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Mademuasel [1]
2 years ago
5

ASAP HURRY!!!!!!!!!!!!!!!!!!!Complete the following statement. 3 ( 20 + 4 ) = a0

Mathematics
2 answers:
Lina20 [59]2 years ago
5 0

Answer:

72

Step-by-step explanation:

3(20+4)=a0

60+12=a0

72=a0

im not quite sure what you want. I solved for a0. Plz tell me if this is correct. If its not, tell me what u need, then I will solve it for you.

siniylev [52]2 years ago
3 0

Answer:

72

Step-by-step explanation:

3 * 20 = 60

3 * 4 = 12

60 + 12 = 72

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In a game, my friend is making 105 dollars per 45 seconds. How much money will he make in an hour?
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Answer:

Step-by-step explanation:

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Good evening ,

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First of all : 3600÷45=80

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3 years ago
Past records indicate that the probability of online retail orders
Tcecarenko [31]

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

Binomial probability distribution

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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

The mean of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

In this problem, we have that:

p = 0.08, n = 20

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

E(X) = np = 20*0.08 = 1.6

Standard deviation

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

P(X \leq 1) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X \leq 1)

In which

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

From itens b and c

P(X \leq 1) = 0.1887 + 0.3282 = 0.5169

P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831

48.31% probability that two or more online retail orders will turn out to be fraudulent.

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a = bx + c

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