If we consider the first half mile to be charged at $0.30 per tenth also, that half-mile costs $1.50 and the charges amount to a fixed fee of $2.00 and a variable fee of $0.30 per tenth mile.
After you subtract the $2 tip and the fixed $2 fee from the trip budget amount, you have $11.00 you can spend on mileage charges. At 0.30 per tenth mile, you can travel
... $11.00/$0.30 = 36 2/3 . . . . tenth-miles
The trip is measured in whole tenths, so you can ride ...
... 36 × 1/10 = 3.6 miles
_____
If you want to see this in the form of an equation, you can let x represent the miles you can travel. Then your budget amount gives rise to the inequality ...
... 3.50 + 0.30((x -.50)/0.10) + 2.00 ≤ 15.00
... 3.50 + 3x -1.50 +2.00 ≤15.00 . . . . . . . eliminate parentheses
... 3x ≤ 11.00 . . . . . . . . . . . . . . . . . . . . . . . . collect terms, subtract 4
... x ≤ 11/3 . . . . . . . . . . . . . . . . . . . . . . . . . . divide by 3
... x ≤ 3.6 . . . . . rounded down to the tenth
Answer:
it's a 0.02% error.
147/150= 0.98% of the folders are there, making the error 0.02%
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
Using the normal distribution, it is found that 25.14% of the batteries will last more than 420 hours.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, we have that the mean and the standard deviation are given, respectively, by:
.
The proportion of the batteries will last more than 420 hours is <u>one subtracted by the p-value of Z when X = 420</u>, hence:


Z = 0.67
Z = 0.67 has a p-value of 0.7486.
1 - 0.7486 = 0.2514.
0.2514 = 25.14% of the batteries will last more than 420 hours.
More can be learned about the normal distribution at brainly.com/question/24663213
#SPJ1