Answer:
given,
Probability of student studying math P(M)=
Probability of student studying physicsP(P) =
Probability of student studying both math and physics together P(M∩P) =
a) student took mathematics or physics
P(M∪P) = P(M) + P(P) - P(M∩P)
= 0.45 + 0.61 - 0.25
= 0.81
b) student did not take either of the subject
P((M∪P)') = 1 - 0.81
= 0.19
c) Student take physics but not mathematics
P(P∩M') = P(P) - P(P∩M)
= 0.61 - 0.25
= 0.36
studying physics and mathematics is not mutually exclusive because we can study both the subjects.
Answer:
Acceleration:
Explanation:
The acceleration of an object is equal to the rate of change of velocity:
where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
For the space probe in this problem, we have:
u = 100 ft/s (initial velocity)
v = 5000 ft/s (final velocity)
t = 0.5 s (time taken)
Therefore, the acceleration is
Answer:
the answers, the correct one is D, Rb₂ = 29.97 ohm
Explanation:
For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.
Serial circuit
(Ra + Rb) is = V
(Ra + Rb) 0.111 = 10
(Ra + Rb) = 10 / 0.111 = 90.09
parallel circuit
R =
\frac{Ra \ Rb}{Ra + Rb} i_p = V
\frac{Ra \ Rb}{Ra + Rb} 0.5 = 10
\frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20
we write and solve our system of equations
Ra + Rb = 90.09
\frac{Ra \ Rb}{Ra + Rb} = 20
we solve for Ra in the first equation
Ra = 90.09 - Rb
RaRb = 20 (Ra + Rb)
we substitute Ra in the second equation
(90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]
90.09 Rb - Rb² = 20 90.09
Rb² - 90.09 Rb + 1801.8 = 0
we solve the quadratic equation
Rb = [90.09 ± ] / 2
Rb = [90.09 ± 30.15] / 2
Rb₁ = 60.12 ohm
Rb₂ = 29.97 ohm
the smallest value is Rb = 30 ohm
When checking the answers, the correct one is D