How to Compute Cubage Box Measurement?
445986
Just multiply the three numbers together - length * width * height.
525014
Let's make this very simple.\n\nTo measure the volume of a cylinder, you find the diameter of the interior portion of the cylinder.  Take that number and divide by two...that's your radius (r).  Now measure the height of the interior of your cylinder...that's your height (h).\n\nNow, take the radius (r) and the height (h) and plug them into the formula below:\n\nV = π*(r)*(r)*(h)\n\n(π equals roughly 3.1516)\n\nNow if your answer is in cubic inches, multiply V by .0036 for British gallons or .00433 for US gallons (yes, they are different).  If V is in cubic centimeters, multiply by .00022 (British) or .00026 (US).\n\nI don't know how to make it any clearer...good luck.
180412
Sure, try kinkos. But, it's relatively easy to do it through ms word (tools->envelopes and labels). Just type in the dimensions of the envelope.
913451
A box and whisker plot!? No problem! \n\nThe Box is from Quartile 1(25th percentile) to Quartile 3(75th percentile),.\n\nThe Whiskers are Q1 minus 1.5 Interquartile ranges and Q3 plus 1.5 Inter Quartile ranges.\n\nIQR (Interquartile ranges) is defined as Q3-Q1\n\nWell say that Q1 is 12 and Q3 is 25 then the IQR is 13, and 1.5IQR is 19.5\n\nTherefore, Anything below -7.5 (12-19.5) and above 44.5 would be considered an outlier. \n\nAs far as drawing the damn thing, just draw a box with the outside edges being Q1 adn Q3, and then draw the whiskers 1.5IQR long.\n\nAdditional info:\n\nThe box and whisker is only 1 criterion for outliers, + or - 3 standard deviations from the mean is another common way, which can get other results.\n\nHope that helps.
201111
You need to measure the distance between the closely spaced lines (J values).  Trans and cis J values will be different
464437
Here is instructions on finding the volume of any prisim.\n\nhttp://www.math.com/school/subject3/lessons/S3U4L2DP.html
408251
Measure the diameter and multiply by pi (3.1415)
22234
Use a high quality digital scale. They sell these at office supply stores like Staples. You can also use a 2-ounce mechanical scale (just make sure it's properly calibrated to 0).
430952
Well, what quality to you want to measure -  size?  The weight?  The density?
909019
You'll get the same strength(within reason) and the same dimensions using this method.I can't believe noone has answered this question yet.Good luck on the other site!
608577
Depends on the shape of the object being measured.
804606
For cubic measure:  L x W x H = CI / 231 = G\nL = Length\nW = Width\nH = Height\nCI = Cubic Inches (Volume)\nG = Gallons\nMeasure the Length, Width, and Height in Inches.\n231 Cubic Inches to one Gallon.\nConvert Volume to Cubic Inches\nLength x Width x Height  = Cubic Inches  / 231 = Gallons\n\nFor Right cylinder measure: Pi x r'2 x H = CI / 231 = G\nPi = 3.14159\nr = radius (half of diameter) '2 = squared\nH = Height (or length)\nCI = Cubic Inches (Volume)\nConvert Volume to Cubic Inches. \nPi x r squared x H = CI / 231 = gallons
872092
It is always understandable when a student feels overwhelmed by "word problems" because they are often presented in an intimidating manner.  If only they could be FUN again ... if YOU wrote the problem, I believe you might find it fun!  :-)\n\nBut let's allow ourselves some joy once we uncover all the unknowns in this problem.  Perhaps we'll get results from converting "written words" into "mathematical expressions."\n\nWhat's most important is that you understand the shape involved (think "shoe box") and how the "openings" are placed (think "windows" on three faces of the shoe box).  All the openings are understood to be identical.\n\nUsing only the letter X -- which stands for the length of an opening -- we can indeed come up with expressions for every other dimension of the container, as well as the openings.\n\nX = Opening LENGTH (given)\nX-2 = Opening HEIGHT ("length is 2 more than height")\n\n2X+8 = Container LENGTH ("8 more than twice X")\n3 = Container HEIGHT (given)\nX+2 = Container WIDTH ("2 more than length of opening")\n\nThe surface area of the container's sides (minus the three openings) is given as 300 square meters.  The equation is also given as 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings), so let's just set these equal:\n\n300 = 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings)\n\nAll we need to do is to write, and then substitute, expressions for the areas in parentheses.\n\nSide area = length * height = 3 (2X+8) = 6X + 24\nFront or back area = width * height = 3 * (X+2) = 3X + 6\nTop or bottom area = length * width = (2X+8) * (X+2)\n____________________________ = 2X^2 + 12X + 16\nOpening area = length * height = X (X-2) = X^2 - 2X\n\nTime to substitute:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n\nNow that this (long) equation has been written, it is necessary to solve for X, since we need that value in order to calculate what the problem asks for:  The area of each element.\n\nCollect like terms:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n300 = 12X + 48 + 6X + 12 + 4X^2 + 24X + 32 - 3X^2 + 6X\n300 = X^2 + 48X + 92\nX^2 + 48X - 208 = 0\n\nWhat we hope is that the trinomial on the right-hand side of this equation is able to be factored easily!  Does it?\nMy approach is to factor '-208' and see whether I can 'build' the linear term's coefficient ('48') from the factors.\n\n-208 = -2 * 104 = -2 * 2 * 52 = -2 * 2 * 2 * 26 = -2 * 2 * 2 * 2 * 13\n\nCan we group these factors into (one positive, one negative) numbers that produce 48 when added together?  Yes!\n\n(-2*2) + (2*2*13) = -4 + 52 = 48\n\nWe got lucky ... the '-4' and '52' provide the perfect pair of factors needed to create a 48 (through addition), so we can now factor.\n\n0 = (X-4)(X+52)\n\nThings are getting better still, because only one of the zeros (some call them 'roots') from the right-hand side will be positive.  The zeros are 4 and -52, and we will only choose the 4, since it represents a physical measurement of length (of an opening).\n\nWe can go back to the expressions above and substitute X=4.\n\nSide area = 6X + 24 = 6(4)+24 = 48 sq. m\nFront/back area = 3X + 6 = 3(4)+6 = 18 sq. m\nTop/bottom area = 2X^2+12X+16 = 2(4^2)+12(4)+16 = 96 sq. m\nOpening area = X^2 - 2X = 4^2-2(4) = 8 sq. m\n\nI hope this helps!
635866
I might be missing something, but the cubic space something occupies is also referred to as it volume, where:\n\nfor a six sides regular box shape volume = heightxbreadthx depth\n\n( make sure all your measurements are in feet eg 1 foot 5 inches is 1 and 5/12 feet. \n\nHowever this formula is not true for a sphere or a trapezoidal box or a pyramid.  For whcih I'd recommend wikipedia or the old maths textbook as its hmmm, err 18 years since I last did all that stuff.
34261
Go to Paragraph in your toolbar and select spacing and then select the box that says single and switch it to double. You should see the sample diagram box go from single to double when you click apply.
493858
this will be  169646.00329384883487698274269709 cubic inches
325473
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)