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Effectus [21]
3 years ago
12

World war 1 lasted from August 4th 1914 until November 11th, 1918. During this time an estimated 10 million lives were lost. Abo

ut how many people were dying per day?
Mathematics
1 answer:
Kryger [21]3 years ago
6 0
The question is so dry, mechanical, and devoid of emotion
that it's terrifying.

There is no way to assign a number to "How many people were
dying per day", and I would prefer not even to think about it in
those terms.

-- The period of time from August 4, 1914 until November 11, 1918 is 1,560 days.

-- The "average", or better, the "unit rate" of 10 million events in 1,560 days
is the quotient

                   (10,000,000 events) / (1,560 days) 

               =  6,410.3 events per day

               =     267.1 events per hour

               =       4.45 events per minute.


Reciprocally, this is a unit rate of 

                  13.48 seconds per event,
                  sustained continuously for 4.274 years !

When will we ever learn ! ?
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3 years ago
The graph shows the cost for renting a post hole digger from a local hardware store. Determine the rate of change or cost per da
kirill115 [55]

The given graph shows the cost for renting a post hole digger from a local hardware store. We have to determine the rate of change or cost per day.

Let us consider any two points.

Let us consider (1,30) and (2,60)

We have to determine the rate of change which is basically termed as "slope".

Rate of change is expressed as a ratio between a change in one variable relative to a corresponding change in another variable.

Therefore, rate of change or cost per day =

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5 0
4 years ago
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Vladimir79 [104]

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Step-by-step explanation:

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Combine like terms

4p-6

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Answer is 32cm^2. See attachment for the working.

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4 years ago
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Find the point on the plane 2x+5y+z=8 that is nearest to​ (2,0,1).
Svetach [21]

Answer:

(2.2, 0.5, 1.1)

Step-by-step explanation:

The parametric equation of the line normal to the plane and through point (2, 0, 1) can be written ...

... L = (2, 0, 1) + t(2, 5, 1)

We want to find the value of t (and the corresponding point) that makes L satisfy the equation of the plane.

... 2(2+2t) +5(0 +5t) +1(1+t) = 8 . . . . . put values from L in for x, y, z in plane

... 5 + 30t = 8 . . . . . simplify

... t = (8 -5)/30 = 0.1 . . . . solve for t (subtract 5, divide by 30)

For this value of t, L is ...

... (2, 0, 1) + 0.1(2, 5, 1) = (2.2, 0.5, 1.1)

5 0
3 years ago
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