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slavikrds [6]
3 years ago
13

Use the ratio table to determine how many people 13 subs would serve. Explain look at the photo

Mathematics
2 answers:
ss7ja [257]3 years ago
8 0

Answer:

13 subs will serve 52 people.    

Step-by-step explanation:

We are given the following information in the question:

Number of subs:     3       5        8

 People served:      12     20     32

We have to find that how many people with be served by 13 subs.

We can solve this question with the help of ratio.

\text{Ratio} = \displaystyle\frac{\text{Number of subs}}{\text{People served}}\\\\= \frac{3}{12} = \frac{1}{4}

The obtained ratio is 1:4, which means that 1 sub can serve 4 peoples.

Let x be the number of people served by 13 subs:

\displaystyle\frac{1}{4} = \frac{13}{x}\\\\\Rightarrow x = \frac{13\times 4}{1} = 52

Hence, 13 subs will serve 52 people.

adoni [48]3 years ago
5 0
I think its 44. The amount of people increase by 12 every time you add more subs. Hope that helps!
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A Riemann Sum is often specified in terms of the overall interval of "integration," the number of divisions of that interval to use, and the method of combining function values.

<u>Example Problem</u>

For the example attached, we are finding the area under the sine curve on the interval [1, 4] using 6 subintervals. We are using a rectangle whose height matches the function at the left side of the rectangle. We say this is a <em>left sum</em>.

When rectangles are used, other choices often seen are <em>right sum</em>, or <em>midpoint sum</em> (where the midpoint of the rectangle matches the function value at that point).

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The second and third attachments show a <em>right sum</em> (r₂) and a <em>midpoint sum</em> (r₃). The latter is the best of these approximations.

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<u>Other Rules</u>

Described above and shown in the graphics are the use of <em>rectangles</em> for elements of the summation. Another choice is the use of <em>trapezoids</em>. For this, the corners of the trapezoid match the function value on both the left and right edges of the subinterval.

Suppose the n subinterval boundaries are at x0, x1, x2, ..., xn, so that the function values at those boundaries are f(x0), f(x1), f(x2), ..., f(xn). Using trapezoids, the area of the first trapezoid would be ...

  a1 = (f(x0) +f(x1))/2·∆x . . . . where ∆x is the subinterval width

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We can see that in computing these two terms, we have evaluated f(x1) twice. We also see that f(x1)/2 contributes twice to the overall sum.

If we collapse the sum a1+a2+...+an, we find it is ...

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That is, each function value except the first and last contributes fully to the sum. When we compute the sum this way, we say we are using the <em>trapezoidal rule</em>.

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<em>Comment on mechanics</em>

As you can tell from the attachments, it is convenient to let a graphing calculator or spreadsheet compute the sum. If you need to see the interval boundaries and the function values, a spreadsheet may be preferred.

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Step-by-step explanation:

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