The answer is 13/25 13 is odd so it can't be evenly divided by anything so it is in its simplest form
Answer: 16
Step-by-step explanation:
8+6+2
Answer:
a) ![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
b) ![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
c) ![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
d) ![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
e) ![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Step-by-step explanation:
Let define some notation:
[L]= represent longitude , [T] =represent time
And we have defined:
s(t) a position function


Part a
If we do the dimensional analysis for v we got:
![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
Part b
For the acceleration we can use the result obtained from part a and we got:
![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
Part c
From definition if we do the integral of the velocity respect to t we got the position:

And the dimensional analysis for the position is:
![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
Part d
The integral for the acceleration respect to the time is the velocity:

And the dimensional analysis for the position is:
![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
Part e
If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:
![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
(13 - 5n) = 29
-5n = 29 - 13
-5n = 16
n = -16/5 or - 3 1/5
There are 9 marbles in the bag. We pick 2 without replacement and get a probability of 1/6.
Each draw of a marble has a probability associated with it. Multiplying these gives 1/6 so let us assume the probabilities are (1/3) and (1/2).
In order for the first draw to have a probability of 1/3 we need to draw a color that has (1/3)(9)=3 marbles. So let's say there are 3 red marbles. The P(a red marble is drawn) = 1/3.
Now that a marble has been drawn there are 8 marbles left. In order for the second draw to have a probability of 1/2 we must draw a color that has (1/2)(8) = 4 marbles. So let's say there are 4 blue marbles out of the 8.
Since there are 9 marbles to start and we have 3 red marbles and 4 blue marbles, the remaining 2 marbles must be a different color. Let us say they are green.
The problem is: There are 3 red marbles, 4 blue marbles and 2 green marbles in a jar. A marble is picked at random, it's color is noted and the marble is not replaced. A second marble is drawn at random and its color noted. What is the probability that the first marble is red and the second blue?