In a circle, what does the ratio A/r^2 represent?

1 answer:

5 0

The ratio A/r² would represent $\pi$ which is the circumference of a circle divided by its diameter.

You can see this because the area of a circle is $\pi$ r² and if you divided that by the radius squared, you are given pi which is the circumference of a circle divided by its diameter.

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QUICKK

VLD [36.1K]

3x + 4(2) = 20
3x + 8 = 20
3x = 12
x = 4
(4,2)
You have y=2 then all you need to do is plug 2 into the y in the equation 3x + 4y = 20 and do the math to find x. Hope this helps

7 0

3 years ago

The difference of two numbers is 1. What is the smallest possible value for the sum of their squares?

Stella [2.4K]

X= 1 number
Y= a number that is 1 less than x
Y= x-1

The sum of squares=

X^2 + (x-1)^2 = x^2 + x^2 - 2x + 1
X^2 + x^2 - 2x + 1 = 2x^2 - 2x + 1

Assuming that integers = numbers. We could try some values

When x = 0, the sum of squares = 1
When x = -1, the sum of squares = 4
When x = 1, the sum of squares = 1

If you put this into a graph, it is a parabola and we see that the minimum value hits at x = 1/2.

To check this value:

2x^2 - 2x + 1 = 2·(1/4) - 2·(1/2) + 1
2·(1/4) - 2·(1/2) + 1 = 1/2 - 1 + 1
1/2 - 1 + 1 = 1/2

When you look back at the question, it asks for 2 answers

X= 1/2
X-1 = -1/2

What is the sum of their squares?

1/4 + 1/4 = 1/2

That fits with graph too.

Answers: 1/2 and -1/2

Hope this helps!!

5 0

4 years ago

Each one of four people will request either Coke D Let X denote the number that request Coke, where the remainder request Sprite

34kurt

Answer:

a) E(x)=2.5 V(x)=1.59

b) E(R)=17 V(R)=6.36

c) The probability that, when the four make their orders, everyone gets what they ordered is P=0.66.

Step-by-step explanation:

a) The mean E(x) can be calculated as:

$E(x)=\sum^4_{i=0} p_ix_i\\\\E(x)=0.09*0+0.15*1+0.18*2+0.33*3+0.25*4\\\\E(x)=0.00+0.15+0.36+0.99+1.00\\\\E(x)=2.5$

The variance V(x) can be calculated as:

$V(x)=\sum^4_{i=0} p_i(x_i-\bar x)^2\\\\V(x)=0.09(0-2.5)^2+0.15(1-2.5)^2+0.18(2-2.5)^2+0.33(3-2.5)^2+0.25(3-2.5)^2\\\\V(x)=0.09*6.25+0.15*2.25+0.18*0.25+0.33*0.25+0.25*2.25\\\\V(x)=0.5625+0.3375+0.045+0.0825+0.5625\\\\V(x)=1.59$

b) The revenue can be expressed as:

$R=5*x+3*(4-x)$

Then, the expected value of R canbe expressed in function of x:

$E(R)=E(5x+3(4-x))=5E(x)+3(4-E(x))\\\\E(R)=5*2.5+3(4-2.5)=12.5+4.5=17$

The expected revenue is $17.

The variance of R is

$V(R)=V(5*x+3(4-x))=V(5x+12-3x)=V(2x+12)\\\\V(R)=V(2x)+V(12)=2^2V(x)+0=4*1.59\\\\V(R)=6.36$

c) In the case there are only 3 Cokes and 3 Sprites available, the only orders that can not be fullfilled are when X=0 (they order 4 Sprites) and X=4 (they order 4 Cokes).

The probability of these events is:

$P(x=0,x=4)=P(x=0)+P(x=4)=0.09+0.25=0.34$