Answer: False, there are actually 900 different three-digit numbers
========================================================
Explanation:
The three digit numbers span from 100 to 999, including both endpoints.
This means we have 999-100+1 = 900 different three-digit numbers.
You subtract the endpoints (large-small) and add 1 to include the lower endpoint.
Here's a smaller example of why this works: say you had the set {1,2,3,4} and we wanted to count the number of items in this set. Clearly there are 4 items. Note how subtracting the endpoints 4-1 gets us 3 instead, so we add on 1 to include that left endpoint.
Answer:
Step-by-step explanThis polynomial has three terms. The first one is 4x2, the second is 6x, and the third is 5. The exponent of the first term is 2. The exponent of the second term is 1
Answer:
(a) 
(b) 100
Step-by-step explanation:
Solving (a):
Given
Shape: Rectangle
Required
Calculate the perimeter.
Area is calculated as:
Where
and 
Substitute 64.8 for Area
Make L the subject:
Perimeter is calculated as:
Substitute 64.8/W for L
To solve further, we take the derivative of P and set it to 0, afterwards.
Set to 0
Collect Like Terms
Cross Multiply:
Divide through by 2
Take square roots
Recall that:
So:
The perimeter is:
Solving (b):
Given
((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)
Required
Solve
So, the expression becomes:
((1 Group of 
) and (1 group of 
))/(6 groups of 
)
This gives:
((1 Group of 
) and (1 group of 
))/(6 groups of 
)
Group means product, so the expression becomes:
And, as used here means addition
Simplify:
Answer:
A number line going from 0 to 4.5 in increments of 0.5
Step-by-step explanation:
This solution makes the most sense because 4.5 and 2.5 both have a decimal of 0.5
Answer:
Step-by-step explanation:
For the random variable 
we define the possible values for this variable on this case ![[1,2,3,4,5]](https://tex.z-dn.net/?f=%20%5B1%2C2%2C3%2C4%2C5%5D)
. We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:
We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable 
like this:
For the distribution of 
we need to take in count that we are finding a conditional distribution. given 
, for this case we see that ![N_2 \in [1,2,...,5-a]](https://tex.z-dn.net/?f=%20N_2%20%5Cin%20%5B1%2C2%2C...%2C5-a%5D)
, so then exist 
ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:
And if we want to find the joint probability we just need to do this:
And if we multiply the probabilities founded we got:
