AAS Postulate
It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.
Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate.
Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!
Answer:
Step-by-step explanation:
Weight decreased = 108 - 88 = 20 Kg
Percentage decrease = 
= 18.52%
Answer:
the 2nd one
Step-by-step explanation:
mecause its new=original
Answer:
a) 2.5% b) 84% c) 95% d) D. The more unusual day is if the stock closed below $185 because it has the largest absolute z-score.
Step-by-step explanation:
For a) b) and c) we will use the empirical rule, so, we can observe the image shown below
a) 211.23 is exactly two standard deviation above the mean, so, the probability that on a randomly selected day in this period the stock price closed above 211.23 is 2.35% + 0.15% = 2.5%
b) 204.11 represents exactly one standard deviation above the mean, so, the probability of being below 204.11 is 50% + 34% = 84%
c) The probability of getting a value between 182.75 and 211.23 is 95%, this because 182.75 is exactly two standard deviations below the mean and 211.23 is exactly two standard deviations above the mean.
d) The z-score related to 208 is 
= (208-196.99)/7.12 = 1.5 and the z-score related to 185 is 
= (185-196.99)/7.12 = -1.7, therefore, the more unusual day is if the stock closed below $185 because it has the largest absolute z-score.
Answer:
(a) ¬(p→¬q)
(b) ¬p→q
(c) ¬((p→q)→¬(q→p))
Step-by-step explanation
taking into account the truth table for the conditional connective:
<u>p | q | p→q </u>
T | T | T
T | F | F
F | T | T
F | F | T
(a) and (b) can be seen from truth tables:
for (a) <u>p∧q</u>:
<u>p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q</u>
T | T | F | F | T | T
T | F | T | T | F | F
F | T | F | T | F | F
F | F | T | T | F | F
As they have the same truth table, they are equivalent.
In a similar manner, for (b) p∨q:
<u>p | q | ¬p | ¬p→q | p∨q</u>
T | T | F | T | T
T | F | F | T | T
F | T | T | T | T
F | F | T | F | F
again, the truth tables are the same.
For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))
