Answer: since safeguarding isn't possible distance and location must be used instead
(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.
<h3>
Weight distribution of the kitten</h3>
In a normal distribution curve;
- 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
- 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
- 1 standard deviation (d) above the mean (M), (M + d) is at 84%
- 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%
M - 2d = 125 g - 2(15g) = 95 g
M - d = 125 g - 15 g = 110 g
95 g is at 2% and 110 g is at 16%
(16% - 2%) = 14%
(110 - 95) = 15 g
14% / 15g = 0.93%/g
From 95 g to 99 g:
99 g - 95 g = 4 g
4g x 0.93%/g = 3.72%
99 g will be at:
(2% + 3.72%) = 5.72%
Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.
<h3>Weight of the kitten in the 90th percentile</h3>
M + d = 125 + 15 = 140 g (at 84%)
M + 2d = 125 + 2(15) = 155 g ( at 98%)
155 g - 140 g = 15 g
14% / 15g = 0.93%/g
84% + x(0.93%/g) = 90%
84 + 0.93x = 90
0.93x = 6
x = 6.45 g
weight of a kitten in 90th percentile = 140 g + 6.45 g = 146.45 g
Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g
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Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
![\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%20%3D%5B%5Cfrac%7Bp2%7D%7Bp1%7D%20%5D%5E%7B%5Cfrac%7By-1%7D%7By%7D%20%7D)
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
![\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.20%7D%7B0.4%7D%20%5D%5E%7B%5Cfrac%7B1.13-1%7D%7B1.13%7D%20%7D)
![\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.2%7D%7B0.4%7D%20%5D%5E%7B0.1150%7D%5C%5C%5Cfrac%7Bt2%7D%7B360%7D%20%3D1.1347)
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. 
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360
cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
Answer:
Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37
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I only know this
sorry for errors
It is uncommon for electrocution to cause crane incidents: False.
<h3>What is
electrocution?</h3>
Electrocution can be defined as a process through which a certain amount of electric current (electricity) passes through the body of a living organism, thereby, leading to death or severe injury.
According to the occupational safety and health administration (OSHA), it is very common for electrocution to cause crane incidents and accidents when adequate safety precautions are not adopted.
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