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Orlov [11]
3 years ago
6

How do you use a strategy of 30%of 900?

Mathematics
1 answer:
Julli [10]3 years ago
3 0

Answer: 2 7 0

Step-by-step explanation:

1. We assume, that the number 900 is 100% - because it's the output value of the task.

2. We assume, that x is the value we are looking for.

3. If 900 is 100%, so we can write it down as 900=100%.

4. We know, that x is 30% of the output value, so we can write it down as x=30%.

5. Now we have two simple equations:

1) 900=100%

2) x=30%

where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:

900/x=100%/30%

6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for what is 30% of 900

900/x=100/30

(900/x)*x=(100/30)*x       - we multiply both sides of the equation by x

900=3.33333333333*x       - we divide both sides of the equation by (3.33333333333) to get x

900/3.33333333333=x

270=x

x=270

now we have:

30% of 900=270

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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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