Answer:
-3<n<=5
Step-by-step explanation:
This is because the coloured circle on the number line means that the number is less than or equal to 5
Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if 
and 
then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.
side note: q = 1-p is the complement of probability p
Answer:
proof below
Step-by-step explanation:
Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)
Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;
(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2
= 2(2a^2 + 2a + 2b^2 + 2b + 1)
Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.
If you need to siplify that radical, you should try to put all the numbers as product of prime factors.
922 = 2*461
2 and 461 are prime numbers. Then the transformation of the square root is √(922) = √(2*461) = (√2)(√461)
So you can use that in any expression to try to simpliy with other numbers whose prime factors include 2 and 461.
B-3/88 because the square trot of 67 is 3
