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3 years ago

Apply the distributive property to create an equivalent expression.

Mathematics

1 answer:

weqwewe [10]3 years ago

6 0

Step-by-step explanation:

$\text{The distributive property:}\ a(b+c)=ab+ac\\\\\dfrac{1}{2}(10x+20y+10z)=\dfrac{1}{2\!\!\!\!\diagup_1}\cdot10\!\!\!\!\!\diagup^5x+\dfrac{1}{2\!\!\!\!\diagup_1}\cdot20\!\!\!\!\!\diagup^{10}y+\dfrac{1}{2\!\!\!\!\diagup_1}\cdot10\!\!\!\!\!\diagup^5z\\\\=5x+10y+5z$

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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.

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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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3 years ago

Reposting because nobody will help me..

Llana [10]

Answer:

Step-by-step explanation:

18 = 2 * 3 * 3 = 2 *3²

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Hint:

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2 years ago

There were 78 girls trying out for dance. They were placed into groups containing d dancers. How many groups were there, in term

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3 years ago

A piece of wire of length 6363 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the

Lerok [7]

Answer:

35.2792 cm from one end (The square)

And 27.7208 cm from the other end (The circle)

b. See (b) explanation below

Step-by-step explanation:

Given

Length of Wire ,= 63cm

Let L be the length of one side of the square

Circumference of a circle = 2πr

Perimeter of a square = 4L

a. To minimise

4L + 2πr = 63 ----- make r the subject of formula

2πr = 63 - 4L

r = (63 - 4L)/2π

r = (31.5 - 2L)/π

Let X = Area of the Square. + Area of the circle

X = L² + πr²

Substitute (31.5 - 2L)/π for r

So,

X² = L² + π((31.5 - 2L)/π)²

X² = L² + π(31.5 - 2L)²/π²

X² = L² + (31.5 - 2L)²/π

X² = L² + (992.25 - 126L + 4L²)/π

X² = L² + 992.25/π - 126L/π +4L²/π ------ Collect Like Terms

X² = 992.25/π - 126L/π + 4L²/π + L²

X² = 992.25/π - 126L/π (4/π + 1)L² ---- Arrange in descending order of power

X² = (4/π + 1)L² - 126L/π + 992.25/π

The coefficient of L² is positive so this represents a parabola that opens upward, so its vertex will be at a minimum

To find the x-cordinate of the vertex, we use the vertex formula

i.e

L = -b/2a

L = - (-126/π) / (2 * (4/π + 1)

L = (126/π) / ( 2 * (4 + π)/π)

L = (126/π) /( (8 + 2π)/π)

L = 126/π * π/(8 + 2π)

L = (126)/(8 + 2π)

L = 63/(4 + π)

So, for the minimum area, the side of a square will be 63/(4 + π)

= 8.8198 cm ---- Approximated

We will need to cut the wire at 4 times the side of the square. (i.e. the four sides of the square)

I.e.

4 * (63/(4 + π)) cm

35.2792 cm from one end.

Subtract this result from 63, we'll get the other end.

i.e. 63 - 35.2792

= 27.7208 cm from the other end

b. To maximize

Now for the maximum area.

The problem is only defined for 0 ≤ L ≤ 63/4 which gives

0 ≤ L ≤ 15.75

When L=0, the square shrinks to 0 and the whole 63 cm wire is made into a circle.

Similarly, when L =15.75 cm, the whole 63 cm wire is made into a square, the circle shrinks to 0.

Since the parabola opens upward, the maximum value is at one endpoint of the interval, either when

L=0 or when L = 15.75.

It is well known that if a piece of wire is bent into a circle or a square, the circle will have more area, so we will assume that the maximum area would be when we "cut" the wire 0, or no, centimeters from the

end, and bend the whole wire into a circle. That is we don't cut the wire at

all.

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