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3 years ago

13

Each character in a password is either a digit [0-9] or lowercase letter [a-z]. How many valid passwords are there with the give

n restriction(s)? Length is 13. No character repeats.

1 answer:

Llana [10]3 years ago

3 0

Answer:

2310789600

Step-by-step explanation:

10 digits + 26 letters = 36

₃₆C₁₃ = 2310789600

Hope this helps, although i am not 100 percent sure its right.

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Lastima Nelson, Inc. is recruiting a new Chief Financial Officer (CFO). The recruiting budget is $44,000. Lastima has spent $10,

Musya8 [376]

The budget is $44000

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The amount left to spend = 44000 - 18500 = $25500

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Let the maximum price of the home be $x$
This value of $x$ will be the 100% before the 6% commission is calculated of it.

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So, the maximum value of the home is $425,000

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3 years ago

The general term for the sequence 2, 4, 8, 16, 32, . . . is

Nuetrik [128]

We are given

sequence is 2 , 4 , 8 ,16 , 32 , .....

Firstly , we will check whether it is geometric sequence

Checking geometric sequence:

We will find common ratio between successive terms

and then we check whether they are equal

r1=(second term)/(first term)

$r_1=\frac{4}{2}=2$

r2=(third term)/(second term)

$r_2=\frac{8}{4}=2$

r3=(fourth term)/(third term)

$r_3=\frac{16}{8}=2$

r4=(fifth term)/(fourth term)

$r_4=\frac{32}{16}=2$

we can see that all four ratios are same

$r_1=r_2=r_3=r_4=2$

so, this is geometric sequence

Calculation of general term:

We got

common ratio is

$r=2$

Let's assume

number of terms is n

first term is 2

$a_1=2$

now, we can use formula

$a_n=a_1 (r)^{n-1}$

we can plug values

and we get

$a_n=2(2)^{n-1}$

$a_n=2^1(2)^{n-1}$

$a_n=(2)^{n-1+1}$

$a_n=(2)^{n}$ ..................Answer

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2 years ago

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Answer:

Step-by-step explanation:

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kumpel [21]

4% of 73 will be 2.92.
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2 years ago

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LiRa [457]

Answer:

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Step-by-step explanation:

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