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liberstina [14]
3 years ago
14

Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected.

How many of the n values can be freely assigned before the remaining values are​ determined? (The result is referred to as the number of degrees of​ freedom.)
Mathematics
1 answer:
Leno4ka [110]3 years ago
6 0

As the problem indicates, degrees of freedom are the number of values ​​that can be independently selected before it is necessary to choose specific values ​​to arrive at the desired result.

The average, on the other hand, results from the sum of a list of values ​​divided by the amount of values ​​in the summed list.

Assume that the mean sought is x and consider that the list is composed of a single element a, in that case no random number can be selected, since the mean x must correspond to that number.

If the list were composed of two elements a and b, one of the two values ​​could be chosen randomly, and according to the chosen value the second should be the one whose sum with the previous one results in 2x, this given that the formula of the average \sum\limits_{i=1}^n \frac{ x_{i}}{n}.

With three values ​​a, b and c, it is possible to select two freely, since the thirteen must be the one that balances the sum of a+b, that is (a + b) + c = 3x.

Thus, in general, with n values, it is possible to select n-1 values ​​freely whose sum must be balanced by the last value so that the whole sum is nx.

Answer

In a list of \bf{n} values ​​you can assign \bf{n-1} values ​​freely, that is, you have n-1<em>degrees of freedom.</em>

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Part 3) (8, 4); The vertex represents the maximum profit

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Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

2x^{2} -12x+1=0

Convert to vertex form

step 1  

Factor the leading coefficient and complete the square

2(x^{2} -6x)+1=0

2(x^{2} -6x+9)+1-18=0

step 2

2(x^{2} -6x+9)+1-18=0

2(x^{2} -6x+9)-17=0

step 3

Rewrite as perfect squares

2(x-3)^{2}-17=0

Part 3) we have

f(x)=-x^{2}+16x-60

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

f(x)+60=-x^{2}+16x

Factor the leading coefficient

f(x)+60=-(x^{2}-16x)

Complete the squares

f(x)+60-64=-(x^{2}-16x+64)

f(x)-4=-(x^{2}-16x+64)

Rewrite as perfect squares

f(x)-4=-(x-8)^{2}

f(x)=-(x-8)^{2}+4

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

-2(x-2)^{2}+5=0

-2(x-2)^{2}=-5

(x-2)^{2}=2.5

square root both sides

(x-2)=(+/-)1.58

x=2(+/-)1.58

x=2(+)1.58=3.58

x=2(-)1.58=0.42

Part 5) we have

f(x)=-x^{2}+50x-264

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

so

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

-x^{2}+50x-264=0

-x^{2}+50x=264

Factor -1 of the leading coefficient

-(x^{2}-50x)=264

Complete the squares

-(x^{2}-50x+625)=264-625

-(x^{2}-50x+625)=-361

(x^{2}-50x+625)=361

Rewrite as perfect squares

(x-25)^{2}=361

square root both sides

(x-25)=(+/-)19

x=25(+/-)19

x=25(+)19=44

x=25(-)19=6

Part 6) we have

-2x^{2}+28x+20

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

-2(x^{2}-14x)+20

Complete the square

-2(x^{2}-14x+49)+20+98

-2(x^{2}-14x+49)+118

Rewrite as perfect square

-2(x-7)^{2}+118

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

Part 7) we have

f(x)=-x^{2}+8x-10

Convert to vertex form

f(x)+10=-x^{2}+8x

Factor -1 the leading coefficient

f(x)+10=-(x^{2}-8x)

Complete the square

f(x)+10-16=-(x^{2}-8x+16)

f(x)-6=-(x^{2}-8x+16)

Rewrite as perfect square

f(x)-6=-(x-4)^{2}

f(x)=-(x-4)^{2}+6

The vertex is the point (4,6)

therefore

The maximum height of the puck is 4 feet.

Part 8) we have

x^{2}+6x+5

Convert to vertex form

Group terms

(x^{2}+6x)+5

Complete the square

(x^{2}+6x+9)+5-9

(x^{2}+6x+9)-4

Rewrite as perfect squares

(x+3)^{2}-4

Part 9) we have

2x^{2}-4x-2=0

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

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Complete the square

2(x^{2}-2x+1)-2-2=0

2(x^{2}-2x+1)-4=0

Rewrite as perfect squares

2(x-1)^{2}-4=0

2(x-1)^{2}=4

The vertex is the point (1,-4)

Part 10) we have

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This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

8(x^{2}-8x)+720

Complete the square

8(x^{2}-8x+16)+720-128

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Rewrite as perfect squares    

8(x-4)^{2}+592

the vertex is the point (4,592)

The population has a minimum at x=4 years ( that is after 4 years since 1998 )

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