A survey of households in a small town showed that in 850 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

Hence the confidence interval is ( 0.6745, 0.7422).

Step-by-step explanation:

Now the given are  

Sample size = n = 1200  

x = 850  

Sample proportion is  

We have to construct 99% confidence interval for the population proportion.  

Formula Used:  

Here E is a margin of error.  

Zc = 2.58  

So confidence interval is ( 0.7083 - 0.0339 , 0.7083 + 0.0339)

                                            = ( 0.6745 , 0.7422).

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}

6 meters

step-by-step explanation: