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AfilCa [17]
3 years ago
15

Kelly went on vacation to the Hamptons. She was curious about whether rich people own a lot of TVs, so she went door to door in

333 neighborhoods, asking people how many TVs they owned. Here's what she found:
The inventor of the flat-screen TV lived in neighborhood AAA. He owned 787878 TVs.
Every home in neighborhood BBB owned 333 TVs.
TV ownership in neighborhood CCC was normally distributed with a median of 666 TVs.
In neighborhood AAA, the median number of TVs will likely be

the mean.
In neighborhood BBB, the median number of TVs will be

the mean.
In neighborhood CCC, the median number of TVs will likely be

the mean.
Mathematics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

In neighborhood AAA, the median number of TVs will likely be < the mean.

In neighborhood BBB, the median number of TVs will be ==equals the mean.

In neighborhood CCC, the median number of TVs will likely be ==equals the mean.

Step-by-step explanation:

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Select the correct answer.
Andreas93 [3]

It would be graph B because it is the only one that is close to it.

If i'm wrong it is ok with me

7 0
2 years ago
Read 2 more answers
Assume that the probability of a defective computer component is 0.02. Components arerandomly selected. Find the probability tha
solmaris [256]

Answer:

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

Step-by-step explanation:

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.

First six not defective, each with 0.98 probability.

7th defective, with 0.02 probability. So

p = (0.98)^6*0.02 = 0.0177

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

Find the expected number and variance of the number of components tested before a defective component is found.

Inverse binomial distribution, with p = 0.02

Expected number before 1 defective(n = 1). So

E = \frac{n}{p} = \frac{1}{0.02} = 50

Variance is:

V = \frac{np}{(1-p)^2} = \frac{0.02}{(1-0.02)^2} = 0.0208

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

5 0
3 years ago
4) This results when you flip the numerator and denominator of a fraction.
Semenov [28]

Answer:

To divide one fraction by another one, flip numerator and denominator of the second one, and then multiply the two fractions. The flipped-over fraction is called the multiplicative inverse or reciprocal.

Step-by-step explanation:

I got this from google.

3 0
3 years ago
Please help If you can! I'm doing Math! and I need some help!
Kisachek [45]

Answer:

x ≈ 370

Step-by-step explanation:

3 0
3 years ago
Fill in the blank. Using the number line below, the plotted number is when rounded to the nearest thousand.​
prisoha [69]

Answer:

5000

Step-by-step explanation:

if its 5 or more round up. If it's 4 or less round down

6 0
3 years ago
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