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STatiana [176]
3 years ago
5

Which expression can be used to change 75 kilometers per hours to meters per minute?

Mathematics
1 answer:
Artemon [7]3 years ago
4 0
The fractions you use must be the correct conversion factors.

1 km = 1000 m, so you can have the fraction (1 km)/(1000 m) or the fraction (1000 m)//(1 km). You must choose the correct one for this problem.

1 hr = 60 min, so you can have the fraction (1 hr)/(60 min) or the fraction (60 min)/(1 hr). Again, you must choose the correct one for this problem.

The important thing is to choose the correct fraction from the choices.
How do you know which conversion fractions to use?
You look at the given units and the units you want.
You must pick the conversion fraction that will cancel out the units you DON'T want, leaving the units you want.

You have a speed in km/hr.
km is in the numerator. To cancel out km, you must pick the conversion fraction that km in the denominator.
hr is in the denominator. To cancel out hr, you must pick the fraction that has hr in the numerator.

75 \frac{km}{hr} \times \frac{1000~m}{1~km} \times \frac{1~hr}{60~min}

Notice that km and km cancel out leaving m.
Also, hr and hr cancel out leaving min.
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Leona [35]

Answer:

The correct option is C

Step-by-step explanation:

A: We know that the lower quartile is 144, and the upper quartile is 129.5 is you add them all up it is going to equal 273.3. So we know that A can now be crossed out.

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3 0
2 years ago
Read 2 more answers
2r2 + 3s3 − r2 + 4t2 − r2 if r = −2, s = −3, and t = 5.
Elza [17]

First, simplify the expression 2r^2 + 3s^3 - r^2 + 4t^2 - r^2. Here you have three terms with r^2, group them in the following way:

2r^2 + 3s^3 - r^2 + 4t^2 - r^2=2r^2-r^2-r^2+3s^3+4t^2=(2r^2-r^2-r^2)+3s^3+4t^2=

=r^2(2-1-1)+3s^3+4t^2=3s^3+4t^2.

Then substitute s=-3 and t=5:

3s^3+4t^2=3\cdot (-3)^3+4\cdot 5^2=3\cdot (-27)+4\cdot 25=-81+100=19.

Answer: 19.

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3 years ago
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