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Zina [86]
3 years ago
13

Paulo is creating a confidence interval based on sample data for the percentage of Americans that own a dog. His work is shown b

elow.
for some reason it won't let me add in the picture as an attachment or as anything. Basically its like this- E = 1.645 • .452(1-.452)/950
And then all of the .452 etc is in a little square root thing.

Based on Paulo’s work, what conclusion can be drawn from the sample data he collected?


Complete the statement to form the conclusion.

It can be said with


90% confidence that between 42.5% and 47.9% of Americans own a dog.

90% confidence that between 45.2% and 54.8% of Americans own a dog.

95% confidence that between 42.5% and 47.9% of Americans own a dog.

95% confidence that between 45.2% and 54.8% of Americans own a dog.

The answer is (A) but if someone wants to explain why it is (A) that would be awesome
Mathematics
2 answers:
Andrews [41]3 years ago
5 0

Answer:

A. 90% confidence that between 42.5% and 47.9% of Americans own a dog.

Step-by-step explanation:

edgu 2020

Gekata [30.6K]3 years ago
4 0

Answer:

Checked on edg. , the answer is A. Thanks!

Step-by-step explanation:

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x=7

Step-by-step explanation:

  • The sum of exterior angles of a polygon is equal to 360 degrees
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3 0
3 years ago
An artificial lake is in the shape of a rectangle and has an area of 9/20 square mile the width of the lake is 1/5 the length of
DIA [1.3K]
Answer:  The dimensions are:   " 1.5 mi.  ×  ³⁄₁₀  mi. " .
_______________________________________________
             { length = 1.5 mi. ;  width =  ³⁄₁₀  mi. } .
________________________________________________
Explanation:
___________________________________________
Area of a rectangle:

A = L * w ; 

in which:  A = Area = (9/20) mi.² ,
                L = Length = ?
                w = width = (1/5)*L = (L/5) = ?
________________________________________
  A = L * w ;  we want to find the dimensions; that is, the values for
                         "Length (L)"  and "width (w)" ; 
_______________________________________
Plug in our given values:
_______________________________________
 (9/20) mi.² = L * (L/5) ;  in which: "w = L/5" ; 
 
     → (9/20) = (L/1) * (L/5) = (L*L)/(1*5) = L² / 5 ;
   
          ↔  L² / 5  = 9/20 ;
 
            →  (L² * ? / 5 * ?) = 9/20 ?    

                →     20÷5 = 4 ;  so; L² *4 = 9 ;
 
                   ↔    4 L² = 9 ; 
 
                   →  Divide EACH side of the equation by "4" ;
           
                   →   (4 L²) / 4 = 9/4 ;
______________________________________
           to get:  →  L² = 9/4 ; 
 Take the POSITIVE square root of each side of the equation; to isolate "L" on one side of the equation; and to solve for "L" ;
___________________________________________          
 
     →   ⁺√(L²)   =   ⁺√(9/4) ;

    →   L  =  (√9) / (√4) ; 

    →  L = 3/2 ; 

    → w = L/5 = (3/2) ÷ 5 = 3/2 ÷ (5/1) = (3/2) * (1/5) = (3*1)/(2*5) = 3/10;
________________________________________________________
Let us check our answers:
_______________________________________
(3/2 mi.) * (3/10 mi.) =? (9/20) mi.² ??

→ (3/2)mi. * (3/10)mi.  =  (3*3)/(2*10) mi.² = 9/20 mi.² ! Yes!
______________________________________________________
So the dimensions are: 

Length = (3/2) mi. ;  write as: 1.5 mi.

width = ³⁄₁₀ mi.
___________________________________________________
or; write as:  " 1.5 mi.  ×  ³⁄₁₀ mi. " .
___________________________________________________
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90. The measure of an inscribed angle is half of that of the measure of its arc, and the measure of the angle corresponding with arc a is 45. 45*2 is 90, so 90 is the measure of arc a.
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