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Oksana_A [137]
3 years ago
5

A. Create a set of 5 points that are very close together and record the standard deviation. Next, add a sixth point that is far

away from the original 5 and record the new standard deviation. What is the impact of the new point on the standard deviation?
Mathematics
1 answer:
defon3 years ago
7 0

Answer: The addition of the new point alters the previous standard deviation greatly

Step-by-step explanation:

Let the initial five points be : 2 3 4 5 and 6. In order to calculate the standard deviation for this data, we will need to calculate the mean first.

Mean = summation of scores/number of scores.

The mean is therefore: (2+3+4+5+6)/5 = 20/5 = 4.

We'll also need the sum of the squares of the deviations of the mean from all the scores.

Since mean = 4, deviation of the mean from the score "2" = score(2) - mean (4)

For score 3, it is -1

For 4, it's 0

For 5 it's 1

For 6 it's 2.

The squares for -2, -1, 0, 1, and 2 respectively will be 4, 1 , 0, 1, 4. Summing them up we have 10 i.e (4+1+0+1+4=10).

Calculating the standard deviation, we apply the formula:

√(summation of (x - deviation of mean)^2)/N

Where N means the number of scores.

The standard deviation = √(10/5) = 1.4142

If we add another score or point that is far away from the original points, say 40, what happens to the standard deviation. Let's calculate to find out.

i.e we now have scores: 2, 3, 4, 5, 6 and 40

We calculate by undergoing same steps.

Firstly mean. The new mean = (2+3+4+5+6+40)/6 = 60/6 = 10.

The mean deviations for the scores : 2, 3, 4, 5, 6 and 40 are -8, -7, -6, -5, -4 and 30 respectively. The squares of these deviations are also 64, 49, 36, 25, 16 and 900 respectively as well. Their sum will then be 1090. i.e. (64+49+36+25+16+900 = 1090).

The new standard deviation is then=

√(1090/6)

= √181.67

= 13.478.

It's clear that the addition of a point that's far away from the original points greatly alters the size of the standard deviation as seen /witnessed in this particular instance where the standard deviation rises from 1.412 to 13.478

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olasank [31]

Answer: 1/30

Step-by-step explanation:

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∫[0,4] 4u cos(u) du

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v = ∫[1,e] π(4-(lnx + 1)^2) dx

Using shells dy

v = ∫[0,1] 2πrh dy

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v = ∫[0,1] 2π(y+1)(e^y-1) dy

v = ∫[0,1] (x-x^2)^2 dx = 1/30

3 0
3 years ago
would someone be able to explain to me how to solve these problems? I don't need the solutions, I just need an explanation. I ne
Leni [432]
The first 3  are examples of the difference of 2 squares so you use the identity 
a^2 - b^2 = (a + b)(a - b)
x^2 - 49 = 0 
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so either x + 7 = 0 or x - 7 = 0
giving x = -7  and 7.
Number 7 reduces to 3x^2 =12, x^2  = 4 so x = +/- 2
Number 8  take out GCf (d) to give 
d(d - 2)  = 0   so d = 0 ,  2
9 and 10 are more difficult to factor
you use the 'ac' method  Google it to get more details
2x^2 - 5x + 2
multiply first coefficient by the constant at the end 
that is 2 * 2 =  4
Now we want  2 numbers  which when multiplied give + 4 and when added give - 5:-    -1 and -4  seem promising so we write the equation as:-

2x^2 - 4x - x + 2 = 0

now factor by grouping 
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(x - 2) is common so

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The last example is solved in the same way.


8 0
3 years ago
Honda Atlas issued 100,000 shares of stock. In 1990, each share of stock was worth Rs. 152.50. In 1993, each share of the stock
noname [10]

Answer:

Rs.13,130 ,000

Step-by-step explanation:

Lets solve first for year 1990

Value of stock in 1990 = Rs. 152.50

No. of stocks issued in 1990 = 100,000

Total value of these stocks in 1990= Value of stock in 1990* no of stocks issued in 1990  = 152.50*100,000 = Rs. 15,250,000

Let solve first for year 1993

Value of stock in 1993 = Rs. 21.20

No. of stocks   1993 = 100,000

Total value of these stocks in 1993 = Value of stock in 1993* no of stocks issued in 1993  = 21.20*100,000 = Rs. 2,120,000

Decrease in value of stock in 1993 in comparison to 1990 = Total value of the stocks in 1990 -Total value of the stocks in 1993

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 = Rs.13,130 ,000

Thus, the 100,000 shares worth Rs.13,130 ,000 lesser in 1993 than in 1990.

6 0
3 years ago
Trapezoid QRST has two right angles. A 5-in.altitude can be drawn dividing QRDT into a rectangle and an isosceles right triangle
Vika [28.1K]

Answer:

The area of the trapezoid is 57.5 square inches

Step-by-step explanation:

we know that

The trapezoid QRST can be divided into a rectangle QRDT and an isosceles right triangle RSD

see the attached figure to better understand the problem

step 1

The area of rectangle is given by the formula

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we have

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QR=9\ in

substitute

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step 2

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we have

RD=DS=5\ in ---> because is an isosceles triangle

substitute

A=\frac{1}{2}(5)(5)=12.5\ in^2

step 3

Adds the areas

A=45+12.5=57.5\ in^2

3 0
4 years ago
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Thus,
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5/6 work in = 1 hour
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3 0
3 years ago
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