A college financial advisor wants to estimate the mean cost of textbooks per quarter for students at the college. For the estimate to be useful, it should have a margin of error of 20 dollars or less. The standard deviation of prices is estimated to be around 100 dollars. How large of a sample size needs to be used to be 95% confident, with the given margin of error?

minimum sample size = 97

Step-by-step explanation:

Margin of error = 20

standard deviation = 100

sample size = n

standard error = 100/sqrt(n)

confidence level, alpha = 95%

Using the standard rule for 95% confidence

standard error <= sample mean 1.96 standard error, or

20 <= 1.96*100 / sqrt(n)

n >= (1.96*100/20)^2 = 9.8^2 = 96.04  

=>

n >= 97

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}