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AfilCa [17]
2 years ago
9

Explain why if a runner completes a 6.2â-mi race in 32 âmin, then he must have been running at exactly 11 âmi/hr at least twice

in the race. Assume theâ runner's speed at the finish line is zero.
Mathematics
1 answer:
MA_775_DIABLO [31]2 years ago
3 0

Answer:

Accelerating to top speed, deaccelerating to finish line.

Step-by-step explanation:

If the runner kept a constant speed of 11 mph for the whole duration of his run (32 minutes), the distance he would have covered is:

d=11*\frac{32}{60}\\d= 5.87\ mi

This means that, in order to run the full 6.2 miles, the runner needs to reach a speed over 11 mph. Assume he starts from rest, while accelerating the runner reaches, and the surpasses, the 11 mph mark. Since his speed at the finish line is zero, the runner has to deaccelerate from his current running speed (which should be higher than 11 mph), passing through 11 mph and reaching zero at the finish line.

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Recall that distance = speed times time. Thus, time = distance / speed.

Here:

time = distance / speed = 0.8 km / 12 km/hr = 1/15 hr.

Note that 1/15 hr = 1/15 (60 min) = 4 minutes

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2 years ago
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Response/ NonResponse: Some students might be too tired to answer the question or in converse,  a student might feel that his or her grade might be reflected on the survey since it is not anonymous, he or she might sugarcoat the answer. (incorrect response bias)
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A tomato grower needs to ship early when prices are high and spoilage is low. She now has 25 tons on hand and by can add two ton
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On first week

Quantity of tomato = 25 tons

profit per ton = $250

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On third week

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2 years ago
A bicycle manufacturer has a stock of 50 frames 70 wheels, and 120 reflectors. if each bicycle manufactured requires 4 reflector
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30 I think but this is probably a trick question so I'm just gonna assume I'm wrong. 

5 0
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Four students wrote sequences during math class. Andre mc011-1.jpg Brenda mc011-2.jpg Camille mc011-3.jpg Doug mc011-4.jpg Which
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<span>A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

</span>The common ration is obtained by dividing the a term by the preceding term.

Given that f<span>our students wrote sequences during math class with
Andre writing -\frac{3}{4} ,\frac{3}{8} ,-\frac{3}{16} ,-\frac{3}{32} , . . .
Brenda writing </span>\frac{3}{4} ,-\frac{3}{8} ,\frac{3}{16} ,\frac{3}{32} , . . .
Camille writing \frac{3}{4} ,\frac{3}{8} ,-\frac{3}{16} ,-\frac{3}{32} , . . .
Doug writing \frac{3}{4} ,-\frac{3}{8} ,\frac{3}{16} ,-\frac{3}{32} , . . .

Notice that the common ratio for the four students is - \frac{1}{2}.

For Andre, the last term is wrong and hence his sequence is not a geometric sequence.
For Brenda, the last term is wrong and hence her sequence is not a geometric sequence.
For Camille, her sequence is not a geometric sequence.
For Doug, his sequence is a geometric sequence with a common ratio of - \frac{1}{2}.

Therefore, Doug wrote a geometric sequence.
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2 years ago
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