Why would someone choose to use a graphing calculator to solve a system of linear equations instead of graphing by hand? Explain

Answer:

Terrence's

Step-by-step explanation:

The cut length of the square corresponds to the box height.

1. a

In Anya's approach: 8.5 - 1.5 = 7, 11 - 1.5 = 9.5; with the height being 1.5, the volume calculates as height x length x width = 1.5 x 9.5 x 7 = 99.75 square inches.

Terrence's method: 8.5 - 3 = 5.5, 11 - 3 = 8. Volume = 5.5 x 8 x 3 = 132 square inches. Clearly, 99.75 < 132 square inches, indicating Terrence's approach would yield a larger volume.

1. b

The dimensions of the box rely on the length, width, and height of the cardboard cutout, thus varying cutting measurements for the same cardboard size can produce different box dimensions.

2. Squares are cut from each corner, thus the combined size of the two squares on the cardboard must be less than the shorter cardboard side. The shorter side is 8.5 inches, which divided by 2 results in 4.25 inches, therefore squares cannot exceed this size. If squares are taken out measuring 4.25 inches, it will leave a strip 2.5 inches wide that cannot form a box.

To cut out squares of 5 inches, how you position them can cause overlap or extend beyond the cardboard. While 5 + 5 equals ten, making the longer side of 11 inches adequate, the 8.5 inches side allows zero room for the 10 inches.

Answer:

The result is 26.4

Step-by-step explanation:

By taking 51.82 and subtracting 26.37, you arrive at 26.45, and when rounding this value to one decimal place, it becomes 26.4.

Answer:

The mean decreases, while the median stays unchanged.

Step-by-step explanation:

A box plot is constructed from the quartiles of the data distribution, incorporating the maximum and minimum values. From this, one can determine the range, median, and interquartile range.

In this scenario, the median maintains its value at $9.5 per hour. The median is represented by the middle line in the box, which remains constant.

Conversely, the range of this data set shrinks from 7 to 3.

Moreover, the mean must decrease because values greater than $11 are no longer represented in the second box plot. The mean is a measure sensitive to such alterations.

Conclusively, the correct finding is The mean decreases, while the median remains unchanged.