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mariarad [96]
3 years ago
15

The Greenbaum family agreed to pay for 3 months of an online TV service in exchange for a $5 credit on the bill each month. If t

he Greenbaums spend a total of $11.85 on the service over the 3 months, what is the normal price of one month of online TV service?
Mathematics
2 answers:
anzhelika [568]3 years ago
6 0
The Greenbaum family agreed to pay for 3 months of an online TV service in exchange for a $5 credit on the bill each month.
If the Greenbaums spend atotal of $11.85 on the service over the 3 months,
Let's solve for the normal price of one month of online TV service.
=> 11.85 dollars / 3 months = 3.95 dollars per month is the monthly service.
blsea [12.9K]3 years ago
4 0

Answer:

$8.95

Step-by-step explanation:

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A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
earnstyle [38]

Answer:

(i). Average speed = 75 km per hour

(ii). Average speed = 75 km per hour

Step-by-step explanation:

This question is incomplete; here is the complete question.

A train traveled from station A to station B at an average speed of 80 kilometers per hour and then from station B to station C at an average of 60 kilometers per hour. If the train did not stop at station b, what was the average speed at which the train traveled from station A to C?

(i). The distance that the train traveled from station A to station B was 4 times the distance that train traveled from station B to station C.

(ii). The amount of time it took to the train to travel from station A to station B is 3 times the amount of time that it took the train to travel from station B to station C.

(i) Let the distance between station B and station C = x km

So the distance between Station A and station B = 4x km

Therefore, time to travel the distance x km with 80 km per hour = \frac{\text{Distance traveled}}{\text{Speed}}

= \frac{4x}{80}

Similarly time taken to travel between station B and station C with speed 60 km per hour = \frac{x}{60} km per hour

Now average speed between station A and station C = \frac{\text{Total distance between station A and station C}}{\text{time taken to travel}}

= \frac{4x+x}{\frac{4x}{80}+\frac{x}{60}}

= \frac{5x}{\frac{4x}{60} }

= \frac{5\times 60}{4}

= 75 km per hour

(ii). Let the train took the time to cover the distance between station B and station C = t hours

Therefore, time taken by the train between station A and B = 3t

Distance between Station A and station B = Speed × time

= 80 × 3t

= 240t km

Similarly distance between station B and station C = 60 × t

= 60t

Now average speed between station A and station C = \frac{240t+60t}{(3t+t)}

= \frac{300t}{4t}

= 75 km per hour

4 0
3 years ago
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