You want the average?
add them all together and then divided by the number of events
"Understand the problem" might rightly consist of
• Perform this step first
• Identify what you are being asked to solve or find.
• Identify the important words or numbers in the problem
• Identify any instructions that you are supposed to follow
_____
One of my professors always insisted we start the solution of any problem by writing down what was Given, and what we had to Find, using those headers for the sections of the paper we turned in. Only after those were listed were we allowed to write the Solution. Solution papers that didn't have that format were tossed in the trash, and no credit was given. Harsh, but effective.
Answer: a)
b) 
<u>Step-by-step explanation:</u>
a) In order to get an even number, you have the 3 different scenarios:
1) Even, Even, Even, Even 
2) Even, Even, Odd, Odd
3) Odd, Odd, Odd, Odd 
<em>Order doesn't matter</em>
Add them up to get your answer: 
b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.

Answer:
30% off
Step-by-step explanation:
63 is 70% of 90 which means that 30 percent of the price is off.
Please mark me brainliest. I need it. Hope this helps
Answer:
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Step-by-step explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.