Answer: A
Explanation: Case study Bayes theorem. Case histories indicate that various diseases produce similar systems. Assume that a particular set of symptoms, which will be denoted as event M, occurs only when any of three diseases, A, B or C, occurs (For simplicity, we will assume that diseases A, B and C are mutually exclusive). Studies show these probabilities of acquiring the three diseases: P (A) = 0.02 P (B) = 0.03 P (C) = 0.4 The probabilities of developing M symptoms, given a specific disease, are: P (M | A) = 0.80 P (M | B) = 0.97 P (M | C) = 0.60 Assuming that a sick person has symptoms M. What is the probability that the person has disease A?
Answer:Vb=-6i-(-0.1ωab+8)j m/s
Explanation:
Va=V0+Va0
Va=V0+(ra0 x ωao)
ω=Angular velocity of link A0
Using r0a=0.1m;
Va=V0+(0.1i x ω0a K)
Va=0
ixk=j
Va=0+0.1ω0aj
Calculating te velocity of using te equation below
Vb=Va+Vba
Vb=Va+ωab x rba
ωab=40rad/s
rab=-0.21i+0.15j
Va=0.1ω0aj
Vb=Va+ωabxrba
Vb=0.1ω0aj+40k x -(0.21i+0.15j)
Vb=0.1ω0aj-8j-6i
Vb=-6i-(-0.1ωab+8)j m/s
Answer:
b)false
Explanation:
Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling is a metal forming process.
We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.
So given statement is wrong.
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1
a) 0.01 μF
b) 0.58 μF
c) 0.060 μF
d) 0.8 μF
Answer:
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
Explanation:
Please refer to the attached Figure 12-1 where three capacitors are connected in series.
We are asked to find out the equivalent capacitance of this circuit.
Recall that the equivalent capacitance in series is given by

Where C₁, C₂, and C₃ are the individual capacitance connected in series.
C₁ = 0.1 μF
C₂ = 0.22 μF
C₃ = 0.47 μF
So the equivalent capacitance is




Rounding off yields

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)