Answer:
We need a sample size of 564.
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 .
Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.
We need a sample size of n
n is found when
So
Rounding up
We need a sample size of 564.
Fixed costs:
$350 + $120 + $170 = $640
Variable costs:
x * ( $6.50 + $3.64 ) = x * 10.14
Sales income ( total ):
x * $36.40
FC + FV - Income = 0
640 + 10.14 x - 36.40 x = 0
640 - 26.26 x = 0
26.26 x = 640
x = 640 : 26.26 = 24.37
Answer:
The minimum number of passengers needed per cruise, so that the cruise company can be sure it will make a profit is 25.
Answer: A) 3(10 - 5y) + 6y =12
Step-by-step explanation:
As per the question , the given system of linear equation :
Here , To solve the system of equations using substitution , Johanna isolates x in the second equation and writes
As per the substitution method , the next step is to substitute the isolated value of 'x' in first equation , thus the next step will become
Hence, the correct option is A) 3(10 - 5y) + 6y =12
C and D are correct because 25 > 35 so that's C.
Its also D because they are not equal.
Hope this helps :)
Answer:
0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 50
Standard Deviation, σ = 1.3
Sample size, n = 12
We are given that the distribution of hardness of pins is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =
P(sample mean hardness for a random sample of 12 pins is at least 51)
Calculation the value from standard normal z table, we have,
0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.