The number sentence that is true is B. 2.9 x 104 < 2.25 x 1011
2.9 x 104 = 301.6
2.25 x 1011 = 2,274.75
Therefore, 301.6 is less than (<) 2,274.75
Completed question:
An initial time study resulted in an average observed time of 2.2 minutes per cycle, and a standard deviation of .3 minutes per cycle. The performance rating was 1.20. What sample size, including the 20 observations in the initial study, would be necessary to have a confidence of 95.44 percent that the observed time was within 4 percent of the true value?
Answer:
47
Step-by-step explanation:
When doing a statistic study, a sample of the total amount must be taken. This sample must be done randomly, and, to be successful, the sample size (n) must be determined, by:
Where Z(α/2) is the value of the standard normal variable associated with the confidence, S is the standard deviation, and E is the precision. The confidence indicates if the study would have the same result if it would be done several times. For a confidence of 95.44, Z(α/2) = 2.
The standard deviation indicates how much of the products deviate from the ideal value, and the precision indicates how much the result can deviate from the ideal. So, if it may vary 4% of the true value (2.2), thus E = 0.04*2.2 = 0.088.
n = [(2*0.3)/0.088]²
n = 46.48
n = 47 observations.
Given:
Number of fish sold in 2005 : 3,320,000,000
Number of fish sold in 2006: 4,220,000,000
The number of fish sold in 2005 and 2006 is the sum of the number of fish sold in 2005 and the number sold in 2006.
Thus:
Hence, the best estimate of the number of fish sold in 2005 and 2006 is 7,540,000,000
Answer:
7,540,000,000 (Option B)
Answer:
option F
Step-by-step explanation:
given,
capacity of tank = 1000 gallon
volume of water leak = 300 gallon
time = 30 min
water is leaking at the rate
formula used
At t = 0 A = 1000 gallons
B = 1000
at t = 30 A = 700
taking ln both side
now,
hence, the correct answer is option F
Answer:
the average number of car(s) in the system is 1
Step-by-step explanation:
Given the data in the question;
Arrival rate; λ = 2.5 cars per hour
Service time; μ = 5 cars per hour
Since Arrivals follows Poisson probability distribution and service times follows exponential probability distribution.
Lq = λ² / [ μ( μ - λ ) ]
we substitute
Lq = (2.5)² / [ 5( 5 - 2.5 ) ]
Lq = 6.25 / [ 5 × 2.5 ]
Lq = 6.25 / 12.5
Lq = 0.5
Now, to get the average number of cars in the system, we say;
L = Lq + ( λ / μ )
we substitute
L = 0.5 + ( 2.5 / 5 )
L = 0.5 + 0.5
L = 1
Therefore, the average number of car(s) in the system is 1