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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.

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:

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

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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 .

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

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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

= 3.5

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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Digiron [165]

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Step-by-step explanation:

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2 years ago

When 23 is subtracted from 5 times a certain number the result is 87

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Answer:

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Step-by-step explanation:

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Can you help me, please ……..

dedylja [7]

Answer: The correct answer is 44.60

Step-by-step explanation:

Let's find the area of the rectangle first, and then add it with the area of the Circle to find the total area of the whole shape.

We know that the area of a Rectangle is L(Length) x W(Width)

So using the given information, 8 x 4 = 32

So the Area of the Rectangle = 32

Now let's find the area of the given circle,

<em>Area of a Circle: πr2 Remember ^ = stands for power</em>

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The diameter of the circle is 4

4/2 = 2 which is our radius, and now let's replace it in the equation.

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Now rounding off to the nearest tenths, would give us 44.60

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2 years ago

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