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lbvjy [14]
3 years ago
15

A motocross rider is in the air for 2.5 seconds. Your camera can take a picture every 0.125 second. Your friend's camera can tak

e a picture every 0.15 second.
a. How much times faster is your camera than your friend's camera?

b. How many more pictures can you take while the rider is in the air?
Mathematics
2 answers:
Tju [1.3M]3 years ago
8 0

Answer: a) Our camera is 1.2 times faster than friend's camera.

b) We can take 20 pictures while the rider is in the air.


Step-by-step explanation:

According to the given problem

a) 1 picture in our camera takes time =0.125 seconds

1 picture in friend's camera takes time =0.15 seconds

The rate of pictures taken by our camera=\frac{1}{0.125}\ picture\ per\ second

The rate of pictures taken by friend's camera=\frac{1}{0.15}\ picture\ per\ second

Now, scale factor by which our camera is faster then friend's camera=

\frac{\frac{1}{0.125}}{\frac{1}{0.15}}=\frac{0.15}{0.125}=1.2

Thus Our camera is 1.2 times faster than friend's camera.

b) Number of pictures we can take while the rider is in the air=\frac{2.5}{0.125}=20 [∵Duration of motocross rider is in the air = 2.5 seconds.]

Thus we can take 20 pictures while the rider is in the air.

alexandr402 [8]3 years ago
6 0
Part a .25 times faster and part b is 20 pictures per second
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Step-by-step explanation:

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Substitute it into the first equality:

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Check whether remaining two values of x and y suit this expression:

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3 years ago
Read 2 more answers
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Step-by-step explanation:

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Step-by-step explanation:

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