Jamal has two bags of marbles. The first bag contains 5 blue marbles, and 3 green marbles. The second bag contains 4 yellow marb
1 answer:
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1)
here, we do the left-hand-side
![\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2 \\\\\\\ [sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)] \\\\\\ 2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2](https://tex.z-dn.net/?f=%5Cbf%20%5Bsin%28x%29%2Bcos%28x%29%5D%5E2%2B%5Bsin%28x%29-cos%28x%29%5D%5E2%3D2%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bsin%5E2%28x%29%2B2sin%28x%29cos%28x%29%2Bcos%5E2%28x%29%5D%5C%5C%5C%5C%2B~%20%5Bsin%5E2%28x%29-2sin%28x%29cos%28x%29%2Bcos%5E2%28x%29%5D%0A%5C%5C%5C%5C%5C%5C%0A2sin%5E2%28x%29%2B2cos%5E2%28x%29%5Cimplies%202%5Bsin%5E2%28x%29%2Bcos%5E2%28x%29%5D%5Cimplies%202%5B1%5D%5Cimplies%202)
2)
here we also do the left-hand-side
![\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x) \\\\\\ \cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)} \\\\\\ \cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2-cos%5E2%28x%29%7D%7Bsin%28x%29%7D%3Dcsc%28x%29%2Bsin%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2-%5B1-sin%5E2%28x%29%5D%7D%7Bsin%28x%29%7D%5Cimplies%20%5Ccfrac%7B2-1%2Bsin%5E2%28x%29%7D%7Bsin%28x%29%7D%5Cimplies%20%5Ccfrac%7B1%2Bsin%5E2%28x%29%7D%7Bsin%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B1%7D%7Bsin%28x%29%7D%2B%5Ccfrac%7Bsin%5E2%28x%29%7D%7Bsin%28x%29%7D%5Cimplies%20csc%28x%29%2Bsin%28x%29)
3)
here, we do the right-hand-side
here, we do the left-hand-side
2)
here we also do the left-hand-side
3)
here, we do the right-hand-side
Answer:
The probability that the test is taken an even number of times is 0.30.
Step-by-step explanation:
The probability that a student passes the driving test at any attempt is,
<em>p</em> = 4/7.
The event of a student passing in any attempt is independent of each other.
The probability that the test is taken an even number of times is:
P (even number of tests) = P (Passing in the 2nd attempt)
+ P (Passing in the 4th attempt)
+ P (Passing in the 6th attempt) ...
If a student passed in the 2nd attempt it implies that he failed in the first.
Then, P (Passing in the 2nd attempt) =
Similarly, P (Passing in the 4th attempt) = , since he failed in the first 3 attempts.
And so on.
Compute the probability of an even number of tests as follows:
P (even number of tests) =
The result follows a Geometric progression for infinite values.
The sum of infinite GP is:
The probability is:
P (even number of tests) =
Thus, the probability that the test is taken an even number of times is 0.30.