C is also the best answer
Answer is a= 51/7 or 7 2/7
Step by step
-2/3 (a + 3) = 5/3a - 19
Distribute to resolve parentheses
-2/3a -2 = 5/3a - 19
Now to move variables to one side
Add 2/3a to both sides
-2/3a + 2/3a -2 = 5/3a + 2/3a -19
Combine like terms
-2 = 7/3a -19
Add 19 to both sides to isolate variable
-2 +19 = 7/3a -19 + 19
17 = 7/3a
Divide both sides by 7/3 to solve for a
(Remember when you divide a fraction, flip it and multiply.)
17 * 3/7 = a
51/7 = a
Picture attached shows steps in order, I think that is what you also needed?
Answer:
'A' is true; theoretically, 50% of the data items reside between the first and third quartiles (40 and 67.5)
Step-by-step explanation:
Range is 84-28 which is 56
Median is 51
1.5 x Interquartile Range (IQR) = 1.5(67.5-40) which equals 41.25
Q1 is 40
Q1 - IQR = -1.25
Low outliers are below -1.25 - 41.25; there are not data items below -42
7^2 + 9^2 = 130
Square root of 130 is about 11.4
Answer:
59.16 units
Step-by-step explanation:
Given that,
The length of rectangle, L = 13
The breadth of the rectangle, B = 6
The perimeter of the shaded region = perimeter of rectangle - 2(perimeter of semicircle)
![P=13\times 6-[2\times \dfrac{2\pi r}{2}]\\\\=78-2\pi r\\\\=78-2\times 3.14\times 3\\\\=59.16](https://tex.z-dn.net/?f=P%3D13%5Ctimes%206-%5B2%5Ctimes%20%5Cdfrac%7B2%5Cpi%20r%7D%7B2%7D%5D%5C%5C%5C%5C%3D78-2%5Cpi%20r%5C%5C%5C%5C%3D78-2%5Ctimes%203.14%5Ctimes%203%5C%5C%5C%5C%3D59.16)
So, the perimeter of the shaded region is 59.16 units.