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lapo4ka [179]
3 years ago
13

5.11.

Mathematics
1 answer:
irinina [24]3 years ago
7 0

Answer:

Following are the response to the given points:

Step-by-step explanation:

For question 5.11:

For point a:

For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.

For point b:

The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last 2 \frac{1}{2} hours. Another sampling approach is a group.

For question 5.12:

This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a \bar{x}line graph when 5 parts were achieved.

This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours 2 \frac{1}{2} . An alternative type of analysis should be a random sample of five consecutive pieces created every 2 \frac{1}{2} hour.

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At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is
Sphinxa [80]

Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. 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.

81% of students own a wireless device.

This means that p = 0.81

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

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

P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487

34.87% probability that all 5 have a wireless device

5 0
3 years ago
Can someone smart help me with this plz:)
agasfer [191]
Bottom right is the correct answer
3 0
3 years ago
Read 2 more answers
2/5(d−10)−23(d+6)25(d−10)−23(d+6) is 
Dimas [21]
Simplify \frac{2}{5} (d - 10) to \frac{2(d - 10)}{5} :
\frac{2(d - 10)}{5} [/tex] - 23(d + 6) x 25(d - 10) - 23(d + 6)

Simplify 23(d + 6) x 25(d - 10) to 575(d + 6)(d - 10) :
\frac{2(d - 10)}{5} [/tex] - 575(d + 6)(d - 10) - 23(d + 6)

Then Expand : 
\frac{2(d - 10)}{5} [/tex] - 575d^{2} + 5750d - 3450d + 34500 - 23d - 138

Now Collect Like Terms :
\frac{2(d - 10)}{5} [/tex] - 575d^{2} + (5750d - 3450d + 34500 - 23d) + (34500 - 138)

Answer :
\frac{2(d - 10)}{5} [/tex] - 575d^{2} + 2277d + 34362



7 0
3 years ago
What’s the correct answer answer asap for brainlist
irga5000 [103]

Answer: B

Step-by-step explanation:

Genes have three regions, the promoter, coding region, and termination sequence.

5 0
2 years ago
Write 3 equivalent ratios for 17:69
Digiron [165]
<h3>♫ :::::::::::::::::::::::::::::: // Hello There ! //  :::::::::::::::::::::::::::::: ♫</h3>

➷  Double the values:

(17 : 69) x 2 = 34 : 138

Multiply both sides by 10:

(17 : 69) x 10 = 170 : 690

Triple the values:

(17 : 69) x 3 = 51 : 207

<h3><u>❄️</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0
4 years ago
Read 2 more answers
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