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3 years ago

A regular pentagon with a perimeter of 21 inches is dilated by a scale factor of

Mathematics

2 answers:

levacccp [35]3 years ago

5 0

<h3>Answer: 35 inches</h3>

Each side of the pentagon is multiplied by 5/3 to get the length of each new side, so the perimeter is multiplied by 5/3 to get the new perimeter.

New perimeter = (scale factor)*(old perimeter)

New perimeter = (5/3)*(old perimeter)

New Perimeter = (5/3)*(21)

New Perimeter = (5/3)*(21/1)

New Perimeter = (5*21)/(3*1)

New Perimeter = 105/3

New Perimeter = 35 inches

Maslowich3 years ago

5 0

Answer:

it is 35 for ttm

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In a multiple regression equation, two independent variables are considered, and the sample size is 30. The regression coefficie

MAXImum [283]

Answer:

Both the variables are important for the regression analysis and cannot be deleted.

Step-by-step explanation:

(1)

The hypothesis for for the testing of coefficient β₁ are:

H₀: β₁ = 0 vs. Hₐ: β₁ ≠ 0

The test statistic is:

$t=\frac{b_{1}}{S_{b_{1}}} =\frac{2.815}{0.75}= 3.753$

It is provided that H₀ is rejected if t > 2.042.

The test statistic value, t = 3.753 > 2.042.

Thus, the null hypothesis is rejected.

Conclusion:

There is a significant relationship between the regression variable and the dependent variable.

(2)

The hypothesis for for the testing of coefficient β₂ are:

H₀: β₂ = 0 vs. Hₐ: β₂ ≠ 0

The test statistic is:

$t=\frac{b_{2}}{S_{b_{2}}} =\frac{-1.249}{0.41}= -3.046$

It is provided that H₀ is rejected if t < -2.042.

The test statistic value, t = -3.046 < -2.042.

Thus, the null hypothesis is rejected.

Conclusion:

There is a significant relationship between the regression variable and the dependent variable.

Thus, both the variables are important for the regression analysis and cannot be deleted.

7 0

3 years ago

Please help thank you

Dvinal [7]

the tale-tell fellow is the number inside the parentheses.

if that number, the so-called "growth or decay factor", is less than 1, then is Decay, if it's more than 1, is Growth.

$\bf f(x)=0.001(1.77)^x\qquad \leftarrow \qquad \textit{1.77 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=2(1.5)^{\frac{x}{2}}\qquad \leftarrow \qquad \textit{1.5 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=5(0.5)^{-x}\implies f(x)=5\left( \cfrac{05}{10} \right)^{-x}\implies f(x)=5\left( \cfrac{1}{2} \right)^{-x} \\\\\\ f(x)=5\left( \cfrac{2}{1} \right)^{x}\implies f(x)=5(2)^x\qquad \leftarrow \qquad \textit{Growth} \\\\[-0.35em] ~\dotfill$

$\bf f(t)=5e^{-t}\implies f(t)=5\left( \cfrac{e}{1} \right)^{-t}\implies f(t)=5\left( \cfrac{1}{e} \right)^t \\\\\\ \cfrac{1}{e}\qquad \leftarrow \qquad \textit{that's a fraction less than 1, Decay}$

now, let's take a peek at the second set.

$\bf f(x)=3(1.7)^{x-2}\qquad \leftarrow \qquad \begin{array}{llll} \textit{the x-2 is simply a horizontal shift}\\\\ \textit{1.7 is more than 1, Growth} \end{array} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3(1.7)^{-2x}\implies f(x)=3\left(\cfrac{17}{10}\right)^{-2x}\implies f(x)=3\left(\cfrac{10}{17}\right)^{2x} \\\\\\ \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill$

$\bf f(x)=3^5\left( \cfrac{1}{3} \right)^x\qquad \leftarrow \qquad \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3^5(2)^{-x}\implies f(x)=3^5\left( \cfrac{2}{1} \right)^{-x}\implies f(x)=3^5\left( \cfrac{1}{2} \right)^x \\\\\\ \textit{that fraction in the parentheses is less than 1, Decay}$