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Cloud [144]
3 years ago
10

Call a household prosperous if its income exceeds $100,000. Call the household educated if the householder completed college. Se

lect an American household at random, and let A be the event that the selected household is prosperous and B the event that it is educated. According to the Current Population Survey, P(A)=0.138, P(B)=0.261, and the probability that a household is both prosperous and educated is P(A and B)=0.082. What is the probability P(A or B) that the household selected is either prosperous or educated?
Mathematics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:  0.317

Step-by-step explanation:

Let A be the event that the selected household is prosperous and B the event that it is educated.

Given : P(A)=0.138,   P(B)=0.261

P(A and B)=0.082

We know that for any events M and N ,

\text{P(M or N)=P(M)+P(N)-P(M or N)}

Thus , \text{P(A or B)=P(A)+P(B)-P(A or B)}

\text{P(A or B)}=0.138+0.261-0.082\\\\\Rightarrow\text{ P(A or B)}=0.317

Hence, the probability P(A or B) that the household selected is either prosperous or educated = 0.317

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Answer:

For a data set of N elements:

{x₁, x₂, ..., xₙ}

The mean is:

M = \frac{x_1 + x_2 + ... + x_n}{N}

The mean absolute deviation is:

MAD = \frac{|x_1 - M| + ... + |x_n - M|}{N}

And the population standard deviation is:

PSD = \frac{\sqrt{|x_1 - M|^2 + ... + |x_n - M|^2} }{\sqrt{N}}

In this case, our set has 5 elements, and the set is:

{2, 4, 6, 9, 14}

The mean of this set is:

M = \frac{2 + 4 + 6 + 9 + 14}{5} = 7

The mean absolute deviation is:

MAD = \frac{|2 - 7| + |4 - 7| + |6 - 7|+ |9 - 7| + |14 - 7| }{5} = 3.6

And the population standard deviation is:

PSD =  \sqrt{\frac{|2 - 7|^2 + |4 - 7|^2 + |6 - 7|^2+ |9 - 7|^2 + |14 - 7|^2 }{5}} = 4.2

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2 years ago
A plant grew 3/10 meter in april in may it grew 27/100 meter in may in june the plant grew 13/100 meter.Which fractions are equi
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What the first four common multiples on 3 like for 2,4,6,8
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PLEASEEEEEEEEEEEEEEEEEE URGENT
jeyben [28]

Option C:

f(x) = 5x + 7 is the function of the input-output table.

Solution:

Option A: f(x) = 6x + 9

Input x = 1 in the above equation.

f(1) = 6(1) + 9

    = 6 + 9

f(1) = 15

But the output is 12 in the table.

So, it is not the function of the table.

Option B: f(x) = 7x + 5

Input x = 1 in the above equation.

f(1) = 7(1) + 5

f(1) = 12

Input x = 5 in the above equation.

f(5) = 7(5) + 5

f(5) = 40

But the output is 32 in the table.

So, it is not the function of the table.

Option C: f(x) = 5x + 7

Input x = 1 in the above equation.

f(1) = 5(1) + 7

f(1) = 12

Input x = 5 in the above equation.

f(5) = 5(5) + 7

f(5) = 32

Input x = 10 in the above equation.

f(10) = 5(10) + 7

f(10) = 57

Input x = 5 in the above equation.

f(15) = 5(15) + 7

f(15) = 82

All outputs are correct for the given input.

Hence it is the function of the table.

Option D: f(x) = 12x + 1

Input x = 1 in the above equation.

f(1) = 12(1) + 1

f(1) = 12

Input x = 5 in the above equation.

f(5) = 12(5) + 1

f(5) = 61

But the output is 32 in the table.

So, it is not the function of the table.

Hence Option C is the correct answer.

f(x) = 5x + 7 is the function of the input-output table.

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