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mihalych1998 [28]
3 years ago
6

If an open box is made from a tin sheet 6 in. square by cutting out identical squares from each corner and bending up the result

ing flaps, determine the dimensions of the largest box that can be made. (Round your answers to two decimal places.)

Mathematics
1 answer:
sesenic [268]3 years ago
8 0

Answer:

4in. square x 4in. square x 1in. square

Step-by-step explanation:

By cutting out identical squares (considering 'x') from each corner and bending up the resulting flaps

The dimensions we have will be (see attachment for the figure)

V= (6-2x) (6-2x) x

As, Volume'V' = length (l) x width(w) x height(h)

V= (6x- 2x²)(6-2x)

V= 36x - 12x²-12x²+ 4x³

V=4x³ - 24x²+ 36x

Next is to find dV/dx,therefore we find the derivative and and set it to zero for the maximum volume

dV/dx = 12x² - 48x + 36

setting it to zero

12x² - 48x + 36 =0

x² - 4x + 3=0

x² -3x -x + 3=0

x(x-3) -1(x-3)=0

Either : x-3=0=> x=3

OR : x-1 =0 => x=1

Now, notice that 'x' cannot be 3 , because if  we cut 3 inch squares out of the original square, there  will be nothing left!

Also,  the volume will be 0 then. That  is the minimum volume, 0, when we cut all the tin away.

So, x=1

Therefore,

height 'x' = 1in. square

length and width = (6-2x) => 4in. square

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Brigitte is shorter than 5 feet. (Brigitte's height=h)

We write this in  inequality form

Brigitte's height is h

So 'h' is shorter than 5 feet

Shorter means less than so we use < symbol. We should not use = symbol because height is shorter than 5 not equal to 5.

Suppose if it is given ' height is more than 5 feet ' then we use greater than symbol (>)

h is less than 5 feet

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3 0
3 years ago
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The random variable x has the following probability distribution: x f(x) 0 .25 1 .20 2 .15 3 .30 4 .10 a. Is this probability di
Reptile [31]

Answer and Explanation:

Given : The random variable x has the following probability distribution.

To find :

a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.

b. Calculate the expected value of x.

c. Calculate the variance of x.

d. Calculate the standard deviation of x.

Solution :

First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.25           0               0                0

1     0.20        0.20             1              0.20

2    0.15          0.3               4             0.6

3    0.30         0.9               9             2.7

4    0.10          0.4               16             1.6

   ∑P(x)=1     ∑xP(x)=1.8               ∑x²P(x)=5.1

a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.25+0.20+0.15+0.30+0.10

\sum P(X)=1

Yes it is a probability distribution.

b) The expected value of x is defined as

E(x)=\sum xP(x)=1.8

c) The variance of x is defined as

V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86

d) The standard deviation of x is  defined as

\sigma=\sqrt{V}

\sigma=\sqrt{1.86}

\sigma=1.136

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3 years ago
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Answer:

-1/5

Step-by-step explanation:

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-1/5

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