Answer:
a. 61.92 in²
b. 21.396 ≈ 21.4%
c. $4.71
Step-by-step explanation:
a. Amount of waste = area of rectangular piece of stock - area of two identical circles cut out
Area of rectangular piece of stock = 24 in × 12 in = 288 in²
Area of the two circles = 2(πr²)
Use 3.14 as π
radius = ½*12 = 6
Area of two circles = 2(3.14*6²) = 226.08 in²
Amount of waste = 288 - 226.08 = 61.92 in²
b. % of the original stock wasted = amount of waste ÷ original stock × 100
= 61.92/288 × 100 = 6,162/288 = 21.396 ≈ 21.4%
c. 288 in² of the piece of stock costs $12.00,
Each cut-out circle of 113.04 in² (226.08/2) will cost = (12*113.04)/288
= 1,356.48/288 = $4.71.
Answer:
y-determinant = 2
Step-by-step explanation:
Given the following system of equation:
Let's represent it using a matrix:
![\left[\begin{array}{ccc}1&2\\1&-3\end{array}\right] = \left[\begin{array}{ccc}5\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C1%26-3%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
The y‐numerator determinant is formed by taking the constant terms from the system and placing them in the y‐coefficient positions and retaining the x‐coefficients. Then:
![\left[\begin{array}{ccc}1&5\\1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%265%5C%5C1%267%5Cend%7Barray%7D%5Cright%5D%20)
y-determinant = (1)(7) - (5)(1) = 2.
Therefore, the y-determinant = 2
Answer:
Step 1
Step-by-step explanation:
they were supposed to multiply 2x by 4 and -1 by 4 but they only multiplied 2x by 4.
The maximum value it can be would be 100%. The only times that it could be this high is when there is only a class on the frequency distribution table. Another way, is that there is only one column variable on the table (& or if there is only one row.) Any other time is will be less, but the maximum stays 100%.
Answer:
hope this helps.............
