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Yakvenalex [24]
2 years ago
11

Alicia took a friend to dinner for her birthday. The total bill for dinner was $30.69 (including tax and a tip). If Alicia paid

a 21.5% tip, what was her bill prior to adding the tip?
Mathematics
1 answer:
tigry1 [53]2 years ago
5 0

Answer:

The original bill was $25.26

Step-by-step explanation:

To find the original bill, you need to divide the amount paid by the percentage paid. Since the tip was 21.5% on top of the full amount already, we know she paid 121.5%. So we can divide the cost by that.

$30.69/121.5% = $25.26

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13.02-6y=8y using the method of solving linear equations
gregori [183]

well add -6y to 6y which equals 0 then add 8y to 6y which equals 14y

now you have 13.02=14y 

now you can divide 13.02 divided into 14 which equals 0.93

0.93 = y

5 0
3 years ago
In a road-paving process, asphalt mix is delivered to the hopper of the paver by trucks that haul the material from the batching
Advocard [28]

Answer:

a) Probability that haul time will be at least 10 min = P(X ≥ 10) ≈ P(X > 10) = 0.0455

b) Probability that haul time be exceed 15 min = P(X > 15) = 0.000

c) Probability that haul time will be between 8 and 10 min = P(8 < X < 10) = 0.6460

d) The value of c is such that 98% of all haul times are in the interval from (8.46 - c) to (8.46 + c)

c = 2.12

e) If four haul times are independently selected, the probability that at least one of them exceeds 10 min = 0.1700

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 8.46 min

Standard deviation = σ = 0.913 min

a) Probability that haul time will be at least 10 min = P(X ≥ 10)

We first normalize/standardize 10 minutes

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (10 - 8.46)/0.913 = 1.69

To determine the required probability

P(X ≥ 10) = P(z ≥ 1.69)

We'll use data from the normal distribution table for these probabilities

P(X ≥ 10) = P(z ≥ 1.69) = 1 - (z < 1.69)

= 1 - 0.95449 = 0.04551

The probability that the haul time will exceed 10 min is approximately the same as the probability that the haul time will be at least 10 mins = 0.0455

b) Probability that haul time will exceed 15 min = P(X > 15)

We first normalize 15 minutes.

z = (x - μ)/σ = (15 - 8.46)/0.913 = 7.16

To determine the required probability

P(X > 15) = P(z > 7.16)

We'll use data from the normal distribution table for these probabilities

P(X > 15) = P(z > 7.16) = 1 - (z ≤ 7.16)

= 1 - 1.000 = 0.000

c) Probability that haul time will be between 8 and 10 min = P(8 < X < 10)

We normalize or standardize 8 and 10 minutes

For 8 minutes

z = (x - μ)/σ = (8 - 8.46)/0.913 = -0.50

For 10 minutes

z = (x - μ)/σ = (10 - 8.46)/0.913 = 1.69

The required probability

P(8 < X < 10) = P(-0.50 < z < 1.69)

We'll use data from the normal distribution table for these probabilities

P(8 < X < 10) = P(-0.50 < z < 1.69)

= P(z < 1.69) - P(z < -0.50)

= 0.95449 - 0.30854

= 0.64595 = 0.6460 to 4 d.p.

d) What value c is such that 98% of all haul times are in the interval from (8.46 - c) to (8.46 + c)?

98% of the haul times in the middle of the distribution will have a lower limit greater than only the bottom 1% of the distribution and the upper limit will be lesser than the top 1% of the distribution but greater than 99% of fhe distribution.

Let the lower limit be x'

Let the upper limit be x"

P(x' < X < x") = 0.98

P(X < x') = 0.01

P(X < x") = 0.99

Let the corresponding z-scores for the lower and upper limit be z' and z"

P(X < x') = P(z < z') = 0.01

P(X < x") = P(z < z") = 0.99

Using the normal distribution tables

z' = -2.326

z" = 2.326

z' = (x' - μ)/σ

-2.326 = (x' - 8.46)/0.913

x' = (-2.326×0.913) + 8.46 = -2.123638 + 8.46 = 6.336362 = 6.34

z" = (x" - μ)/σ

2.326 = (x" - 8.46)/0.913

x" = (2.326×0.913) + 8.46 = 2.123638 + 8.46 = 10.583638 = 10.58

Therefore, P(6.34 < X < 10.58) = 98%

8.46 - c = 6.34

8.46 + c = 10.58

c = 2.12

e) If four haul times are independently selected, what is the probability that at least one of them exceeds 10 min?

This is a binomial distribution problem because:

- A binomial experiment is one in which the probability of success doesn't change with every run or number of trials. (4 haul times are independently selected)

- It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure. (Only 4 haul times are selected)

- The outcome of each trial/run of a binomial experiment is independent of one another. (The probability that each haul time exceeds 10 minutes = 0.0455)

Probability that at least one of them exceeds 10 mins = P(X ≥ 1)

= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 1 - P(X = 0)

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 4 haul times are independently selected

x = Number of successes required = 0

p = probability of success = probability that each haul time exceeds 10 minutes = 0.0455

q = probability of failure = probability that each haul time does NOT exceeds 10 minutes = 1 - p = 1 - 0.0455 = 0.9545

P(X = 0) = ⁴C₀ (0.0455)⁰ (0.9545)⁴⁻⁰ = 0.83004900044

P(X ≥ 1) = 1 - P(X = 0)

= 1 - 0.83004900044 = 0.16995099956 = 0.1700

Hope this Helps!!!

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