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riadik2000 [5.3K]

3 years ago

6

An auction house sells raffle tickets for $1. There are 10 prizes, each worth $10. If 100 raffle tickets are sold, what is the e

xpected value of each ticket? $0 $1 -$1 $10

2 answers:

zaharov [31]3 years ago

8 0

Each prize is $10, 100 tickets for a dollar and then it's a $100, divide it by a total of ten prizes and it's 10 dollars a prize

allsm [11]3 years ago

7 0

1 dollar is the expected value

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A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has eight ident

Fudgin [204]

Answer:

a) 0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b) 0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either they last more than 1,000 hours, or they do not. Components operate independently, which means that the binomial probability distribution is used to solve this question.

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.

One subsystem has eight identical components, each with a probability of 0.1 of failing in less than 1,000 hours.

So 1 - 0.1 = 0.9 probability of working for more, which means that $p = 0.9$

a. exactly two of the four components last longer than 1000 hours.

This is P(X = 2) when $n = 4$ . So

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

$P(X = 2) = C_{4,2}.(0.9)^{2}.(0.1)^{2} = 0.0486$

0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b. the subsystem operates longer than 1000 hours.

The subsystem has 8 components, which means that $n = 8$

It will operate if at least 4 components are working correctly, so we want:

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

So

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

$P(X = 0) = C_{8,0}.(0.9)^{0}.(0.1)^{8} \approx 0$

$P(X = 1) = C_{8,1}.(0.9)^{1}.(0.1)^{7} \approx 0tex][tex]P(X = 2) = C_{8,2}.(0.9)^{2}.(0.1)^{6} \approx 0$

$P(X = 3) = C_{8,3}.(0.9)^{3}.(0.1)^{5} = 0.0004$

Then

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0 + 0 + 0.0004 = 0.0004$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0004 = 0.9996$

0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

5 0

3 years ago

Find the area of the shaded region. Round to the nearest hundredth when necessary.

dusya [7]

Answer: 423.11 m^2

Step by step explanation:

Area of a circle formula pi*r^2

First solve the area of the big circle

3.14*21^2=1384.74 m^2

Now find the are of the smaller circle

r=21-3.5=17.5
3.14*(17.5)^2=961.63 m^2

Now we can minus the small area off the big area to get the shaded area.

1384.74-961.63=423.11 m^2

So the area of the shaded region is 423.11 m^2

3 0

3 years ago

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Lapatulllka [165]

Answer:

8) answer=5 9)answer=15

Step-by-step explanation:

all you have to do is replace in the following formula

$\sqrt{(x2-x1)^{2}+ (y2-y1)^{2}} =d$

Where

X1: is the first number of the first point

Y1: is the second number of the first point

X2: is the first number of the second point

Y2: is the second number of the second point

you replace in the formula and solve

I attach solution

4 0

3 years ago

Which expression below is equivalent to the expression -3/4(-12x-20Y)

Brut [27]

The answer would be a

6 0

3 years ago

X<br> Combine like terms:<br> 10k - 2k + 7s2 - 32

Free_Kalibri [48]

Answer:

$7s^{2}$ + 8k - 32

4 0

3 years ago