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3 years ago

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Question 1: The Millers have hired an interior designer to decorate their family room. they want to have a 22" by 30" oil painti

ng frame and incorporated in the design plan. if the designer chooses molding that is 4" Wide and is priced at $1.45 per inch, how much will the molding cost (Before Tax)

1 answer:

____ [38]3 years ago

6 0

Answer: Total molding cost is $197.20.

Step-by-step explanation:

Since we have given that

Length of oil painting frame = 22 inch

Width of oil painting frame = 30 inch

Width for molding = 4 inches

So, Length after molding would be

4+22+4 = 30 inches

Width after molding would be

4+30+4 = 38 inches

So, Perimeter of molding would be

$2(L+B)\\\\=2(30+38)\\\\=2\times 68\\\\=136\ inches$

Cost per inch = $1.45

So, Total molding cost would be

$1.45\times 136\\\\=\$197.20$

Hence, Total molding cost is $197.20.

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Step-by-step explanation:

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The line of best fit is a straight line that can be used to predict the

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Correct responses:

The equation of best fit is; $\underline{ \hat Y = 1,042 - 4.9 \cdot X_i}$

The correlation coefficient is; r ≈<u> -0.969</u>

<h3>Methods by which the line of best fit is found</h3>

The given data is presented in the following tabular format;

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|}Cost, (dollars), x&20&21&22&24&25&27&28&30\\Daily attendance, y&940&935&940&925&920&905&910&890\end{array}\right]$

The equation of the line of best fit is given by the regression line

equation as follows;

$\hat Y = \mathbf{b_0 + b_1 \cdot X_i}$

Where;

$\hat Y$ = Predicted value of the<em> i</em>th observation

b₀ = Estimated regression equation intercept

b₁ = The estimate of the slope regression equation

$X_i$ = The <em>i</em>th observed value

$b_1 = \mathbf{\dfrac{\sum (X - \overline X) \cdot (Y - \overline Y) }{\sum \left(X - \overline X \right)^2}}$

$\overline X$ = 24.625

$\overline Y$ = 960.625

$\mathbf{\sum(X - \overline X) \cdot (Y - \overline Y)} = -433.125$

$\mathbf{\sum(X - \overline X)^2} = 87.875$

Therefore;

$b_1 = \mathbf{\dfrac{-433.125}{87.875}} \approx -4.9289$

Therefore;

The slope given to the nearest tenth is b₁ ≈ -4.9

$b_0 = \mathbf{\dfrac{\left(\sum Y \right) \cdot \left(\sum X^2 \right) - \left(\sum X \right) \cdot \left(\sum X \cdot Y\right)} {n \cdot \left(\sum X^2\right) - \left(\sum X \right)^2}}$

By using MS Excel, we have;

n = 8

∑X = 197

∑Y = 7365

∑X² = 4939

∑Y² = 6782675

∑X·Y = 180930

(∑X)² = 38809

Therefore;

$b_0 = \dfrac{7365 \times 4939-197 \times 180930}{8 \times 4939 - 38809} \approx \mathbf{1041.9986}$

The y-intercept given to the nearest tenth is b₀ ≈ 1,042

The equation of the line of best fit is therefore;

$\underline{\hat Y = 1042 - 4.9 \cdot X_i}$

The correlation coefficient is given by the formula;

$\displaystyle r = \mathbf{\dfrac{\sum \left(X_i - \overline X) \cdot \left(Y - \overline Y \right)}{ \sqrt{\sum \left(X_i - \overline X \right)^2 \cdot \sum \left(Y_i - \overline Y \right)^2} }}$

Where;

$\sqrt{\sum \left(X - \overline X \right)^2 \times \sum \left(Y - \overline Y \right)^2} = \mathbf{446.8121}$

$\sum \left(X_i - \overline X \right) \times \left(Y - \overline Y\right) = \mathbf{-433.125}$

Which gives;

$r = \dfrac{-433.125}{446.8121} \approx \mathbf{-0.969367213}$

The correlation coefficient given to the nearest thousandth is therefore;

<u>Correlation coefficient, r ≈ -0.969</u>

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