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Studentka2010 [4]
3 years ago
9

Find the mode(s) of the number of calories

Mathematics
1 answer:
alexdok [17]3 years ago
6 0
A mode is the most repeated data point
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What is the name of the place value of 9 in 199
Sliva [168]

The name of place value of 9 in 199 is ones.

The second 9 has a value of tens.

The number 1 has a place value of hundreds.

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3 years ago
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Sam is filling up his pool. The water hose filled the 315 gallon pool in 45 minutes. At what rate did the hose fill the pool in
Serhud [2]

Answer:

7 gallons per minute

Step-by-step explanation:

315/45 = 7

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A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot
Olegator [25]

Answer:

The ratio of the drag coefficients \dfrac{F_m}{F_p} is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, c_{m} = The drag coefficient of the prototype, c_{p}

The medium of the test for the model, \rho_m = The medium of the test for the prototype, \rho_p

The drag force is given as follows;

F_D = C_D \times A \times  \dfrac{\rho \cdot V^2}{2}

We have;

L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m

Therefore;

\dfrac{L_p}{L_m}  = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2

\dfrac{L_p}{L_m}  =\dfrac{17}{1}

\therefore \dfrac{L_p}{L_m}  = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2

\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2

\dfrac{F_p}{F_m}  = \dfrac{c_p \times A_p \times  \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times  \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}

\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2

\dfrac{F_p}{F_m}  = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3

\dfrac{F_m}{F_p}  = \left( \left\dfrac{1}{17} \right)^3= (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients \dfrac{F_m}{F_p} ≈ 0.0002.

5 0
3 years ago
If you dont answer all I won't give brain list...(If you do I will!)
Goryan [66]

Answer:

dude bace times hight 4×2×3

3 0
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How do we calculate the rate of change for a given domain
SashulF [63]

Answer:The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

Step-by-step explanation:

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