A sample of n = 25 n=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of σ = $ 4 σ=$4 . We obtain a 95% confidence interval as ( 14.43 , 17.57 ) (14.43,17.57) . Which choice correctly interprets this interval?

( 14.43 , 17.57 ) correctly interprets this interval.

Step-by-step explanation:

mean = x`= $ 16

Standard deviation= S= $4

sample size= n= 25

Using the formula

x`± zₐ/₂  S/ √n

= 16 ± (1.96) 4/ √25

= 16 ± (1.96) 4/5

= 16 ± (1.96)0.8

= 16 ± (1.568)

= 16+ 1.568         ,       16-1.568

= 17.568             , 14.432

From 14.43 to 17.57

( 14.43 , 17.57 ) correctly interprets this interval.

If two samples are taken  and two standard deviations are considered then the answer would be completely different from 13.78 to 18.22

step-by-step explanation:

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}