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sashaice [31]
3 years ago
7

Ned has a coupon for one fourth off the cost of any item. Today all items are on sale for one third off. Does it matter whether

the store applies the sale first or the coupon first? Explain.
Mathematics
1 answer:
Aleksandr [31]3 years ago
4 0

Answer:

Step-by-step explanation:

Let the cost of the item is C.

Case I :

first sale is applied

so, the price of item  = one third = C - C/3 = 2C/3

Now coupon is applied

so, the price of item = \frac{2C}{3}-\frac{1}{4}\times \frac{2C}{3} = C/2

case II :

First coupon is applied

so, the price of item = C - C/4 = 3C/4

Now sale is applied

the price of item = \frac{3C}{4}-\frac{1}{3}\times \frac{3C}{4} = C/2

So, it does not matter that either coupon is applied first or the sale is applied first the final cost of the item remains same.  

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3 years ago
what are the extraneous variables eliminated by randomly selecting schools into the experiment and control groups?
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6 0
1 year ago
The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and me
Dovator [93]

Answer:

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

Step-by-step explanation:

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 time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

There are 5 weekdays, with a mean of 0.1 calls per day.

The weekend is 2 days long, with a mean of 0.2 calls per day.

So:

\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286

If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?

This is P(X = 2). So:

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

P(X = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

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