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vfiekz [6]
2 years ago
8

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 18 m

inutes and a standard deviation of 4 minutes. Using the empirical rule, what percentage of customers have to wait between 10 minutes and 26 minutes?
Mathematics
1 answer:
anyanavicka [17]2 years ago
3 0

Answer:

95% of the customers  have to wait between 10 minutes and 26 minutes

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 18 minutes

Standard Deviation, σ = 4 minute

We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.

Empirical Formula:

  • Almost all the data lies within three standard deviation from the mean for a normally distributed data.
  • About 68% of data lies within one standard deviation from the mean.
  • About 95% of data lies within two standard deviations of the mean.
  • About 99.7% of data lies within three standard deviation of the mean.

Now, we can write:

10 = 18-2(4) = \mu - 2(\sigma)\\26 = 18+2(4) = \mu + 2(\sigma)

Thus, by empirical formula, 95% of the data lies within two standard deviations of the mean.

Thus, 95% of the customers  have to wait between 10 minutes and 26 minutes

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2 years ago
Use the distributive property of multiplication to find 5×45.<br> Hint: 45=5+40.
alexgriva [62]

Answer:

Step-by-step explanation:

Distributive property: a(b +c) = (a*b) + (a*c)

45 = 40 + 5

5* 45 = 5* (40 + 5)

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2 years ago
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Ainat [17]
We can say at month 0 she was 43 inches tall.

And at month 18 she was 52 inches tall.

We can say that x is the number of months and y is her height.

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y2-y1/x2-x1
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5 0
3 years ago
The sum of two positive numbers is 16 and their difference is 2 . the larger number is?​
Aleonysh [2.5K]

Answer:

9

Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
(HELP ME 15 PTS)
kap26 [50]

Answer:   \bold{\dfrac{27x^2y^3}{16}}

<u>Step-by-step explanation:</u>

\dfrac{(3x^2y^5)^3}{(2xy^3)^4}\\\\\\=\dfrac{3^3\cdot x^{2\cdot 3}\cdot y^{5\cdot 3}}{2^4\cdot x^4\cdot y^{3\cdot 4}}\\\\\\=\dfrac{27\cdot x^6\cdot y^{15}}{16\cdot x^4\cdot y^{12}}\\\\\\=\dfrac{27\cdot x^{6-4}\cdot y^{15-12}}{16}\\\\\\=\dfrac{27\cdot x^2\cdot y^3}{16}

8 0
2 years ago
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