The dot plot shows the frequency or length of the tadpoles
The total length of the tadpoles that are 1/8-inch and 1/4-inch long is 7/8 inches
<h3>How to determine the total length?</h3>
From the dot plot, we have the following parameters:
- 3 tadpoles have a length of 1/8
- 2 tadpoles have a length of 1/4
So, the total length is:
Total = 3 * 1/8 + 2 * 1/4
Evaluate the product
Total = 3/8 + 1/2
Rewrite as;
Total = 3/8 + 4/8
Evaluate the sum
Total = 7/8
Hence, the total length of the tadpoles that are 1/8-inch and 1/4-inch long is 7/8 inches
Read more about dotplots at:
brainly.com/question/24309209
Answer:
2.67 x 10^3
Step-by-step explanation
The exponent is equal to how many numbers are after the decimal point. So, since there are 3 numbers after the decimal point, the exponent would be 3. Hope this helps :)
Answer:
![\large \boxed{\text{A. }\dfrac{13}{7}\text{; B. }\dfrac{17}{14} \text{; C. 507 in}^{2}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7BA.%20%7D%5Cdfrac%7B13%7D%7B7%7D%5Ctext%7B%3B%20B.%20%7D%5Cdfrac%7B17%7D%7B14%7D%20%5Ctext%7B%3B%20C.%20507%20in%7D%5E%7B2%7D%7D)
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 $\large \boxed{\mathbf{ \dfrac{17}{14}}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7BEF%7D%7BAB%7D%20%26%20%3D%20%26%20%5Cdfrac%7B13%7D%7B7%7D%5C%5C%5C%5C%5Cdfrac%7BEF%7D%7B%5Cfrac%7B17%7D%7B26%7D%7D%20%26%20%3D%20%26%20%5Cdfrac%7B13%7D%7B7%7D%5C%5C%5C%5CEF%20%26%20%3D%20%26%20%5Cdfrac%7B13%7D%7B7%7D%5Ctimes%5Cdfrac%7B17%7D%7B26%7D%5C%5C%5C%5C%20%26%20%3D%20%26%5Cdfrac%7B1%7D%7B7%7D%5Ctimes%5Cdfrac%7B17%7D%7B2%7D%5C%5C%5C%5C%20%26%20%3D%20%26%20%5Cmathbf%7B%5Cdfrac%7B17%7D%7B14%7D%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20length%20of%20EF%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B%20%5Cdfrac%7B17%7D%7B14%7D%7D%7D%24%7D)
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}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B13%7D%7B7%7D%20%5Cright%29%5E%7B2%7D%20%3D%20%5Cdfrac%7B169%7D%7B49%7D)
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 $\large \boxed{\textbf{507 in}^{\mathbf{2}}}$}](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20EFGH%7D%20%3D%20%5Ctext%7B147%20in%7D%5E%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B169%7D%7D%7B%5Ctext%7B49%7D%7D%20%3D%203%20%5Ctimes%20169%5Ctext%7B%20in%7D%5E%7B2%7D%20%3D%20507%20%5Ctext%7B%20in%7D%5E%7B2%7D%5C%5C%5C%5C%5Ctext%7BThe%20area%20of%20EFGH%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B507%20in%7D%5E%7B%5Cmathbf%7B2%7D%7D%7D%24%7D)
There are 4,000 Grams in 4 Kilograms