Let's attack this problem using the z-score concept. The sample std. dev. here is (0.25 oz)/sqrt(40), or 0.040. Thus, the z score representing 3.9 oz. is
3.9 - 4.0
z = -------------- = -2.5
0.040
In one way or another we must find the area under the std. normal curve that lies to the left of z = -2.5. Use a table of z-scores or a calculator with built-in statistics functions. According to my TI-83 Plus calculator, that area is
0.006. One way of interpreting this that with so small a standard deviation, most volumes of coffee put into the jars are very close to the mean, 4 oz.
Answer:
The Alans's average for the course is 84.5
Step-by-step explanation:
We are given
There are 4 tests, A Term paper and a Final examination.
Score = 92, 78, 82, 90.
Term paper score = 80
Final examination score = 86
Weighted mean = ∑ w.f/∑w
also the sum of all the weights is 100% = 1
weighted mean = 15%*92 + 15%*78+15%*82+15%*90+20%*80+20%*86/1
= 51.3 + 33.2
= 84.5
Therefore the Alans's average for the course is 84.5
Obviously it will be 0.25.
Because this number is a rational number not a integer number....
Answer:
11.01 - 11.09 , 11.21 - 11.29, 11.31 - 11.39, 11.41 - 11.49
Hope this is what you were looking for :)
Answer:
94.5 days
Step-by-step explanation:
We are asked to calculate the time.
The formula to calculate the time for half life =
t = t½ × In (Nt/No)/-In2
t½ = half life = 16 days
No = Initial substance = 120mg
Nt = Amount of substances after time t = 2 mg
t = 16 × In (2/120)/-In2
t = 94.510249529736 days
Approximately = 94.5 days
Therefore, the time it would take for the substance to decay from 120mg to 2 mg is 94.5 days.
