$\displaystyle \binom{6}{3}\cdot2+\binom{6}{2}=\dfrac{6!}{3!3!}\cdot2+\dfrac{6!}{2!4!}=\dfrac{4\cdot5\cdot6}{2\cdot3}\cdot2+\dfrac{5\cdot6}{2}=40+15=55$

8 0

3 years ago

What is the diameter of a circle with a radius of 4.375"? 2.187" or 2.375" or 8.375" or 8.750"

AURORKA [14]

DONT CLICK ON THE LINKS! It’s 8.750 BY THE WAY the radius is half the diameter so just add the radius by itself or multiply by 2

3 0

3 years ago

Cargo ships arrive at a loading dock at a rate of 2 per day. The dock has the capability of handling 3 arrivals per day. How man

Vsevolod [243]

Answer:

4 days

Step-by-step explanation:

We need to use the probability functions of each of the intervals to know the Probability number and then use it in the expected value.

P(x>3 cargo ships) = 1-P(x<=3)

P(x>3) = 1-P(x<=3)

P(x>3) = 1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)]

$P(x>3) = 1 - [\frac{e^{-2}2^0}{01}+\frac{e^{-2}2^1}{11}+\frac{e^{-2}2^2}{21}+\frac{e^{-3}2^3}{31}]$

P(x>3) = 1- [0.1353(1+2+2+1.33)]

P(x>3) = 1-0.856

P(x>3) = 0.1431

Als n=30, expected number is

E(x) =30*P(x>3)

E(x) = 30*0.1431

E(x) = 4.293

I expect 4.293 or 4 days per month the block being unable to hanlde all arriving ships

6 0

3 years ago