Answer:
Step-by-step explanation:
Line q will be graphed on the same grid. The only solution to the system of linear equations formed by lines n and q occurs when x = and y = 0.
Now, as x = is a solution of the equation, y = f(x) = 0, so,
will be a factor of the linear function y = f(x).
Therefore, we can write the possible equation for the line q will be
. (Where k is any constant} (Answer)
Answer:
Question 13: For age groups y=1 and y=1.3 response is 8 microseconds.
Question 14: The club was making a loss between 11.28 and 4.88 years.
Step-by-step explanation:
Question 13:
The age group y for which the response rate R is 8 microseconds is given by the solution of the equation
We graph this equation and find the solutions to be
Since only positive solutions for y are valid in the real world we take only those.
Thus only for age groups y=1 and y=1.3 the response is 8 microseconds.
Question 14:
The footbal club is making a loss when
Or
We graph this inequality and find the solutions to be
and
Since in the real world only positive values for t are valid, we take the the second solution to be true.
Thus the club was making a loss in years
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:
a ≅ 12