The university bought a total of 2200 computers they bought three kinds of computers HP IBM and dell the ratio of the number of HP computers to IBM to dell is 2/3 and 6 each HP computer cost $750 each. Computer cost 90% of the price of the HP computer if the university spent a total 1650000 what is the cost of each IBM computer

Complete question :

The University but a total of 2200 computers they bought three kinds of computers HP, IBM and dell. The ratio of theNumber of HP computers to IBM to dell is 2:3:6 each HP computer cost $750 each dell computer cost 90% of the price of an HP computer if the University spent a total of $1,650,000 what is the total cost of each IBM computer

$900

Step-by-step explanation:

Given that :

Total amount of computers purchased = 2200

Number of brands purchased = 3;

HP, IBM and DELL

The ratio of theNumber of HP computers to IBM to dell is 2:3:6

Cost per HP = $750

Cost per Dell = 90% of HP = 0.9 × $750 = $675

Total amount spent = $1,650,000

2:3:6 = (2+3+6) = 11

Number of IBM computers purchased :

(3/11) * 2200 = 600 IBM

(2/11) * 2200 = 400 HP

(6/11) * 2200 = 1200 DELL

TOTAL Cost of HP = (400 × $750) = $300,000

TOTAL cost of Dell = (1200 * $675) = $810,000

TOTAL cost of IBM :

$1,650,000 - ($300,000 + $810,000)

$1,650,000 - $1,110,000 = $540,000

Cost of each IBM:

540,000 / 600 = $900

the circle one

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm