The answer is none of those because you find the median of 51 which is Q2 then the lower median is 34.5 and the upper median is 73.4 then you subtract 73.5-34.5=39
Answer:
StartRoot 2 squared + 6 squared EndRoot
Step-by-step explanation:
we have
A(4,3) and B(-2,1)
we know that
the formula to calculate the distance between two points is equal to
substitute the given values
therefore
StartRoot 2 squared + 6 squared EndRoot
Answer:
C. option is right
Step-by-step explanation:
I hope it's your helpful
Part A
Answer: The common ratio is -2
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Explanation:
To get the common ratio r, we divide any term by the previous one
One example:
r = common ratio
r = (second term)/(first term)
r = (-2)/(1)
r = -2
Another example:
r = common ratio
r = (third term)/(second term)
r = (4)/(-2)
r = -2
and we get the same common ratio every time
Side Note: each term is multiplied by -2 to get the next term
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Part B
Answer:
The rule for the sequence is
a(n) = (-2)^(n-1)
where n starts at n = 1
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Explanation:
Recall that any geometric sequence has the nth term
a(n) = a*(r)^(n-1)
where the 'a' on the right side is the first term and r is the common ratio
The first term given to use is a = 1 and the common ratio found in part A above was r = -2
So,
a(n) = a*(r)^(n-1)
a(n) = 1*(-2)^(n-1)
a(n) = (-2)^(n-1)
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Part C
Answer: The next three terms are 16, -32, 64
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Explanation:
We can simply multiply each previous term by -2 to get the next term. Do this three times to generate the next three terms
-8*(-2) = 16
16*(-2) = -32
-32*(-2) = 64
showing that the next three terms are 16, -32, and 64
An alternative is to use the formula found in part B
Plug in n = 5 to find the fifth term
a(n) = (-2)^(n-1)
a(5) = (-2)^(5-1)
a(5) = (-2)^(4)
a(5) = 16 .... which matches with what we got earlier
Then plug in n = 6
a(n) = (-2)^(n-1)
a(6) = (-2)^(6-1)
a(6) = (-2)^(5)
a(6) = -32 .... which matches with what we got earlier
Then plug in n = 7
a(n) = (-2)^(n-1)
a(7) = (-2)^(7-1)
a(7) = (-2)^(6)
a(7) = 64 .... which matches with what we got earlier
while the second method takes a bit more work, its handy for when you want to find terms beyond the given sequence (eg: the 28th term)
