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melomori [17]
1 year ago
11

Elizabeth brought a box of donuts to share. There are two-dozen (24) donuts in the box, all identical in size, shape, and color.

6 are jelly-filled, 10 are lemon-filled, and 8 are custard-filled. You randomly select one donut, eat it, and select another donut. Find the probability of selecting two jelly-filled donuts in a row. Write your answer as a fraction in simplest form.
Mathematics
1 answer:
pashok25 [27]1 year ago
8 0

The probability of selecting two jelly-filled donuts in a row is 1/ 22

<h3>What is probability?</h3>

Probability is simply the chance that a given event will occur.

It is also the ratio of the number of outcomes in a set of equally likely outcomes that produce a given event to the total number of possible outcomes.

From the information given, we have that the donuts are:

  • 6 are jelly-filled
  • 10 are lemon-filled
  • 8 are custard-filled

Note that two jelly - filled donuts were taken without replacement

Probability( 2 jelly -filled donuts) = 6 / 24 × 4/ 22 = 24/ 24 × 22 = 1/ 22

Thus, the probability of selecting two jelly-filled donuts in a row is 1/ 22

Learn more about probability here:

brainly.com/question/24756209

#SPJ1

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Answer:

(D)-2(28z-14) \: and\:\\(E) -7(8z-4)

Step-by-step explanation:

We are to determine which of these expressions are equivalent to: -56 z + 28

(A)\frac{1}{2}\cdot(-28z+14) \\(B)(-1.4z+0.7)40\\(C)(14-7z)\cdot(-4)\\(D)(8z-4)(-7)\\(E)-2(-28z-14)

-56 z + 28

  • If we factor out -2

-56 z + 28=-2(\frac{ -56 z}{-2} + \frac{28}{-2})=-2(28z-14)

  • If we factor out -7

-56 z + 28=-7(\frac{ -56 z}{-7} + \frac{28}{-7})=-7(8z-4)

Therefore, the two equivalent expressions are:

-2(28z-14) \: and\: -7(8z-4)

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If you want to be 95​% confident of estimating the population mean to within a sampling error of plus or minus3 and the standard
Aleksandr [31]

Answer:

The sample size required is at least 171

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We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

What sample size is​ required?

A samle size of at least n is required, in which n is found when M = 3, \sigma = 20

So

M = z*\frac{\sigma}{\sqrt{n}}

3 = 1.96*\frac{20}{\sqrt{n}}

3\sqrt{n} = 20*1.96

\sqrt{n} = \frac{20*1.96}{3}

(\sqrt{n})^{2} = (\frac{20*1.96}{3})^{2}

n = 170.7

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The sample size required is at least 171

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Answer:

L = 60

Step-by-step explanation:

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