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3 years ago

Roberto is a systems analyst. His annual salary is $165,125. What is his monthly salary?

Mathematics

2 answers:

Evgesh-ka [11]3 years ago

8 0

Step-by-step explanation:

We have,

Annual salary = $165,125

→ Annual Salary = Months in a year × Monthly salary

Let the monthly salary be x.

→ $165,125 = 12x

→ $165,125 ÷ 12 = x

→ $13,760.4167 = Monthly Salary [Answer]

lilavasa [31]3 years ago

4 0

Answer: it is 13,760.416

Step-by-step explanation:

If you divide his annual salary by 12 months you get your answer which is $13,760.416

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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 $P(X \geq 2)$

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

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.

4 0

4 years ago