Answer: 0.0516
Step-by-step explanation:
Given : The number of defective components produced by a certain process in one day has a Poisson distribution with a mean of
.
The probability mass function for Poisson distribution:-
For x= 15 and
, we have

Hence, the probability that exactly 15 defective components are produced in a particular day = 0.0516
The corresponding answers for each student plan are:
plan A: 15 years
plan B: 5 years
plan C: 10 years
plan D: 9 years
Thus, first, the graph corresponding to each of the plans must be analyzed. The Y-axis corresponds to money and the x-axis corresponds to the time in years.
Thus, in-plane A, it can be observed that to reach the desired value, the graph will be projected in exponential, so it will be close to 15 years. In plan B, this projection will be reached around 5 years. In plans C and D, this projection will be 10 years and 9 years respectively.
Learn more: brainly.com/question/14355665
Answer:
okay first one is ,3
Step-by-step explanation:
Answer:
Bet
Step-by-step explanation:
It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
Answer:
A sample size of 79 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

The margin of error is:
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?
A sample size of n is needed.
n is found when M = 0.09. So






Rounding up to the nearest whole number.
A sample size of 79 is needed.