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Mice21 [21]
2 years ago
6

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width 8.75 inches, length 8.75 i

nches, height 23 inches.
The hot coffee has a density of 5.8 oz/in3 and the cold coffee has a density of 3.5 oz/ in3

(A) What is the volume of the coffee in each tank?
Show all math work needed to complete this problem.

(B) What is the mass of the coffee in each the tank?
Show all math work needed to complete this problem.

(C) How many tanks can be placed on a table which can support a maximum weight of 800 lb?
Explain why or why not.
Show all math work needed to complete this problem.
Mathematics
1 answer:
Blababa [14]2 years ago
3 0

The volume of the coffee in each tank will be 1760. in³. The mass of cold coffee and hot coffee will be 6163.29 oz and 10213.45 oz. Then the number of the tank placed on a table is one only.

<h3>How to find the density and volume?</h3>

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width of 8.75 inches, a length of 8.75 inches, and a height of 23 inches.

The hot coffee has a density of 5.8 oz/in³ and the cold coffee has a density of 3.5 oz/ in³.

Then the volume of the coffee in each tank will be

We know that the dimension of both tanks is equal. Then the volume will remains equal.

Volume = L × B × H

Volume = 8.75 × 8.75 × 23

Volume = 1760.9375 ≅ 1760.94 in³

Then  the mass of the coffee in each the tank will be

The mass of hot coffee will be

Mass = Density × Volume

Mass = 5.8 × 1760.94

Mass = 10213.45 oz

The mass of cold coffee will be

Mass = Density × Volume

Mass = 3.5 × 1760.94

Mass = 6163.29 oz

Then the number of the tank placed on a table is one only.

More about the density and volume link is given below.

brainly.com/question/952755

#SPJ1

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Answer:

E. 99.7%

Step-by-step explanation:

Calculate each z-score.

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3 years ago
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Answer:

Volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

Step-by-step explanation:

Given: The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s.

To find: The rate at which volume of the cone changing when the radius is 134 in. and the height is 136 in.

Solution:

We have,

\frac{dr}{dt} =1.9 \:\text{in/s}, \frac{dh}{dt}=-2.2\:\text{in/s}, r=134 \:\text{in}, h=136\:\text{in}

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Differentiate with respect to t.

\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]

Now, on substituting the values, we get

\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left (  136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]

\frac{dv}{dt} =\frac{1}{3}\pi\left [  -39503.2+69251.2 \right ]  

\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]

\frac{dv}{dt} =9916\pi \frac{in^3}{s}

Hence, the volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

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During 2002 and 2003, a total of 9316 people were infected with a specific virus. In 2003, 2.35 times as many people were infect
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X = 2002, y = 2003

x + y = 9316
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