The formula of the line is
y=mx + b
The likes goes through the point (0,0)
that means that y-int = 0, b=0
y=mx
So we need to find only slope.

0.99 is the answer it greater than 0.1
6/11=x/33 is not a proportion because they aren't equal to each other.
10 is the default base for a logarithm
log of 50 is equal to log base 10 of 50

1) Since this is a Continuously Compounded operation in a 10 yrs period, then we can write out the following equation:

2) Plugging into the equation the given data and since Otto is 20 yrs old and he plans to get $4,000 in ten years, we can write out:

3) Thus the rate Otto needs is

Note that since the 0.1386 the six here is greater than 5 then we can round up to the next thousandth, in this case: 0.139. For the 0.1386 is closer to 0.139 (0.004) than to 0.138 (0.006).
Or 13.9%