<span>n = 5
The formula for the confidence interval (CI) is
CI = m ± z*d/sqrt(n)
where
CI = confidence interval
m = mean
z = z value in standard normal table for desired confidence
n = number of samples
Since we want a 95% confidence interval, we need to divide that in half to get
95/2 = 47.5
Looking up 0.475 in a standard normal table gives us a z value of 1.96
Since we want the margin of error to be ± 0.0001, we want the expression ± z*d/sqrt(n) to also be ± 0.0001. And to simplify things, we can omit the ± and use the formula
0.0001 = z*d/sqrt(n)
Substitute the value z that we looked up, and get
0.0001 = 1.96*d/sqrt(n)
Substitute the standard deviation that we were given and
0.0001 = 1.96*0.001/sqrt(n)
0.0001 = 0.00196/sqrt(n)
Solve for n
0.0001*sqrt(n) = 0.00196
sqrt(n) = 19.6
n = 4.427188724
Since you can't have a fractional value for n, then n should be at least 5 for a 95% confidence interval that the measured mean is within 0.0001 grams of the correct mass.</span>
Answer:
Answer is 100 sq m
Step-by-step explanation:
A = 
A = 
A = 100 sq m
HOPE IT HELPS
<u><em>(FROM CROSS)</em></u>
Answer:
17/13
Step-by-step explanation:
9514 1404 393
Answer:
1.63 cm (across the centerline from release)
Step-by-step explanation:
If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...
h(t) = -7cos(2πt/4)
h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm
That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.
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If it is released from <em>above</em> the resting position, it will be 1.63 cm <em>below</em> the resting position at t=1.15 seconds.