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Elanso [62]
3 years ago
15

Enter a positive common favor (other than 1) of the fractions numerator and denominator

Mathematics
2 answers:
Romashka [77]3 years ago
6 0

Answer:

7

Step-by-step explanation:

You can divide both 14 and 21 by seven and get a whole number.

Ket [755]3 years ago
4 0

Answer:

7

Step-by-step explanation:

hope this helps

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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
Solve for x please and thank you​
Anarel [89]

Answer:

2x -5(x-3) = -4 +5x -29

2x-5x+15 = -4 +5x -29

-3x +15 = 5x -33

-8x= -48

x=6

6 0
3 years ago
Help! My birthday is today and no one decorated or got me a cake or got me presents.! Help. If a dog has two apples and I have t
Ahat [919]

Answer:

I'm pretty sure the total would be 5 apples

Step-by-step explanation: If your birthday actually is today then happy birthday if not then.........i dont know

7 0
2 years ago
Read 2 more answers
Please help, i'm horrible @ math :)
katovenus [111]
   
\displaystyle\\
1 \frac{2}{5} +\Big(-5 \frac{1}{2} \Big) = ? \\  \\ 
1 \frac{2}{5} = \frac{1\times 5+2}{5}=\boxed{\frac{7}{5}} \\  \\ 
-5 \frac{1}{2} =-5 -\frac{1}{2} = \frac{-5\times2-1}{2} =\frac{-10-1}{2}=\frac{-11}{2}= \boxed{-\frac{11}{2} }\\  \\ \texttt{OR} \\  \\ 
-5 \frac{1}{2} = -\Big(5 \frac{1}{2} \Big)= -\Big( \frac{5\times2+1}{2} \Big)=-\Big( \frac{11}{2} \Big)= \boxed{-\frac{11}{2} }


\displaystyle\\
\Longrightarrow ~~1 \frac{2}{5} +\Big(-5 \frac{1}{2} \Big) =\frac{7}{5} -\frac{11}{2} = \frac{7\times 2}{5\times 2} -\frac{11\times 5}{2\times 5} = \\  \\ 
= \frac{14}{10} -\frac{55}{10} = \frac{14-55}{10} =\frac{-41}{10} = -\frac{41}{10}=-\frac{40+1}{10}=\boxed{\boxed{-4\frac{1}{10}}}



7 0
3 years ago
Need help on the mean and the median 1-10 pleaaaase for brainlest answer and thanks !!!!!!!!!!
kati45 [8]
For number one the media is 22 and 5

8 0
3 years ago
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