Can someone help me with this problem.
1 answer:
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121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range
The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.
The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:
So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range
That's probably the answer they're looking for.
2x+5y=20
X=10-2.5y
-2(10-2.5y)+20y=-90
-20+5y+20y=-90
25y=-70
Y=2.8
2(2.8)+5y=20
5.6+5y=20
5y=14.4
Y=2.88
CHECK:
2(2.8)+5(2.88)=20
X=10-2.5y
-2(10-2.5y)+20y=-90
-20+5y+20y=-90
25y=-70
Y=2.8
2(2.8)+5y=20
5.6+5y=20
5y=14.4
Y=2.88
CHECK:
2(2.8)+5(2.88)=20