In a survey taken in july 1985, of the 50 households randomly surveyed from across the united states, the proportion of households having a personal computer was determined to be 11%. the u. s. government census data taken in august 1985 suggested that 11.7% of households in the united states had a personal computer. which statement(s) below support the small discrepancy between the sample proportion of households having a personal computer in july 1985 and the population proportion of households having a personal computer in august 1985?

 I.The sample size was too small.

II.The sample was not truly random.

III.The data from July 1985 to August 1985 changed substantially.

IV.The sample statistic is nearly equal to the population parameter and will never be exactly equal.

A. Only statements I, II, and III support the discrepancy.

B. Only statements I and III support the discrepancy.

C. Only statements I and IV support the discrepancy.

D. Only statement IV supports the discrepancy.

I think the answer is letter d. only statement IV supports the discrepancy. In the problem, it shows there that the survey happened in July 1985 was just a sample proportion of the family that uses a computer at home. So, the sample survey is nearly equal to the result of in August 1985 data, but will never be exactly equal

i am chris from the united states i have skill in math and i am here to evryone

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}