Hello.<span><span> </span><span>
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Let 2003 be the zero year; then 2005 is the three year, and 2008 the 5 year.
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P = ab^x
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P(3) = ab^3 = 800000
P(0) = ab^0 = 900000
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a = 900000
Solve for "b"::
b^3 = 8/9
b = 2/cbrt(9)
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Equation::
P(x) = 900000^x
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Ans: P(5) = 900000
Have a nice day</span></span></span></span>
It's actually legit 50 pounds because you sound up if it's 5 or more but if it's less it stays the same
Answer:
It is in the tenths place
Step-by-step explanation:
In the number 9.365, 3 is in the tenths place.
Answer:
492 0/10
Step-by-step explanation:
We have that the perimeter is 4 times the length of the side, now we know that this side "l" is given by:
l = 4 in * 24 + 23 * 5 ft
l = 96 in + 115 ft
Now, we pass the inches to pes, knowing that 1 foot equals 12 inches, so
96 in * 1 ft / 12 in = 8 ft
therefore replacing it remains:
l = 8 ft + 115 ft
l = 123 ft
now the perimeter:
p = 123 * 4
p = 492
to change to a mixed number:
4920/10 = 492 0/10
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
