if we have a geometric series with u1 = 2.1 and r= 1.06, what would the least value of n be such that Sn > 5543?

  88

Step-by-step explanation:

Write the expression for the sum in the relation you want.

  Sn = u1(r^n -1)/(r -1) = 2.1(1.06^n -1)/(1.06 -1)

  Sn = (2.1/0.06)(1.06^n -1) = 35(1.06^n -1)

The relation we want is ...

  Sn > 5543

  35(1.06^n -1) > 5543 . . . . substitute for Sn

  1.06^n -1 > 5543/35 . . . .  divide by 35

  1.06^n > 5578/35 . . . . . . add 1

  n·log(1.06) > log(5578/35) . . . take the log

  n > 87.03 . . . . . . . . . . . . . . divide by the coefficient of n

The least value of n such that Sn > 5543 is 88.

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

i think the length is 12 and the width is 17