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3 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

6 0

I you think about it 12a with the 2 on the rope is actually 144 because it’s multiplied by 12 so maybe that helps?

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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

Please help I really do not understand this yeah I can be weird sometimes but please help will mark the correct answers brainies

NISA [10]

Answer:

Step-by-step explanation:

1.0.39

2.0.33

3.Loose

4.0.14

I did the first four... hope that helps

3 0

3 years ago

Number 14 please and thank you

Fed [463]

You can do a notebook a door , a frame or a mirror, mostly anything that has angles

4 0

3 years ago

Two opposing opinions were shown to a random sample of 1,500 buyers of a particular political news app in the United States. The

kifflom [539]

Answer:

A) 98% Confidence interval for the proportion of all US buyers of this particular app who would have chosen Opinion B

= (0.51, 0.57)

This means that the true proportion of all thay would chose opinion B can take on values between the range of (0.51, 0.57)

B) For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion of all US buyers of this particular app who would have chosen Opinion B = 0.54

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 98% confidence interval for sample size of 1500 is obtained from the z-tables.

Critical value = 2.33

Standard error = σₓ = √[p(1-p)/n]

p = sample proportion = 0.54

n = sample size = 1500

σₓ = √(0.54×0.46/1500) = 0.0128685664 = 0.01287

98% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.54 ± (2.33 × 0.01287)

CI = 0.54 ± 0.02998)

98% CI = (0.51, 0.57)

98% Confidence interval = (0.51, 0.57)

B) The number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B are both greater than 10. Why must this inference condition be met?

For this confidence interval to be obtained using sample data, a couple of conditions are necessary to be satisfied. They include;

- The sample data must have been obtained using a random sampling technique.

- The sampling distribution must be normal or approximately normal.

- The variables of the sample data must be independent of each other.

On the second point, the condition for a binomial distribution to approximate a normal distribution is that

np ≥ 10

and n(1-p) ≥ 10

The quantity np is the actual sample mean which is the actual number of buyers that chose Opinion B while n(1-p) is the number of buyers that did not chose Opinion B.

For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Hope this Helps!!!

8 0

4 years ago

Someone solve this for me please

Butoxors [25]

Answer:

x=2; y=1

Step-by-step explanation:

Since the triangle is equilateral, which means three sides have the same length.

So we have：4x-y=2x+3y=7

$4x-y=2x+3y\\4x-y+y=2x+3y+y\\4x=2x+4y\\2x=4y\\x=2y\\$

Since 2x+3y=7 and plug in x=2y

we have

$2x+3y=7\\2(2y)+3y=7\\4y+3y=7\\7y=7\\y=1$

and since y=1

x=2

5 0

3 years ago