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Nezavi [6.7K]
2 years ago
8

The tile along the edge of a triangular community pool needs to be replaced. A right triangles, where the hypotenuse is labeled

8 x squared, the height is labeled 4 x squared 15, and the base is labeled 8 x 10. Which expression represents the total perimeter of the pool edge? 12x2 15 20x2 25 12x2 8x 25 24x2 16x 50.
Mathematics
1 answer:
jolli1 [7]2 years ago
3 0

Perimeter of a closed figure is the sum of length of its outline. The perimeter of the consider triangular pool is: 12x^2 + 8x + 25 units.

<h3>How to calculate perimeter of a triangular figure?</h3>

Perimeter of a triangle = Sum of lengths of all its 3 sides.

For the given case, we have:

Length of first side(hypotenuse) = 8x^2 units

Height = 4x^2 + 15 units

Base is of length 8x + 10

Thus, its perimeter is calculated as:

Perimeter of a triangle = Sum of lengths of all its 3 sides.

Perimeter =  8x^2 + 4x^2 + 15 + 8x + 10 = (8+4)x^2 + 8x + 25 = 12x^2 + 8x + 25 units

Perimeter = 12x^2 + 8x + 25 units

Thus,

The perimeter of the consider triangular pool is: 12x^2 + 8x + 25 units.

Learn more about perimeter here:

brainly.com/question/10466285

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Answer:

x=2,x=4

So, the 6-meter side will be 2 centimeters and the 12-meter side will be 4 centimeters in the drawing.

Step-by-step explanation:

Solve for x by using cross multiplication

6=3x

x=2

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In the diagram, A||B and B||C. Find the values of X.
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Answer:

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Step-by-step explanation:

3x + 12 and 105 are alternate exterior angles because A || B and B || C.

If the two or more lines are parallel, then the alternate exterior angles are equal.

3x + 12 = 105

Substract 12 on both sides,

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What is 43.09 to the nearest whole number​
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43

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Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning." These words begin a
IgorLugansk [536]

Answer: -8.16 to 15.84

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval in which we are a percentage sure the true mean is in the interval.

A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be

x_{1}-x_{2} ± t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }

where

x₁ and x₂ are sample means

t is t-score

S_{p} is estimate of standard deviation

n₁ and n₂ are the sample numbers

The estimate of standard deviation is calculated as

S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }

where

s₁ and s₂ are sample standard deviation of each sample

Degrees of freedom is:

df=n_{1}+n_{2}-2

df = 12 + 9 - 2

df = 19

Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.

The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.

The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.

Calculating estimate of standard deviaton:

S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }

S_{p}=\sqrt{\frac{299.07}{19} }

S_{p}= 15.74

The difference between means is

x_{1}-x_{2} = 17.5 - 13.66 = 3.84

Calculating the interval:

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 27.21\sqrt{\frac{21}{108} }

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 27.21\sqrt{0.194}

t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} } = 12

Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:

lower limit: 3.84 - 12 = -8.16

upper limit: 3.84 + 12 = 15.84

The interval is from -8.16 to 15.84.

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