A pitcher's earned-run average e varies directly as the number r of earned runs allowed and inversely as the number i of innings pitched. a certain pitcher had an earned-run average of 2.92, giving up 85 earned runs in 262 innings. how many earned runs would the pitcher have given up having pitched 330 innings with the same average?

The pitcher would have 107 earned runs

Step-by-step explanation:

- The earned-run average is E

- The number of earned runs is R

- The number of innings pitched is I

- E varies directly with R ⇒ E ∝ R

- E is varies inversely with I ⇒ E ∝

- The earned average is 2.92 when giving up 85 earned runs in

  262 innings

- We need to find how many earned runs would the pitcher have given

  up having pitched 330 innings with the same average

∵ E ∝ R and E ∝

∴ , where k is the constant of variation

- To find the constant of variation k we must use the initial values of E,

  R, and I

∵ E = 2.92 , R = 85 and I = 262

- Substitute these values in the equation above

∴

- Multiply both sides by 262

∴ 765.04 = 85 k

- Divide both sides by 85

∴ k ≅ 9

- Substitute the value of k in the equation

∴ The equation is

∵ E = 2.92 (same average)

∵ I = 330 innings

- We need to find the number of earned runs R

∴  

- Multiply both sides by 330

∴ 963.6 = 9 R

- Divide both sides by 9

∴ R = 107.07 ≅ 107

* The pitcher would have 107 earned runs

pretty sure 750 each month!

step-by-step explanation:

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages