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coldgirl [10]
3 years ago
14

Danielle constructs a scale model of a building with a rectangular base. Her model is 2 inches in length and 1 inch in width. Th

e scale on the model is 1 inch = 47 feet. What is the actual area, in square feet, of the base of the building
Mathematics
2 answers:
IRINA_888 [86]3 years ago
7 0

The area of the base is of the building  4,418 square feet

NeTakaya3 years ago
6 0

Answer:

Answer:

Area=4418\,feet^{2}

Step-by-step explanation:

We have that in the model the length is 2 inches and 1 inch in width. As we know 1 inch is 47 feet, then:

For the length we have: 2\,inches\times\dfrac{47\,feet}{1\,inch}=94\,feet

For the width we have: 1\,inches\times\dfrac{47\,feet}{1\,inch}=47\,feet

To know the area we just need to multiply length by width.

Area=94\,feet\times47\,feet=4418\,feet^{2}

then the area of the building is: 4418ft^{2}

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Likurg_2 [28]

Answer:

1.25 cups

Step-by-step explanation:

Since you have 5 5/8 cups = 4 1/2 water bottles

you can make it

5.625  cups = 4.5 water bottles

Then, all you have to do is divide both sides of this equation by 4.5, so

5.625 / 4.5 = 1 water bottle

1.25 cups = 1 water bottle

7 0
3 years ago
Customers experiencing technical difficulty with their Internet cable hookup may call an 800 number for technical support. Itake
Paladinen [302]

Answer:

(A) The percent of the problems takes more than 5 minutes to resolve is 60.87%.

(B) The problem solving time will be 50% as long as the difference between the two end points is 5.75.

(C) <em>a</em> = 0.50 minutes, <em>b</em> = 12 minutes.

(D) Mean = 6.25 minutes, Standard deviation = 3.32 minutes

Step-by-step explanation:

Let the random variable <em>X</em> represent the time it takes the technician to resolve the problem.

The random variable <em>X</em> follows an Uniform distribution with parameters <em>a</em> =  0.50 minutes and <em>b</em> = 12 minutes.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

(A)

Compute the probability that the problems takes more than 5 minutes to resolve as follows:

P(X>5)=\int\limits^{12}_{5}{\frac{1}{12-0.50}}\, dx

               =\frac{1}{11.50}\cdot \int\limits^{12}_{5}{1}\, dx \\\\=\frac{1}{11.50}\cdot [x]\limits^{12}_{5}\\\\=\frac{1}{11.50}\cdot [12-5]\\\\=\frac{7}{11.50}\\\\=0.608696\\\\\approx 0.6087

Thus, the percent of the problems takes more than 5 minutes to resolve is 60.87%.

(B)

Let the middle 50% of the problem-solving times be between <em>u</em> and <em>v</em>.

Then,

P (<em>u</em> < X < <em>v</em>) = 0.50

\int\limits^{v}_{u}{\frac{1}{12-0.50}}\, dx=0.50\\\\\frac{1}{11.50}\cdot \int\limits^{v}_{u}{1}\, dx=0.50\\\\\frac{v-u}{11.50}=0.50\\\\(v-u)=5.75

Thus, the problem solving time will be 50% as long as the difference between the two end points is 5.75.

(C)

The interval in which the technician can solve the problem is 30 seconds to 12 minutes.

So, the values of <em>a</em> and <em>b</em> in minutes is:

<em>a</em> = 30 seconds = 0.50 minutes

<em>b</em> = 12 minutes.

(D)

The mean time is:

\mu=\frac{a+b}{2}=\frac{0.50+12}{2}=6.25\ \text{minutes}

The standard deviation of the time is:

\sigma=\sqrt{\frac{(b-a)^{2}}{12}}=\sqrt{\frac{(12-0.50)^{2}}{12}}=3.3197\approx 3.32\ \text{minutes}

5 0
3 years ago
Simplify: 13.6 – 4.1 – 3.8 + 6.7
blsea [12.9K]

Answer:

12.4 this is the answer

13.6-4.1=9.5

-3.8+6.7=2.9

9.5+2.9=12.4

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topjm [15]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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