Answer:
The second number after the decimal point- the 6
Hello!
When finding the perimeter of a rectangle, you have to consider the properties of a rectangle. A rectangle has two pairs of equal sides where one is the width, while the other one is the length.
Now looking back at your question, it says "... a rectangle that is x units wide" ⇒ you let the width = x ; this is the same with "y units long" ⇒ length = y. Perimeter can just be : P = 24.
Therefore,
The equation would be:
x + x + y + y = P.
2x + 2y = P.
(Sub in P = 24)
∴ 2x + 2y = 24. (This should be your answer.)
:) Good luck (Message me if you have any problem)
Just put the numbers under one square root symbol and try to simplify.
Answer:
your finding the absolute value, I I this means absolute value which means if its negative it automatically is its own number you get rid of the negative sign if it's positive just keep it because it's already in it's normal number.
For example: I -4 I = 4
For example: I -5 I = 5
For example: I 3/4 I = 3/4
For example: I 3 I = 3
For example: I 4 I = 4
Answer:
(a) Shown below
(b) There is a positive relation between the number of assemblers and production.
(c) The correlation coefficient is 0.9272.
Step-by-step explanation:
Let <em>X</em> = number of assemblers and <em>Y</em> = number of units produced in an hour.
(a)
Consider the scatter plot below.
(b)
Based on the scatter plot it can be concluded that there is a positive relationship between the variables <em>X</em> and <em>Y</em>, i.e. as the value of <em>X</em> increases <em>Y</em> also increases.
(c)
The formula to compute the correlation coefficient is:
![r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^{2}-(\sum X)^{2}][n\sum Y^{2}-(\sum Y)^{2}]}} }](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bn%5Csum%20XY-%5Csum%20X%5Csum%20Y%7D%7B%5Csqrt%7B%5Bn%5Csum%20X%5E%7B2%7D-%28%5Csum%20X%29%5E%7B2%7D%5D%5Bn%5Csum%20Y%5E%7B2%7D-%28%5Csum%20Y%29%5E%7B2%7D%5D%7D%7D%20%7D)
Compute the correlation coefficient between <em>X</em> and <em>Y</em> as follows:
![r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^{2}-(\sum X)^{2}][n\sum Y^{2}-(\sum Y)^{2}]}} }=\frac{(5\times430)-(15\times120)}{\sqrt{[(5\times55)-15^{2}][(5\times3450)-120^{2}]}} =0.9272](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bn%5Csum%20XY-%5Csum%20X%5Csum%20Y%7D%7B%5Csqrt%7B%5Bn%5Csum%20X%5E%7B2%7D-%28%5Csum%20X%29%5E%7B2%7D%5D%5Bn%5Csum%20Y%5E%7B2%7D-%28%5Csum%20Y%29%5E%7B2%7D%5D%7D%7D%20%7D%3D%5Cfrac%7B%285%5Ctimes430%29-%2815%5Ctimes120%29%7D%7B%5Csqrt%7B%5B%285%5Ctimes55%29-15%5E%7B2%7D%5D%5B%285%5Ctimes3450%29-120%5E%7B2%7D%5D%7D%7D%20%3D0.9272)
Thus, the correlation coefficient is 0.9272.
