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2 years ago

A password contains exactly 7 letters. How many passwords are possible if letters cannot be used more than once?

Mathematics

2 answers:

DerKrebs [107]2 years ago

8 0

Answer:

Step-by-step explanation:

This is a combination problem. For the first letter, there are 26 possible letters in the alphabet to use. For the second, we cannot use the letter used in the first, so there are only 25 possible letters. Repeat this for the remaining 5 letters.

1st - 26

2nd - 25

3rd - 24

4th - 23

5th - 22

6th - 21

7th - 20

To find the number of possible passwords, simply multiply all these numbers together. 26*25*24*23*22*21*20 = 3,315,312,000. Therefore B is the correct answer.

aliya0001 [1]2 years ago

6 0

Answer:

Step-by-step explanation:

The answer is B

you can find answers for questions like these on quizlet.

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Pleaseee help

Licemer1 [7]

Answer:

D : -5 ≤ x < 1

R : -4 ≤ y < 7

Step-by-step explanation:

In a graph, domain is always represented by the x-axis while the range is represented by the y-axis.

In this diagram, at x-axis, the dots are connected by -5 and 1. At y-axis, the dots are connected by -4 and 7.

● represents smaller than or equal to

○ represents smaller than

3 0

4 years ago

To examine the potential effects on the transmission of dengue, Bian et al. (2010) infected mosquitoes from both the WB1 (labora

nekit [7.7K]

Answer:

reject H0

the viral titer is lower in WB1

Step-by-step explanation:

yeah

6 0

4 years ago

HELP WITH 4 QUESTIONS!!!!

m_a_m_a [10]

When dilation is about the origin, as it is here in every case, the image point coordinates are the original (pre-image) coordinates multiplied by the scale factor.

1. Multiply every coordinate value by 5:

... W' = (-5, 10), X' = (-15, -5), Y' = (25, -5), Z' = (15, 10)

2. Multiply every coordinate value by 1/3:

... A' = (-2, 5), B' = (0, 5/3), C' = (1, 10/3)

3. A' = (2, 8), B' = (6, 2), C' = (2, 2)

4. The image coordinates are 5 times the original coordinates, so ...

... the scale factor of the dilation is 5.

5 0

3 years ago

Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p

scZoUnD [109]

Answer:

The probability that they will teach different courses is $\frac{2}{3}$ .

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here x represents the course taught by the first part-time instructor and y represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events E is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Compute the probability that they will teach different courses as follows:

$P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}$

Thus, the probability that they will teach different courses is $\frac{2}{3}$ .

3 0

3 years ago

PLEASE HELP WITH #14 AND #7 PLEASE A.S.A.P !!!

kodGreya [7K]

Answer:

7.) 16.7

Step-by-step explanation:

7). First you have to find the length of the line with the triangle that is at a 90 angle with the longest side of the triangle

Using the pythagrorem theorem a² + b² = c²

4² + b² = 5²

b² = 5² - 4²

b² = 25 - 16

b² = 9

√b² =√9

b = 3

next you do the same thing to find the missing segment of the longest side

3² + b² = 13²

b² = 169 - 9

b² = 160

b = √160 or ≈ 12.7

12.7 + 4 = 16.7

so 16.7 is the length of that side

3 0

4 years ago