A recent survey asked UF students if they had a job or not, and if they were underclassmen (Freshmen and Sophomores) or uppercla
ssmen (Juniors and Seniors). The answers appear on the table below.
JOB NoJob
Underclassmen 40 37 77
Upperclassmen 42 39 81
82 76 158
Let A be the event that a student has a job, and let B be the event that a student is an underclassmen. Find P(AUB).
JOB NoJob
Underclassmen 40 37 77
Upperclassmen 42 39 81
82 76 158
Let A be the event that a student has a job, and let B be the event that a student is an underclassmen. Find P(AUB).
1 answer:
Following the table and knowing that the total number of students interviewed were 158 ( we can see this by looking adding either the total number of upperclassment or adding the total number of people with jobs or no jobs, this value is at the bottom right of the table in the figure attached).
Recall that:
In the figure provided each of these terms is highlighted in a different color. To convert these values to their matching probabilities we have to divide each by the total number of students, this is due to the fact that the probability is the number of favorable cases (in this case a group matching the qualities we seek) divided by the total amount of cases ( that is the total number of people interviewed). In the figure the answer is provided. For the intersection of the two events we're looking for people that is both an undercalssman and also has a job.
Recall that:
In the figure provided each of these terms is highlighted in a different color. To convert these values to their matching probabilities we have to divide each by the total number of students, this is due to the fact that the probability is the number of favorable cases (in this case a group matching the qualities we seek) divided by the total amount of cases ( that is the total number of people interviewed). In the figure the answer is provided. For the intersection of the two events we're looking for people that is both an undercalssman and also has a job.
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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.
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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 ;
__________________________________________________
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: .
_________________________________________
______________________________
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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