Both calculated the slope as 2/3
![a\ slope=\dfrac{\Delta y}{\Delta x}\\\\MEG:\dfrac{2}{3}\\\\TANIA:\ \dfrac{4}{6}]=\dfrac{2}{3}](https://tex.z-dn.net/?f=a%5C%20slope%3D%5Cdfrac%7B%5CDelta%20y%7D%7B%5CDelta%20x%7D%5C%5C%5C%5CMEG%3A%5Cdfrac%7B2%7D%7B3%7D%5C%5C%5C%5CTANIA%3A%5C%20%5Cdfrac%7B4%7D%7B6%7D%5D%3D%5Cdfrac%7B2%7D%7B3%7D)
Answer:
TRUE
Step-by-step explanation:
i just know it :>
The vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)
<h3>How to determine the new vertices?</h3>
The vertices are given as:
a (-6,-7)
b(-3,-10)
c(-5,2)
The transition rule is given as
(x,y)-->(x,y-3)
So, we have
a' = (-6, -7 - 3)
a' = (-6, -10)
b' = (-3, -10 - 3)
b' = (-3, -13)
c' = (-5, 2 - 3)
c' = (-5, -1)
Hence, the vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)
Read more about translation at:
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Answer:
Type I: 1.9%, Type II: 1.6%
Step-by-step explanation:
given null hypothesis
H0=the individual has not taken steroids.
type 1 error-falsely rejecting the null hypothesis
⇒ actually the null hypothesis is true⇒the individual has not taken steroids.
but we rejected it ⇒our prediction is the individual has taken steroids.
typr II error- not rejecting null hypothesis when it has to be rejected
⇒actually null hypothesis is false ⇒the individual has taken steroids.
but we didnt reject⇒the individual has not taken steroids.
let us denote
the individual has taken steroids by 1
the individual has not taken steroids.by 0
predicted
1 0
actual 1 98.4% 1.6%
0 1.9% 98.1%
so for type 1 error
actual-0
predicted-1
therefore from above table we can see that probability of Type I error is 1.9%=0.019
so for type II error
actual-1
predicted-0
therefore from above table we can see that probability of Type I error is 1.6%=0.016
