Answer:
Do the data from this shipment indicate statistical control: No
Step-by-step explanation:
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
= (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10
= 0.044/10
= 0.0044
Calculating the lower control limit (LCL) using the formula;
LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n
= 0.0044 - 3*√(0.0044(1-0.0044))
= 0.0044- (3*0.0042)
= 0.0044 - 0.01256
= -0.00816 ∠ 0
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
= 0.0044 + 3*√(0.0044(1-0.0044))
= 0.0044+ (3*0.0042)
= 0.0044 + 0.01256
=0.01696∠ 0
Do the data from this shipment indicate statistical control: No
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment do not indicate statistical control.
Answer:
Step-by-step explanation:
Let f(x) = 4x3 + 3x2 − 20x − 15 and g(x) = 4x + 3. Find f of x over g of x.
Answer:
32π is the volume of the larger sphere
Step-by-step explanation:
The volume of a sphere is V = 4/3πr³. The smaller sphere is V = 4π. This means r³ divided by 3 was 1. So r³ = 3. For the larger sphere, substitute r = 2r into the volume formula.
V = 4/3πr³
V = 4/3π(2r)³
V = 4/3π8r³
We know from the previous problem that r³ = 3.
So the volume formula simplifies to V = 4π(8) = 32π
The period of oscillation of a pendulum is
Where L is the length of the pendulum and
the gravitational acceleration.
Re-arranging the formula and using T=1.0 s, we find the length of the pendulum:
Which of the following pairs of equations will never break even?
<u><em>B: I and IV</em></u>
