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max2010maxim [7]
3 years ago
15

The area of a playground is 160 yd2. The width of the playground is 6 yd longer than its length. Find the length and width of th

e playground.
Mathematics
2 answers:
snow_tiger [21]3 years ago
5 0

Answer:

Length of the playground is 10 yd and the Width of the playground is 16 yd .

Step-by-step explanation:

Formula

Area of a rectangle = Length × Width

As given

The area of a playground is 160 yd².

The width of the playground is 6 yd longer than its length.

Let us assume that the length be x .

Thus

Width = x + 6

Put all the values in the formula

160 = x × (x + 6)

160 = x² + 6x

x² + 6x - 160 = 0

x² + 16x - 10x - 160 = 0

x (x + 16)-10 (x + 16) = 0

(x -10)(x + 16) = 0

Thus

Either (x - 10) = 0 or (x + 16 ) = 0

x = 10 or x = -16

(Negative value is neglected because sides cannot be negative.)

Thus

x = 10

Length = 10 yd

Width = 10 + 6

             = 16 yd

Therefore the length of the playground is 10 yd and the width of the playground is 16 yd .

Sonbull [250]3 years ago
4 0

Answer:

length=10\ yd,  width=16\ yd

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Answer:

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\mu_{{\overline}{X}} = 3.5

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Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

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∀ i = 1(1)10 and j = 1(1)20

Then,

E(X_{ij}) = \frac {1 + 2 + 3 + 4 + 5 + 6}{6}

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and,

E(X^{2}_{ij} = \frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}

                                = \frac {1 + 4 + 9 + 16 + 25 + 36}{6}

                                = \frac {91}{6}

                                \simeq 15.166667

so, Var(X_{ij} = (E(X^{2}_{ij} - {(E(X_{ij})}^{2})

                                    \simeq 15.166667 - 3.5^{2}

                                    = 2.91667

   and \sigma_{X_{ij}} = \sqrt {2.91667}[/tex                                            [tex]\simeq 1.7078261036

Now we get that,

 Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}

We get that Y_{j}'s are iid RV's ∀ j = 1(1)20

Let, {\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}

      So, we get that E({\overline}{Y}) = E(Y_{j})

                                                                 = E(X_{ij}  for any i = 1(1)10

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and,

       \sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}}                                             = \frac {\sigma_{X_{ij}}}{\sqrt {20}}                                             = \frac {1.7078261036}{\sqrt {20}}                                            [tex]\simeq 0.38

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10  

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2 years ago
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