PLEASE HELP 30 cm
30 cm
76 cm
Casa wames to put a layer of sand at the bottom of the tank. The sand layer is to be 3 cm deep. If the density of sand is 2.65 9 what mass of sand does Carla need to purchase? Round to the nearest gram.
cm
6810
18. 136
25.11

cm
6810
18. 136
25.11

PLEASE HELP 30 Cm30 Cm76 CmCasa Wames To Put A Layer Of Sand At The Bottom Of The Tank. The Sand Layer

Answers

Answer 1

The mass of Sand that Carla needs to purchase for the sand tank is gotten to be; 25811 g

What is the mass required?

Formula for density is;

Density = mass/volume

From the box we are given, it's volume is;

Volume = 30 * 30 * 76

Volume = 68400 cm³

Now, from conversion;

1 cm³ = 1g. Thus;

68400 cm³ = 68400 g

Volume = Mass/density

Volume = 68400/2.65

Volume = 25811 g

Read more about mass required at; https://brainly.com/question/1762479


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Answers

Answer:

Step-by-step explanation:

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Standard deviation     0.9798

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in 6-month distribution, the probability of the negative value of the cash position is as follows.

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[tex]z = \dfrac{x-\mu}{\sigma}[/tex]

[tex]z = \dfrac{0 - 2.6}{0.9798}[/tex]

[tex]z = -2.6536[/tex]

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[tex]z = \dfrac{x-\mu}{\sigma} \\ \\ z = \dfrac{0-3.2}{1.3856} \\ \\ z = -2.3094[/tex]

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c) To determine the time period over which the likelihood of achieving a negative cash condition is highest, it's necessary to examine the z-score more closely. Essentially, the z-score measures the difference between a given value(x) and the mean of all potential values [tex](\mu)[/tex], expressed in terms of the total set's standard deviation [tex](\sigma)[/tex]

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[tex]z = \dfrac{x-\mu}{\sigma} \\ \\ z = \dfrac{0-(initial \ value + \alpha T)}{b \sqrt{T}} \\ \\ z = \dfrac{-initial \ value }{b\sqrt{T}}-\dfrac{a \sqrt{T}}{b} \\ \\[/tex]

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[tex]\dfrac{dz}{dT}= \dfrac{initial \ value \times T^{-\dfrac{3}{2}}}{2b} - \dfrac{aT^{-\dfrac{1}{2}}}{2b}[/tex]

Now, figure out the value of T at which this derivative is equal to zero by substituting all values as follows:

[tex]0 = \dfrac{2.0 \times T^{-\dfrac{3}{2}}}{2\times 0.4}- \dfrac{0.1 \times T^{-\dfrac{1}{2}}}{2 \times 0.4} \\ \\ \\ 0.1 \times T^{-\dfrac{1}{2}}= 2.0 \times T^{-\dfrac{3}{2}} \\ \\ \\T = \dfrac{2}{0.1} \\ \\ \\ T = 20[/tex]

As a result, the time period in which achieving a negative cash condition is = 20 months.

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