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Mandarinka [93]

4 years ago

13

A website requires a 7 digit password. How many different passwords can be made using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and

9 if each number can only be used once?
A. 120
B. 70
C. 5,040
D. 604,800

Instructions/formula would be sooo great

1 answer:

olga_2 [115]4 years ago

4 0

B because 10 x 7 = 70

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

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b) 6₁₀

c) 7₁₀

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i) 181₁₀

Step-by-step explanation:

<u>Binary to Decimal Conversion (Positional Notation Method)</u>

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.
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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

<u>Question (a)</u>

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

Therefore, 11₂ = 3₁₀

<u>Question (b)</u>

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

Therefore, 110₂ = 6₁₀

<u>Question (c)</u>

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

Therefore, 111₂ = 7₁₀

<u>Question (f)</u>

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

Therefore, 10011₂ = 19₁₀

<u>Question (i)</u>

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

Therefore, 10110101₂ = 181₁₀

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