Answer:
124984638540
Step-by-step explanation:
just use a calculator
Answer:
The answer is B) 0.57.
Step-by-step explanation:
In this problem we have to apply queueing theory.
It is a single server queueing problem.
The arrival rate is
and the service rate is
.
The proportion of time that the server is busy is now as the "server utilization"and can be calculated as:

where c is the number of server (in this case, one server).
For synthetic division:
1) Write the coefficients of the polynomial under the division symbol, in order. Make sure you include coefficients that are 0, so they may not be written out in the polynomial!
The coefficients are 7, 1, 0 (the coefficient of x), and -4 in that order
2) The divisor is in the form of x - a. The number "a" goes outside the bracket.
a = 5
Since 7, 1, 0, -4 are under the bracket and 5 is outside the bracket, the answer is C.
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Answer: C
If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.
Given 95% confidence level, standard deviation=0.05.
We know that margin of error is the range of values below and above the sample statistic in a confidence interval.
We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data near the mean are more frequent in occurence than data far from mean.
We know that margin of error for a confidence interval is given by:
Me=
α=1-0.95=0.05
α/2=0.025
z with α/2=1.96 (using normal distribution table)
Solving for n using formula of margin of error.

n=
=96.4
By rounding off we will get 97.
Hence the sample size required will be 97.
Learn more about standard deviation at brainly.com/question/475676
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The given question is incomplete and the full question is as under:
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?