What is 3.56 x 10^-5 in standard form?
1 answer:
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Answer:
D) 54 cm
Step-by-step explanation:
We can use the Centroid Theorem to solve this problem, which states that the centroid of a triangle is of the distance from each of the triangle's vertices to the midpoint of the opposite side.
Therefore, is
of the distance from
to
, since the latter is the midpoint of the side opposite to
. We know this because
belongs to
, so
must be
's midpoint due to the fact that by definition, the centroid of a triangle is the intersection of a triangle's three medians (segments which connect a vertex of a triangle to the midpoint of the side opposite to it).
We can then write the following equation:
Substituting into the equation gives us:
Solving for , we get:
(Multiply both sides of the equation by
to get rid of
's coefficient)
(Simplify)
(Symmetric Property of Equality)
Therefore, the answer is D. Hope this helps!
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: