The reduction in cholesterol level after eating a certain brand of oatmeal for breakfast for one month in people with cholesterol levels over 200 is Normally distributed, with mean (in milligrams) μ and standard deviation σ = 8 . Although the brand advertises that eating their oatmeal for breakfast daily for one month will produce a mean decrease in cholesterol of more than 20 points for people with cholesterol levels over 200, you believe that the mean decrease in cholesterol is actually less than advertised. To explore this, you test the hypotheses H0 : μ = 20 , H a : μ < 20 at the 0.05 level. You will randomly select 144 people with cholesterol levels over 200. After they eat this brand of oatmeal for one month, you will compute the mean decrease in cholesterol levels of the subjects.

If you observe a sample mean cholesterol reduction of ¯x = 18.52 , the P ‑value obtained is:

A. less than 0.01.
B. between 0.025 and 0.5.
C. between 0.05 and 0.1.
D. between 0.01 and 0.025.

After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.

Step-by-step explanation:

To find a percent of a number, change the percent to a decimal and multiply by the number. To find 90% of a number, change 90% to a decimal and multiply by the number.

90% = 0.9

He first walks 1 km. The next day, he walks 90% of 1 km. To find 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km. For the next day, he walks 90% of 0.9 km, which is 0.9 * 0.9 km = 0.81 km. To find how much he walks each day, multiply what he walked on the previous day by 0.9.

Now you can find out how much he walks each day until you see he walks less than 0.5 km.

Day 1: 1 km

Day 2: 1 km * 0.9 = 0.9 km

Day 3: 0.9 km * 0.9 = 0.81 km

Day 4: 0.81 km * 0.9 = 0.729 km

Day 4: 0.729 km * 0.9 = 0.6561

Day 5: 0.6561 km * 0.9 = 0.59049 km

Day 6: 0.59049 km * 0.9 = 0.531441 km

Day 7: 0.531441 km * 0.9 = 0.4782969 km

On the 6th day, he is still walking more than 0.5 km, but by the 7th day, he is walking less than 0.5 km.

After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.