The answer to your question is: Yes, someone undoubtedly can.
Although you haven't asked to be told or shown how to solve it, I'm here
already, so I may as well stick around and go through it with you.
The sheet is telling you to find the solutions to two equations, AND THEN
DO SOMETHING WITH THE TWO SOLUTIONS. But you've cut off the
instructions in the pictures, so all we have are the two equations, and
you'll have to figure out what to do with their solutions.
<u>First equation:</u>
(2/5) x - 6 = -2
Add 6 to each side:
(2/5) x = 4
Multiply each side by 5:
2x = 20
Divide each side by 2 :
<u>x = 10</u>
<u>Second equation:</u>
-3y + 1/4 = 13/4
Subtract 1/4 from each side:
-3y = 12/4
Multiply each side by 4 :
-12 y = 12
Divide each side by -12 :
<u> y = -1</u>
What you want is P(6∩1) or P(1∩6) or P(2∩5) or P(5∩2) or P(3∩4) or P(4∩3).
The events of rolling the dice are independent (i.e. they don't affect one another) so:
E.g.
P(6∩1) = P(6) * P(1)
P(2∩5) = P(2) * P(5)
The probability of getting a given number on a roll is 1/6 for both dice.
So:
P(6∩1) = 1/6 * 1/6 = 1/36
This is the same for any arrangement of numbers you could get from rolling two dice.
So, we can see that there are 6 arrangements of numbers that will give a sum of 7 and so that is 6 * 1/36 = 6/36 = 1/6
4x = 157.2
4x/4 = 157.2/4
x = 39.3
6 sin 2x = 6 cos x Using the identity sin 2x = 2 sin x cos x:-
12 sin x cos x = 6 cos x
6 cos x ( 2 sin x - 1) = 0
either 6 cos x = 0 or 2 sin x - 1 = 0 so sin x = 1/2
x = pi/2, 3pi/2 , pi/6, 5pi / 6
answer is ( pi/6, pi/2, 5pi/6, 2pi/2)
