Step-by-step explanation:
an² = 182
When a = 2, n² = 182/2 = 91, n = 9.539...
When a = 5, n² = 182/5 = 36.4, n = 6.033...
Since 2 <= a <= 5, we have 6.033 <= n <= 9.539.
Therefore our possible values of n are 7, 8 and 9.
Answer: Marianne made an inference that is true based on the data. More than half of the people surveyed in each sample chose oranges as their favorite fruit. Since most people in each sample chose oranges, it is likley that oranges are the favorite fruit of the entire population.
Step-by-step explanation:
If you would like to find the final score, you can do this using the following steps:
Team A: 6 points * 3 + 3 points * 2 + 2 points * 1 + 1 point * 0 = 6 * 3 + 3 * 2 + 2 * 1 + 1 * 0 = 18 + 6 + 2 + 0 = 26 points
Team B: 6 points * 4 + 3 points * 4 + 2 points * 1 + 1 point * 2 = 6 * 4 + 3 * 4 + 2 * 1 + 1 * 2 = 24 + 12 + 2 + 2 = 40 points
The final score of team A is 26 points and the final score of team B is 40 points.
Answer:
The answer is below
Step-by-step explanation:
A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?
Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

a) P(x > 5) = 
b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.
That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757
c) Let b be the amount of raw sugar should be stocked for the plant each day.
P(x > a) = 
But P(x > a) = 0.05
Therefore:
![e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98](https://tex.z-dn.net/?f=e%5E%7B-0.25a%7D%3D0.05%5C%5Cln%5Be%5E%7B-0.25a%7D%5D%3Dln%280.05%29%5C%5C-0.25a%3D-2.9957%5C%5Ca%3D11.98)
a ≅ 12