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amm1812
3 years ago
9

Working on Summer Vacation. An Adweek/Harris (July 2011) poll found that 35% of U.S. adults do not work at all while on summer v

acation. In a random sample of 10 U.S. adults, let x represent the number who do not work during summer vacation
a. For this experiment, define the event that represents a "success"

b. Explain why x is (approximately) a binomial random variable

c. Give the value of p for this binomial experiment

d. Find P(x = 3)

e. Find the probability that 2 or fewer of the 10 U.S. adults do not work during summer vacation
Mathematics
1 answer:
Alex Ar [27]3 years ago
6 0

A becuase i worked it out and i got that so im really confident

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a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

5 0
3 years ago
PLS HELP ME I WILL GIVE YOU BRAINLIEST!
dezoksy [38]
I think its 180 i think
5 0
3 years ago
Two data sets have the same mean,
nirvana33 [79]

For Data Set B, we see that the data is more varied. The absolute deviations are 4, 3, 2, 5. The average of these absolute deviations is 3.5. MAD_B = (4+3+2+5)/4 =3.5 M ADB

Hence, The average of these absolute deviations is 3.5.

5 0
2 years ago
Please help thank you
Keith_Richards [23]

Answer:

5

alternates

Step-by-step explanation:

The series is for k = 0 to 4, so k = 0, 1, 2, 3, 4.

That is 5 terms.

-1/2 is inside the parentheses and is raised to 0, 1, 2, 3, 4.

A negative number raised to an even integer is positive.

A negative number raised to an odd integer is negative.

The sign of each term in the series alternates.

7 0
3 years ago
Answer Fast Please! Will give Brainlist!! What common fraction is one-third of the way from 1/4 to 1/3
Brut [27]

Answer:

I think it’s 7/24’s I looked it up on Quora

Step-by-step explanation:

4 0
2 years ago
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