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.

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5