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Novosadov [1.4K]
3 years ago
7

An architect drew the blueprint for a new office building. He used a scale in which 1 inch represents 7.5 feet. The floor of an

office in the building will have actual dimensions of 18 feet by 16 feet. What will the dimensions on the blueprint be?
Mathematics
1 answer:
VMariaS [17]3 years ago
7 0
We know that
scale factor=1 in/7.5 ft
scale factor=measurements   on the blueprint/measurements <span> in the actual
</span>measurements on the blueprint=[measurements  in the actual*scale factor]
so

for 18 ft
measurements on the blueprint=[18 ft*(1 in/7.5 ft)-----> 2.4 in

for 16 ft
measurements on the blueprint=[16 ft*(1 in/7.5 ft)-----> 2.1 in

the dimensions on the blueprint are
2.4 in x 2.1 in
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<h3>Given</h3>

regular paper costs $3.79 per ream

recycled paper costs $5.49 per ream

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<h3>Find</h3>

the numbers of reams of each type that were purchased

<h3>Solution</h3>

Let r and g represent the numbers of reams of regular and recycled ("green") paper, respectively.

... r + g = 116 . . . . . . . . 116 reams were purchased

... 3.79r + 5.49g = 582.44 . . . . this is the total cost of the purchase

Solve the first equation for r and substitute that into the second equation.

... r = 116 - g

... 3.79(116 - g) + 5.49g = 582.44 . . . . . use the expression for r

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... r = 116 -84 = 32 . . . . . . . . . . . . . . . . . use the equation for r

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E = Occurrence of 1 on first die

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Are E and F independent

First, we need to list the sample space of a roll of a die

Event\ 1 = \{1,2,3,4,5,6\}

Next, we list out the sample space of F

Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\

6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}

In (1): the sample space of E is:

E = \{1\}

So:

P(E) = \frac{n(E)}{n(Event\ 1)}

P(E) = \frac{1}{6}

In (2): the sample space of F is:

F = \{5,5,5,5\}

So:

P(F) = \frac{n(F)}{n(Event\ 2)}

P(F) =\frac{4}{36}

P(F) =\frac{1}{9}

For E and F to be independent:

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Substitute values for P(E) and P(F)

This gives:

P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}

P(E\ and\ F) = \frac{1}{54}

However, the actual value of P(E and F) is 0.

This is so because E = \{1\} and F = \{5,5,5,5\} have 0 common elements:

So:

P(E\ and\ F) = 0

Compare P(E\ and\ F) = \frac{1}{54} and P(E\ and\ F) = 0.

These values are not equal.

Hence: the two events are not independent

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