<h3>
Answer: 127</h3>
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Explanation:
The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.
The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.
Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.
So we have these scores
- first exam = 70
- second exam = 80
- third exam = 43
- fourth exam = x+5
Adding up these scores and dividing by 4 will get us the average
(sum of scores)/(number of scores) = average
(70+80+43+x+5)/4 = 80
(x+198)/4 = 80
x+198 = 4*80
x+198 = 320
x = 320 - 198
x = 122
So the student got a score of x+5 = 122+5 = 127 on the fourth exam.
Answer:
(12.1409, 14.0591
Step-by-step explanation:
Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.
Group Group One Group Two
Mean 87.600 74.500
SD 1.000 1.500
SEM 0.316 0.433
N 10 12
The mean of Group One minus Group Two equals 13.100
standard error of difference = 0.556
90% confidence interval of this difference:

t = 23.5520
df = 20
9n - 2 - 6n - 3
(9n + -6n) + (-2+-3)
=3n -5