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KonstantinChe [14]
4 years ago
5

Jim wants to build a rectangular parking lot along a busy street but only has 2400 feet of fencing available. if no fencing is r

equired along the​ street, find the maximum area of the parking lot.
Mathematics
1 answer:
PSYCHO15rus [73]4 years ago
7 0
<span>Length = 1200, width = 600
   First, let's create an equation for the area based upon the length. Since we have a total of 2400 feet of fence and only need to fence three sides of the region, we can define the width based upon the length as: W = (2400 - L)/2
   And area is: A = LW
    Substitute the equation for width, giving: A = LW A = L(2400 - L)/2
    And expand: A = (2400L - L^2)/2 A = 1200L - (1/2)L^2
    Now the easiest way of solving for the maximum area is to calculate the first derivative of the expression above, and solve for where it's value is 0. But since this is supposedly a high school problem, and the expression we have is a simple quadratic equation, we can solve it without using any calculus. Let's first use the quadratic formula with A=-1/2, B=1200, and C=0 and get the 2 roots which are 0 and 2400. Then we'll pick a point midway between those two which is (0 + 2400)/2 = 1200. And that should be your answer. But let's verify that by using the value (1200+e) and expand the equation to see what happens:
   A = 1200L - (1/2)L^2
 A = 1200(1200+e) - (1/2)(1200+e)^2
 A = 1440000+1200e - (1/2)(1440000 + 2400e + e^2)
 A = 1440000+1200e - (720000 + 1200e + (1/2)e^2)
 A = 1440000+1200e - 720000 - 1200e - (1/2)e^2
 A = 720000 - (1/2)e^2
   And notice that the only e terms is -(1/2)e^2. ANY non-zero value of e will cause this term to be non-zero and negative meaning that the total area will be reduced. Therefore the value of 1200 for the length is the best possible length that will get the maximum possible area.</span>
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Answer:

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Step-by-step explanation:

a. H0: μ = 26.6

b. Ha: μ > 26.6

c. Let x = the mean age for online students at De Anza College.

d. Student’s t-distribution

e. 9.98

f. p-value = 0.0000

g. Check student’s solution.

h. i. Alpha: 0.01        

ii. Decision: Reject the null hypothesis.        

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iv. There is sufficient evidence to conclude that the mean age of online students at De Anza College is greater than 26.6 years.i. (28.8, 30.0)

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3 years ago
A small family home in tucson arizona has a rooftop area of 1967 square feet, and it is possible to capture rain falling on abou
strojnjashka [21]

Answer:

<em>Volume = 36.4 m^3 or 9616.6 gallon</em>

<em>weight = 18025.47 N or 80213.33 lbs</em>

<em></em>

Step-by-step explanation:

The complete question is

A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about 56% of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water weigh when the tank was full (in N and in lbf)?

The area of the roof = 1967 ft^2

Area of water to be possibly captured = 56% of 1967 ft^2 = 0.56 x 1967 = 1101.52 ft^2

Rainfall depth = 14 inches = 14/12 ft = 1.167 ft

Volume of water = (area of water) x (depth of water) = 1101.52 ft^2 x 1.167 ft = <em>1285.47 ft^3</em>

<em></em>

35.315 ft^3 = 1 m^3

1285.47 ft^3 = x m^3

==> 1285.47/35.315 = <em>36.4 m^3</em>

<em></em>

1 ft^3 = 7.481 gallon

1285.47 ft^3 = x gallon

==> 1285.47 x 7.481 = <em>9616.6 gallon</em>

<em></em>

Weight of water W = ρv

where ρ is the density of water = 62.4 lbs/ft^3

v is the volume of water = 1285.47 ft^3

substituting values,

W = 62.4 x 1285.47 = <em>80213.33 lbs</em>

but,

1 lb = 4.45 N

therefore,

80213.33 lbs = 80213.33/4.45 = <em>18025.47 N</em>

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3 years ago
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3 years ago
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Answer:

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Step-by-step explanation:

Given data

Principal=$4000

Rate=1.5%

Time= 20years

The expression for simple interest is

SI= PRT/100

Substitute

SI= 4000*1.5*20/100

SI=120000/100

SI=$1200

Hence the interest is $1200

3 0
3 years ago
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