<span>During the distribution of bonbons to her grandchildren, Grandma Claudia realized that if she distributed 11
chocolates for each of them, 10 candies would remain. In the meantime, distribute 13 chocolates
each of his grandchildren, would lack 6.
Based on this information, it is CORRECT to state that the difference between the number of candies
and the number of grandchildren is
A) 90.
B) 30.
C) -90.
D) -30.
Let n=number of candies,
and c=number of grandchildren
The story expressed in modulo gives the two following equations:
then
mod(n,11)=10 (ten left if each given 11).............(1)
mod(n,13)=-6, or
mod(n,13)=7 (7 left if each given 13)
By enumerating (1), we have
10,21,32,43,54,65,76,87,98,109...
=> 8*11+10=98
By enumerating (2), we have
7,20,33,46,59,72,85,98,101...
=> 8*13-6=98
Hence there are 8 grandchildren and 98 candies.
The difference of the number of candies and the number of grandchildren is therefore 98-8=90.
Another way to get the number of grandchildren is
c=(10-(-6))/(13-11)=8
</span>
Answer:
The expected cost is 152
Step-by-step explanation:
Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y
if 
and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)
Recall that, by definition
Also, we are given that 
. Recall the following properties of the expected value. If X,Y are random variables, then
Then, using this property we have that ![E[C] = 50+3E[Y]+ 9E[Y^2]](https://tex.z-dn.net/?f=E%5BC%5D%20%3D%2050%2B3E%5BY%5D%2B%209E%5BY%5E2%5D)
.
Thus, we must calculate E[Y] and E[Y^2].
Using the definition, we get that
![E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%20%5Cint_%7B1%7D%5E%7B5%7D%5Cfrac%7By%7D%7B4%7D%20dy%20%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft%5Cfrac%7By%5E2%7D%7B2%7D%5Cright%7C_%7B1%7D%5E%7B5%7D%20%3D%20%5Cfrac%7B25%7D%7B8%7D-%5Cfrac%7B1%7D%7B8%7D%20%3D%203)
![E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}](https://tex.z-dn.net/?f=E%5BY%5E2%5D%20%3D%20%5Cint_%7B1%7D%5E%7B5%7D%5Cfrac%7By%5E2%7D%7B4%7D%20dy%20%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft%5Cfrac%7By%5E3%7D%7B3%7D%5Cright%7C_%7B1%7D%5E%7B5%7D%20%3D%20%5Cfrac%7B125%7D%7B12%7D-%5Cfrac%7B1%7D%7B12%7D%20%3D%20%5Cfrac%7B31%7D%7B3%7D)
Then
Answer:
2.6 (6 is recurring)
Step-by-step explanation:
2/3 x 4
= 2.6
Answer:
254,251,200
Step-by-step explanation:
This is a combination question, since the order doesn't matter, the formula for combinations is n!/(n-r)! n is the amount of things we can choose from but r is the amount of things (employees in this case) we actually select. n = 50 and r = 5. This we get 50!/(50-5)! or 50!/45!, using a calculator, we can find that 50!/45! is equal to 254,251,200. That is our final answer for the amount of combinations available.
