You toss a coin twice. which calculation proves that landing on tails for the first toss and heads on the second toss are independent events?

If a coin is tossed twice than the sample space of outcomes are:

{ HH, HT, TH, TT}

Now let A denote the event that the tail comes up in the first toss

and B denote the event that head comes up on the second toss.

Let P denote the probability of an event.

Hence,

P(A)=1/2  ( As probability of tail is half during  single toss)

Similarly P(B)=1/2 ( Since , Probability of obtaining a head is half on a single toss)

Also,

P(A∩B)=1/4

( Since the outcomes is TH )

Now, as:

P(A∩B)=P(A)×P(B)

Hence,  landing on tails for the first toss and heads on the second toss are independent events.

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6