Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : 
Standard deviation : 
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula,
, the z-value corresponds to 2.39 will be :-

z-value corresponds to 2.60 will be :-

Using the standard normal table for z, we have
P-value = 

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
Answer: r = 0.8081; s = -0.07071
Explanation:
A = (150i + 270j) mm
B = (300i - 450j) mm
C = (-100i - 250j) mm
R = rA + sB + C = 0i + 0j
R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j
R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j
Equating the i and j components;
150r + 300s - 100 = 0
270r - 450s - 250 = 0
150r + 300s = 100
270r - 450s = 250
solving simultaneously,
r = 0.8081 and s = -0.07071
QED!
Answer:
true because BCD used 6 bits to represent a symbol .
Explanation:
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The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.
<h3>How does kVp impact the exposure to digital receptors?</h3>
The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.
<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>
Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.
To know more about kVp visit:-
brainly.com/question/17095191
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Answer:
Enthalpy at outlet=284.44 KJ
Explanation:


We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation

m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation

By putting the values

So h=284.44 KJ