A) it is left skewed
B) the median is 5
C) the mean is 5.15
D) the mean would be more affected (a change of 1.05 versus a change of 0.5).
The majority of the data is to the right of the graph; this means it is left skewed.
To find the median, write all of the data values out:
2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7
The middle value is 5.
We find the sum of this set of values and divide by 13, the number of data points, to find the mean:
2+3+4+4+5+5+5+6+6+6+7+7+7 = 67/13 = 5.15
If we added an additional data value at 20, the new median would be 5.5. The new mean would be (67+20)/14 = 6.2. The mean changes more than the median.
4 x 10^-7 (7 x 10^-9) (1 x 10^4) (2 x 10^4)
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
Answer:
53.04
Step-by-step explanation:
