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.
40 degrees because i just got it right, and shout out to cash carti!
Tammy's recipe makes a sweeter fruit punch. This is because 3/4 (Kim's recipe) equals 18/24 (Tammy's recipe) and 5/6 equals 20/24 and since 20/24 is the bigger ratio, Tammy's recipe will make a sweeter fruit punch. I hope this helps. (:
Answer:
For this case the correct answer is the product
Step-by-step explanation:
When we have two events that occur at the same time, let's assume A and B the two events, and we want to find this probability:
And we want that he probability of each occurrence yielding the same result is the of the probability of each separate event since "The probability of two independent events occurring together is the of the probability of each event occurring separately". So for this case we need to use the independence condition like this:
And then the correct option is product rule.
