sin(30°)cos(45°) + cos(30°)sin(45°)

Answer:

I'm not sure would love to help but dont know the answer

Answer:

The ratio of the drag coefficients [tex]\dfrac{F_m}{F_p}[/tex] 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, [tex]c_{m}[/tex] = The drag coefficient of the prototype, [tex]c_{p}[/tex]

The medium of the test for the model, [tex]\rho_m[/tex] = The medium of the test for the prototype, [tex]\rho_p[/tex]

The drag force is given as follows;

[tex]F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}[/tex]

We have;

[tex]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[/tex]

Therefore;

[tex]\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[/tex]

[tex]\dfrac{L_p}{L_m} =\dfrac{17}{1}[/tex]

[tex]\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[/tex]

[tex]\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2[/tex]

[tex]\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}[/tex]

[tex]\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2[/tex]

[tex]\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[/tex]

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

The ratio of the drag coefficients [tex]\dfrac{F_m}{F_p}[/tex] ≈ 0.0002.

Answer:

4.25.                                            

8 divides into 34 4 times, which is 32, then do 34−32, which is 2 (make sure to add your decimal) , then drop the 0 and it goes into 20 twice, subtract and get 4, drop a 0, then subtract 40 by 40 again and get 4.25

Answer:

He would need a score of 0.8 = 80%.

Step-by-step explanation:

Average score:

The average score is the sum of all scores divided by the number of tests.

What score would he need on his next test if the test scores he has so far are: 70%, 105%, 75%, 90%, 90%?

Combining with the next text, there will be 6, and the sum will be: 0.7 + 1.05 + 0.75 + 0.9 + 0.9 + x = 4.3 + x.

Joe wants his overall test average to be 85%.

So

[tex]\frac{4.3 + x}{6} = 0.85[/tex]

[tex]4.3 + x = 5.1[/tex]

[tex]x = 5.1 - 4.3 = 0.8[/tex]

He would need a score of 0.8 = 80%.

Answer:

the distance formula is square root (x2-x1)^2 + (y2-y1)^2

7^2 + 4^2 = 49+16 = 65

squarroot of 65 = 8.06

Step-by-step explanation:

Step-by-step explanation:

The distance formula is [tex]\sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2} }[/tex]. I don't provide answers for problems like these but will give the equations.

Answer:

1/2

Step-by-step explanation:

2/5 = 4/10

4/10 + 1/10 = 5/10

5/10 = 1/2

Answer:

decimal form-0.5, fraction form-1/2

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

F(-2)= 4(-2)-1= -9

G(-2)= (-2)^2-3= 1

f(-2)-(g-2)= -9-1= -10

Answer:

A = 36.8°

B = 23.2°

a = 7.6

Step-by-step explanation:

Given:

C = 120°

b = 5

c = 11

Required:

Find A, B, and a.

Solution:

✔️To find B, apply the Law of Sines

[tex] \frac{sin(B)}{b} = \frac{sin(C)}{c} [/tex]

Plug in the values

[tex] \frac{sin(B)}{5} = \frac{sin(120)}{11} [/tex]

Cross multiply

Sin(B)*11 = sin(120)*5

Divide both sides by 11

[tex] sin(B) = \frac{sin(120)*5}{11} [/tex]

[tex] sin(B) = \frac{sin(120)*5}{11} [/tex]

Sin(B) = 0.3936

B = [tex] sin^{-1}(0.3936) [/tex]

B = 23.1786882° ≈ 23.2° (nearest tenth)

✔️Find A:

A = 180° - (B + C) (sum of triangle)

A = 180° - (23.2° + 120°)

A = 36.8°

✔️To find a, apply the Law of sines:

[tex] \frac{sin(A)}{a} = \frac{sin(B)}{b} [/tex]

Plug in the values

[tex] \frac{sin(36.8)}{a} = \frac{sin(23.2)}{5} [/tex]

Cross multiply

a*sin(23.2) = 5*sin(36.8)

Divide both sides by sin(23.2)

[tex] a = \frac{5*sin(36.8)}{sin(23.2) [/tex]

a = 7.60294329 ≈ 7.6 (nearest tenth)

Answer: 19 odd numbers: 11 , 13,  15,  17 , 19,  23,  25,  27,  29,  31,  33 , 35,  37, 39,  41,  43,  45, 47,  49

Hope that helped! :)

Answer:

x = -3

Step-by-step explanation:

2x + 9 = 3

2x = 3 - 9

2x = -6

x = -3

Answer:

- 3

Step-by-step explanation:

given

2x + 9 = 3

2x = 3 - 9

2x = - 6

x = - 6/2

x = - 3

hope it helps :)

Answer:

im not 100 percent sure but i think its

140 inches squared

Step-by-step explanation:

Answer:

No, I have not heard. Give me brainliest plz, I neeeeeeeeeeed it to level up

Step-by-step explanation:

Answer:

your answer is 7 meters

I hope it's helps you ❤️

Answer:

A

Step-by-step explanation:

since mean is the most annoying to calculate I went through and did everything else, the range is the same as both charts go 0-6. The median is is the middle number which means store 1 has a higher number of 3' s but our option says store 2 is greater and B is the mode, the number that shows up the most.

