Mr. Donley drove 504 km in each direction on a vacation

Mathematics

1 answer:

mestny [16]2 years ago

6 0

Answer:

Data that we know:

He drove 504km in each direction.

Going, the average speed was 14km/h more than returning.

The total trip lasted 21 hours.

Ok, to solve this first:

Let's define the variables:

Sg = Average speed when going.

Sr = Average speed when returning.

Then we can write one of our relationships as:

Sg = Sr + 14km/h.

Now, you can recall the relation:

Time = Distance/speed.

Then we can write the equation that represents the total time of the travel as:

504km/Sg + 504km/Sr = 21h

Now we can replace the Sg by Sr + 14km/h, and get:

504km/(Sr + 14km/h) + 504km/Sr = 21h.

Now we must multiplacate by each denominator, in order to remove them:

504km*Sr + 504km*(Sr + 14km/h) = 21h*Sr*(Sr + 14km/h)

Sr*1008km + 7056km^2/h = 21h*Sr^2 + 294km*Sr

Now we can write a cuadratic equation, where i will ignore the units so it is easier to read, as:

21*Sr^2 -714*Sr - 7056 = 0

The solutions are given by the Bhaskara formula:

$Sr = \frac{714 +- \sqrt{714^2 - 4*21*(-7056)} }{2*21} = \frac{714 +- 1050}{42}$

We have two solutions, one negative and one positive, and because Sr represents an average speed, it must be a positive number, then we choose the positive solution:

Sr = (714 + 1050)/42 km/h = 42km/h

Now we have the average speed for the returning, with this we can find Sg.

Sg = Sr + 14km/h = 42km/h + 14km/h = 56km/h.

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The time it takes to deliver a pizza (from time of phone call to delivery at door) follows a normal distribution with a mean (µ)

Yuki888 [10]

Answer:

99.87% of the store’s total delivery orders will be delivered to consumers with charge

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:

$\mu = 45, \sigma = 5$

If a pizza store’s policy is, "Orders delivered within one hour or they’re free!", what percentage of the store’s total delivery orders will be delivered to consumers with charge?

Within one hour, which is 60 minutes. So this is the pvalue of Z when X = 60.

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

$Z = \frac{60 - 45}{5}$