. A computer rework center has the capacity to rework 300 computers per day. The expected number of computers needing to be reworked per day is 225. The center is paid $26 for each computer reworked. The fixed cost of renting the reworking equipment is $250 per day. Work space rents for $150 per day. The cost of material is $18 per computer and labor costs $3 per computer. What is the break-even number of computers reworked per day?

The break-even number of computers reworked per day is 80.

Step-by-step explanation:

The break-even number of computers reworked per day represents the number of computers that makes the total revenue equals the total expenditure.

Let the break-even number of computers reworked per day be y

Revenue for each computer reworked = $26

Total revenue for y computers reworked = $26y

Fixed cost of renting the reworking equipment per day = $250

Work space rent per day = $150

Cost of material per computer = $18

Cost of material for y computers = y × $18 = $18y

Labor cost per computer = $3

Labor cost for y computers = $3y

Total expenditure = $250 + $150 + $18y + $3y = $400 + $21y

Total revenue = total expenditure

$26y = $400 + $21y

$26y - $21y = $400

$5y = $400

y = $400/$5 = 80

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)