3 years ago

Does anyone know the answer?

1 answer:

5 0

For which ones I’m confused ?

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m_a_m_a [10]

Number of possible outcome for tossing N coins = $2^N$

Solution:

Possible outcomes when tossing one coin = {H, T}

Number of possible outcomes when tossing one coin = 2 $=2^1$

Possible outcomes when tossing two coins = {HH, HT, TH, TT}

Number of possible outcomes when tossing two coins = 4 $=2^2$

Possible outcomes when tossing three coins

= {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

Number of possible outcomes when tossing three coins = 8 $=2^3$

Therefore, the sequence obtained is $2^1, 2^2, 2^3$ .

If continue this sequence, we can obtain number of possible outcome for tossing N coins is $2^N$ .

7 0

2 years ago

andreyandreev [35.5K]

Answer:

12 dimes and 21 nickles

Step-by-step explanation:

4 0

2 years ago

Pavlova-9 [17]

You just need to find distance between two points
d= $\sqrt{(y2-y1)^2+(x2-x1)^2}$
d=6

6 0

3 years ago

Dima020 [189]

Only 2 forces are out of balance - the 22N and the 14N
So the resultant force is 22-14 = 8N.

7 0

3 years ago

Troyanec [42]

Answer:

Hope this helps

7 0

3 years ago