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Kipish [7]
3 years ago
15

A music industry researcher wants to estimate, with a 90% confidence level, the proportion of young urban people (ages 21 to 35

years) who go to at least 3 concerts a year. Previous studies show that 35% of those people (21 to 35 year olds) interviewed go to at least 3 concerts a year. The researcher wants to be accurate within 2% of the true proportion. Find the minimum sample size necessary.
Choose one • 10 points

2185

1539

2401

8740
Mathematics
1 answer:
Maslowich3 years ago
6 0

Answer:

1,539

Step-by-step explanation:

Using Simple Random Sampling in an infinite population (this is such a large population that we do not know the exact number) we have that the sample size should be the nearest integer to

\large \frac{Z^2pq}{e^2}

where

<em>Z= the z-score corresponding to the confidence level, in this case 90%, so Z=1.645 (this means that the area under the Normal N(0,1) between [-1.645,1.645] is 90%=0.9) </em>

<em>p= the proportion of young urban people (ages 21 to 35 years) who go to at least 3 concerts a year= 35% = 0.35 </em>

<em>q = 1-p = 0.65 </em>

<em>e = the error proportion = 2% = 0.02 </em>

Making the calculations

\large \frac{Z^2pq}{e^2}=\frac{(1.645)^2*0.35*0.65}{(0.02)^2}=1,539.09

So, the sample size should be 1,539 young urban people (ages 21 to 35 years)

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F:[1/2,infinity)—&gt;[3/4,infinity) <br> f(x)=x^2-x+1 <br> find the inverse of f(x);please explain
dexar [7]

Answer:

f^{-1}(x)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}

Step-by-step explanation:

y=x^2-x+1

We want to solve for x.

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y-1=x^2-x

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This allows me to write the right hand side as a square.

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Now remember we are solving for x so now we square root both sides:

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The problem said the domain was 1/2 to infinity and the range was 3/4 to infinity.

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That is we want:

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Add 1/2 on both sides:

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8 0
3 years ago
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. T
Kaylis [27]

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

6 0
3 years ago
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