Simplify: 19w5+ (-3075)
1 answer:
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The confidence interval is
. This means that we can be 99% confident that the mean number of books people read lies between 9.15 and 11.85.
To find the confidence interval, we first find the z-score associated with it:
Convert 99% to a decimal: 0.99
Subtract from 1: 1-0.99=0.01
Divide by 2: 0.01/2 = 0.005
Subtract from 1: 1-0.005 = 0.995
Using a z-table (http://www.z-table.com) we see that this is associated with a z-score between 2.57 and 2.58. Since both are equally far from this value we will use 2.575.
We calculate the margin of error using
This means that the confidence interval is
The lower limit is given by 10.5-1.35 = 9.15.
The upper limit is given by 10.5+1.35 = 11.85
To find the confidence interval, we first find the z-score associated with it:
Convert 99% to a decimal: 0.99
Subtract from 1: 1-0.99=0.01
Divide by 2: 0.01/2 = 0.005
Subtract from 1: 1-0.005 = 0.995
Using a z-table (http://www.z-table.com) we see that this is associated with a z-score between 2.57 and 2.58. Since both are equally far from this value we will use 2.575.
We calculate the margin of error using
This means that the confidence interval is
The lower limit is given by 10.5-1.35 = 9.15.
The upper limit is given by 10.5+1.35 = 11.85
The answer is: " 
" .
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Given: 0.1212121212..... repeating ; write that value as a fraction;
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In other words; we are given: "0.1212121212..... repeating infinitely" ;
→ that is to say; "0.12 ... ; {the "12" decimal portion repeats infinitely} ;
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→ We write this value, as a fraction, as: "12/99" ;
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→Explanation:
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Note: "0.99999999...... repeating infinitely; = "1" .
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→Since:
Let us say that we have:
"x = 0.999999 ; repeating infinitely;
In order words, let us say we have: "x = 0.9.... ; the "9" decimal repeats infinitely;
_____________________________________________________
Then "10x" ; (that is: "10" multlipled by "x"; or "10*x" or "10x" ); is equal to:
"10" * (0.999999.....) = 9.99999999...... (the "9" decimal repeats infinitely);
in other words: 10x = 9.99999999....
Divide each side by "10" ;
to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;
But if you have: "10x = 10" ; divide each side of the equation by "10" ;
you get: "x = 1" .
____________________
Also, if "x = 0.9999...(repeating infinitely);
then: 10x = 9.99999.
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10x = 9.999999999999999......
− x = 0.999999999999999.......
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9x = 9.00000000000000000000.....
→ 9x = 9 ;
Divide each side of the equation; to get;
9x /9 = 9/ 9 ; to get: x = 1 ; and we have: x = 0.9999.... ; so
x= 0.99999.... = 1 ;
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So, if the numbers "12" is repeating, we divie "12" by "99" ;
that is; we divide "12" by "two 9's" ; since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
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So; 0.12121212.....(the "12" is the decimal that repeats infinitely);
= 12/99 ; which can be simplified;
Divide each side (both the numerator AND the denominator); by "3" ;
_________________________________________
" 12/99 " = "(12÷3) / (99÷3) = 4/33 " .
_________________________________________
The answer is: .
_________________________________________
______________________________
Given: 0.1212121212..... repeating ; write that value as a fraction;
______________________________________________________
In other words; we are given: "0.1212121212..... repeating infinitely" ;
→ that is to say; "0.12 ... ; {the "12" decimal portion repeats infinitely} ;
_______________________________________________________
→ We write this value, as a fraction, as: "12/99" ;
__________________________________________
→Explanation:
__________________________________________
Note: "0.99999999...... repeating infinitely; = "1" .
_________________________________________
→Since:
Let us say that we have:
"x = 0.999999 ; repeating infinitely;
In order words, let us say we have: "x = 0.9.... ; the "9" decimal repeats infinitely;
_____________________________________________________
Then "10x" ; (that is: "10" multlipled by "x"; or "10*x" or "10x" ); is equal to:
"10" * (0.999999.....) = 9.99999999...... (the "9" decimal repeats infinitely);
in other words: 10x = 9.99999999....
Divide each side by "10" ;
to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;
But if you have: "10x = 10" ; divide each side of the equation by "10" ;
you get: "x = 1" .
____________________
Also, if "x = 0.9999...(repeating infinitely);
then: 10x = 9.99999.
_______________________________________________
10x = 9.999999999999999......
− x = 0.999999999999999.......
_____________________________________
9x = 9.00000000000000000000.....
→ 9x = 9 ;
Divide each side of the equation; to get;
9x /9 = 9/ 9 ; to get: x = 1 ; and we have: x = 0.9999.... ; so
x= 0.99999.... = 1 ;
__________________________________________________
So, if the numbers "12" is repeating, we divie "12" by "99" ;
that is; we divide "12" by "two 9's" ; since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
________________________________________
So; 0.12121212.....(the "12" is the decimal that repeats infinitely);
= 12/99 ; which can be simplified;
Divide each side (both the numerator AND the denominator); by "3" ;
_________________________________________
" 12/99 " = "(12÷3) / (99÷3) = 4/33 " .
_________________________________________
The answer is:
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