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Ann [662]
3 years ago
14

Suppose about 900,000 people live in an area of 1,800 square miles. What is the best estimate for the population density?

Mathematics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

The Population Density is 500\ People/mi^2.

Step-by-step explanation:

Given,

Total number of People = 900,000

Total Land Area = 1800 sq. mi.

Solution,

For calculating the population density, we have to divide the total number of people by the area of the land.

This can be framed in equation form'

Population\ Density=\frac{Total\ Number\ of\ People}{Land\ Area}

Now putting the given values, we get;

Population\ Density=\frac{900,000}{1800\ mi^2}=500\ People/mi^2

Hence The Population Density is 500\ People/mi^2.

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A $60,000 piece of machinery depreciates in value at a rate of 11% per year. About what will its value be in 5 years?
kicyunya [14]

Answer:

$27000

Step-by-step explanation:

11% of 60000 is 3300 per year. You need the value for five years so multiply 3300 * 5 to get 33000. 33000 is the total depreciation of the five years. Then subtract 33000 from 60000 to get 27000.

3 0
3 years ago
Read 2 more answers
7/2 inches by 3/2 inches?
miskamm [114]

Answer:

<u><em>5.25 inches or </em></u>5 \frac{1}{4}<u><em> inches</em></u>

Step-by-step explanation:

<u><em>I believe that the question is what do they multiply to..? If so, we start by </em></u><u><em>making them both in decimal form:</em></u>

<u><em>7 / 2 = 3.5</em></u>

<u><em>3 / 2 = 1.5</em></u>

<u><em>Now, we </em></u><u><em>multiply</em></u><u><em>:</em></u>

<u><em>3.5 * 1.5 = 5.25</em></u>

<u><em>If you want fraction form, we can turn .25 to 1/4.</em></u>

<u><em>Fraction form: </em></u>5 \frac{1}{4}

6 0
2 years ago
Emily has a coupon for 20 percent off her purchase. She finds a backpack she likes on the discount rack. It’s original price is
Bas_tet [7]
She’s incorrect I’m pretty sure u divide 20 and 60.
4 0
3 years ago
Solve the unknown viable<br><br> 7c/12 + 5c/16=15/48 +c
Mashutka [201]
<span>deconstructing
48 : 2 = 24
24 : 2 = 12
12 :2 = 6
6 : 2 = 3
3 : 3 = 1
2 * 2 * 2 * 2 * 3 = 48 
...................................
16 : 2 = 8
8 : 2 = 4
4 : 2 = 2
2 : 2 = 1
2 * 2* 2 * 2 = 16
..................................
12 : 2 = 6</span>\frac{7c}{12}  +  \frac{5c}{16}  =  \frac{15}{48}  + c \\  \\  \frac{7c*4}{12*4}  +  \frac{5c*3}{16*3}  -  \frac{15}{48}  = c \\  \\  \frac{28c + 15c}{48}  -  \frac{15}{48}  = c \\  \\ 48c =  43c - 15 \\  \\ 48c - 43c = -15 \\  \\ 5c =- 15 \\  \\ c =  -3&#10;<span>
6 : 2 = 3
3 : 3 = 1
</span><span>common denominator 48</span>
5 0
3 years ago
A uniform beam of length L = 7.30m and weight = 4.45x10²N is carried by two ovorkers , Sam and Joe - Determine the force exert e
Mama L [17]

Answer:

Effort and distance = Load  x distance

7.30 x 4.45x10^2N = 3.2485 X 10^3N

We then know we can move 3 points to the right and show in regular notion.

= 3248.5

Divide by 2 = 3248.5/2 = 1624.25 force

Step-by-step explanation:

In the case of a Second Class Lever as attached diagram shows proof to formula below.

Load x distance d1 = Effort x distance (d1 + d2)

The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.

If the wheelbarrow is held steady (i.e. in Equilibrium) then the moment of the effort must be equal to the moment of the load :

Effort x its distance from wheel centre = Load x its distance from the wheel centre.

This general rule is expressed as clockwise moments = anti-clockwise moments (or CM = ACM)

 

This gives a way of calculating how much force a bridge support (or Reaction) has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!

The moment of the load on the beam (F) must be balanced by the moment of the Reaction at the support (R2) :

Therefore F x d = R2 x D

It can be seen that this is so if we imagine taking away the Reaction R2.

The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up.

The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example attached.

We know from our experience that a lighter person will have to sit closer to the end of the see-saw to balance a heavier person - or two people.

So if CM = ACM then F x d = R2 x D

from our kitchen scales example above 2kg x 0.5m = R2 x 1m

so R2 = 1m divided by 2kg x 0.5m

therefore R2 = 1kg - which is what the scales told us (note the units 'm' cancel out to leave 'kg')

 

But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated.

Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams (or bridges) have several loads acting upon them.

In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.e. F1=60kg, F2=40kg, d1=6m and d2=9m

 

CM = ACM

(F1 x d1) + (F2 x d2) = R2 x Length of beam

(60kg x 6m) + (40Kg x 9m) = R2 x 12m

(60kg x 6m) + (40Kg x 9m) / 12m = R2

360kgm + 360km / 12m = R2

720kgm / 12m = R2

60kg = R2 (note the unit 'm' for metres is cancelled out)

So if R2 = 60kg and the total load is 100kg (60kg + 40kg) then R1 = 40kg

4 0
2 years ago
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