Nylah buys a pack of 20 markers for $6.00. How much does a pack of 30<br> markers cost? *

Expand the expression (3-2y)^6 using binomial theorem

Nitella [24]

She should have added 2 to both sides of the equation instead of subtracting 2.

2x -2 = 14

Add 2 to both sides:

2x = 16

Divide both sides by 2:

x = 8

5 0

3 years ago

Put the following in order from the smallest to the largest 0.2 1/2 2%

aliya0001 [1]

Answer:

2%, 0.2, 1/2

Step-by-step explanation:

0.2 = 0.20 = 20%

1/2 = 0.50 = 50%

2% = 0.02

3 0

3 years ago

Read 2 more answers

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

Factor the equation.

Kryger [21]

Hi!

The fully factored form of this equation is (3x+2)²

5 0

3 years ago

PLEASE HELP! A campground rents bicycles by the hour. The total cost y to rent a bicycle, including deposit, is presented by the

Lelu [443]

The domain and the range of the function are all possible values of the x and y, respectively, that function can take.
The range is given by:

$y_{o}= \frac{3}{3} +12 \\ y_{o}=12 \\ \\ y_{e}= \frac{27}{3} +12 \\ y_{e}=21 \\ \\ \boxed {Im=(y\in R \mid 12\leq y \leq 21)}$

The domain is given by:

$\boxed {D=(x\in R \mid 3\leq y \leq 27)}$

4 0

3 years ago

Nylah buys a pack of 20 markers for $6.00. How much does a pack of 30 markers cost? *