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OpenStudy (anonymous):

In the morning, Alan drove 100 miles at the rate of 60 miles per hour; in the afternoon, he drove another 100 miles at the rate of 40 miles per hour. What was his average rate of speed, in miles per hour, for the day?

OpenStudy (anonymous):

k...

OpenStudy (anonymous):

d=100

OpenStudy (anonymous):

anything?

OpenStudy (anonymous):

ahh

OpenStudy (anonymous):

60+40/2=50

OpenStudy (anonymous):

derp

OpenStudy (anonymous):

C.

OpenStudy (anonymous):

no

OpenStudy (anonymous):

B.

OpenStudy (anonymous):

48

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

that'swrong

OpenStudy (anonymous):

it's 48

OpenStudy (anonymous):

heck no

OpenStudy (anonymous):

oh i mean heck no to 50

OpenStudy (anonymous):

average speed is not the average of the speeds

OpenStudy (anonymous):

oh man, you are a guru

OpenStudy (anonymous):

you are even higher than DA PANIC!!!!!!

OpenStudy (anonymous):

Oh, sorry...just been a while since I've done these...

OpenStudy (anonymous):

figure out how long it took to get there, then figure out how long it took to get back that is the total time divide the total distance (200 miles ) by the total time to get the average speed

OpenStudy (anonymous):

2.5+1.666666667

OpenStudy (anonymous):

divided by 200

OpenStudy (anonymous):

equal 48

OpenStudy (anonymous):

that it?

OpenStudy (anonymous):

200/4.166666667?

OpenStudy (anonymous):

=48

OpenStudy (anonymous):

100/60=5/3

OpenStudy (anonymous):

100/40=2.5

OpenStudy (anonymous):

time going it \(\frac{100}{60}\) time returning is \(\frac{100}{40}\) total time is \[\frac{100}{60}+\frac{100}{40}=\frac{25}{6}\] and \[200\div \frac{25}{6}=200\times \frac{6}{25}=8\times 6=48\] as luisz said

OpenStudy (anonymous):

5/2+5/3=15/6+10/6=25/6

OpenStudy (anonymous):

200 divided by 25/6=48

OpenStudy (anonymous):

ty

OpenStudy (anonymous):

you nothing about truth tables, guru?

OpenStudy (anonymous):

now redo the problem replacing 100 by \(m\) and see that the answer is still 48 mph

OpenStudy (anonymous):

by friend, aripotta needs help with them?

OpenStudy (anonymous):

yes, we can do them

OpenStudy (anonymous):

go ahead and post

OpenStudy (anonymous):

its not me who needs help with them

OpenStudy (anonymous):

its aripotta

OpenStudy (anonymous):

I'm posting a different problem

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