Answer:
Explanation:
A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than
1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with ? = 68. Let ? denote the true average compressive strength.
(b) Let
X
denote the sample average compressive strength for n = 11 randomly selected specimens. Consider the test procedure with test statistic X and rejection region x ? 1331.26. What is the probability distribution of the test statistic when H0 is true? (Round your standard deviation to three decimal places.)
(c) What is the probability distribution of the test statistic when
? = 1350? (Round your standard deviation to three decimal places.)
Using the test procedure of part (b), what is the probability that the mixture will be judged unsatisfactory when in fact
? = 1350 (a type II error)? (Round your answer to four decimal places.)
(d) How would you change the test procedure of part (b) to obtain a test with significance level 0.05? (Round your answer to two decimal places.) Replace 1331.26 KN/m2 wit
What impact would this change have on the error probability of part (c)? (Round your answer to four decimal places.)
The probability that the mixture will be judged unsatisfactory when in fact ? = 1350 will change to .
(e) Consider the standardized test statistic
Z = (X ? 1300)/(?/n).
What are the values of Z corresponding to the rejection region of part (b)? (Round your answer to two decimal places.)
True since they’re engineers they don’t mess with the actual petroleum
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:-
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Answer:
(a) 16.27 Vpk
(b) 48.7%
Explanation:
The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.
__
<h3>(a)</h3>
The peak voltage will be 0.7 V less than the transformer secondary peak voltage:
((120 V)√2)/10 -0.7 V ≈ 16.27 V
__
<h3>(b)</h3>
The fraction of the amplitude for which the diode is non-conducting is ...
0.7/(12√2) ≈ 0.041248
The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:
arccos(0.041248) ≈ 87.64°
As a fraction of half the cycle, this is ...
conduction fraction ≈ 87.64°/180° ≈ 48.7%
Answer:
a) 84.034°C
b) 92.56°C
c) ≈ 88 watts
Explanation:
Thickness of aluminum alloy fin = 12 mm
width = 10 mm
length = 50 mm
Ambient air temperature = 22°C
Temperature of aluminum alloy is maintained at 120°C
<u>a) Determine temperature at end of fin</u>
m = √ hp/Ka
= √( 140*2 ) / ( 12 * 10^-3 * 55 )
= √ 280 / 0.66 = 20.60
Attached below is the remaining answers