Answer:

The answer is: B.)32.0–32.1 ounces

Step-by-step explanation:

Got it right on my test

The true statement is The median weight for both types is between 32.0 and 32.1 ounces.

Data is referred to as a collection of information gathered by observations, measurements, research, or analysis. It may comprise facts, figures, numbers, names, or even general descriptions of things. Data can be organized in the form of graphs, charts, or tables for ease in our study.

Given:

Brand A and brand B of frozen green beans both list a weight of 32 ounces on their bags.

The median weight for brand B is less than the median weight of brand A.

So, if both brands weigh 32 ounces then the median weight for both types is between 32.0 and 32.1 ounces.

Learn more about data here:

https://brainly.com/question/10980404

#SPJ6

Answer:

Part a:

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)=  0.6628

Step-by-step explanation:

The null and alternate hypothesis are

H0: μ1 ≥ μ2   against the claim Ha: μ1 < μ2

This is left tailed test

The critical region for this test at 0.10 ∝ is z < ± 1.28

The test statistic

z= x1-x2/ s/√n

z= 66-61/ 3/√24

z= 5/0.61238

z= 8.16486

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)  = P ( accept H0/ H0 is false)

z ( ∝)= X-u/s/√n

1.28= X-61/0.61238

X= 61.7838

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 61.7838

 The probability of a Type II error is given by

P (z> 61.7838-66/24)

P (z>- 0.42161)

From the tables

P (z>- 0.42161)=  0.6628

Answer:

Step-by-step explanation:

When looking at a graph, the greatest speed will be the steepest part. Even though one line is further than another, does not mean it is faster.

Answer:

There was a increase of 1.7% over these three​ years

Step-by-step explanation:

Multipliers:

For a decrease of a%, we multiply by: [tex]\frac{100-a}{100}[/tex]

For a increase of a%, we multiply by: [tex]\frac{100+a}{100}[/tex]

What was the percentage​ increase/decrease of groundhogs over these three​ years?

Decrease of 12%(multiplication by 0.88).

Increase of 6%(multiplication by 1.06).

Increase of 9%(multiplication by 1.09).

After these three years:

0.88*1.06*1.09 = 1.017

1.017 - 1 = 0.017

0.017*100% = 1.7%

There was a increase of 1.7% over these three​ years

Answer:

56m

Step-by-step explanation:

5+7+5+11+7+16+5

= 56

Answer:

2( x + 1) = 18

2x + 2 = 18

2x = 16

x = 16/2 = 8

Answer:

D

Step-by-step explanation:

Just divide by 2/3

Answer:

[tex]PC = 0.333[/tex]

Step-by-step explanation:

Given

[tex]AB = 5[/tex]

[tex]AP = 4[/tex]

[tex]DP =12\\[/tex]

See attachment

Required

Find PC

When two chords intersect, the product of the segments are always equal.

In chord DC, the segments are DP and PC

In chord AB, the segments are AP and PB

So:

[tex]DP * PC = AP * PB[/tex]

This gives:

[tex]12 * PC = 4 * PB[/tex] --- (1)

In chord AB, we have:

[tex]AB = AP + PB[/tex]

Make PB the subject

[tex]PB = AB - AP[/tex]

This becomes

[tex]PB = 5 - 4[/tex]

[tex]PB = 1[/tex]

[tex]12 * PC = 4 * PB[/tex] --- (1) becomes

[tex]12 * PC = 4 * 1[/tex]

[tex]12 * PC = 4[/tex]

Make PC the subject

[tex]PC = \frac{4}{12}[/tex]

[tex]PC = 0.333[/tex]

Answer:

the answer gonna be 3

Answer: I the answer is 3 100%

Step-by-step explanation: I took the test

Answer:

Step-by-step explanation:

Number of blue marbles = 10

Number of white marbles = 7

Number of Red marbles = 6

Number of Green marbles = 2

Total marbles = 10 + 7 + 6 + 2 = 25

Probability of selecting the blue marble = 10/25 = 2/5

Answer:

3.294

Step-by-step explanation:

Answer:

The answer is A

Step-by-step explanation:

x to the power of 1 is the lowest x and when you multiply -5 and -5 you get 5 so the answer is A.

Answer:

A sample of 383 would be required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In an earlier study, the population proportion was estimated to be 0.3.

This means that [tex]\pi = 0.3[/tex]

80% confidence level

So [tex]\alpha = 0.2[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.2}{2} = 0.9[/tex], so [tex]Z = 1.28[/tex].

How large a sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 80% confidence level with an error of at most 0.03?

A sample of n is needed.

n is found when M = 0.03. So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.03 = 1.28\sqrt{\frac{0.3*0.7}{n}}[/tex]

[tex]0.03\sqrt{n} = 1.28\sqrt{0.3*0.7}[/tex]

[tex]\sqrt{n} = \frac{1.28\sqrt{0.3*0.7}}{0.03}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.28\sqrt{0.3*0.7}}{0.03})^2[/tex]

[tex]n = 382.3[/tex]

Rounding up:

A sample of 383 would be required.

Answer:

[tex] log_{a}(a) \ + log_{a}( {10}^{ - 4} ) = 1 + \frac{ - 4}{ log_{10}(a) } = 1 + \frac{ - 4}{b} = \frac{b - 4}{b} [/tex]

................................................................................................................................................................................

................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................