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Nat2105 [25]
3 years ago
7

Mia spins two spinners. Each spinner has sections labeled A, B, and C. If each spinner lands in one of the lettered sections, wh

ich outcome is missing from the sample space represented by the table?
A B C
A A A A B A C
B B B B C
C C A C B C C
Select one:

B B

A B

C A

B A
Mathematics
1 answer:
VMariaS [17]3 years ago
7 0

Answer:

BB

Step-by-step explanation:

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The real solution for x2 = 25
Westkost [7]

Answer:

x = 12.5 or 12 1/2 (depends if you want decimal or fractions)

Step-by-step explanation:

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3 years ago
Which set of ordered pairs represents a function?
worty [1.4K]

Answer:

b. {(2, 1), (3, 1) (4, 3), (5, 3)}

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What is the value of AAA when we rewrite 4^{31x}4 31x 4, start superscript, 31, x, end superscript as A^{ x}A x A, start supersc
Yuri [45]

Answer:

A = 4^31

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
In international morse code, each letter in the alphabet is symbolized by a series of dots and dashes: the letter a, for example
Ostrovityanka [42]
Morse code is essentially the same as binary. That is, there are two "digits", a dot or a dash.

There are 26 letters in the English alphabet. Clearly, we can't just use one dot or dash, since that could only encode 2 letters at the most. We can't use two symbols because that could only encode 4 letters at the most. Similarly, 3 symbols means 2^3=8 letters at most.

We have to select the smallest power of 2 that exceeds or is equal to 26. In this case, 2^5=32>26, so we would have to use up to 5 symbols to encode each letter in the alphabet.
5 0
3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
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