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saul85 [17]
4 years ago
14

The perimeter of a rectangular toddler play area is 100 feet. The length is 10 feet more than 3 times the width. Find the length

and width of the play area.
Mathematics
1 answer:
Mekhanik [1.2K]4 years ago
8 0

Answer:

length and width = 40 and 10 ft.

Step-by-step explanation:

The perimeter of a rectangular toddler play area is 100 feet.

Perimeter = 2(length + width)

Let the width = w

The length is 10 feet more than 3 times the width. Length = 10+(3w)

now put the values

100 = 2(10+3w + w)

100 = 2w + 20 + 6w

100 = (2w + 6w) + 20

100 = 8w + 20

8w = 100 - 20

w = \frac{80}{8}

w = 10 feet.

Now Length = 10+ (3w)

                     = 10+( 3 × 10)

                     = 10 + 30

                     = 40 feet

Therefore, length = 40 feet and width = 10 feet

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Can you explain how you got the answer as well please?
Vikki [24]

Answer:

A. 35

Step-by-step explanation:

The total amount of movies that the critic rates that we are given is 20. Now they want to figure us to figure it out with 100 movies. 20 times 5 is one hundred. We are supposed to look at the 3 star reviews, which is 7 movies. 7 times 5 is 35. Hope this helps!

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The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d
jek_recluse [69]

Answer:

50%

Step-by-step explanation:

68-95-99.7 rule

68% of all values lie within the 1 standard deviation from mean (\mu-\sigma,\mu+\sigma)

95% of all values lie within the 1 standard deviation from mean  (\mu-1\sigma,\mu+1\sigma)

99.7% of all values lie within the 1 standard deviation from mean  (\mu-3\sigma,\mu+3\sigma)

The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4.

\mu = 55\\\sigma = 4

68% of all values lie within the 1 standard deviation from mean (\mu-\sigma,\mu+\sigma) = (55-4,55+4)= (51,59)

95% of all values lie within the 2 standard deviation from mean  (\mu-1\sigma,\mu+1\sigma)= (55-2(4),55+2(4))= (47,63)

99.7% of all values lie within the 3 standard deviation from mean  (\mu-3\sigma,\mu+3\sigma)= (55-3(4),55+3(4))= (43,67)

Refer the attached figure

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Hence The approximate percentage of light bulb replacement requests numbering between 43 and 55 is 50%

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