The 3rd term and 6th term of ageometric progression are
1 answer:
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Answer:
C) 515 hours.
D) 500 hours
c) sample 3
Step-by-step explanation:
1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours
Sample Service Life (hours)
1 2 3
495 525 470
500 515 480
505 505 460
<u>500 515 470 </u>
<u>∑2000 2060 1880</u>
x1`= ∑x1/n1= 2000/4= 500 hours
x2`= ∑x2/n2= 2060/4= 515 hours
x3`= ∑x3/n3= 1880/4= 470 hours
2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using μ ± σ , μ±2 σ or μ ± 3 σ.
In this question the limits are determined by using μ ± σ .
3. Upper control limit = UCL = 520 hours
Lower Control Limit= LCL = 480 Hours
Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours
Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours
Sample 3 mean = x3`= ∑x3/n3= 1880/4= 470 hours
This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.
We see that the sample 3 mean is lower than the LCL. The other two means are within the given UCL and LCL.
This can be shown by the diagram.
Answer:
see explanation
Step-by-step explanation:
Calculate the slopes between pairs of the 3 points using the slope formula
m =
with (x₁, y₁ ) = (- 9, 3) and (x₂, y₂ ) = (- 3, 9)
m = =
=
= 1
Repeat with (x₁, y₁ ) = (5, 1) and (x₂, y₂ ) = (- 3, 9)
m = =
= - 1
If lines are perpendicular then the product of their slopes = - 1 , then
1 × - 1 = -1
Thus there is a right angle between the 2 lines
Then triangle is right- angled