Answer:
Step-by-step explanation:
Average speed = distance/time
Trip to dry dock
Let x be the velocity
Time = 2hours
Distance = average speed ×time
Distance = x×2
Distance = 2x
Return trip
Velocity = x-5 (5 miles per hour faster on the trip there than on the return trip)
Time = 3hrs
Distance = 3(x-5)
Since the distance is the same both to and fro;
2x = 3(x-5)
2x = 3x -15
2x-3x = -15
-x = 15
x = 15
Hence the aircraft carrier's average speed on the outbound trip is 15miles/hour
Answer:
Step-by-step explanation: it cant
Answer:
50 degrees
Step-by-step explanation:
40+50=130
130-180
50
I did it on paper just now. Sorry for the sloppy handwriting, but you should be able to answer the two questions with the info I got. If you don't know what five number summary is, just google it.
Basically, arrange the list of numbers in order, and then find the minimum, 1st quartile, median, 3rd quartile, and maximum.
Answer: y=4/3x+2
Step-by-step explanation:
-4x-3y=6
+4x +4x
-3y=4x+6
-3/-3=4/3+6/3
y=4/3x+2