assignment8-QA

course Mth 151

February 11 around 10:00

assignment8-QA

q001:

I didn't get it myself but i understood how your strategy works.

q002:

We would pair the first number with the last as we get toward to the middle and we get 1000 pairs and they all add up to 2001. So 1000X2001=2,001,000

*3 on confidence scale

q003:

We can use the same strategy as the previous one but I have to keep in mind that the middle number is left out. I would use the previous strategy to get 250X502 and then add 251(the middle number) to =125,000+251=125,751

*2 on confidence scale

q004:

Every pair adds up to 1534 and we would multiply that by 766.5(number of pairs) to get 1,175,811

*2 on confidence scale

q005:

pair them up and we have to add 1 to 890 because the 55 is included to be added. so 891 divided by 2=445.5. So we would multiply the number of pairs, 1000 by 445.5 to equal 445,000

*2 on confidence scale

q006:

The pairs equal up to 904. so the numbers ""jump"" by 4. We would again have to add 1 and so we get 225 numbers. 225 divided by 2 is 112.5 and times that by 904 to get 101,700

*2 on confidence scale

q007:

We would pair 1 and n, 2 and n-1, 3 and n-2, etc. This would give us the sum of n+1. So n divided by 2Xn+1 would give you the total

*1 on confidence scale

&#Please let me know if you have questions. &#

*&$*&$