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dybincka [34]
3 years ago
8

What is the standard deviation of the data set?. 14 34 23 50 25 16 48. . a. about 35.6. . b. about 26.8. . c. about 14.5. . d. a

bout 13.4. my answer: c. about 14.5. or is it d?? idk please help i got confused... on the scientific calculator, is Sx the symbol for standard deviation? or is it something else?.
Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0
The standard deviation of the data set is D. ABOUT 13.4

14, 34, 23, 50, 25, 16, 48

Arrange in ascending order: 14, 16, 23, 25, 34, 48, 50

mean = (14 + 16 + 23 + 25 + 34 + 48 + 50) / 7 = 210/7 = 30

|14 - 30| = 16 ⇒ 16² = 256
|16 - 30| = 14 ⇒ 14² = 196
|23 - 30| = 7 ⇒ 7² = 49
|25 - 30| = 5 ⇒ 5² = 25
|34 - 30| = 4 ⇒ 4² = 16
|48 - 30| = 18 ⇒ 18² = 324
|50 - 30| = 20 ⇒ 20² = 400

(256 + 196 + 49 + 25 + 16 + 324 + 400) / 7 = 1,266/7 = 180.85  variance
√180.85 = 13.45 standard deviation.

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Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the
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Summation notation:

\frac{1}{2}\sum_{k=1}^6f((.5k))

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\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

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We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have  a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

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The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)

or by factoring out the 1/2 part:

\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

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1

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-2.5^2+2(2.5)+4

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-1.5^2+2(1.5)+4

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-1^2+2(1)+4

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To find f(.5) replace x in -x^2+2x+4 with .5:

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-.25+1+4

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\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))

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I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

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