Answer: Hello!
Let's start with the word TRISKAIDEKAPHOBIA wich has 17 letters (some of them repeat, but it does not matter in this problem)
We want to know how many permutations we can do with 17 letters: then think this way, Lets compose a word. The first letter of this word has 17 options, the second letter of the word has 16 options (you already took one of the set) the third letter of the word has 15 options, and so on.
The total number of permutations is the product of the number of options that you have for each letter, this is:
17*16*15*14*....*3*2*1 = 17! = 3.6e+14
(b) FLOCCINAUCINIHILIPILIFICATION now we have 30 letters in total, using the same reasoning as before, here we have 30! permutations; this is
30! = 2.65e+32
(c) PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS: now there are 47 letters.
then P = 47! = 2.59e+59
(d) DERMATOGLYPHICS: here are 18 letters, then:
p = 18! = 6.4e+15
Answer:
Thank god im virtual
Step-by-step explanation:
In x^3=64, X is equal to 4
So you want to set up an equation for a weighted average. You know the final is 30% of the grade, so everything else is 70%. This gives you:
(Final)(.30) + (other grades)(.70) = course grade
The best grade the student can get would be if they get a hundred on the final, since that’s the best score you can make on the final. Then,
(100)(.30) + (82)(.70) = course grade
30 + 57.4 = course grade = 87.4 Which, If you round, the student would get an 87.
For the last part, we use the same equation, just filling in different parts.
(Final)(.30) + (other grades)(.70) = course grade
This time, we don’t know the grade for the final, but we know the course grade.
(Final)(.30) + (82)(.70) = 75
(Final)(.30) + 57.4 = 75
(Final)(.30) = 17.6
Final = (17.6)/(.30)
Final = 58.667 Which is approx a 59.
