Can someone answer and explaine this to me
1 answer:
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Answer:
Step-by-step explanation:
The sample space is shown in the attached photo.
Probability = number of favourable outcomes/total number of outcomes.
Looking at the diagram, the total possible outcomes is 36.
A) from the diagram, the total number of events in which the sum of the numbers is odd is 18. Therefore, the probability that the sum of the numbers is odd is
18/36 = 1/2
B) the total number of events in which the sum of the numbers is 10 or more is 6. Therefore, the probability that the sum of the numbers is 10 or more is
6/36 = 1/6
C) the total number of events in which 3 appears on each of the two dice is 1. Therefore, the probability that 3 appears on each of the two dice is
1/36
Well, the cos(∅) is negative in the second and third quadrants. If you are solving for theta, you would use the inverse of the cos or arccos
take the inverse of both sides to get:
x = arccos(-2/3) now evaluate the right
x = 131.8103149 degrees
to find your second solution, subtract your reference angle from 360 degrees.
360 - 131.8103149 = 228.1896851 degrees
Now the period of cos is 2π or 360 degrees. So if you want to consider all possible solutions, you would need to add/subtract 360n to both solutions above..
Not sure if this is what you're looking for, but thought I would leave it here for you just in case... As a side note, you could do this problem in radian measurement as well.
x = arccos(-2/3) now evaluate the right
x = 131.8103149 degrees
to find your second solution, subtract your reference angle from 360 degrees.
360 - 131.8103149 = 228.1896851 degrees
Now the period of cos is 2π or 360 degrees. So if you want to consider all possible solutions, you would need to add/subtract 360n to both solutions above..
Not sure if this is what you're looking for, but thought I would leave it here for you just in case... As a side note, you could do this problem in radian measurement as well.