Answer:
Answer:18+4.5
Step-by-step explanation:
This shape is a triangle with a semicircle connected to it
and this triangle is a right triangle so side A=B
that means the other leg is 6.Knowing that we can solve
The formula for the area of a triangle is (base×height)÷2
so that means 6×6 equals 36 and if you divide that by 2 you get the 18.
Now we will deal with the semicircle. We know that both of the legs are 6 so that means the diameter is 6 and now we solve 6 divided by 2 equals 3 and we will have to square that and we will 9 and since it is a semicircle we have to divide it by 2 and that will give us 4.5 and since we have to express it in terms of pi it will be 4.5(pi) and then we add both of the areas
giving us 18+4.5(pi)
and do you go to RSM
<u>Given</u>:
Let x represents the radius of the cylinder.
Given that the height of the cylinder is twice the radius of its base.
The height of the cylinder is 2x.
We need to determine the volume of the cylinder.
<u>Volume of the cylinder:</u>
The volume of the cylinder can be determined using the formula,
Substituting r = x and h = 2x, we get;
Simplifying, we get;
Thus, the expression that represents the volume of the cylinder is 2πx³ cubic units.
Hence, Option b is the correct answer.
Answer:
(a) 0.0686
(b) 0.9984
(c) 0.0016
Step-by-step explanation:
Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).
Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.
So, from the given data:
P(A)=0.90, P(B)=0.03, and P(C)=0.07.
Let E be the event that the part is disregarded by the inspection machine.
As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.
So,
Now, from the conditional probability,
This is the probability of disregarding the defect-free parts by inspection machine.
Similarly,
and
This is the probability of disregarding the partially defective parts by inspection machine.
and
This is the probability of disregarding the defective parts by inspection machine.
(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine
[from equation (iii)]
(b) The total probability that the parts produced get disregarded by the inspection machine,
So, the total probability that the part produced get shipped
The probability that the part is good (either defect free or slightly defective)
So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped
(c) The probability that the 'bad' (defective} parts get shipped
=1- the probability that the 'good' parts get shipped
=1-0.9984
=0.0016