Consider the endothermic reactions in the table below. Final Temperature Conditions Heat Absorbed (cal)
Below Target Above Target
266 K 13 40
271 K 44 17
274 K 56 36
Let denote the event that a reaction's final temperature is 271 K or less. Let denote the event that the heat absorbed is above target. Determine the following probabilities.
A) P(AUB).
B) P(A'UB').
C) Use the total probability rule to determine P(A).

0.277

0.723

0.553

Step-by-step explanation:

Given the data:

Final temp condtn__ below target __above

266K 13 40

271 K 44 17

274 K 56 36

A = final temp ≤ 271K : {44, 40, 13, 17}

Total = {13 + 40 + 44 + 17 + 56 + 36}= 206

B = heat absorbed is above target : {40, 17, 36}

P(AuB) :

AuB = {40, 17} = (40 + 17) = 57

P(AuB} = (AuB) / total = 57 /206 = 0.277

A' = {36, 56,}

B' = {13, 44, 56}

(A'uB') = {36 + 56 + 13 + 44} = 149

P(A'uB') = 149 / 206 = 0.723

P(A) = (44 + 40 + 13 + 17) / 206 = 0.553

i got 86 for this

hope this !

if you need an explanation just ask

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}