Answer:
I think b, but I may be wrong
Let x be the distance (in feet) along the road that the car has traveled and h be the distance (in feet) between the car
and the observer.
(a) Before the car passes the observer, we have dh/dt < 0; after it passes, we have dh/dt > 0. So at the instant it passes the observer we have
dh/dt = 0, given that dh/dt varies continuously since the car travels at a constant velocity.
Answer: 72%
Step-by-step explanation: Well, we want to know what percent 608 is of 845, so to solve that, let's call 845 100%, as we're solving it terms of 845. 845 is 100%, then it follows that 8.45 is 1%.
Now we can see how many 8.45's go into 608, which comes out to 71.95266...%, which can be rounded nicely to 72%
Mean:
E[Y] = E[3X₁ + X₂]
E[Y] = 3 E[X₁] + E[X₂]
E[Y] = 3µ + µ
E[Y] = 4µ
Variance:
Var[Y] = Var[3X₁ + X₂]
Var[Y] = 3² Var[X₁] + 2 Covar[X₁, X₂] + 1² Var[X₂]
(the covariance is 0 since X₁ and X₂ are independent)
Var[Y] = 9 Var[X₁] + Var[X₂]
Var[Y] = 9σ² + σ²
Var[Y] = 10σ²
Ii.First ,expand: 5(x-2)=32
5x-10=32
bring 10 to the other side (add it to 32)
5x=42
divide both sides by 5 to get x
x=42/5
x=8.4
iii.Expand 5+2(x+3)=21
5+2x+6=21
bring 5 and 6 to the other side
2x=21-5-6
2x=10 divide both sides by 2
x=5