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Arturiano [62]
3 years ago
13

An appliance store reduced the price of a refrigerator by 20% and then raised the prices by 10% from the lower price. What was t

he original price of the refrigerator, if the price was $70.49 ?
Mathematics
1 answer:
Romashka [77]3 years ago
8 0

Answer: $80

Step-by-step explanation:

Let the original price of the refrigerator be represented by y.

We are told that an appliance store reduced the price of a refrigerator by 20%. This will be:

= y - (20% of y)

= y - (20/100 × y)

= y - (0.2 × y)

= y - 0.2y

= 0.8y

We are then told that the store then raised the prices by 10% from the lower price. This will be:

= 0.8y + (10% of 0.8y)

= 0.8y + (0.1 × 0.8y)

= 0.8y + 0.08y

= 0.88y

Since we told that now the price was $70.49, this means that we have to find the value of y using the above expression. This will be:

0.88y = $70.49

y = $70.49/0.88

y = $80.10

y = $80 approximately

The original price of the refrigerator is $80.

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The number of daily sales of a product was found to be given by S = 600xe−x2 + 600 x days after the start of an advertising camp
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Answer:

a. 20520

b. 12600

Step-by-step explanation:

Given the function S = 600xe^-x² + 600.

a. To find the average, we have to find the definite integral of the function because average, as it is known, is the sum of data points divided by the size of its dataset, this can be used for discrete data. Integral formula is just the continuous form of average, so we are using integral because we were given an interval of x= 0 to X = 30.

Let's integrate 100xe^x² first. Let –x² = u, this means –2xdx = du and we have dx = –du/2x. Also, when x = 0, u = –(0)² = –0 and when x = 30, u = –(30)² = –900. When we make our substitutions we have:

–600(xe^udu)/2x = –600(e^udu)/2 upon integrating that we have –600(e^u)/2. Applying our interval we have

–600[(e^900)/2 – (e^0)/2] ≈ – [– 3.7 – (1/2)] = –600 (–4.2) = 600 x 4.2 = 2520

Now let's integrate 600, with the interval x = 0 to x = 30 (we are using this interval here because the substitution we made didn't affect this).

We have, upon integrating:

600x.

Substituting our intervals we have:

600(30 – 0) = 600 x 30 = 18000.

Adding that up we have: 2520 + 18000 = 20520.

b. The same method is needed, just difference of interval.

Therefore, after integrating the first component with intervals u = 900 to u = 2500 (from x² = u) we have:

–600[(e^2500)/2 – (e^900)/2] ≈ –600[2.7 – 3.7] = –600(–1) = 600.

Then for the second component:

600x using x = 30 to x = 50

600(50 – 30) = 600 x 20 = 12000.

Adding that up we have:

12000 + 600 = 12600.

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