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3 years ago

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There are six households in a rural community. Each household earns $40,000 per year. Suppose that a new resident builds a mansi

on in the community and that the income in the new household is $4 million per year. After the new resident arrives, the median household income has _____, and the mean household income has _____. increased; increased not changed; increased not changed; not changed increased; not changed

1 answer:

dusya [7]3 years ago

7 0

Answer:

Median income has not changed

Mean income has increased

Step-by-step explanation:

Initial number of houses = 6

Earning per household = $40,000

The initial mean :

$(40,000 * 6) / 6 = $40000

Initial median :

40000, 40000, 40000, 40000, 40000, 40000

= 1/2(n+1)th term,

Median = 40000

A 7th household earns 4,000,000

Mean = ((40000*6) + 4,000,000) / 7 = $605,714

Median = 1/2(n+1)th term

40000, 40000, 40000, 40000, 40000, 40000, 4000000

Median = 4th term = 40000

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-- The perimeter of the whole thing is (33 + 6 + 21 + 15 + 12 + 15 + 6) = 108

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1 year ago

Read 2 more answers

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution 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.

In this problem we have that:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

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

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897$

$P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339$

$P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669$

$P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211$

$P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148$

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

$E(X) = np$

In this question, we have $n = 22$

So

$E(X) = 22*0.75 = 16.5$

The closest integer to 16.5 is 17.

7 0

3 years ago

Of the cars sold during the month of July, 89 had air conditioning, 99 had an automatic transmission, and 74 had power steering.

DedPeter [7]

Answer:

There are a total of 23 cars with air conditioning and automatic transmission but not power steering

Step-by-step explanation:

Let A be the cars that have Air conditioning, B the cars that have Automatic transmission and C the cars that have pwoer Steering. Lets denote |D| the cardinality of a set D.

Remember that for 2 sets E and F, we have that

$|E \cup F| = |E| + |F| - |E \cap F|$

Also,

|E| = |E ∩F| + |E∩F^c|

We now alredy the following:

|A| = 89

|B| = 99

|C| = 74

$|A \cap B \cap C| = 5$

|(A \cup B \cup C)^c| = 24

|A \ (B U C)| = 24 (This is A minus B and C, in other words, cars that only have Air conditioning).

|B \ (AUC)| = 65

|C \ (AUB)| = 26

$|B \cap C| = 11$

We want to know |(A∩B) \ C|. Lets calculate it by taking the information given and deducting more things

For example:

99 = |B| = |B ∩ C| + |B∩C^c| = 11 + |B∩C^c|

Therefore, |B∩C^c| = 99-11 = 88

And |A ∩ B ∩ C^c| = |B∩C^c| - |B∩C^c∩A^c| = |B∩C^c| - |B \ (AUC)| = 88-65 = 23.

This means that the amount of cars that have both transmission and air conditioning but now power steering is 23.

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2 years ago

Select the sets that are not functions.

Temka [501]

D and E are the ones that aren’t functions because they use the x-values more then once

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2 years ago