Answer and Step-by-step explanation:
we have the following data:
Point estimate = sample mean = \ bar x = 12.39
Population standard deviation = \ sigma = 3.7
Sample size = n = 177
a) the margin of error with a 90% confidence interval
α = 1 - 90%
alpha = 1 - 0.90 = 0.10
alpha / 2 = 0.05
Z \ alpha / 2 = Z0.05 = 1,645
Margin of error = E = Z \ alpha / 2 * (\ sigma / \ sqrtn)
we replace:
E = 1.645 * (3.7 / \ sqrt177)
Outcome:
E = 0.46
b) margin of error with a 99% confidence interval
α = 1-99%
alpha = 1 - 0.99 = 0.01
alpha / 2 = 0.005
Z \ alpha / 2 = Z0.005 = 2,576
Margin of error = E = Z \ alpha / 2 * (\ sigma / \ sqrtn)
we replace:
E = 2,576 * (3.7 / \ sqrt177)
Outcome:
E = 0.72
c) A larger confidence interval value will increase the margin of error.
Answer:
Below.
Step-by-step explanation:
About x3 bigger then figure b.
Assuming 5 is the base. I'm going to leave that out for now.
2log(5x^3) + (1/3)log(x^2+6)
power rule
log(5^2 x^3*2) + log((x^2 + 6)^(1/3))
log(25x^6) + log((x^2 + 6)^(1/3))
quotient rule
log(25x^6 / (x^2 + 6)^(1/3))
Answer:
the students that brought a lunch box is 28
Step-by-step explanation:
The computation of the students that brought a lunch box is shown below:
= Entire school students × students that carry a lunch box ÷ entering students
= 84 students × 8 ÷ 24 students
= 28 students
Hence, the students that brought a lunch box is 28
Answer:
I think that its 58 degrees