So we are given the mean and the s.d.. The mean is 100 and the sd is 15 and we are trying the select a random person who has an I.Q. of over 126. So our first step is to use our z-score equation:
z = x - mean/s.d.
where x is our I.Q. we are looking for
So we plug in our numbers and we get:
126-100/15 = 1.73333
Next we look at our z-score table for our P-value and I got 0.9582
Since we are looking for a person who has an I.Q. higher than 126, we do 1 - P. So we get
1 - 0.9582 = 0.0418
Since they are asking for the probability, we multiply our P-value by 100, and we get
0.0418 * 100 = 4.18%
And our answer is
4.18% that a randomly selected person has an I.Q. above 126
Hopes this helps!
Not sure if that's a multiplication or a composition. It doesn't matter. Both f and g have a domain of all real numbers. The domain of
is the same as the domain of g(x), all real numbers
Answer: all real numbers
Answer:
TC (A) = 40x , TC (B) = 500 + 20x
Step-by-step explanation:
Let the number of students be = x
Hall A Total Cost
Relationship Equation, where TC (A) = f (students) = f (x) 40 per person (student) = 40x
Hall B Total Cost
Relationship Equation, where TC (B) = f (students) = f (x) 500 fix fee & 20 per person (student) = 500 + 20x
You had 12 the next day and the day after that you had 18
4.29 + 97.2 + 0.687 = 102.177
In adding decimal numbers, make sure that the decimal points are aligned. Since each number has different counts of numbers after the decimal point, use 0 to pad the missing places.
4.290
97.200
<u> 0.687
</u> 102.177
The count of numbers after the decimal point is the same count of number of the decimal who has the greatest count of number after the decimal point.
4.29 only has 2 counts of places after the decimal point
97.2 only has 1 count of place after the decimal point
0.687 has 3 counts of places after the decimal point.
The sum of the decimals must also have 3 counts of places after the decimal point.
