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frosja888 [35]
3 years ago
10

Tiffani works in a baby shop in which she prints personalized bibs. She uses a probability model to predict that the next custom

er ordering a white bib would be 30%. For one day, Tiffani gathers data by tallying the the number of customers who order each color in this table.
Color Tally
white 21
grey 9
blue 15
pink 26

Based on her experiment, which statement is true?

Tiffani's prediction is valid. The probability of the next customer ordering a white bib is about 30%.

Tiffani's prediction is not valid. The probability of the next customer ordering a white bib should be about 21% because 21 people ordered white in her data.

Tiffani's prediction is not valid. The next customer ordering a white bib would be the same as any color, so the probability should be about 25%.

There is not enough information to determine the validity of her prediction.
Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
4 0

Answer:

Tiffani's prediction is valid. The probability of the next customer ordering a white bib is about 30%.

Step-by-step explanation:

First you add all of your amounts together

21 + 9 + 15 + 26 = 71

Then to find the probability of the next customer buying a white bib you divide 21 by 71

21/71 = 0.295

Which is approximately 30%

I hope this helped!

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Find (3) if<br> f(x)=x^3+2x^2-x-1
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Answer:

f(3)=3³+2×2²-2-1=27+8-3=32

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3 years ago
Help please!!!!!!!I don’t know this! Tyy
aleksley [76]
<h3>Answer: Choice C.  4*sqrt(6)</h3>

====================================================

Explanation:

Each cube has a side length of 4. Placed together like this, the total horizontal side combines to 4+8 = 8. This is the segment HP as shown in the diagram below. I've also added point Q to form triangle HPQ. This is a right triangle so we can find the hypotenuse QH

Use the pythagorean theorem to find QH

a^2 + b^2 = c^2

(HP)^2 + (PQ)^2 = (QH)^2

8^2 + 4^2 = (QH)^2

(QH)^2 = 64 + 16

(QH)^2 = 80

QH = sqrt(80)

Now we use segment QH to find the length of segment EH. Focus on triangle HQE, which is also a right triangle (right angle at point Q). Use the pythagorean theorem again

a^2 + b^2 = c^2

(QH)^2 + (QE)^2 = (EH)^2

(EH)^2 = (QH)^2 + (QE)^2

(EH)^2 = (sqrt(80))^2 + (4)^2

(EH)^2 = 80 + 16

(EH)^2 = 96

EH = sqrt(96)

EH = sqrt(16*6)

EH = sqrt(16)*sqrt(6)

EH = 4*sqrt(6), showing the answer is choice C

-------------------------

A shortcut is to use the space diagonal formula. As the name suggests, a space diagonal is one that goes through the solid space (rather than stay entirely on a single face; which you could possibly refer to as a planar diagonal or face diagonal).

The space diagonal formula is

d = sqrt(a^2+b^2+c^2)

which is effectively the 3D version of the pythagorean theorem, or a variant of such.

We have a = HP = 8, b = PQ = 4, and c = QE = 4 which leads to...

d = sqrt(a^2+b^2+c^2)

d = sqrt(8^2+4^2+4^2)

d = sqrt(96)

d = sqrt(16*6)

d = sqrt(16)*sqrt(6)

d = 4*sqrt(6), we get the same answer as before

The space diagonal formula being "pythagorean" in nature isn't a coincidence. Repeated uses of the pythagorean theorem is exactly why this is.

4 0
3 years ago
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antiseptic1488 [7]
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3 years ago
based on the simulation, what is the probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to
Triss [41]

Using the binomial distribution, supposing that 0.3 of the callers have to wait more than 8 minutes to have their calls answered, it is found that there is a 0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

For each caller, there are only two possible outcomes, either they have to wait more than 8 minutes to have their calls answered, or they do not. The probability of a caller having to wait more than 8 minutes is independent of any other caller, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 10 callers, hence n = 10
  • Suppose that 0.3 of them have to wait more than 8 minutes, hence p = 0.3

The probability that <u>at most 2</u> of the next 10 callers will have to wait more than 8 minutes is:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

Then

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

P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.0282

P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.1211

P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.2335

Then:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0282 + 0.1211 + 0.2335 = 0.3828

0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

A similar problem is given at brainly.com/question/25537909

3 0
3 years ago
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