Question

can someone explain steps to calculate the following: 9 (2/3)^1 times (1/3)^9 ===

Answer 1

\[9 \times{\left(\frac23\right)}^1 \times \left(\frac 13\right)^9\]
?

Answer 2

how do you perform this math? can you explain?

Answer 3

so i have the right equation?

Answer 4

use these

\[x^1=x\]
\[\left(\frac ab\right)^n=\frac{a^n}{b^n}\]

Answer 5

Supposely the result is .0004 for the problem ... why?

Answer 6

how did you get that

Answer 7

My instructor
That is why I am trying to figure out and can't

Answer 8

what number are you calculating?

Answer 9

9 times (2/3)^1 times (1/3)^9 = .0004

Answer 10

\[9 \times{\left(\frac23\right)}^1 \times \left(\frac 13\right)^9\]\[=9\times\frac 23\times\frac 1{3^9}\]\[=\]

Answer 11

\[\neq0.0004\]

Answer 12

I tried to do this manually right... I come up to (9/1) (2/3) (1/19683) = 18/59049 then is totally different result, right?

Answer 13

that is what i am too,

\[=9\times\frac 23\times\frac 1{3^9}=\frac {18}{3^{10}}\approx0.0003048\]

Answer 14

are we solving the right equation?

Answer 15

i am getting*

Answer 16

\[(9+\frac23)\times\frac1{3^9}\approx0.0004911\]

Answer 17

so that is where is coming from, i see now

Answer 18

but 0.0004911=0.0005 to one significant figure

Answer 19

the part drives me nuts is that there is a difference when you round it, right?

Answer 20

okay so I am allow to use a calculator that does not store or have memory, riight for the exam... I have a Casio FX-115s that will be allow right ... do you have any recommendations

Answer 21

ah, i have a Casio fx-82MS
very similar , should be allowed in to exam

Answer 22

im not sure if your instructor made a rounding error
or if the equation is different

Answer 23

I believe he made a mistake... it is a few other mistakes he had done