Answer:
1a) 93ten
1b) 135eight
2a) 565ten
2b) 769ten
Step-by-step explanation:
1a)
1011101
7 digits long the first digit (the one that is most far right) is the ones digit or 2^0 digit and so the last digit (the most far left) is 2^6.
So to put the number in base 10 we do:
1(2^6)+0(2^5)+1(2^4)+1(2^3)+1(2^2)+0(2^1)+1(2^0)
=64+0+16+8+4+0+1
=64+16+8+4+1
=(64+16)+(8+4+1)
=80+13
=93
1b)
93
What is the hightest power of 8 in 93?
8^x=93
8^1=8
8^2=64
8^3 would be a factor of 8 too much
So how many 8^2 's are in 93?
64 goes into 93 how many times?
It's once with a remainder of 93-64=29.
So now 29...what is the highest power of 8 that goes into 29? 8^1=8.
How many times? 3(8)=24 with remainder 5.
So 93=1(8^2)+3(8^1)+5(8^0).
93=135eight
2a) 2341six
2(6^3)+3(6^2)+4(6^1)+1(6^0)
2(216)+3(36)+4(6)+1
432+108+24+1
540+25
565
2b) 30001four
3(4^4)+0(4^3)+0(4^2)+0(4^1)+1(4^0)
3(256)+0+0+0+1
768+1
769
I believe the range is the middle number so start from 63 then 46 then 53 then go back and forth until you get the middle number now after you get the middle number for the IQR add up both sides of the middle and / by the all the number on the left and right of the middle number
The answer would be 60
The first think you would need to do would simply 144. Which would be 12.
Then you are going to want to multiply-5 by 12.
Which would be 60.
With 12 as a positive number, being greater or to than 60 then will become a positive.
If you put it in fraction form and use PEMDAS to find the answer
here is the answer its in the picture and it also has the solution in picture 2
