What cycle time would match capacity and demand if demand is 173 units a day, there are 2 shift(s) of 458 minutes each, and workers are given three 35 minute breaks during each shift, one of which is for lunch or dinner? (round your answer to 2 decimal places.) cycle time 0 minutes per cycle

so cycle time is 4.39 min

Step-by-step explanation:

given data

demand d = 173 units

shift s  = 2

time t1 = 485 min

break b = 3

break time t2 = 35 min

to find out

cycle time

solution

first we find total time available that is

time available = t1 × s - s × b × t2   1

put here all these value which is given

time available = 485 × 2 - 2 × 3 × 35

time available = 760

so here cycle time is express as

cycle time = time available / demand   2

put here value

cycle time = 760 / 173

so cycle time is 4.39 min

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204