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garik1379 [7]
3 years ago
14

Anna enjoys having dinner at a restaurant in a town where the sales tax on meals is 10%. She leaves 15% tip before the sales tax

is applied, and the tax is calculated on the pre-tip amount. She spends a total of $27.50 on her dinner. What was the cost of her dinner without tax or tip?
Mathematics
1 answer:
vitfil [10]3 years ago
5 0

Answer:

The cost of her dinner without tax or tip is <u>$22</u>.

Step-by-step explanation:

Given:

Anna enjoys having dinner at a restaurant in a town where the sales tax on meals is 10%.

She leaves 15% tip before the sales tax is applied, and the tax is calculated on the pre-tip amount.

She spends a total of $27.50 on her dinner.

Now, to find the cost of her dinner without tax or tip.

Let the cost of dinner without tax or tip be x.

So, the tax is:

10\%\ of\ x.

=\frac{10}{100} \times x

=0.10\times x

=0.10x.

And, the tip is:

15\%\ of \ x.

=\frac{15}{100} \times x

=0.15\times x

=0.15x.

<u><em>As given, she spends a total $27.50 on her dinner.</em></u>

Now, to get the cost of her dinner without tax or tip:

x+0.15x+0.10x=27.50

x+0.25x=27.50

1.25x=27.50

<em>Dividing both sides by 1.25 we get:</em>

x=22

x=\$22.

Therefore, the cost of her dinner without tax or tip is <u>$22</u>.

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Answer:

a) P( X < 24 ) =  0.9772

b) P ( X > 18 ) =0.8413

c) P ( 14 < X < 26) = 0.9973

d)  P ( 14 < X < 26)  = 0.9973

e) P ( 16 < X < 20)  = 0.4772

f) P ( 20 < X < 26)  =  0.4987

Step-by-step explanation:

Given:

- Mean of the distribution u = 20

- standard deviation sigma = 2

Find:

a. P ( X  < 24 )

b. P ( X  > 18 )

c. P ( 18 < X  < 22 )

d. P ( 14 < X  < 26 )

e. P ( 16 < X  < 20 )

f. P ( 20 < X  < 26 )

Solution:

- We will declare a random variable X that follows a normal distribution

                                   X ~ N ( 20 , 2 )

- After defining our variable X follows a normal distribution. We can compute the probabilities as follows:

a) P ( X < 24 ) ?

- Compute the Z-score value as follows:

                                   Z = (24 - 20) / 2 = 2

- Now use the Z-score tables and look for z = 2:

                                   P( X < 24 ) = P ( Z < 2) = 0.9772

b) P ( X > 18 ) ?

- Compute the Z-score values as follows:

                                   Z = (18 - 20) / 2 = -1

- Now use the Z-score tables and look for Z = -1:

                    P ( X > 18 ) = P ( Z > -1) = 0.8413

c) P ( 18 < X < 22) ?

- Compute the Z-score values as follows:

                                   Z = (18 - 20) / 2 = -1

                                   Z = (22 - 20) / 2 = 1

- Now use the Z-score tables and look for z = -1 and z = 1:

                   P ( 18 < X < 22)  = P ( -1 < Z < 1) = 0.6827

d) P ( 14 < X < 26) ?

- Compute the Z-score values as follows:

                                   Z = (14 - 20) / 2 = -3

                                   Z = (26 - 20) / 2 = 3

- Now use the Z-score tables and look for z = -3 and z = 3:

                   P ( 14 < X < 26)  = P ( -3 < Z < 3) = 0.9973

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- Compute the Z-score values as follows:

                                   Z = (16 - 20) / 2 = -2

                                   Z = (20 - 20) / 2 = 0

- Now use the Z-score tables and look for z = -2 and z = 0:

                   P ( 16 < X < 20)  = P ( -2 < Z < 0) = 0.4772

f) P ( 20 < X < 26) ?

- Compute the Z-score values as follows:

                                   Z = (26 - 20) / 2 = 3

                                   Z = (20 - 20) / 2 = 0

- Now use the Z-score tables and look for z = 0 and z = 3:

                   P ( 20 < X < 26)  = P ( 0 < Z < 3) = 0.4987

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The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

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A. P(x=6)

P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\

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