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umka2103 [35]
3 years ago
15

Who can help me out with this?

Mathematics
2 answers:
kirill115 [55]3 years ago
8 0
To solve this, I will use proportions:

The defected ones have a proportion of 1:400

So just make this proportion:

\frac{1}{400} =  \frac{x}{2000}

Once you solve it, you get a value of 5
Ne4ueva [31]3 years ago
8 0
Make a fraction 399/400 are not defected. How many are in 2000 not defected? We divide 2000 by 400 and the answer is 5. Then you multiply 5 x 399, which will give you 1995 as your answer.
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Express the given integral as the limit of a riemann sum but do not evaluate: the integral from 0 to 3 of the quantity x cubed m
WARRIOR [948]
We will use the right Riemann sum. We can break this integral in two parts.
\int_{0}^{3} (x^3-6x) dx=\int_{0}^{3} x^3 dx-6\int_{0}^{3} x dx
We take the interval and we divide it n times:
\Delta x=\frac{b-a}{n}=\frac{3}{n}
The area of the i-th rectangle in the right Riemann sum is:
A_i=\Delta xf(a+i\Delta x)=\Delta x f(i\Delta x)
For the first part of our integral we have:
A_i=\Delta x(i\Delta x)^3=(\Delta x)^4 i^3
For the second part we have:
A_i=-6\Delta x(i\Delta x)=-6(\Delta x)^2i
We can now put it all together:
\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\
\sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]
We can also write n-th partial sum:
S_n=(\frac{3}{n})^4\cdot \frac{(n^2+n)^2}{4} -6(\frac{3}{n})^2\cdot \frac{n^2+n}{2}

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Step-by-step explanation:

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HELP!!!
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Step-by-step explanation:

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