<span>h = (19 - sqrt(97))/6, which is approximately 1.525190366
The volume of the box will be
V = lwh
And l will be
a - 2h
And w will be
b - 2h
So using the above, the volume of the box will be
V = lwh
V = (a - 2h)(b - 2h)h
V = (11 - 2h)(8 - 2h)h
V = (88 - 22h -16h + 4h^2)h
V = (88 - 38h + 4h^2)h
V = 88h - 38h^2 + 4h^3
Since you're looking for a maximum, that screams "First derivative" So let's calculate the first derivative of the function and solve for 0.
V = 88h^1 - 38h^2 + 4h^3
V' = 1*88h^(1-1) - 2*38h^(2-1) + 3*4h^(3-1)
V' = 1*88h^0 - 2*38h^1 + 3*4h^2
V' = 88 - 76h + 12h^2
We now have a quadratic equation. So using the quadratic formula with A=12, B=-76, and C=88, calculate the roots as:
(19 +/- sqrt(97))/6
which is approximately 1.525190366 and 4.808142967
We can ignore the 4.808142967 value since although it does indicate a slope of 0, it produces a negative width and is actually a local minimum of the volume function.
So the optimal value of h is (19 - sqrt(97))/6, which is approximately 1.525190366</span>
Answer:
m<11=5.75
Step-by-step explanation:
<u>since you already know what m<4 is </u>
<u>we just have to divide the result (63) by 4</u>
<u>63%4=15.75 </u>
therefore m<1= 15.75
Integers are whole numbers that are either positive or negative. Operation of signs is important when dealing with integers.
In this case, -3 and -2 are integers that are being added.
if we add -3 by -2 we get -6 because when a negative number is added to a negative number the negative sign does not change
thus, (-3)-2 = -5
Answer:
We want to construct a confidence interval at 99% of confidence, so then the significance level would be 
and the value of 
. And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.
And if we find the critical value in the normal standard distribution or excel and we got:
And we can use the following excel code:
"=NORM.INV(0.005,0,1)"
Step-by-step explanation:
For this case we have the following info given:
We want to construct a confidence interval at 99% of confidence, so then the significance level would be and the value of . And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.
And if we find the critical value in the normal standard distribution or excel and we got:
And we can use the following excel code:
"=NORM.INV(0.005,0,1)"
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