Answer:
The margin of error that corresponds to a 95% confidence interval is 4.96.
Step-by-step explanation:
We have the standard error(which is the same as the standard deviation of the sample), so we use the students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. There are 6 days, so
df = 6 - 1 = 5
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of ![1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{0.975}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.975%28%5Btex%5Dt_%7B0.975%7D)
). So we have T = 2.5706
The margin of error is:
M = T*s = 2.5706*1.93 = 4.96
The margin of error that corresponds to a 95% confidence interval is 4.96.
Answer:
Step-by-step explanation:
-Let x be the sample size and n the population size
-The conditions for a one-sample proportion z-test are:
-The sample is randomly selected from the population.
-The sample size is greater than or equal to ten times the population size:
-The expectation np is greater than or equal to 10:
The answer to ur question is 4
Answer:
9 years
Step-by-step explanation:
Let the present age = x
Age 6 years ago = x - 6
Thrice my age 6 years ago = 3*(x -6) =3x - 3*6 = 3x - 18
Twice my present age = 2 *x = 2x
Thrice my age 6 years ago be subtracted from
twice my present age = 2x - (3x - 18)
2x - (3x - 18) = x
2x - 3x + 18 = x
-x + 18 = x {Add x to both sides}
18 = x + x
18 = 2x {Divide both sides by 2}
18/2 = x
x = 9 years
