Exercise 155 Page 539

a If m∠ ACB=90^(∘), the shaded region occupies 14 of the circle.

This means per spin, the probability of winning is 14. This means that we have a 14 chance of winning $1000 and 34 chance of winning nothing ($0). The expected value for one spin is the sum of the winnings multiplied by their probabilities. E = $1000 1/4 + $0 3/4 = $250 The expected value for one spin is $250.

b If the shaded region is 1^(∘), it makes up 1360 of the spinner. This means that we have a 1360 chance of winning $1000 and 359360 chance of winning nothing ($0). The expected value for one spin is the sum of the winnings multiplied by their probabilities.

E = $1000 1/360 + $0 359/360 ≈ $2.77 The expected value for one spin is approximately $2.77.

c We want to find the probability of winning and the measure of the shaded angle based on the given expectation value of winnings.

Since unless we win $1000 we get nothing, the expectation value E of winnings is $1000 multiplied by the probability of winning it. E = $1000 * P( winning $1000)Let's substitute the known value of the expectation value for E and solve the equation above for P( winning $1000).

E = $1000 * P( winning $1000)

Substitute

E= $200

$200 = $1000 * P( winning $1000)

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Solve for P( winning $1000)

DivEqn

.LHS /$1000.=.RHS /$1000.

0.2 = P( winning $1000)

WriteAsFrac

Write as a fraction

1/5 = P( winning $1000)

RearrangeEqn

Rearrange equation

P( winning $1000) = 1/5

The probability of winning is 15. Therefore, the ratio of the central angle and 360^(∘) must also equal 15. m∠ ACB/360^(∘)=1/5 Let's solve for m∠ ACB.

m∠ ACB/360^(∘)=1/5

MultEqn

LHS * 360^(∘)=RHS* 360^(∘)

m∠ ACB=1/5* 360^(∘)

MoveRightFacToNumOne

1/b* a = a/b

m∠ ACB=360^(∘)/5

CalcQuot

Calculate quotient

m∠ ACB=72^(∘)