The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;
- Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.
- Maximum volume of the box is approximately 1048.6 in.³
<h3>How can the dimensions and volume of the box be calculated?</h3>
The given dimensions of the cardboard are;
Width = 18 inches
Length = 35 inches
Let <em>x </em>represent the side lengths of the cut squares, we have;
Width of the box formed = 18 - 2•x
Length of the box = 35 - 2•x
Height of the box = x
Volume, <em>V</em>, of the box is therefore;
V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x
By differentiation, at the extreme locations, we have;
Which gives;
6•x² - 106•x + 315 = 0
Therefore;
x ≈ 4.55, or x ≈ -5.55
When x ≈ 4.55, we have;
V = 4•x³ - 106•x² + 630•x
Which gives;
V ≈ 1048.6
When x ≈ -5.55, we have;
V ≈ -7450.8
The dimensions of the box that gives the maximum volume are therefore;
- Width ≈ 18 - 2×4.55 in. = 8.89 in.
- Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.
- The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³
Learn more about differentiation and integration here:
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Answer:a) 1,2,6,24,120
Step-by-step explanation:
The formula for the nth term of the sequence is expressed as
n=(n+1)!/n+1
For the first term, n = 1. Therefore,
(1+1)!/(1+1) = 2!/2 = (2×1)/2 = 1
For the second tern, n = 2. Therefore,
(2+1)!/2+1 = 3! /3 = (3×2×1)/3 = 6/3 = 2
For the third term, n = 3, therefore
(3+1)!/3+1 = 4!/4 =(4×3×2×1)/4 = 6
For the fourth term, n = 4. Therefore
(4+1)!/4+1 = 5!/5 = (5×4×3×2×1)/5 = 24
For the fifth term, n = 5. Therefore
(5+1)!/5+1 = 6!/6 = 6× 5×4×3×2×1)/6 = 120
Answer:
it would be 14
Step-by-step explanation:
you have to simplify the expressions
Given the table below which contains data from 200 endothemic reactions involving <span>sodium bicarbonate:
![\begin{tabular} {|c|c|} Final Temperature Conditions&Number of Reactions\\[1ex] 266 K&48\\ 271 K&60\\ 274 K&92 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0AFinal%20Temperature%20Conditions%26Number%20of%20Reactions%5C%5C%5B1ex%5D%0A266%20K%2648%5C%5C%0A271%20K%2660%5C%5C%0A274%20K%2692%0A%5Cend%7Btabular%7D)
We find the probability distribution function as follows:
Let X denote the final temperature, then
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Answer:
B. There is an association because the value 0.15 is not similar to the value 0.55
Step-by-step explanation:
Based on the above picture, for the nutritionist to determine whether there is an association between where food is prepared and the number of calories the food contains, there must be an association between two categorical variables.
The conditions that satisfy whether there exists an association between conditional relative frequencies are:
1. When there is a bigger difference in the conditional relative frequencies, the stronger the association between the variables.
2. When the conditional relative frequencies are nearly equal for all categories, there may be no association between the variables.
For the given conditional relative frequency, we can see that there exists a significant difference between the columns of the table in the picture because 0.15 is significantly different from 0.55 and 0.85 is significantly different from 0.45
We can conclude that there is an association because the value 0.15 is not similar to the value 0.55
