The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. however, records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes. suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. on a typical saturday, the oil-change facility will perform 40 oil changes between 10 a. m. and 12 p. m. treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? this will be the goal established by the manager.

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!