Question

question 5 (1 point) a car is brought in for new spark plugs, filters, and a tune - up. when the car comes in initially the engine is producing 290 hp running at 2000 rpm. after the tune - up is completed the engine produces 385 hp at 2000 rpm. what is the percent increase in the engines efficiency. round your answer to the nearest tenth.

Explanation:

Step1: Identify initial and final values

Initial power $P_1 = 290$ Hp, final power $P_2=385$ Hp

Step2: Use percent - increase formula

The formula for percent increase is $\text{Percent Increase}=\frac{P_2 - P_1}{P_1}\times100\%$. Substitute $P_1 = 290$ and $P_2 = 385$ into the formula: $\frac{385 - 290}{290}\times100\%=\frac{95}{290}\times100\%$.

Step3: Calculate the result

$\frac{95}{290}\times100\%=\frac{9500}{290}\approx 32.8\%$

Answer:

$32.8$