Answer:
<h3>{44-55}</h3><h3>{52- 59 } are the last two interval s</h3>
Step-by-step explanation:
<h2>To get the class boundaries deduct 0.5 from the first number and add 0.5 to the second number = 4-0.5 = 3.5 </h2><h2> = 11.5 + 0.5 = 11.5 </h2><h2> = 3.5 -11.5 as in the fish class boundary .</h2><h2>same for all other boundaries</h2>
Let's define the vectors:
U = (4.4)
V = (3.1)
The projection of U into V is proportional to V
The way to calculate it is the following:
Proy v U = [(U.V) / | V | ^ 2] V
Where U.V is the point product of the vectors, | V | ^ 2 is the magnitude of the vector V squared and all that operation by V which is the vector.
We have then:
U.V Product:
U.V = (4,4) * (3,1)
U.V = 4 * 3 + 4 * 1
U.V = 12 + 4
U.V = 16
Magnitude of vector V:
lVl = root ((3) ^ 2 + (1) ^ 2)
lVl = root (9 + 1)
lVl = root (10)
Substituting in the formula we have:
Proy v U = [(16) / (root (10)) ^ 2] (3, 1)
Proy v U = [16/10] (3, 1)
Proy v U = [1.6] (3, 1)
Proy v U = [1.6] (3, 1)
Proy v U = (4.8, 1.6)
Answer:
the projection of (4,4) onto (3,1) is:
Proy v U = (4.8, 1.6)
To find the slope, you must this formula: (Y2-Y1)/(X1-X2)
Now let's input the numbers in the formula.
(3-0)/(3-0)=3/3
Now simply the slope if possible.
3/3=1/1=1
So, the slope is equal to 1 or 1/1
<h2>
Answer:</h2><h2>-16 -6.</h2><h2 /><h2>
-2(5+3)(-2+4)</h2><h2>
-2(8) (-6)</h2><h2>
Then multiply the one's outside with the one's in the bracket.</h2><h2>
Making it...</h2><h2>
-16 -6</h2>
I believe it’s the second one
