Find the greatest common factor of 36 and 60.
2 answers:
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Answer:
15 = 2x - 3y
Step-by-step explanation:
We have the two points P1(3,-3) and P2(9,1).
First, calculate the slope between those points:
slope
Now simply move 3 steps from (3,-3) towards the y-axis to get the intersection with the y-axis (each step reduces x by 1 and applies the slope to the y-value):
(3,-3)
---> (2, -3 - (2/3)) = (2,-3 2/3)
---> (1, (-3 2/3) - 2/3) = (1, -4 1/3)
---> (0, (-4 1/3) - 2/3) = (0,-5)
This tells us, that our line intersects the y-axis at (0,-5).
The general form of a line is f(x) = <SLOPE> * x + <Y-VALUEATXEQUALSZERO>.
In our case: f(x) = y = (2/3)x -5.
To get it into the correct form, we simply subtract y from both sides and get:
0 = (2/3)x - y - 5
To get only integers just multiply everything with 3 and add 15 and get:
15 = 2x - 3y, which fulfils the form asked for in the problem.
Answer:
NOT 100% sure! If its not this I think its
Answer:
95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].
Yes, this data suggest that the tensile strength was changed after the adjustment.
Step-by-step explanation:
We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.
A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.
Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;
P.Q. = ~ N(0,1)
where, = sample mean strength of 50 items = 74.28
= population standard deviation = 2.15
n = sample of items = 50
= population mean tensile strength after machine was adjusted
<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>
So, 95% confidence interval for the population mean, is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of
significance are -1.96 & 1.96}
P(-1.96 < < 1.96) = 0.95
P( <
<
) = 0.95
P( <
<
) = 0.95
<u><em>95% confidence interval for</em></u> = [
,
]
= [ ,
]
= [73.68 kg/mm2 , 74.88 kg/mm2]
Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].
<em>Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.</em>
Answer:
Step-by-step explanation:
