The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
- Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
- Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.
Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'
D = 180t + 10 + 160t
D = 340t + 10
<h3>Number of minutes before they are 1870 feet</h3>
Making t subject of the formula, we have
t = (D - 10)/340
Since they are 1870 feet apart after t minutes, D = 1870 feet.
t = (D - 10)/340
t = (1870 - 10)/340
t = 1860/340
t = 5.47 minutes
So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
Learn more about minutes of distance apart here
brainly.com/question/8783264
The probability of one head must be 3/8. I believe there is a 50% chance of getting a toss of more then 1 head in 3 tosses.
Answer:
squares
Step-by-step explanation:
Lantana needs $1776 to reach its goal of $14000
Answer:
Step-by-step explanation:
x is the number of students
(x+5) is the number of students and chaperons
$2.50·(x+5) is the amount of money the number of students and chaperons will pay for the trip, and that number should be less or equal than $90 because those are all the money they have
2.50(x+5) ≤ 90 which is the same as 90 ≥ 2.50(x+5)
The inequality "90 greater-than 2.50 (x + 5) "
90 > 2.50(x+5) is an error becase it excludes the possibility that <u>the trip can cost exact $90</u> so we need not just greater than > , yet greater and equal than ≥ sign
90 ≥ 2.50(x+5)
