Exams are approaching and Helen is allocating time to studying for exams. She feels that with the appropriate amount of studying, she has an 80% chance of getting an A in Marketing. She also feels that she has a 60% chance of getting an A in Spanish with the appropriate amount of studying. Given the demands on her time, she feels that she has only a 45% chance of getting an A in both classes. What is the probability that Helen does not get an A in either class?

0.05

Step-by-step explanation:

Let M = Event of getting an A in Marketing class.

S = Event of getting an A in Spanish class,

i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45

Required probability = P(neither M nor S)

= P(M'∩S')

= P(M∪S)'                                 [∵P(A'∩B')=P(A∪B)']

=1- P(M∪S)                               [∵P(A')=1-P(A)]

= 1- (P(M)+P(S)- P(M∩S))   [∵P(A∪B)=P(A)+P(B)-P(A∩B)]

= 1- (0.80+0.60-0.45)

= 1- 0.95

= 0.05

hence, the probability that Helen does not get an A in either class= 0.05

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}

answer: 30

step-by-step explanation: its 30 because im a basketball player and stephen curry hits most of his shots so if you put 30 you might make the shot goodluck