To solve the problem: "What is 3/4 of 12," you would ____

Five divided sixty eight

Schach [20]

5/68, or about 0.07

8 0

3 years ago

TRUE OR FALSE

Zielflug [23.3K]

Answer:

true

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

14n = -126<br> Need Hell ASAP<br> Algebra Quiz

sweet-ann [11.9K]

14n=-126

/14 /14 divide by 14 on both sides of the

equation

n=-9

5 0

3 years ago

Read 2 more answers

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago

In a survey of 1000 eligible voters selected at random, it was found that 100 had a college degree. Additionally, it was found t

o-na [289]

Answer:

A. 8%

B. 39.6%

C. 58.4%

D. 41.6%

Step-by-step explanation:

Computation to determine the probability of eligible voter selected at random

First step is to Draw up a contingincy table which will include Rows = Degree/No degree

and Columns= Vote/Not vote

..............Vote..No vote

Degree 80...20...100

(80%*100=80)

(100-80=20)

No Degree 504..396..900

(1000-100=900)

(56%*900=504)

(504-900=396

Totals 584..416...1000

(80+504=584)

(20+396=416)

(900+100=1,000)

Summary

..............Vote..No vote

Degree 80...20...100

No Degree 504..396..900

Total Totals 584..416...1000

A. Calculation to determine the probability of The voter had a college degree and voted in the last presidential election.

P = 80/1,000

P=0.08*100

P=8%

Therefore the probability of The voter had a college degree and voted in the last presidential election will be 8%

B. Calculation to determine the probability of The voter did not have a college degree and did not vote in the last presidential election.

P =396/1000

P=0.396*100

P=39.6%

Therefore the probability of The voter did not have a college degree and did not vote in the last presidential election will be 39.6%

C. Calculation to determine the probability if The voter voted in the last presidential election.

P = 584/1,000

P=0.584*100

P=58.4%

Therefore the probability if The voter voted in the last presidential election will be 58.4%

D. Calculation to determine the probability if The voter did not vote in the last presidential election.

P = 416/1000

P=0.416*100

P=41.6%

Therefore the probability if The voter did not vote in the last presidential election will be 41.6%

8 0

3 years ago

To solve the problem: "What is 3/4 of 12," you would _____ .