A coin will be tossed twice, and each toss will be recorded as heads (H) or tails (1)

Give the sample space describing all possible outcomes.

Then give all of the outcomes for the event that the second toss is heads.

Use the format HI to mean that the first toss is heads and the second is tails.

If there is more than one element in the set, separate them with commas.

a.  {HH, H1, 1H, 11}

b.  {HH, 1H}

Step-by-step explanation:

A coin has two surfaces referred to as either the head (H) or tail (T). Thus when a coin is tossed, the out could be a head or a tail. In this case, both outcome have the same probability.

Given the conditions that:

Head - H and tail - 1, and  H1 mean the first toss is head and the second is tail, then:

i. Tossing a coin twice would give the following sample space for possible outcome;

      {HH, H1, 1H, 11}

This implies that since the coin would be tossed twice, the outcome of the event could be either; the first toss is head and the second toss is head (HH), the first toss is head and the second toss is tail (H1), the first toss is tail and the second toss is head (1H), or the first toss is tail and the second toss is tail (11).

ii. All outcomes for the event that the second toss is head is;

     {HH, 1H}

the answer is b; false. 2 dimensional things are a lot more different from 3 dimensional things.

i hope this ! : d

"stay and stay proud! " - zalgo

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!