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Gre4nikov [31]
3 years ago
12

Which table correctly shows all the sample spaces for tossing a coin and rolling an odd number on a six-sided number cube? Coin

Number cube Outcome Heads 1 Heads, 1 Heads 3 Heads, 3 Heads 5 Heads, 5 Tails 1 Tails, 1 Tails 3 Tails, 3 Tails 5 Tails, 5 Coin Number cube Outcome Heads 1 Heads, 1 Heads 2 Heads, 2 Heads 3 Heads, 3 Tails 4 Tails, 4 Tails 5 Tails, 5 Tails 6 Tails, 6 Coin Number cube Outcome Heads 2 Heads, 2 Heads 4 Heads, 4 Heads 6 Heads, 6 Tails 2 Tails, 2 Tails 4 Tails, 4 Tails 6 Tails, 6 Coin Number cube Outcome Heads 2 Heads, 1 Heads 4 Heads, 3 Heads 6 Heads, 5 Tails 2 Tails, 2 Tails 4 Tails, 4 Tails 6 Tails, 6
Mathematics
2 answers:
antiseptic1488 [7]3 years ago
8 0
Do you do flvs, and are you in 7th grade? If yes for both what math teacher do you have? The answer is B, I think.
slavikrds [6]3 years ago
7 0

Answer with explanation:

When a coin is tossed , possible outcomes are Head (H) , and Tail (T).

When ,a die is thrown , possible outcomes are {1,2,3,4,5,6}.

Among six numbers, odd numbers are {1,3,5}.

So, it is an arrangement of two things of a kind and three things of other kind.

And each of them is independent of another.

Total Possible outcome = 2 × 3=6

Which are , 1 H, 1 T, 3 H,3 T, 5 H, 5 T.

Option A: Coin ,Number cube→ Outcome, Heads→ 1 Heads,  3 Heads,  5 Heads,1 Tails,3 Tails, 5 Tails.

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4 0
3 years ago
getting home from trick or treat celia and emma counted their candies. half of celias candies is equal to 2/3 of emmas candies.
blagie [28]

Candies with celias and emmas is 60 and 45 respectively.

<u>Solution:</u>

Given, Getting home from trick or treat celia and emma counted their candies.  Half of celias candies is equal to 2/3 of emmas candies.  

They had a total of 105 candies altogether.  

We have to find how many candies did each of them have.

Let the number of candies with celias be n, then number of candies with emma will be 105 – n.

Now according to given condition.

\begin{array}{l}{\frac{1}{2} \times \text { celias candies count }=\frac{2}{3} \times \text { emmas candies count }} \\\\ {\rightarrow \frac{1}{2} \times n=\frac{2}{3} \times(105-n)} \\\\ {\rightarrow 3 \times n=2 \times 2 \times(105-n)} \\\\ {\rightarrow 3 \times n=2 \times 2 \times(105-n)} \\\\ {\rightarrow 3 n=4(105-n)} \\\\ {\quad \rightarrow 3 n=420-4 n} \\\\ {\rightarrow 3 n=420-4 n} \\\\ {\rightarrow 3 n+4 n=420} \\\\ {\rightarrow 7 n=7 \times 60} \\\\ {\rightarrow n=60}\end{array}

Hence, candies with celias and emmas is 60 and 45 respectively.

4 0
3 years ago
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