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2 years ago

14

How many different passwords are there that contain only digits and lower-caseletters and satisfy the given restrictions?(iii) L

ength is 7 and the password must contain at least one digit.

1 answer:

seraphim [82]2 years ago

6 0

Answer:

36⁷- 26⁷ = 70332353920 passwords.

Step-by-step explanation:

Number of digits 0-9 = 10

Number of lowercase letters a-z = 26

We have to find different passwords of length 7 that contain only digits and lower-case letters.

Total number of characters = 26 lowercase letters + 10 digits

= 36

The total number of length 7 passwords without restrictions = 36⁷

= 78364164096

The number of length 7 passwords with no digits (only lowercase letters)

= 26⁷

= 8031810176

Now the restriction here is that the password must contain at least 1 digit. This can be computed by subtracting the number of length 7 passwords with no digits from total number of length 7 passwords.

Number of length 7 passwords with restriction (at least one digit) = 36⁷- 26⁷

= 36⁷- 26⁷

= 78364164096 - 8031810176

= 70332353920

So the number of different passwords that contain only digits and lower-case letters and satisfy the given restrictions that length is 7 and the password must contain at least one digit = 70332353920

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Arlecino [84]

Answer:

12.36

Step-by-step explanation:

the standard deviation is 12.36123 or 12.36 rounded.

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1 year ago

How can you make 0.40 into a fraction??

melamori03 [73]

<h3>2/5 is your answer. </h3><h3>▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓</h3>

☞ First, 0.40 is the same as 40/100.

But we are not done.

<u>Now, doing this you need to find the GCF (Greatest common factor) of 40, 100.</u>

2 x 2 x 5 = 20.

☞ So our GCF is 20.

Divide both the numerator and the denominator by the GCF.

40 ÷ 20 / 100 ÷ 20.

<h2>Simplify to get ☞ 2/5.</h2><h2>▬▬ι══════════════ι▬▬</h2>

Best of luck on your assignment!

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- Emacathy | Top Answerer ♫♪.ılılıll|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̲̅̅●̲̅̅|llılılı.♫♪

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2 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

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= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

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2 years ago

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nika2105 [10]

Easy here
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2 years ago

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madam [21]

Answer:

The three unknown angles X, Y , and Z are:

X = 40, Y = 20, and Z = 120

Step-by-step explanation:

Let's name X the measure of the first angle, Y the measure of the second one, and Z that of the third one.

Then we can create the following equations:

X = 2 Y

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and

X + Y + Z = 180

So we use the first equation and the second one to substitute for the variable X and Z in the thrid equation:

2 Y + Y + (100 + Y) = 180

4 Y + 100 = 180

4 Y = 80

Y = 80/4 = 20

Then X = 40, Y = 20, and Z = 120

7 0

2 years ago