Simplify the expression. 8(5k + 6m)

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

3 years ago

In a week, a man received a paycheck for $826.60. The withholding for taxes was $91.40. If he worked 34 hours in that week,

sleet_krkn [62]

Answer:

Step-by-step explanation:

3 0

2 years ago

A research firm tests the miles-per-gallon characteristics of three brands of gasoline. Because of different gasoline performanc

Anni [7]

Answer:

A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.

B.)The advantage of attempting to remove the block effect is

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

Step-by-step explanation:

Using Two-way ANOVA method

Given problem

Observation I II III Row total (xr)

A 18 21 20 59

B 24 26 27 77

C 30 29 34 93

D 22 25 24 71

E 20 23 24 63

Col total (xc) 114 124 129 367

∑x²=9233→(A)

∑x²c/r

=1/5(114²+124²+129²)

=1/5(12996+15376+16641)

=1/5(45013)

=9002.6→(B)

∑x²r/c

=1/3(59²+77²+93²+71²+67²)

=1/3(3481+5929+8649+5041+4489)

=1/3(27589)

=9196.3333→(C)

(∑x)²/n

=(367)²/15

=134689/15

=8979.2667→(D)

Sum of squares total

SST=∑x²-(∑x)²/n

=(A)-(D)

=9233-8979.2667

=253.7333

Sum of squares between rows

SSR=∑x²r/c-(∑x)²/n

=(C)-(D)

=9196.3333-8979.2667

=217.0667

Sum of squares between columns

SSC=∑x²c/r-(∑x)²/n

=(B)-(D)

=9002.6-8979.2667

=23.3333

Sum of squares Error (residual)

SSE=SST-SSR-SSC

=253.7333-217.0667-23.3333

=13.3333

ANOVA table

Source Sums Degrees Mean Squares

of Variation of Squares of freedom

SS DF MS F p-value

B/ w SSR=217.0667 4 MSR=54.2667 32.56 0.0001

rows

B/w SSC=23.3333 c-1=2 MSC=11.6667 7 0.01

columns

Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667

Total SST=253.7333 rc-1=14

Conclusion:

1. F for between Rows

The critical region for F(4,8) at 0.05 level of significance=3.8379

The calculated F for Rows=32.56>3.8379

Therefore H0 is rejected

2. F for between Columns

The critical region for F(2,8) at 0.05 level of significance=4.459

We see that the calculated F for Colums=7>4.459

therefore H0 is rejected,and concluded that there is significant differentiating between columns

Part B:

To analyze the data for completely randomized designs click on anova two factor without replication in the data analysis dialog box of the excel spreadsheet.

The following table is obtained

Source DF Sum Mean F Statistic

(df1,df2) of Square (SS) Square (MS) P-value

Factor A 1 1496.5444 1496.5444 769.6514 0.001297

Rows

Factor B - 2 19.4444 9.7222 5 0.1667

Columns

Interaction

AB 2 3.8889 1.9444 0.1013 0.9045

Error 12 230.4 19.2

Total 17 1750.2778 102.9575

Factor - A- Rows

Since p-value < α, H0 is rejected.

Factor - B- Columns

Since p-value > α, H0 can not be rejected.

The averages of all groups assume to be equal.

Interaction AB

Since p-value > α, H0 can not be rejected.

The advantage of attempting to remove the block effect is

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

3 0

3 years ago

Pen are shipped to the office supply store in boxes of 12 each. Write a function rule to calculate the total number of pens when

Tom [10]

p = 12x is the function rule to calculate the total number of pens when you know the number of boxes

Solution:

Given that,

Pen are shipped to the office supply store in boxes of 12 each

To find: Write a function rule to calculate the total number of pens when you know the number of boxes

Let "p" be the total number of pens

Let "x" be the number of boxes

Number of pens in each box = 12

Then from given information,

Total number of pens = Number of boxes multiplied by number of pens in each box

$p = x \times 12\\\\p = 12x$

Thus the function rule is found

6 0

4 years ago

Please help right away

andrew-mc [135]

Answer:

100

Step-by-step explanation:

Mixed candy question... Skittles jar... to be filled with Jelly beans.

Let's first calculate the volume of the jar. We'll assume it's a regular cylindrical prism jar, unlike the one on the photo which is narrower on top.

V = π * r² * h = π * (3.5)² * 11.5 = 140.875 π = 442.6 cubic cm

Now, we don't have the precise measurement of a jelly bean, but we know it's roughly 2-3 cubic cm. The precision isn't needed to answer this question, just to have a rough idea... it's no 300 cu cm per jelly bean.

So, let's assume a 3 cu cm per jelly bean (2 cu cm wouldn't the final answer)....

442.6 / 3 = 147.5 jelly beans, approximately.

So, can they fit 100,000? No

Can we fit 10,000 in there? No

Can we fit 100? Yes.

Can we fit 1? Certainly

The most reasonable lower-limit would then be 100.

4 0

3 years ago