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1 year ago

Giving Brainliest to whoever solves it correctly w an explanation first!

Mathematics

1 answer:

mojhsa [17]1 year ago

4 0

The recommended dimensions for the garden is as follows:

Length 1 = Length 2 = 20 feet

width 1 = width 2 = 5 feet

depth 1 = 1.5 feet

depth 2 = 3 feet

Given data

The garden should be at least 200 square feet for the for the plants

The volume should be less than 500 cubic feet

<h3>How to get the are at least 200 cubic feet</h3>

The area of the garden available should be the top of both levels

say the top has equal widths of 5 and the length is 20

therefore area of the two levels will be

= area of level 1 + area of level 2

= L1 * B1 + L2 * B2

= 20 * 5 + 20 * 5

= 100 + 100

= 200 square feet

condition one: the garden should be at least 200 square feet for the plants is maintained.

volume of the platform

The platform has two levels representing two depths say 3 for the higher level and 2 for the lesser level.

Using the same division used for area and adding them together we find the volume as follows:

= volume of level 1 + volume of level 2

= L1 * B1 * D1 + L2 * B2 * D2

= 20 * 5 * 1.5 + 20 * 5 * 3

= 150 + 300

= 450 cubic feet

condition two: the garden should be less than 500 cubic feet for the platform is maintained

Read more on volume and area calculation here: brainly.com/question/463363

#SPJ1

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3 years ago

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3 years ago

use a graphing calculator or other technology to answer the question which quadratic regression equation best fits the data set

Greeley [361]

Answer:

Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

Step-by-step explanation:

x y

3 470

4 416

5 403

Analyzing Option A:

Considering the equation

$y=32.86\:\left(x\right)^2+379.14\left(x\right)-1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2+379.14\left(3\right)-1369.14\:$

$y=64.02$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2+379.14\left(4\right)-1369.14\:\:$

$y=673.18$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2+379.14\left(5\right)-1369.14\:$

$y=1348.06$

Analyzing Option B:

$y=32.86\:\left(x\right)^2-379.14\left(x\right)$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)$

$y=-841.68$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)$

$\:y=-990.8$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)$

$\:y=-1074.2$

Analyzing Option C:

Considering the equation

$y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)+1369.14\:$

$y=527.46$

So, the approximately result is (3, 527)

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)+1369.14\:$

$y=378.34$

So, the approximately result is (4, 378)

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)+1369.14\:\:\:$

$y=294.94$

So, the approximately result is (5, 295)

Analyzing Option D:

Considering the equation

$y=-1369.14\:\left(x\right)^2-379.14\left(x\right)+32.86$

From (3, 470), putting x = 3

$y=-1369.14\:\left(3\right)^2-379.14\left(3\right)+32.86\:\:$

$y=-13426.82$

From (4, 470), putting x = 4

$y=-1369.14\:\left(4\right)^2-379.14\left(4\right)+32.86$

$y=-23389.94$

From (5, 403), putting x = 5

$y=-1369.14\:\left(5\right)^2-379.14\left(5\right)+32.86\:\:$

$y=-36091.34$

Therefore, from the above calculations and analysis, we conclude that Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.