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Aleksandr [31]
4 years ago
12

A group of 85 people were asked whether or not they prefer a bagel or a muffin for breakfast and whether or not they prefer milk

or orange juice. The results of the poll showed people prefer orange juice over milk, regardless of whether or not they prefer a bagel or a muffin for breakfast. W. X. Y. Z. Which two-way table is a possible representation of the information collected from the poll? A. X B. Z C. Y D. W
Mathematics
1 answer:
gladu [14]4 years ago
5 0

Answer:

The answer(s) would be WY/WZ, or WYZ (If this is multiple choice)

Step-by-step explanation:

The reasoning is that in the "IRL Equation" the polls showed that orange juice was prefered over milk, and didn't matter whether they wanted a bagel or muffin. All that matters in this equation is that it is one sided (For the 1st poll that is) and now you are able to narrow down the possibilities of the answer(s).

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Yuri computes the mean and standard deviation for the sample data set 12, 14, 9, and 21. he finds the mean is 14. his steps for
german
The complete question in the attached figure N 1

we know that
The formula for Sample Standard Deviation is indicated in the attached figure N2
n in this problem is 4 <span>and Yuri had to divide in the formula by 3  </span>

therefore

the answer is
 Yuri divided by n instead of n -1

8 0
3 years ago
Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili
lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}

Use the Bayes' theorem to compute the value of P (B | D) as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

                                                                                     =0.3333\times 0.3333\\=0.11108889\\\approx 0.1111

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0
3 years ago
Mr. Gonzalez drove 30 miles in March. He drove 5 times as many miles in March as he did in January. He drove 2 times as many mil
Deffense [45]
<h2>Answer:</h2>

<h2>12 miles</h2>

Step-by-step explanation:

Given information :

Mr. Gonzalez drove 30 miles in March. In January, she drove 1/5 as much as in March.

30 ÷ 5 = 6 miles in January

in JanuaryShe drove 2 times as many miles in February as she did in January. In January, she drove 30 miles.

6 x 2 = 12 miles in February

8 0
3 years ago
What multiplies to give you 108 and adds up to give you -31?
meriva
4 x 27  = 108 and 4 + 27 = 31... I don't think there is any pair of numbers that would equal -31. So the answer would be 4 and 27. Good luck! Hope this helped!!
4 0
3 years ago
The side lengths of a right triangle are 24 , 45 , and 51 . Determined which is the length of the hypotenuse
HACTEHA [7]
The side length 51 is the hypotenuse.
24*24+45*45=2601
The square root of 2601 is 51.
Please note that the hypotenuse is most likely the longest side.
5 0
4 years ago
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