Aradiator contains 10 quarts of fluid, 30% of which is antifreeze. how much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?

1 3/7 quarts should be drained off and replaced with pure antifreeze.

1 3/7 ≈ 1.4286

Current amount of antifreeze in quarts is -

30/ 100 × 10 = 3

40% ---> 4 quarts

Let the amount drained of and replaced with antifreeze be x-

The amount left after draining off is 10 − x.

The amount of antifreeze  is 30/ 100 (10−x).

30/100(10-x)+x=4

3-3/10x+x=4

3+x(1-3/10)=4

x=1*10/7=1 3/7 quarts

check;

10- 1 3/7 = 8 4/7

=(30/100*8 4/7)+1 3/7

=(3/10 * 60/7) + 10/7

=3*6/7 + 10/7

=28/7

=4

4 liters of pure antifreeze is mixed into 10 quarts.

0.6

39mph

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.