The complete version of question:
<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>
Answer:
Step-by-step explanation:
As the description of the statement is:
'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>
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As
- <em>Five times the sum of a number and 27 </em>is written as:
- <em>greater than or equal </em>is written as:
- <em>six times the sum of that number and 26' </em>is written as: 6(x + 26)
so lets combine the whole statement:
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solving
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Therefore,
We first determine the z-scores for the given x-values of 64 and 96.
For x = 64: z = (64 - 80) / 8 = -2
For x = 96: z = (96 - 80) / 8 = 2
Therefore we find the probability that -2 < z < 2, which is around 0.95. Therefore, out of 100 students, approximately 100(0.95) = 95 students will weigh between 64 and 96 pounds.
I'm assuming that the diagonals intersect at E. Since diagonals of a parallelogram bisect each other, BE=ED.
- 7x-2=x²-10
- x²-7x-8=0
- (x-8)(x+1)=0
- x = -1, 8
As distance must be positive, we reject the negative case, so x=8.
Thus, BE=ED=54.
B is the answer because first number is 2 and common ratio is - 5