solution:
P(passed)=P(Dr. J and passed)+P(Dr. S and passed) =0.4*0.8+(1-0.4)*0.6=0.68
hence probability that she was enrolled in Dr. J's section given passed
=P(Dr. J and passed)/P(passed)=0.4x0.8/0.68=0.4706
Basically, all you have to do is substitue the terms given at the top.
x = -3
y = 6
z = -4
<em>The first one will be set up like this:</em>
(-3) + 6 + (-1) = 2
<em>The second one will be set up like this: </em>
-15 + (- -3) + 6 = -6
<em>The third one will be set up like this:</em>
3(-3) - 2(6) + (-4) = -25
<em>The fourth one will be set up like this:</em>
2(-4) - 3 (6) + 4(-3) = -38
<em>The fifth one will be set up like this: </em>
3 ( -3) + 2 (6) + (-4) = -1
<em>The sixth will be set up like this: </em>
(-3)(-4)
---------- = -1
-2 (6)
<em>The seventh will be set up like this:</em>
2 (6) (-4) - 3 (-3) = -39
Hope this helps! Have a good day/night! c:
Answer:
This graph has been translated. Translation is when the image on the graph has been shifted along one or both of the axis. In this case, the graph has been shifted up two units as shown by the operation of addition on the y-value. Remember that translations are rigid movements, so the new image will be congruent to the pre-image.
So, since 100% = 128, x should equal 96, and we are trying to find x.
Multiply both sides of the equation by x.
(100/x) * x = (128/96) * x
Cancel out the x's on the left side so
100 = 1.333(x)
75 = x
96 is 75% of 128
Answer:
in my ipinion...
Step-by-step explanation:
let a no.be x
1/3*x+77=33
1/3*x=-44
x=-44*3
x=-132
