I believe it would be the third option
To complete all 42 laps it will take him 10.5 pounds of gas. so if he only has 5.92 pounds then he wont finish.
So we are starting out with 30 messages.
If 1/5 of the messages Lily receives are picture messages, then take 1/5 x 30 to find out how many that is.
1/5 x 30 = 30/5 = 6. So Lily receives 6 picture messages.
If 7/8 of the remaining messages are texts, then we need to find the number of remaining messages. If 6 are pictures and she receives 30 overall, then 30 - 6 = 24, the remaining number of messages. Since we know 7/8 of <em>this </em>number is the number of texts Lily received, let's do some more multiplication:
7/8 x 24 = (7 x 24)/8 = 168/8 = 21.
So Lily received 21 text messages, 6 picture messages, and 3 of something else!
Answer:
1050 g < weight ≤ 1150 g
Step-by-step explanation:
Let w represent the weight of the package in grams. The the number of 100-gram increments after the first 250 grams is given by ...
⌈(w-250)/100⌉ . . . . . . . where ⌈ ⌉ signifies the <em>ceiling</em> function
and the charges for a package exceeding 250 grams will be ...
0.65 + 0.10⌈(w -250)/100⌉ = 1.55
0.10⌈(w -250)/100⌉ = 0.90 . . . . . . . . subtract 0.65
⌈(w -250)/100⌉ = 9 . . . . . . . . . . . . . . . divide by 0.10
8 < (w-250)/100 ≤ 9 . . . . . . . . . . . . . . meaning of ceiling function
800 < w -250 ≤ 900 . . . . . . . . . . . . . multiply by 100
1050 < w ≤ 1150 . . . . . . . . . . . . . . . . . add 250
The weight in grams could be greater than 1050 and at most 1150 for a charge of $1.55.
If the car is traveling at 36ft/s per converted mph divide the amount of travel length the person will travel and subtract is by 1 this being the second tracked which is 36ft/s