I’ll help you out, but what are the numbers in between? Like does it go boy 0, .3, .6, .9, to 1? Or is there a different pattern
So she has 400 and 500 so far.
now if she were to have an average of 2300 for the next 5 years, that means some years she made more, some she made less, but in the 5 years run, she made 2300 * 5 or 11500 total.
Neverminding the lows and highs, if we bundle up the 11500 bucks and divide by 5, that'd 2300, which is what's called the average, the average amount doesn't take into account that some numbers are higher than others, it kinda just flattens them out.
so in total she would have made 11500 in each account.
so in the account with 400, she needs 11100 more, how much will it be 1100 for the next three years on average? 11100/3 or 3700.
so if she earns 3700 each year, for the next 3 years in the account with 400 already, she'd have 11500 bucks in 5 years with an average of 2300.
now let's look at the 500 account, she needs 11000 more, how much will it be for 11000 for the next three years on average? 11000/3 or about 3666.67.
so if she earns 3666.67 each year, for the next 3 years, she'd have 11500 bucks in 5 yeas with an average of 2300.
Answer:
a. 1.75
Step-by-step explanation:
r values, or the correlation coefficient must be between -1 and 1, inclusive
-1 is a perfect negative correlation
0 is no correlation
+1 is a perfect positive correlation
1.75 is not between -1 and 1
Answer:
√3 is irrational
Step-by-step explanation:
The location of the third point of a triangle can be found using a rotation matrix to transform the coordinates of the given points.
<h3 /><h3>Location of point C</h3>
With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)
<h3>Equilateral triangle</h3>
The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.
The math of it is this. For rotation of (x, y) counterclockwise 60° about the origin, the transformation matrix is ...
![\left[\begin{array}{cc}\cos(60^\circ)&\sin(60^\circ)\\-\sin(60^\circ)&\cos(60^\circ)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}x'\\y'\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ccos%2860%5E%5Ccirc%29%26%5Csin%2860%5E%5Ccirc%29%5C%5C-%5Csin%2860%5E%5Ccirc%29%26%5Ccos%2860%5E%5Ccirc%29%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%27%5C%5Cy%27%5Cend%7Barray%7D%5Cright%5D)
Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...
cos(60°)·x + sin(60°)·y
will be irrational.
As in the attached diagram, if one of the coordinates of the rotated point (B) is zero, then one of the coordinates of its image (C) will be rational. The other image point coordinate cannot be rational.