We know that
The formula for combinations is
C=n!/[(n-r)!*r!]
where
n is the total number of objects you choose from
r is the number that you choose to arrange
in this problem
n=15 students
r=4 students
C=15!/[(15-4)!*4!]-----> C=15!/[11!*4!]---> (15*14*13*12*11!)/(11!*4*3*2*1)
C=(15*14*13*12)/(24)----->C=1365
the answer is
1365
From trigonometry we know that:
if
then, (where is an integer)
This can be rewritten in degrees as:
.............(Equation 1)
Now, in our case,
Therefore, (Equation 1) can be written as:
..........(Equation 2)
Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of that match.
For n=2, (Equation 2) gives us: .
Thus,
Now, we know that:
Let n=-1, then:
Thus,
Likewise,
Only the last option will never match because no integral value of will ever give
Thus the last option is the correct option.
Answer:
PLEASE BRAINLIEST
Step-by-step explanation:
400-2a =332
400-332 =2a
2a=68
a=34
The answer is letteer C. You just have to multiply 9 and pi
Answer:
a. P(x = 0 | λ = 1.2) = 0.301
b. P(x ≥ 8 | λ = 1.2) = 0.000
c. P(x > 5 | λ = 1.2) = 0.002
Step-by-step explanation:
If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:
a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:
b. What is the probability of finding eight or more defective pens in a carton?
This can be calculated as one minus the probablity of having 7 or less defective pens.
c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?
We can calculate this as we did the previous question, but for k=5.