You invested $6000 in two accounts paying 7% and 9% annual interest. at the end of the year, the total interest from these investments was $510. how much was invested at each rate?

Step-by-step explanation:

Let x represent the amount invested at 9%. Then 6000-x is the amount invested at 7%, and the total interest earned is ...

  0.09x + 0.07(6000-x) = 510

  0.02x + 420 = 510 . . . simplify

  0.02x = 90 . . . . . . . . . subtract 420; next, divide by 0.02

  x = 4500 . . . . . . . . . . . the amount invested at 9%

  6000-x = 1500 . . . . . .the amount invested at 7%

$4500 was invested at 9%; $1500 was invested at 7%.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

b and c

step-by-step explanation: