A proper fraction is a fraction such as 3/5, or 2/4, any fraction where the numerator is less than the denominator. Then it says "the whole is divided into an even number of equal parts, and an odd number of those parts exists." Such as 1 3/5. From my understanding, you then divide by 2, which would give you 0.8, which could then be turned into 0.4 if that's what your teacher wants you to do. In the end, there will be no odd number.
There are 36 possible ways of throwing two dices. The following lists are shown in ordered pair:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
a) a pair of 3's
The probability is 1/36 because there is only one pair which is (3,3).
b) any pair
The probability is 6/36 or 1/6 because the number of pairs are (1,1) (2,2) (3,3) (4,4) (5,5) and (6,6).
c) at least one 4
The probability is 10/36 or 5/18 because there would be (4,1) (4,2) (4,3) (4,5) (4,6) (1,4) (2,4) (3,4) (5,4) (6,4) for any combination.
d) a total of 9
The probability is 4/26 or 2/13 because (4,5) (5,4) (3,6) (6,3) are the totals of 9.
5) a total greater than 8
The probability is 10/36 or 5/18. Just find the ordered pair having a sum greater than 8.
Hope this helps.
Answer:
The Total number of Professors = 48
Step-by-step explanation:
Given as,
The total number of Faculty = 136
And the Ratio of Professors to Lecturers = 6 Ratio 11
Now According to question
Professors : Lecturers = 6 : 11
So, Let the Number of Professors = 6x
And The Number of Lecturers = 11x
∴ 6x + 11x = 136
Or, 17x = 136
Or, x = = 8
Hence, the number of Professors = 6 × 8 = 48 Answer
Answer:
Step-by-step explanation:
we would like to solve the following trigonometric equation:
the left hand side can be rewritten using <u>angle </u><u>sum </u><u>indentity</u><u> </u><u>of </u><u>sin </u>which is given by:
therefore Let
Thus substitute:
simplify addition:
keep in mind that <u>sin(</u><u>t)</u><u>=</u><u>sin(</u><u>π-t)</u><u> </u>saying that there're two equation to solve:
take inverse trig and that yields:
add π to both sides of the second equation and that yields:
sin function has a period of <u>2</u><u>n</u><u>π</u><u> </u>thus add the period:
divide I equation by 4 and II by -4 which yields:
recall that,<u>-</u><u>½</u><u>(</u><u>nπ)</u><u>=</u><u>½</u><u>(</u><u>nπ)</u><u> </u>therefore,
by using a calculator we acquire:
hence,
the general solution for: for the trig equation are