Answer:
The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.
Step-by-step explanation:
The three different assembly lines are: A₁, A₂ and A₃.
Denote <em>R</em> as the event that a component needs rework.
It is given that:
Compute the probability that a randomly selected component needs rework as follows:
Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:
Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
We can proceed in solving the problem since all information are given such as 2*4*5costheta.
we have a=4, b=5, and C=theta
let us solve for "c" using Pythagorean
c²=a²+b²
c²=4²+5²
c=6.4
Solving for theta or C
c²=a²+b²-2abcosC
6.4²=4²+5²-2*4*5*cosC
C=90
Answer:
y = 50 + 35x we dont know how many hours they worked
Step-by-step explanation:
<h2>
Answer:</h2>
7
<h2>
Step-by-step explanation:</h2>
4 1/5 ÷ 3/5 = 7
Kate cut 7 pieces of rope.
<h2>#BrainliestAnswer</h2><h2>#CarryOnLearning</h2>
