Can someone help me on this

Abby is registering at a Web site and must select a six-character password.The password can contain either letters or digits.

zaharov [31]

Answer:

a. 2176782336 with repeated characters. 1402410240 with no repeated characters.

b. 7964171460 when all letters can be repeated, but not numbers. 118813760 when only one number and letters can be repeated.

I believe that the password with all letters able to repeat and numbers not being able to is the most secure password, because if someone were to guess the password there is a 1/7964171460 chance of guessing the password.

Step-by-step explanation:

A. Since there are 26 different letters and 10 different numbers, there are 36 characters we can type. If characters can be repeated then, there is 2176782336 different passwords, since on all six number there are 36 possibilities each. So, 36 x 36 x 36 x 36 x 36 x 36 or 36^6 is to evaluated to find the answer. If characters are not to be repeated, there are 1402410240 different passwords, since on the first number there are 36, second has 35, third has 34, fourth has 33, fifth has 32, and sixth has 31. So, 36 x 35 x 34 x 33 x 32 x 31 is to be evaluated to solve this.

B. Since all characters that are letters can be repeated, then there are 26 letters to use forever and 10 numbers you can use with a limit. So, 36 x 35 x 34 x 33 x 32 x 31 which is to be solved if only numbers were to be used, which is 1402410240. Then, you add that with 36 x 35^5 and 36 x 35 x 34^4 and 36 x 35 x 34 x 33^3 and 36 x 35 x 34 x 33 x 32^2. which will be 7964171460. If the password must contain one digit, then you must multiply 10 with 26^5. Since, there is 10 different digits to use for the first number and 26 letters to choose from for the other five. So, it will be 26^5 times 10 which is 118813760.

6 0

3 years ago

True or False:

lara [203]

The answer would be true.

6 0

3 years ago

2 MATH K.

Ann [662]

Answer:

94

Step-by-step explanation:

Given the addition problem:

58 + 36

We can form a table :

5 _______ 8

+

3 _______ 6

9 4

Hence, (58 + 36) = 94

4 0

3 years ago

A jar contains 8 red marbles numbered 1 to 8 and 7 blue marbles numbered 1 to 7. A marble is drawn at random from the jar. Find

Shkiper50 [21]

Answer:

Therefore the probability that the marble is blue or even numbered is $\frac{11}{15}$

Step-by-step explanation:

Probability: The ratio of favorable outcomes to the total outcomes.

It is denoted by P.

$Probability= \frac{\textrm{favorable outcomes}}{\textrm{Total outcomes}}$

Given that a jar contains 8 red marbles and 7 blue marbles.

Total number of marbles = (8+7) = 15

Let A = Event of getting a blue marble

B= Event of getting of even marble.

Even number blue marbles are 2, 4,6

Even number red marbles are 2, 4,6,8

The number of even marbles are =(3+4)=7

The probability of getting a blue marble is P(A)

$=\frac{\textrm{Total number of blue marbles}}{\textrm{Total number of blue marbles}}$

$=\frac{7}{15}$

The probability of getting a even marble is P(B)

$=\frac{\textrm{The number of even number marbles}}{\textrm{Total number of marbles}}$

$=\frac{7}{15}$

The probability of getting a even numbered blue marble P(A∩B)

$=\frac{3}{16}$

P(blue marble or even- numbered)

=P(A∪B)

=P(A)+P(B)-P(A∩B)

$=\frac{7}{15} +\frac{7}{15}-\frac{3}{15}$

$=\frac{11}{15}$

Therefore the probability that the marble is blue or even numbered is $\frac{11}{15}$

3 0

4 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago