$Use:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\a^n:a^m=a^{n-m}\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\a^{-n}=\left(\frac{1}{a}\right)^{n}\\==================================\\\frac{(xy)^{-3}}{(x^{-5}y)^3}=\frac{x^{-3}y^{-3}}{(x^{-5})^3y^3}=\frac{x^{-3}y^{-3}}{x^{-15}y^3}=x^{-3-(-15)}y^{-3-3}=\boxed{x^{12}y^{-6}=\frac{x^{12}}{y^6}}$

6 0

3 years ago

20 earth pounds on the moon

tatiyna

20 earth pounds on the moon would equate to 3.31 pounds.

4 0

3 years ago

Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine

Paladinen [302]

Answer:

a) 117.4

b) 117.9

c) Option A) When there is rounding or grouping, the median can be highly sensitive to small change

Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

n = 9

a) Median of the reported blood pressure values

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4

Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4$

b) New median of the reported values

Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

$\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9$

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A) When there is rounding or grouping, the median can be highly sensitive to small change

7 0

3 years ago