9514 1404 393
Answer:
(c) dZ/dt = 11/√544, so moving away at about 0.47 m/s
Step-by-step explanation:
The (x, y) coordinates of the car are related by the fact that they are on a circle centered at (12, 7) with a radius of 5. Then their rates of change are related by the derivative of the circle equation with respect to time.
(x -12)^2 + (y -7)^2 = 25
2(x -12)x' +2(y -7)y' = 0
y' = -(x -12)/(y -7)x'
At the time and point of interest, we have ...
y' = -(15 -12)/(11 -7)(1) = -3/4
__
The distance (z) to the observer is given by the Pythagorean theorem:
z^2 = (x -0)^2 +(y -2)^2
and its rate of change with time is ...
2z·z' = 2(x-0)x' +2(y -2)y'
The distance d at the point of interest is ...
z = √((15 -0)^2 +(11 -2)^2) = √(225 +81) = √306 = 3√34
So, the rate of change of distance to the observer at the time and point of interest is ...
z' = (x(x') +(y -2)y')/z
z' = ((15)(1) +(11-2)(-3/4))/(3√34) = (33/4)/(3√34)
z' = 11/√544 ≈ 0.47 . . . m/s
140 you divide 490 by 3.5.
Answer:
option c
x + 4 = 64
and
option e
x = 60
Step-by-step explanation:
Given in the question an expression
3∛x+4 = 12
∛x + 4 = 12/3
∛x + 4 = 4
Take cube on both sides of the equation
∛(x + 4)³ = 4³
x + 4 = 64
x = 64 - 4
x = 60
Answer:
the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds is 0.05
Step-by-step explanation:
Given the data in the question;
μ_x = 10 pound bags
standard deviation s_x = 0.24 pounds
sample size n = 4
The bag weights are normally distributed so;
p( x' less than 9.8 ) will be;
p( (x'-μ_x' / s_x') < (9.8-μ_x' / s_x') )
we know that;
μ_x' = μ_x = 10
and s_x' = s_x/√n = 0.24/√4
so; we substitute
p( z < ( (9.8 - 10) / (0.24/√4) )
p( z < -0.2 / 0.12 )
p( z < -1.67 )
{ From z-table }
⇒ p( z < -1.67 ) = 0.0475 ≈ 0.05
Therefore, the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds is 0.05