Answer:
27
Step-by-step explanation:
Answer:
Zabato is thinking of 6
Step-by-step explanation:
I first set it up as an equation based on the word problem you have given me
(n + 10)3 = 8n
Distribute the 3
3n + 30 = 8n
Isolate n
3n - 8n +30 - 30 = 8n - 8n - 30
-5n = -30
-5n/-5 = -30/-5
n = 6
To check replace 6 with n
(6 + 10)3 = 8 (6)
(16)3 = 48
48 = 48
The English alphabet contains 26 letters (a, b, c, ...y, z).
The digits from 0 to 9 are a total of 10.
A keycode contains 2 letters, and 3 numbers, for example:
AB 598; MM 139; NT 498; ...
So there are 26 possible choices for the first letter, which can combined with any of the 26 possible choices for the second letter, so there are a total of
26*26=676 possible pairs of letters.
Similarly, the 10 possible choices for the first number, which can be combined with the 10 possible choices for the second number, and the 10 possible choices for the third number make a total of :
10*10*10=1,000 possible triples of numbers.
Any of the 676 possible pairs of letters can be combined with any of the possible 1,000 triples of numbers. This makes a total of
676*1,000=676,000 keycodes.
Answer: 676,000
Answer: 62
Step-by-Step Explanation:
First Term (a) = 7
Common Difference (d) = 12 - 7 = 5
Term to Find (n) = 12th
Therefore, finding the 12th Term :-
=> a+(n-1)d
= 7 + (12 - 1)5
= 7 + (11)5
= 7 + 55
=> 62
Hence, 12th Term of this AP is 62
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
