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tangare [24]
3 years ago
6

A worker at a landscape design center uses a machine to fill bags with potting soil. Assume that the quantity put in each bag fo

llows the continuous uniform distribution with low and high filling weights of 14.1 pounds and 16.2 pounds, respectively.Calculate the expected value and the standard deviation of this distribution.
Mathematics
1 answer:
ddd [48]3 years ago
4 0

Answer:

Mean = 15.15

Standard deviation = 0.6062

Step-by-step explanation:

We are given a uniform distribution of  quantity to be put in each bag.

Low and high filling weights:

14.1 pounds and 16.2 pounds

a = 14.1

b = 16.2

a) Mean:

\mu = \displaystyle\frac{a+b}{2}\\\\\mu = \frac{14.1+16.2}{2} = 15.15

b) Standard Deviation:

\sigma = \sqrt{\displaystyle\frac{(b-a)^2}{12}}\\\\= \sqrt{\dfrac{(16.2-14.1)^2}{12}} = \sqrt{0.3675} = 0.6062

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Answer:

you have to find common denominators first.

Step-by-step explanation:

24 is their common

8x2=16/24

3x3=9/24

your first fraction is larger!

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Answer:

22.222222%

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The segments shown below could form a triangle ac=9. cb=8. ba=17
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Examplelt: A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall
harina [27]

Answer :

  • 4167 bricks.

Explanation :

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here,

{\qquad \dashrightarrow{ \sf{Length=10 \: m=1000 \: cm}}}

\qquad \dashrightarrow{ \sf{Thickness=24 \: cm}}

\qquad \dashrightarrow{ \sf{Height=4 m=400 \: cm}}

Therefore,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: the  \: wall = length \times breadth \times height}}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: the  \: wall = 1000 \times 24 \times 400 \:  {cm}^{3} }}}

Now, each brick is a cuboid with Length = 24 cm, Breadth = 12 cm and height = 8 cm.

So,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: each  \: brick = length \times breadth \times height}}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: each  \: brick = 24 \times 12 \times 8 \:  {cm}^{3} }}}

So,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: bricks  \: required =  \dfrac{volume \: of \: the \: wall}{volume \: of \: each \: brick} }}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: bricks  \: required =  \dfrac{1000 \times 24 \times 400}{24 \times 13 \times 8} }}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: bricks  \: required =   \bf \: 4166.6} }}

Therefore,

  • <u>The wall requires 4167 bricks. </u>

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24 cars 8 times 3 hope this helps
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3 years ago
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