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First pear Second pear probability
green (3/7)---------green (2/6) (3/7)*(2/6) = 1/7
---------red (4/6) (3/7) *(4/6) = 2/7
red (4/7)-------------green (3/6) (4/7)*(3/6) = 2/7
-------------red (3/6) (4/7)*(3/6) = 2/7
-------------------------
1/7 + 2/7 + 2/7 + 2/7 = 1
The probailities that two are green are: 1/7
The probabilities that one is green and the other is red are:2 /7 + 2/7 = 4/7
The probabilities that two are red are: 2/7
you have the hypotenuse and the opposite side so you will use sin
sin58= (30/q)
multiply both sides by q
q(sin58)=30
divide by sin58 in your calculatior
q=(30/sin58)
^ I don't have a calculator with me so you will need to plug that in
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
