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ahrayia [7]
3 years ago
7

The​ _______ of a discrete random variable represents the mean value of the outcomes.

Mathematics
2 answers:
USPshnik [31]3 years ago
3 0

Answer:

expected value

Step-by-step explanation:

artcher [175]3 years ago
3 0

Answer:

Expected value

Step-by-step explanation:

Expected value- Any discrete variable's expected value is probability-weighted average of all of its possible values.

In simple words, any possible value that can be inferred by the any given variable is compounded by its probability of occurrence and the resulting products are added up to deliver the expected value.

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Nothing times itself will equal 29, but in decimal form only is 5.38
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For Madeline's lemonade recipe, 3 lemons are required to make 6 cups
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Answer:

1/2 lemon per 1 cup

Step-by-step explanation:

6/6=1

3/6=1/2= .5 lemon

6 0
3 years ago
Vacation Property Rentals purchased 50 television sets for their rental units at a cost of $250 per television. The televisions
Alex787 [66]
Under A straight line basis which is a method of computing depreciation and amortization by dividing the difference between an asset's cost and its expected salvage value by the number of years it is expected to be used. Also known as straight line depreciation or straight line amortization, this is the simplest depreciation method. But instead of that find The rate of depreciation
100/5 years=20% depreciation rate per year
Total cost 250×50=12,500
Salvage value 40×50=2,000
Subtract the salvage value from the total cost of televisions
12,500−2,000=10,500
In the first year the depreciation is
10,500×0.2=2,100
Book value
12,500−2,100=10,400
In the second year the depreciation is
10,500×0.2=2,100
Book value
10,400−2,100=8,300
In the third year the depreciation is
10500×0.2=2100
Book value
8300-2100=6200

the book value for all of the televisions at the end of the third year is 6200
8 0
4 years ago
​41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the
lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, 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.

​41% of U.S. adults have very little confidence in newspapers.

This means that p = 0.41

You randomly select 10 U.S. adults.

This means that n = 10

(a) exactly​ five

This is P(X = 5). So

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

P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

(b) at least​ six

This is:

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209

P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480

P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125

P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019

P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001

Then

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

(c) less than four.

This is:

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

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

P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051

P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355

P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111

P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058

So

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0
3 years ago
Distributive property<br> 1/2(6x-14)
stich3 [128]

Answer:

The answer is 3x−7

Step-by-step explanation:

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4 years ago
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