Answer:
\mu = 14.5\\
\sigma = 5.071\\
k = 1.084
Step-by-step explanation:
given that a statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.
i.e. his findings with respect to probability are
Recall Chebyshev's inequality that
Comparing with the Ii equation which is appropriate here we find that
Next what we find is
Thus from the given information we find that
Step-by-step explanation:
which expression there is no expression here
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.
Answer:
180-146=34 146+34=180
Step-by-step explanation:
Step-by-step explanation:
45.50
Converting decimal
45.50 x 100 = 4550
