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3 years ago

How can the effect of an outlier be overcome in an experiment that examines a random sample of a population?

1 answer:

3 0

To overcome and identify outliers, the data obtained can be arranged into from the highest to lowest and be oberved on which values have the highest or lowest value.

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-6 (-2/3)<br> How is it solved

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You multiply -6 and -2/3. Which is 4

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Please help this is timed

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3 years ago

Find the volume. Round to the nearest tenth.

sergey [27]

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3 years ago

A manufacturer wants to make open tin boxes from pieces of tin with dimensions 8 in. by 15 in. by cutting equal squares from the

Stells [14]

Answer:

$\frac{5}{3}$ in

Step-by-step explanation:

Let x be the side of square.

Length of box=8-2x

Width of box=15-2x

Height of box=x

Volume of box= $L\times B\times H$

Substitute the values then we get

Volume of box=V(x)= $(8-2x)(15-2x)x=(15x-2x^2)(8-2x)$

$V(x)=120x-30x^2-16x^2+4x^3$

$V(x)=4x^3-46x^2+120x$

Differentiate w.r.t x

$V'(x)=12x^2-92x+120$

$V'(x)=0$

$12x^2-92x+120=0$

$3x^2-23x+30=0$

$3x^2-18x-5x+30=0$

$3x(x-6)-5(x-6)=0$

$(x-6)(3x-5)=0$

$x-6=0\implies x=6$

$3x-5=0\implies x=\frac{5}{3}$

Again differentiate w.r.t x

$V''(x)=24x-92$

Substitute x=6

$V''(6)=24(6)-92=52>0$

Substitute x=5/3

$V''(5/3)=24(5/3)-92=-52$

Hence, the volume is maximum at x= $\frac{5}{3}$

Therefore, the side of the square , $x=\frac{5}{3}$ in cutout that gives the box the largest possible volume.

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3 years ago