Answer:
i would pick line c and e before i would do a and e and b fnhwbh bfrhb bdh hejwhjd febhrew
Step-by-step explanation:
The length of the rectangle is 20 inches, and the width of the rectangle is 4 inches if the table runner has an area of 80 square inches. The length and width of the table runner are whole numbers. The length is 5 times greater than the width.
<h3>What is the area of the rectangle?</h3>
It is defined as the space occupied by the rectangle which is planner 2-dimensional geometry.
The formula for finding the area of a rectangle is given by:
Area of rectangle = length × width
The area of the table runner = 80 square inches
Let's assume the length of the rectangle is L and the width is W
Then L = 5×W ...(1)
L×W = 80 ...(2)
Put the value of L in the equation (2)
5W(W) = 80
5W² = 80
W² = 16
W = ±4
Width cannot be negative.
W = 4 inches putting this value in the equation (1)
L = 5(4) = 20 inches
Thus, the length of the rectangle is 20 inches, and the width of the rectangle is 4 inches if the table runner has an area of 80 square inches. The length and width of the table runner are whole numbers. The length is 5 times greater than the width.
Learn more about the area here:
brainly.com/question/14383947
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X=0 because when you have something to the power of zero your answer is 1.
x=0 since 1 to the power of zero equals 1.
hope im not too late
Answer:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ![\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D)
Step-by-step explanation:
Given that
3 white, 2 blue and 5 gray shirts are there.
To find:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?
Solution:
Here, total number of shirts = 3+2+5 = 10
First of all, let us learn about the formula of an event E:
![P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}](https://tex.z-dn.net/?f=P%28E%29%20%3D%20%5Cdfrac%7B%5Ctext%7BNumber%20of%20favorable%20cases%7D%7D%7B%5Ctext%20%7BTotal%20number%20of%20cases%7D%7D)
![P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}](https://tex.z-dn.net/?f=P%28First%5C%20White%29%20%3D%20%5Cdfrac%7B%5Ctext%7BNumber%20of%20white%20shirts%7D%7D%7B%5Ctext%20%7BTotal%20number%20of%20shirts%20left%7D%7D)
![P(First\ White) = \dfrac{3}{10}](https://tex.z-dn.net/?f=P%28First%5C%20White%29%20%3D%20%5Cdfrac%7B3%7D%7B10%7D)
Now, this shirt is set aside.
So, total number of shirts left are 9 now.
![P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }](https://tex.z-dn.net/?f=P%28First%5C%20White%5C%20and%5C%20second%5C%20gray%29%20%3D%20P%28First%20White%29%20%5Ctimes%20P%28Second%5C%20Gray%29%5C%5C%5CRightarrow%20P%28First%5C%20White%5C%20and%5C%20second%5C%20gray%29%20%3D%20P%28First%20White%29%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7BNumber%20of%20gray%20shirts%7D%7D%7B%5Ctext%7BTotal%20number%20of%20shirts%20left%7D%7D%5C%5C%5C%5C%5CRightarrow%20P%28First%5C%20White%5C%20and%5C%20second%5C%20gray%29%20%3D%20%5Cdfrac%7B3%7D%7B10%7D%20%5Ctimes%20%5Cdfrac%7B5%7D%7B9%7D%5C%5C%5CRightarrow%20P%28First%5C%20White%5C%20and%5C%20second%5C%20gray%29%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5CRightarrow%20P%28First%5C%20White%5C%20and%5C%20second%5C%20gray%29%20%3D%20%5Cbold%7B%5Cdfrac%7B1%7D%7B4%7D%20%7D)
So, the answer is:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ![\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D)