On a given school day, the probability that nick oversleeps is 48% and the probability he has a pop quiz is 25%. assuming these two events are independent, what is the probability that nick oversleeps and has pop quiz on the same day?

0.12

Step-by-step explanation:

Independent events are when probability of one event happening doesn't affect the probability of the other event happening.

Also,

In probability "AND" means "multiplication" and "OR" means "addition".

The question asks for probability that Nick oversleeps "AND" has pop quiz. So we need to multiply the individual probabilities (since independent, they have no affect on each other).

P(Nick oversleeps) = 0.48

P(pop quiz) = 0.25

Thus,

P(Nick oversleeps AND pop quiz) = 0.48 * 0.25 = 0.12

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}

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation: