Yup! We're all pretty good here. :)
Answer:
μ = 5.068 oz
Step-by-step explanation:
Normal distribution formula to use the table attached
Z = (x - μ)/σ
where μ is mean, σ is standard deviation, Z is on x-axis and x is a desired point.
98% of 6-oz. cups will not overflow means that the area below the curve is equal to 0.49; note that the curve is symmetrical respect zero, so, 98% of the cases relied between the interval (μ - some value) and (μ + some value)].
From table attached, area = 0.49 when Z = 2.33. From data, σ = 0.4 oz and x = 6 oz (maximum capacity of the cup). Isolating x from the formula gives
Z = (x - μ)/σ
2.33 = (6 - μ)/0.4
μ = 6 - 2.33*0.4
μ = 5.068
This means that with a mean of 5 oz and a standard deviation of 0.4 oz, the machine will discharge a maximum of 6 oz in the 98% of the cases.
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
That would be 8 because the answer is 8.21 and you round down to 8
Answer:
r=-3
Step-by-step explanation: