The binomial distribution is given by,
P(X=x) =

q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) =

where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) =

+

+

+

+

+

+

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.
We want to find the mean of two elements in a set, given that we know the other elements of the set and the mean of the whole set.
The answer is: -490
-----------------------------
For a set with N elements {x₁, x₂, ..., xₙ} the mean is given by:

Here we know that:
- The mean of the set is 0.
- The set has 1000 elements.
- 998 of these elements are ones, the other two are A and B.
We want to find the mean of the values of A and B.
First, we can start by writing the equation for the mean:

We can rewrite this as:

And we have 998 ones, then:

Now we have B isolated.
With this, the mean of A and B can be written as:

So we can conclude that the mean of the other two numbers is -490.
If you want to learn more, you can read:
brainly.com/question/22871228
Answer:
Step-by-step explanation:
m = 5 + 10d
The variable "d" is the independent variable because its value does not depend on another variable. The variable "m" is the dependent variable because its value depends on the value of the variable "d"
Answer:
Integers are always closed under subtraction, addition and multiplication, not division though. Hope this helped! :D
Step-by-step explanation:
Answer: Graph D
Step-by-step explanation: Got it right on plato