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Brilliant_brown [7]
3 years ago
11

Quadrilateral ABCD is similar to Quadrilateral EFGH. Diagonal AC has length 7 and diagonal EG has length 13. What is the scale f

actor that describes a dilation from BC to FG? Give the exact scale factor and state whether the dilation is an expansion or a contraction.
If side AB has length 17/26 what is the length of side EF? Give the exact, un-rounded value.




PLEASE HELP ASAP

If the area of ABCD is 147 square inches, what is the area of EFGH? Give the exact answer.

Mathematics
1 answer:
olasank [31]3 years ago
8 0

Answer:

\large \boxed{\text{A. }\dfrac{13}{7}\text{; B. }\dfrac{17}{14} \text{; C. 507 in}^{2}}

Step-by-step explanation:

A. Scale factor

When you dilate an object by a scale factor, you multiply its line lengths by the same number.

If EF/AB = 13/7, the scale factor is 13/7.

B. Length of EF

\begin{array}{rcl}\dfrac{EF}{AB} & = & \dfrac{13}{7}\\\\\dfrac{EF}{\frac{17}{26}} & = & \dfrac{13}{7}\\\\EF & = & \dfrac{13}{7}\times\dfrac{17}{26}\\\\ & = &\dfrac{1}{7}\times\dfrac{17}{2}\\\\ & = & \mathbf{\dfrac{17}{14}}\\\end{array}\\\text{The length of EF is $\large \boxed{\mathbf{ \dfrac{17}{14}}}$}

C. Area of EFGH

If the lengths in a shape are all multiplied by a scale factor, then the areas will be multiplied by the scale factor squared.

ABCD is dilated by a scale factor of 13/7, so its area is dilated by a scale factor of

\left(\dfrac{13}{7} \right)^{2} = \dfrac{169}{49}

The area of its dilated image EFGH is

\text{Area of EFGH} = \text{147 in}^{2} \times \dfrac{\text{169}}{\text{49}} = 3 \times 169\text{ in}^{2} = 507 \text{ in}^{2}\\\\\text{The area of EFGH is $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}

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Answer:

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Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

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The expected number in the sample that treats hazardous waste on-site is given by

$ E(X) =  \frac{n \times r}{N} $

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The probability is given by

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For the given case, we have

N = 209

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$ P(x = 4) =  \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} $

$ P(x = 4) =  \frac{70 \times 84944276340}{35216131179263320}

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