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3 years ago

Arif and keisha go to a restaurant fot dinner. their meals total $13.75 and the tax is 5%

1 answer:

5 0

The total cost of the dinner meals taken by Arif and Keisha = $13.75
Percentage of tax that need to be given = 5%
Then
Amount of tax that needs to be given = (5/100) * 13.75 dollars
   = 13.75/20 dollars
   = 0.687 dollars
   = 0.69 dollars
So
The total amount that needs to be
paid for dinner including tax = (13.75 + 0.69) dollars
   = 14.44 dollars
Percentage of tips that they both want to leave = 15%
Then
Amount of tips that need to be given = (15/100) * 14.44
   = 216.6/100 dollars
   = 2.17 dollars
So
The total amount that they should leave = (14.44 + 2.17) dollars
   = 16.61 dollars
So they should leave a total of $16.61 in the restaurant.

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Answer:

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Step-by-step explanation:

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Read 2 more answers

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

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4 years ago

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3 years ago