a_1$

So the sum is

$a_1+ra_1+r^2a_1+\cdots+r^6a_1=a_1(1+r+r^2+\cdots+r^6)=a_1\times\dfrac{1-r^7}{1-r}$

You know that $a_1=-11$ and $r=-4$ , so the sum is equal to

$-4\times\dfrac{1-(-4)^7}{1-(-4)}=-13108$

8 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 trials before q successes 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