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3 years ago

How many different combinations are possible if a number cube is tossed three times?

2 answers:

7 0

If you toss it the first time there are 6 outcomes

For each of these outcomes, the second toss could yield another 6 outcomes - total 6 x 6 = 6² outcomes 

For each of the these outcomes, the third toss could yield another 6 outcome - total = 6² x 6 = 6³ outcomes. 

Extending this pattern you see that for x tosses, the total number of outcomes is 6^x.
That's what I got, hope it helps you.

Nataly_w [17]3 years ago

6 0

It would be 6 x 6 x 6, since the number cube has six sides and it was thrown 3 times.

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Three more than the product of 16 and Goran's savings

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16x+3

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3 years ago

Which time is 4 hours and 34 minutes later than 00:15?

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Answer:

4:49

Step-by-step explanation:

00:15 + 4 hours = 4:15

4:15 + 34 minutes = 4:49

Overall the answer is 4:49

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2 years ago

Solve 6x = 4x^2 using the quadratic formula

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3 years ago

1)Convert the following binary numbwers into decimal number(only a,b,c,f,i)

kozerog [31]

Answer:

a) 3₁₀

b) 6₁₀

c) 7₁₀

f) 19₁₀

i) 181₁₀

Step-by-step explanation:

Binary to Decimal Conversion (Positional Notation Method)

Multiply each digit by the base (2) raised to the power dependent upon the position of that digit in the binary number.
Sum all the values obtained for each digit.
Express the number as a decimal number by placing subscript 10 after it.

For a binary number with 'n' digits:

The right-most digit is multiplied by 2⁰
The left-most digit is multiplied by $\sf 2^{n-1}$

For example, to convert the binary number 111001₂ into a decimal:

$\begin{array}{ c c c c c c}1 & 1 & 1 & 0 & 0 & 1\\\downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 2^0\\\end{array}$

Multiply each digit by the base (2) raised to the power as indicated above and sum them:

$=(1 \times 2^5)+(1 \times 2^4)+(1 \times 2^3)+(0 \times 2^2)+(0 \times 2^1)+(1 \times 2^0)$

$= 32+16+8+0+0+1$

$= 57$

Finally, express as a decimal number ⇒ 111001₂ = 57₁₀

Question (a)

$\begin{aligned}\implies 11_2 & = (1 \times 2^1)+(1 \times 2^0)\\& = 2+1\\& = 3\end{aligned}$

Therefore, 11₂ = 3₁₀

Question (b)

$\begin{aligned}\implies 110_2 & = (1 \times 2^2)+(1 \times 2^1)+(0 \times 2^0)\\& = 4+2+0\\& = 6\end{aligned}$

Therefore, 110₂ = 6₁₀

Question (c)

$\begin{aligned}\implies 111_2 & = (1 \times 2^2) +(1 \times 2^1)+(1 \times 2^0)\\& =4+2+1\\& = 7\end{aligned}$

Therefore, 111₂ = 7₁₀

Question (f)

$\begin{aligned}\implies 10011_2 & =(1 \times 2^4)+(0 \times 2^3)+ (0 \times 2^2) +(1 \times 2^1)+(1 \times 2^0)\\& =16+0+0+2+1\\& = 19\end{aligned}$

Therefore, 10011₂ = 19₁₀

Question (i)

$\phantom{)))}10110101_2 \\\\=(1 \times 2^7)+(0 \times 2^6)+(1 \times 2^5)+(1 \times 2^4)+(0 \times 2^3)+ (1 \times 2^2) +(0 \times 2^1)+(1 \times 2^0)\\\\=128+0+32+16+0+4+0+1\\\\= 181$

Therefore, 10110101₂ = 181₁₀

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2 years ago

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andre [41]

what yung papi said

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3 years ago