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2 years ago

Find the percent of discount. Original price: $95 Sale price: $61.75

2 answers:

6 0

To find the percentage, we divide the sale price by the original price, and then multiply it by 100 to change the answer from a decimal to a percentage.

61.75 / 95 = 0.65
0.65 x 100 = 65%

This tells us that $61.75 is 65% of $95

To find the percentage decrease, we subtract this answer from 100%.

Answer: 100 - 65 = 35%

So the discount was 35%

Schach [20]2 years ago

4 0

1.538 i took the test

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A. The number of 10-boards Peter bought is equal to n divided by 10. Then, each of the 10-boxes will get two boxes of nails. The number of boxes of nails that Peter will have after buying n boards will be,

N = (2)(n/10)

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b. If the number of boards are 90 then,

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2 years ago

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2 years ago

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1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,

crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

$PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}$

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

$PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111$

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

$PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666$

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

$PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222$

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2 years ago

A firm has a revenue function that can be represented by r (x)=700x-0.35x^2, where r (x) is the total revenue (in dollars) and x

Fudgin [204]

Answer:

How many units need to be sold to produce the maximum revenue? 1000 units

How many in dollars is the maximum revenue when the maximum of units are sold? $350,000

Step-by-step explanation:

We get max value of a function if we differentiate it and set it equal to 0.

We need to differentiate r(x) and set it equal to 0 and solve for x.

That would be number of units sold to get max revenue.

Then we take that "x" value and substitute into r(x) to get the max revenue amount.

Before differentiating, we see the rules shown below:

$f(x)=ax^n\\f'(x)=n*ax^{n-1}$

Where

f'(x) is the differentiated function

Now, let's do the process:

$r (x)=700x-0.35x^2\\r(x)=700-2*0.35x\\r(x)=700-0.7x\\0=700-0.7x\\0.7x=700\\x=1000$

So, 1000 units need to be sold for max revenue

Now, substituting, we get:

$r (x)=700x-0.35x^2\\r(1000)=700(1000)-0.35(1000)^2\\r(1000)=350,000$

The max revenue amount is $350,000

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2 years ago