This is a problem of Permutations. We have 3 cases depending on the number of B's. Since no more than three B's can be used we can use either one, two or three B's at a time.
Case 1: Five A's and One B
Total number of letters = 6
Total number of words possible = 
Case 2: Five A's and Two B's
Total number of letters = 7
Total number of words possible = 
Case 3: Five A's and Three B's
Total number of letters = 8
Total number of words possible = 
Total number of possible words will be the sum of all three cases.
Therefore, the total number of words that can be written using exactly five A's and no more than three B's (and no other letters) are 6 + 21 + 56 = 83
Answer:
Step-by-step explanation:
<u>a)</u>
- Given that ; X ~ N ( µ = 65 , σ = 4 )
From application of normal distribution ;
- Z = ( X - µ ) / σ, Z = ( 64 - 65 ) / 4, Z = -0.25
- Z = ( 66 - 65 ) / 4, Z = 0.25
Hence, P ( -0.25 < Z < 0.25 ) = P ( 64 < X < 66 ) = P ( Z < 0.25 ) - P ( Z < -0.25 ) P ( 64 < X < 66 ) = 0.5987 - 0.4013
- P ( 64 < X < 66 ) = 0.1974
b) X ~ N ( µ = 65 , σ = 4 )
From normal distribution application ;
- Z = ( X - µ ) / ( σ / √(n)), plugging in the values,
- Z = ( 64 - 65 ) / ( 4 / √(12)) = Z = -0.866
- Z = ( 66 - 65 ) / ( 4 / √(12)) = Z = 0.866
P ( -0.87 < Z < 0.87 )
- P ( 64 < X < 66 ) = P ( Z < 0.87 ) - P ( Z < -0.87 )
- P ( 64 < X < 66 ) = 0.8068 - 0.1932
- P ( 64 < X < 66 ) = 0.6135
c) From the values gotten for (a) and (b), it is indicative that the probability in part (b) is much higher because the standard deviation is smaller for the x distribution.
Find out the how many kilobytes have been downloaded so far .
To prove
As given
While waiting for a video game to download.
notice that 30 percent of 32000 kilobytes have been downloaded so far .
30% is written in the decimal form.
= 0.30
kilobytes uses in downloaded = 0.30 × 32000
= 9600 kilobytes
Therefore 9600 kilobytes have been downloaded so far .
Answer:
the answer is 160.37
Step-by-step explanation:
174.69 -15%==148.49
148.49×8% 11.88 add that back to 148.49 =160.37
