Answer:
H0 : μ = 4.50
H1 : μ > 4.50
Test statistic = 1.525
we conclude that there is no sufficient evidence to conclude that the mean wait time for customers is longer than 4.50.
Step-by-step explanation:
H0 : μ = 4.50
H1 : μ > 4.50
Test statistic :
(xbar - μ) ÷ s/sqrt(n)
(4.94 - 4.50) ÷ 2.10/ sqrt(53)
0.44 / 2.10/ sqrt(53)
= 1.525
α = 1 - 98% = 0.02
Decision region :
Reject H0 ; if Pvalue < α
Pvalue = p(Z < 1.525) = 0.936
Pvalue = 0.936
Since Pvalue > α ; We fail to reject H0
Hence, we conclude that there is no sufficient evidence to conclude that the mean wait time for customers is longer than 4.50.
Yes it would be 7.1 is greater than 7
Answer: the velocity and position vector is given by;
v(t) = 3i + 2j + (1-32t)k
r(t) = (3t)i + (5-2t)j + (2+t-16t^2)k
The position at time t=2; r(2)
r(2) = 6i + j - 60k
Step-by-step explanation:
Shown in the attachment.
Answer:
I think total of is right answer
Surface area: S
radius: r
diameter: e
time: t
[dS/dt] = [dS/de]*[de/dt] => de/dt = [dS/dt] / [dS/de]
S = 4pi*r^2 = 4p*i(e/2)^2 = pi*e^2 => dS/de =2pi*e
dS/dt = 7cm^2 / min
de / dt = [2pi*e] / [7cm^2/min] =[7cm^2/min] / [2pi*10cm] = 0.11 cm/min
