So, the best way to solve this problem is by creating a equation for which we can substitute any number of days. For this problem that equation would be W=5+3(t-1)
W=number of words
T=time or number of days
-1= This is here because we are already accounting for the first day of school so we need to have 1 less day than the number of days they have been in school.
So in 20 days the students had received
5+3(19) or 62words.
Answer: 62 words
If the mean score is 55% and the s.d. is 15%, then a test score of at least
55% + 1.5 × 15% = 77.5%
falls in the A range.
The next cutoff score is 0.5 s.d. from the mean, which is
55% + 0.5 × 15% = 62.5%
so any test score between 62.5% and 77.5% is in the B range.
The next cutoff is
55% - 0.5 × 15% = 47.5%
so a score between 47.5% and 62.5% gets a C.
The lower limit of the next range is
55% - 1.5 × 15% = 32.5%
so 32.5% - 47.5% gets a D.
Anything lower than 32.5% is an F.
The answer is (d) where x = -1 & 9
45 is what percent of 190
45/190 = 0.2368
Make it into a percentage
0.2368 * 100 = 23.7
23.7% of the food is from fat
