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hichkok12 [17]
3 years ago
15

Leslie has a rectangular patio. She measures it and finds out it is 2135 feet long by 1115 feet wide. She wants to know how many

square feet of tile she will need to completely cover the patio. Draw the patio, and label the measurements. How much tile will Leslie need to cover the patio? If each square foot of tile costs 75 cents, how much will she have to pay to cover her patio?

Mathematics
1 answer:
larisa86 [58]3 years ago
7 0

Answer: (1) Leslie will need 2,380,525 square tiles (1 foot by 1 foot each)

(2) She will have to spend $1,785,393.75 to cover her patio.

Step-by-step explanation: Please refer to the picture attached.

The rectangular patio has been designed and as shown in the diagram has on side measuring 2135 ft and the other side measuring 1115 ft. This means, in order to cover the entire patio, she would have to cover an entire area defined as 2,135 feet into 1,115 feet. If one tile measures 1 foot long by 1 foot wide, the total number of tiles to cover the patio can simply be derived as;

Area = L x W

Where the length is 2135 and the width is 1115,

Area = 2135 x 1115

Area = 2380525

**Having in mind that one tile measures 1 ft by 1 ft, the area of each tile is given as

Area = L x W

Area = 1 x 1

Area = 1 square foot**

(1) The above calculation shows that Leslie would be using up a total of 2,380,525 square tiles to completely cover her patio.

(2) Having been told that each square foot of tile costs 75 cents ($0.75), the total amount spent to cover her patio would be calculated as follows;

Cost = Area x cost per tile

Cost = 2380525 x 0.75

Cost = 1785393.75

The total cost therefore is $1,785,393.75

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Answer:

25.63694268 = r

Step-by-step explanation:

The circumference is

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3 years ago
. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar
yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are  K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

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(b)

The probability of selecting three 40-W bulbs is

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The probability of selecting three 60-W bulbs is

\frac{5*4*3}{455}=0.1318=13.18\%

The probability of selecting three 75-W bulbs is

\frac{6*5*4}{455}=0.2637=26.37\%

Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

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(d)

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As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is \frac{9}{15}=0.6

Since there are no replacement, the probability of taking a second non 75-W bulb is now \frac{8}{14}=0.5714

Following this procedure 5 times, we find the probabilities

\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}

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