Let 
denote the value on the 
-th drawn ball. We want to find the expectation of 
, which by linearity of expectation is
![E[S]=E\left[\displaystyle\sum_{i=1}^5B_i\right]=\sum_{i=1}^5E[B_i]](https://tex.z-dn.net/?f=E%5BS%5D%3DE%5Cleft%5B%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E5B_i%5Cright%5D%3D%5Csum_%7Bi%3D1%7D%5E5E%5BB_i%5D)
(which is true regardless of whether the 
are independent!)
At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.
and so
![E[X_i]=\displaystyle\sum_{i=1}^{10}x\,P(X_i=x)=\frac1{10}\frac{10(10+1)}2=5.5](https://tex.z-dn.net/?f=E%5BX_i%5D%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7B10%7Dx%5C%2CP%28X_i%3Dx%29%3D%5Cfrac1%7B10%7D%5Cfrac%7B10%2810%2B1%29%7D2%3D5.5)
Then the expected value of the total is
![E[S]=5(5.5)=\boxed{27.5}](https://tex.z-dn.net/?f=E%5BS%5D%3D5%285.5%29%3D%5Cboxed%7B27.5%7D)
-6r+5
the two like terms are the ones with R. combine -4r and -2r to get -6r
Answer:
(-3,10) (0,5) (3,0) (6,-5)
Step-by-step explanation:
x intercept = 3
y intercept =5
If brainiest is earned its greatly appreciated
Answer: 0.87400mg of caffeine.
Step-by-step explanation:
You have
N(t)=N0(e^−rt)(1)
as a general Exponential decay equation where N0 is the amount at t=0, N(t) is the amount remaining at time t and r is the exponential decay constant. You're specifically given that after 10 hours, the decay factor is 0.2601, i.e.,
N(10)/N(0)=N0(e^−10r)/N0(e^0)= e^−10r=0.2601 . .(2)
Taking the last 2 parts of (2) to the power of 0.1t gives
e^−rt=0.2601^.1t . .(3)
This means that
N(t)=N0(e^−rt)=N0(0.2601^.1t). .(4)
Also,
N(2.56)N(1.56)=N0(0.2601.1(2.56))N0(0.2601.1(1.56))=0.2601.1(2.56−1.56)=0.2601^.1
= 0.87400mg of caffeine.
$1.00 because the first ounce is 0.40 then you have to times the other 3 ounce by 0.20 then u would have $1.00
