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Lady bird [3.3K]
3 years ago
10

3. A researcher randomly selects a sample of 61 former student leaders from a list of graduates of UNCG who had participated in

leadership positions while a student. She discovered that it has taken an average of 4.97 years for these student leader graduates to finish their degrees, with a standard deviation of 1.23. The average for the entire student body is 4.56 years. Is the difference between student leaders and the entire student population statistically significant at the alpha
Mathematics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

not statistically significant at ∝ = 0.05

Step-by-step explanation:

Sample size( n )  = 61

Average for student leader graduates to finish degree ( x') = 4.97 years

std = 1.23

Average for student body = 4.56 years

<u>Determine if the difference between the student leaders and the entire student population is statistically significant at alpha</u>

H0( null hypothesis ) : u = 4.56

Ha : u ≠ 4.56

using test statistic

test statistic ; t = ( x' - u ) / std√ n

                        = ( 4.97 - 4.56 ) / 1.23 √ 61

                        = 2.60

let ∝ = 0.05 , critical value = -2.60 + 2.60

Hence we wont fail to accept  H0

This shows that the difference between the student leaders and the entire student population is not statistically significant at ∝ = 0.05

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sukhopar [10]

Answer:

Step-by-step explanation:

Given the mass is m =16kg, and 1N force will stretch the spring 1 m.

That is, F =1N,Z =1m. Now find the spring constant k:

F = kL = 1 = k(1) = k= 1N/m.

The damping force is 8times the instantaneous velocity, this means β = 8,

and the external force is f(t) = 0

Initially the object compressed 0.6m above equilibrium position,

with the downward velocity is 2m/s.

The differential equation for a spring mass system with

damping force and extemal force is: mx" + βxt + kx = f(t).

so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.

Now solve the DE:

The auxilary equation for the homogeneous equation is 16x"+8x'+x=0

solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.

Then the general solution for the homogenous system is: x(t)=c_1e^{-t/4} +c_2te^{-t\4}.

Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:

x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85.

Hence, x(t) =-0.6e^{-t/4}+1.85te^{-t/4}.

5 0
3 years ago
Can someone please explain how I would figure this out
skad [1K]
[1], [3] => a = 49,000/2 = 24,500 => b +c = 24,500 => c = 24,500 - b;
[2] => 25*24,500 + 20b + 15c = 1,052,000 => 20b + 15c = 1,027,500;
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5 0
3 years ago
Solve for x.
Ivanshal [37]

Answer:

x = -3.25

Step-by-step explanation:

2/5 = 0.4

0.4x - 0.2 = -2(3/4)

-2(3/4) = -2/1 * 3/4 = -6/4

0.4x - 0.2 = -6/4

Convert to fraction

2/5x - 1/5 = -6/4

Add 1/5 to both sides

-6/4 + 1/5 = -30/20 + 4/20 = -26/20

2/5x = -26/20

Convert to decimal

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-1.3/0.4 = -3.25

x = -3.25

8 0
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Answer:

Break-even point in units= 20,000

Step-by-step explanation:

Giving the following information:

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<u>To calculate the break-even point in units, we need to use the following formula:</u>

Break-even point in units= fixed costs/ contribution margin per unit

Break-even point in units= 314,800 / (29.99 - 14.25)

Break-even point in units= 20,000

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