6x2 +10x−1 discriminant
1 answer:
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Observe the figure below.
Statement Reason
1. AC and BD bisect each other Given
2. AE = EC and BE = ED Definition of bisection
3. Vertical angle theorem
Vertical angle theorem states " When two lines intersect each other, the vertically opposite angles are always equal".
4. SAS criterion for congruence
5. Corresponding angles of congruent triangles are congruent
6. Converse of alternate interior angle theorem
7. Vertical angle theorem
8. SAS criterion for congruence
9. Converse of alternate interior angle theorem
As, , so
as corresponding parts of corresponding triangles are equal. As these angles are alternate interior angles, so the lines BC and AD are parallel by "Converse of alternate interior angle theorem".
Answer:
The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.
The probability that a randomly chosen plate contains the sub-string HI is 0.002548 approximately.
Step-by-step explanation:
Consider the provided information.
Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
Part (A)
Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.
Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.
The numbers of ways in which three letters followed by three digits can be chosen is:
The numbers of ways in which two letters followed by four digits can be chosen is:
Hence, the total number of ways are 17576000 + 6760000 = 24336000
Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.
Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are:
The probability that a randomly chosen plate contains the number 2222 is:
Part (B)
The number of ways in which chosen plate contains the sub-string HI:
If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:
If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:
Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:
From part (A) we know that the total number of ways to chose a number plate is 24336000.
The probability that a randomly chosen plate contains the sub-string HI is: