you want to make five-letter codes that use the letters a ke r. and mwithout repeating any letter. what is the probability that a randomly
chosen code starts with mand ends with e?

0.05

Step-by-step explanation:

There are five letters to choose from. Since there is no repetition, you can choose one of 5 for the first letter, one of four for the second letter, etc. The total number of five-letter codes is

5 * 4 * 3 * 2 * 1 = 120

Now we need to find out the number of codes that start with M and end with E.

The first position is occupied by M. There is one single letter that can be placed there. The last position is occupied by E. There is a single letter that goes there. There are 3 letters left.

1 * 3 * 2 * 1 * 1 = 6

There are 6 codes that start with M and end with E

p(code that starts with M and ends with E) = 6/120 = 0.05

5%

Step-by-step explanation:

First, find the total number of combinations.

There are 5 options for the first letter.  Since no letter can be repeated, there are 4 options for the second letter, 3 options for the third letter, and so on.

5 × 4 × 3 × 2 × 1 = 120

Now find the number of combinations that start with M and end with E.

There's 1 option for the first letter.  3 options for the second letter.  2 options for the third letter.  1 option for the fourth letter.  And 1 option for the fifth letter.

1 × 3 × 2 × 1 × 1 = 6

So the probability is:

6 / 120 = 5%

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles

but there’s no any figure there!