Answer:
a) 28 stations
b) Rp = 21.43
E = 0.5
Explanation:
Given:
Average downtime per occurrence = 5.0 min
Probability that leads to downtime, d= 0.01
Total work time, Tc = 39.2 min
a) For the optimum number of stations on the line that will maximize production rate.
Maximizing Rp =minimizing Tp
Tp = Tc + Ftd
At minimum pt. = 0, we have:
dTp/dn = 0
Solving for n²:
The optimum number of stations on the line that will maximize production rate is 28 stations.
b) 
Tp = 1.4 +1.4 = 2.8
The production rate, Rp =
The proportion uptime,
Answer: Last week, Nate and I counted all the inventory.
Explanation: all other choices are passive voices
this sentence follows a clear subject + verb + object construct that's why it is an active voice. In fact, sentences constructed in the active voice add impact to your writing. but on the other hand With passive voice, the subject is acted upon by the verb.
Ape x
Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:
- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:
- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .
- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate
Answer:
A)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
Explanation:
Given data :
45.00 % mole of methane
55.00 % of ethane
attached below is a detailed solution
A) calculate - Q(kw)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B ) calculate - Q ( KJ )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
since combustion takes place in a constant-volume batch reactor
Answer:
correct option is (B) 315,500
Explanation:
given data
volume = 320,000 yd³ = 8640000 ft³
dry unit weight = 106 pcf
moisture content = 18.2%
total unit weight = 122 pcf
moisture content = 16.7%
to find out
volume of borrow lyd needed
solution
first we get here weight of material that is
weight = volume × unit weight
weight = 8640000 × 122
weight = 1054080000 lb
that weight is weight of water + weight of solid so
0.167 × weight is weight of water + weight of solid ) = 1054080000 lb
and weight of solid = 
weight of soil solid is = 903239075 pound
and weight of water = 150840925 pound
so volume of soil = 903239075 ÷ 106 lb/ft³ = 8521123.34 ft³
and volume required = 8521123.34 ft³ ÷ 27 ft³ = 315597.161 yd³
volume required = 315500 yd³
so correct option is (B) 315,500
