Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Battery A: . Battery B: . Assume equal variances and that we calculated . Perform a .05 level test to determine if Battery B outlasts Battery A by more than 2 hours.

Gdjehhjjjbb bvggcddfghg bbgg

F(x) =2.96.(1.04)ˣ, where x is the number of monthsprice at the origin, when x = 0→ f(x) = 2.96(1.04)⁰  →f(x) = 2.96(remember any number with exponent 0 gives 1).price after 1 month (x=1)  → f(x) =2.96.(1.04)¹ = 3.0784increase of cost = (3.0784-.96)/2.96 = 0.04 or 4% increase.however you could have noticed it from the formula: f(x) =2.96.(1.04)ˣ = 2.96(1+0.04)ˣ (and the 0.04 or 4% is the increase over 1.sorry for the 2nd part, i am confused with your "table" . i have to see a proper table in order to you