Answer:
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error:

For this problem, we have that:

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is
We need a sample size of at least n, in which n is found M = 0.04.







With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
Answer:
the answer is 6
Step-by-step explanation:
6×3=18
So 6×5=30, 6×2=12
30-12=18
18=18
600/800 descansos porque mencionaste que se toma un descanso cada 3 cm pero se desliza 1 cm cada vez que se toma un descanso, así que solo cambia los 2 ma cm, que es 200 veces eso a 3 o 4 y obtendrás el número de descansos (acabo de adivinar)
Figure how much percent she reads daily. Then add on till 7 until ur done... For example 7= 35% 14(days)= 70