Find the m∠2 90° 67° 157° 57°

A rectangular box has length x and width 3. The volume of the box is given by y = 3x(8 – x). The greatest x-intercept of the gra

CaHeK987 [17]

Consider rectangular box with

length x units (x≥0);
width 3 units;
height (8-x) units (8-x≥0, then x≤8).

The volume of the rectangular box can be calculated as

$V_{box}=\text{length}\cdot \text{width}\cdot \text{height}.$

In your case,

$V_{box}=3\cdot x\cdot (8-x).$

Note that maximal possible value of the height can be 8 units (when x=0 - minimal possible length) and the minimal possible height can be 0 units (when x=8 - maximal possible length).

From the attached graph you can see that the greatest x-intercept is x=8, then the height will be minimal and lenght will be maximal.

Then the volume will be V=0 (minimal).

Answer: correct choices are B (the maximum possible length), C (the minimum possible height)

7 0

3 years ago

My Algebra 1 teacher is teaching us slope. Any tips and tricks?

Nuetrik [128]

Slope Intercept form is: y=mx+b
Standard Form is: Ax+By=C
Point Slope form is: y-y1=m(x-x1)

6 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$