Answer:
So, The possible values of h and r are
1. r = 3 , h = 3
2. r = 1, h = 27
3. r = 2, h = 6.75
Step-by-step explanation:
P.S - The exact ques is -
As we know that Volume of cone , V =
As given,
V =
⇒
So,
The possible values of h and r are
1. r = 3 , h = 3
2. r = 1, h = 27
3. r = 2, h = 6.75
In Case I-
When r = 3, h = 3
(3)² ×3 = 9×3 = 27
Satisfied
Case II-
r = 1, h = 27
(1)²×27 = 1×27 = 27
Satisfied
Case III-
r = 2, h = 6.75
(2)²×6.75 = 4×6.75 = 27
Satisfied
The fraction of a defective item getting by both inspectors is =
Step-by-step explanation:
Step 1; Assume that the probability of the first inspector missing a defective part is P(A) and the probability of the second inspector missing those that do get past the first inspector is P(B).
Step 2; It is given that P(A) = 0.1, we convert this into a fraction so that the final probability will be a fraction and not a decimal.
P(A) = 0.1 = .
It is given that the second inspector misses 5 out of 10 that get past the first inspector, so P(B) = .
Step 3; To calculate the probability of both inspectors missing a defective part, we multiply both the probabilities.
P(A and B happening) = P(A) × P(B) = ×
=
=
= 0.05%. So there is a 0.05% chance of both inspectors missing a defective part.