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3 years ago

8

An isosceles triangle has two sides of equal length, a, and a base, b. The perimeter of the triangle is 15.7 inches, so the equa

tion to solve is 2a + b = 15.7

1 answer:

lana [24]3 years ago

8 0

Answer:

15.7 - 2a

Step-by-step explanation:

Perimeter = 15.7 inches

2a + b = 15.7

b = 15.7 - 2a

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Tamiku [17]

Answer:

The answer to your question is Extras charges = $999

Step-by-step explanation:

Data

Total price = $ 11,325

State Tax 8%

License plates = $53

Emissions inspection = $40

Process

1.- Calculate the 8% of the price of the car

11325 ---------------- 100%

x --------------- 8%

x = (8 x 11325)/100

x = 90600/100

x = $ 906 state tax

2.- Calculate the total charges

Total charges = $906 + $53 + $40

= $ 999

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Step-by-step explanation:

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In 2000, U.S. Airways burned about 115 litres of fuel per passenger. In 2014 that dropped

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Answer:

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Step-by-step explanation:

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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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4 years ago