Calculate a 15% Tip for $12.68

The wholesale price for a dozen pairs of socks is $4.30. The retail price for the same dozen pairs of socks is $9.25. What is th

Mandarinka [93]

Answer:

the mark up used by the retailer is 115.12% of the wholesale price

Step-by-step explanation:

Mark-up is the percentage increase of the retail price over the wholesale price

Mark-up = (difference between retail and whole sale price / retail price ) x 100

difference between retail and whole sale price = $9.25 - $4.30 = $4.95

(4.95 / 4.3) x 100 = 115.12%.

5 0

3 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

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

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

6 0

3 years ago

Based on the experimental results, how many rolls of 2 would you expect to get in a new experiment with 75 trials

Mila [183]

Answer =37.5 rows of two
Explanation= you can do 75 divid by 2 to get rows of two each of 75.

3 0

3 years ago

Read 2 more answers

What is the circumference of a circle with a radious of 7?

TEA [102]

Answer:

to get the circumference 3.14×14=43.96

radious is doubled because it is half the diameter

8 0

3 years ago

He can either pay a $150 joining fee and a $10 monthly fee or he can pay a $50 joining fee and a $30 monthly fee. On what month

larisa86 [58]

Answer:it will take 5 months for the cumulative costs of the plans to be equal and the total cos is $200

Step-by-step explanation:

Let x represent the number of months that for which the cumulative costs of the plans will be equal.

Let y represent the total cost of using plan A for x months.

Let z represent the total cost of using plan B for x months.

He can either pay a $150 joining fee and a $10 monthly fee. This means that the total cost of using plan A would be

y = 150 + 10x

For plan B, he can pay a $50 joining fee and a $30 monthly fee. This means that the total cost of using plan B would be

y = 50 + 30x

To determine the number of hours for which the cumulative costs of the plans will be equal, we would equate y to z. It becomes

150 + 10x = 50 + 30x

30x - 10x = 150 - 50

20x = 100

x = 100/20 = 5 months

The total cost would be

150 + 10 × 5 = 150 + 50 = $200

6 0

3 years ago