A taxi driver provides service in two zones of a city. Fares picked up in zone A will have destinations in zone A with a probability of 0.6 or in zone B with a probability of 0.4. Fares picked up in zone B will have destinations in zone A with a probability of 0.3 or in zone B with a probability of 0.7. The driver's expected profit for a trip entirely in zone A is 6; for a trip entirely in zone B is 8, and for a trip that involves both zones is 12. Find the taxi driver's average profit per trip.

The taxi driver's average profit per trip is 4.4.

Step-by-step explanation:

Probability of fares picked up in zone A with destinations in zone A = 0.6

Probability of fares picked up in zone A with destinations in zone B = 0.4

Probability of fares picked up in zone B with destinations in zone A = 0.3

Probability of fares picked up in zone B with destinations in zone B = 0.7

The driver's expected profit for a trip entirely in zone A = 6

The driver's expected profit for a trip entirely in zone B = 8

The driver's expected profit for a trip involving both zones is 12

The taxi driver's average profit per trip:

zone A with destinations in zone A = 0.6 * 6 =  3.6

zone A with destinations in zone B = 0.4 * 12 = 4.8

zone B with destinations in zone A = 0.3 * 12 = 3.6

zone B with destinations in zone B = 0.7 * 8 =  5.6

Total profit expected for 4 trips =                     17.6

Average profit per trip = 4.4 (17.6/4)

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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