In order to solve this problem you have to add the distance Harriet traveled away from her starting point which is 2 1/4+3/4. Then you should end up with 3 miles. After that it says she went back on the same route so you have to double the distance. So in total she rode 6 miles.
Answer:
Median for a: 50 because it is in the middle of the points
Median for b: 35 because it is in the middle of the points.
Step-by-step explanation:
What i do is i write all the numbers out as shown in graph and cross one out on both sides untill there is one left.
IQR A: 15
IQR B: 15
Find the 25 percentile the. the 75 pecentile then subtract the numbers off from eachother.
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The reciprocal of 3/8 would be 8/3 (basically reversing the numbers).
When dividing fractions, you can just multiply the first fraction by the reciprocal of the second fraction instead of dividing by the second fraction.
3/4 ÷ 3/8
= 3/4 × 8/3
Simplifying gives us the final answer:
= 2
Let me know if you need any clarifications, thanks!
Answer:
y = 5/6x - 5.167
Or
y = 5/6x - 31/6
Step-by-step explanation:
The equation of a line
y = mx + c
Step 1: find the slope
m = y2 - y1 / x2 - x1
Given points
( 0 , -7) ( 5 , -1)
x1 = 0
y1 = -7
x2 = 5
y2 = -1
Insert the values into the equation
m = -1 - (-7) / 5 - 0
m = -1 + 7 / 5
m = 6/5
y = 6/5x + c
Step 2: sub any of the two points given into the equation
y = 5/6x + c
( 5 , -1)
x = 5
y = -1
-1 = 5/6(5) + c
-1 = 5*5/6 + c
-1 = 25/6 + c.
c = -1 - 25/6
LCM = 6
c = -6 - 25 / 6
c = -31/6
c = -5.167
Step 3: sub c into the equation
y = 5/6x + c
y = 5/6x - 31/6
Or
y = 5/6x - 5.167
The equation of the line is
y = 5/6x - 5.167
Answer:
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)
Step-by-step explanation:
For this case we can use the moment generating function for the normal model given by:
![\phi(t) = E[e^{tX}]](https://tex.z-dn.net/?f=%20%5Cphi%28t%29%20%3D%20E%5Be%5E%7BtX%7D%5D)
And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:
And we have that the moment generating function can be write like this:
And we can write this as an infinite series like this:
And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:
![E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%20E%5B1%2B%20tX%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28tX%29%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn%21%7D%28tX%29%5En%20%2B....%5D)
![E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%201%2B%20E%5BX%5Dt%20%2B%5Cfrac%7B1%7D%7B2%7DE%5BX%5E2%5Dt%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn1%7DE%5BX%5En%5D%20t%5En%2B...)
and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:
![\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%282k%29%21%7D%20E%5BX%5E%7B2k%7D%5D%20t%5E%7B2k%7D%3D%5Cfrac%7B1%7D%7Bk%21%7D%20%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%5Ek%20%3D%5Cfrac%7B1%7D%7B2%5Ek%20k%21%7D%20t%5E%7B2k%7D)
And then we have this:
![E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...](https://tex.z-dn.net/?f=E%5BX%5E%7B2k%7D%5D%3D%5Cfrac%7B%282k%29%21%7D%7B2%5Ek%20k%21%7D%2C%20k%3D0%2C1%2C2%2C...)
And then we can find the ![E[X^2]](https://tex.z-dn.net/?f=E%5BX%5E2%5D)
And we can find the variance like this :
![Var(X^2) = E[X^4]-[E(X^2)]^2](https://tex.z-dn.net/?f=Var%28X%5E2%29%20%3D%20E%5BX%5E4%5D-%5BE%28X%5E2%29%5D%5E2)
And first we find:
![E[X^4]= \frac{4!}{2^2 2!}= 3](https://tex.z-dn.net/?f=E%5BX%5E4%5D%3D%20%5Cfrac%7B4%21%7D%7B2%5E2%202%21%7D%3D%203)
And then the variance is given by:
