Answer:
a= -4/15
Step-by-step explanation:
Step one: 4+5a=8+20a
Step two: 5a-20a=8-4
Step three: -15a=4 then divide both sides by 15 to get your answer :)
We will use the following formula to work out the confidence interval
Upper limit = μ + z* (σ/√n)
Lower limit = μ - z* (σ/√n)
We have
μ = 79.3
σ = 7.8
n = 12
z* is the z-score for 95% confidence level = 1.96
Substitute these into the formula, we have
Upper limit = 79.3 + 1.96 (7.8/√12) = 83.7
Lower limit = 79.3 - 1.96 )7.8/√12) = 74.9
Aye I have the same computer
Answer:
(72460, 97540)
Step-by-step explanation:
Given:-
- The sample size, n = 8
- The sample parameters of normal distributed data:
mean u = $85000
standard deviation s.d = $15000
Find:-
Construct a 95% confidence interval for the average starting salary of all computer scientists in gainesville.
Solution:-
- A random variable (X) denotes average starting salary of all computer scientists in gainesville has a sample normally distribution as follows:
X ~ N ( $85000 , $15000^2 )
- The general formulation for CI for known standard deviation with significance α = 1 - 0.95 = 0.05:
( u - Z_a/2*(s.d/√n) < X < u + Z_a/2*(s.d/√n) )
- The Z_a/2 = Z_0.025 = 1.96:
( 85,000 - 1.96*(15,000/√8) < X < 85,000 + 1.96*(15,000/√8) )
(72460, 97540)
Answer:
(e) The women on the Maryland field hockey team are not a random sample of all female college field
hockey players. Similarly, the women on the Maryland basketball team are not a random sample of all female
college basketball players. However, for the purposes of this question, suppose that these two groups can be
regarded as random samples of all female college field hockey players and all female college basketball
players, respectively. If these were random samples, would you think that female college basketball players
are typically taller than female college field hockey players? Explain your decision using answers to the
previous questions and/or additional reasoning.
