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Select the example which illustrates the distributive property. (7 x 39), 39 x 7 7(30 + 9) = 7(30) + 7(9) (7 + 3) (7 + 9) (7 x 3

Andreyy89

Answer is 7(30+9)=7(30)+7(9)

4 0

2 years ago

Trevor took one of his friends out to lunch. The lunches cost $248.40 and he paid 12.5% sales tax. If Trevor left a 15% tip on t

bixtya [17]

He paid ...

-- The cost of the lunches (100% = 1.00 of it)

-- 15% of the cost as a tip (15% = 0.15 of it)

-- 12% of the cost as sales tax (12% = 0.12 of it)

Total that he paid = (1.00 + 0.15 + 0.12) = 1.27 of $248.40

= $315.47 .

Trevor is one generous guy !

4 0

3 years ago

Evaluate 3b2 when b= 4 and when b= - 4.<br> When b=4, 3b2

nydimaria [60]

Answer:

24

Step-by-step explanation:

3 times 4 times 2

3 times 4 is 12

12 times 2 is 24

Hope this helps you

4 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

2 years ago

1. When Lucas dumped the coins out of his piggy bank, he noticed that there were only nickels and dimes, and the number of nicke

vaieri [72.5K]

1- The total value of the Lucas coins dumped out of his piggy bank was $ 5.

2- Franco dumped out 36 coins from his piggy bank (16 quarters and 20 nickels).

3- There is no possible solution, as neither combination gives a total of $ 10, which is the combined total of Lucas and Franco's coins.

1- Given that when Lucas dumped the coins out of his piggy bank, he noticed that there were only nickels and dimes, and the number of nickels was twice the number of dimes, to determine, if there were 50 nickels, what was the total value of the Lucas coins dumped out of his piggy bank, the following calculation must be performed:

Nickel = 0.05
Dime = 0.10
N + D = X
N + 1 / 2N = X
50 + 25 = 75
50 x 0.05 + 25 x 0.1 = X
2.50 + 2.50 = X
5 = X

Therefore, the total value of the Lucas coins dumped out of his piggy bank was $ 5.

2- Since when Franco dumped the coins out of his piggy bank, he noticed that there were only nickels and quarters, and the number of quarters was six more than half the number of nickels, to determine, if the total value of the coins Franco and Lucas dumped out of each piggy bank was the same, how many coins did Franco dump out of his piggy bank, the following calculation must be performed:

Nickel = 0.05
Quarter = 0.25
Q = 2N + 6
4 x 0.25 + 6 x 0.25 = 1 + 1.5 = 2.5
8 x 0.05 = 0.40
2.5 + 0.4 = 2.9
8 x 0.25 + 6 x 0.25 = 2 + 1.5 = 3.5
16 x 0.05 = 0.8
3.5 + 0.8 = 4.3
10 x 0.25 + 6 x 0.25 = 2.5 + 1.5 = 4
20 x 0.05 = 1
4 + 1 = 5
10 + 6 + 20 = 36

Therefore, Franco dumped out 36 coins from his piggy bank (16 quarters and 20 nickels).

3- Since when Callie dumped the coins out of her piggy bank, she noticed that there were only nickels, dimes and quarters, and the number of dimes was one more than twice the number of quarters, and there were half as many quarters as nickels, y the coins dumped out of Callie's piggy bank had a total value equal to that of the coins dumped out of Franco's and Lucas' piggy banks combined, to determine what was the total value of the nickels and dimes dumped out of Callie's piggy bank, the following calculation should be performed:

Nickel = 0.05
Dime = 0.10
Quarter = 0.25
(D + 1) + 1 / 2Q + N = 10
1 x 0.1 + 20 x 0.1 + 10 x 0.25 + 20 x 0.05 = 2.1 + 2.5 + 1 = 5.6
1 x 0.1 + 36 x 0.1 + 18 x 0.25 + 36 x 0.05 = 3.6 + 4.5 + 1.8 = 9.9
1 x 0.1 + 38 x 0.1 + 19 x 0.25 + 38 x 0.05 = 3.9 + 4.75 + 1.9 = 10.55

Therefore, there is no possible solution, as neither combination gives a total of $ 10, which is the combined total of Lucas and Franco's coins.

Learn more in brainly.com/question/11096797

6 0

3 years ago