What is the next letter? Y B W D U F

Answers

Answer 1

Answer: The correct answer is S. The sequence is each alternate letter, changing from one end of the alphabet to the other.

Z Y X W V U T S R Q P L M N O K J I H G F E D C B A
Since we started at the end of the alphabet and then went back to the beginning, there are two letters in the middle that aren't part of the pattern. We only need to worry about which letter is next at the end of the alphabet, though, and the next highlighted letter after U is S.

Hope this helps!

Related Questions

What other information do you need to prove triangle GHK is congruent to triangle KLG by SAS

Answers

The angles of the shapes
shape
size

Here are the possible outcomes when a pair of fair dice is rolled.1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

In a roll of a pair of fair dice, what is the probability of the outcome being either a multiple of 3 or an even number? Are these events mutually exclusive?
, mutually exclusive
, not mutually exclusive
, mutually exclusive
, not mutually exclusive

Answers

I don't understand the big table of numbers at the top at all,
and don't see how it relates to the question.

There are 36 possible outcomes for the roll of a pair of dice.
According to your specifications, the successes are:

-- Multiples of 3:
   1 ... 2 (3)
   2 ... 1
   3 ... 3 (6)
   1 ... 5
   5 ... 1
   2 ... 4
   4 ... 2
   3 ... 6 (9)
   6 ... 3
   4 ... 5
   5 ... 4
   6 ... 6 (12)
(12 different outcomes)

-- Even numbers:
   1 ... 1 (2)
   1 ... 3 (4)
   3 ... 1
   2 ... 2
   3 ... 3 (6)
   1 ... 5
   5 ... 1
   2 ... 4
   4 ... 2
   2 ... 6 (8)
   6 ... 2
   3 ... 5
   5 ... 3
   4 ... 4
   4 ... 6 (10)
   6 ... 4
   5 ... 5
   6 ... 6    (12)
(18 different outcomes)

The events are NOT mutually exclusive.
A roll of 6 (5 ways) or 12 (1 way) meets both requirements.

Successful outcomes:
Multiples of 3 . . . . 12
Even numbers . . . 18
       Duplicates . . . . 6

So there are 24 different successful outcomes.

Probability = (24) / (36) = 2/3 = (66 and 2/3) %

it takes 1 gallon of diesel 1 fuel for a cruise ship to trarel 6 inches how many gallons will it take for the cruise ship to travel 1 mile

Answers

considering that there are 63360 inches in one mile I can divide that by 6 inches to get 10560 gallons of diesel fuel

Identify all of the expressions that simplify to −7/8.2 1/2 ÷ -7/20
______________

-1 2/5÷ -1 3/5
_____________
4 ÷ −4 4/7
_____________
−7/10 ÷ 4/5

Show your work pls

Answers

5/2 ÷ 7/10
5/2 x 10/7
50/14; 25/7; Answer: 3 4/7

-7/5 ÷ -8/5
-7/5 x 5/8
-35/40
-(-35/40)
-(-7/8). Answer: 7/8

-7/10 x 5/4
-35/40
-7/8. Answer: -7/8

Hoped this helped

Weather records show that the probability of rainfall in Florida in June is 0.86. What is the probability that it rains there in June at least 9 times in a decade?

Answers

0.5816 = 58.16% probability that it rains there in June at least 9 times in a decade.

For each June 9 in Florida, there are only two possible outcomes. Either it rains, or it does not. The probability of raining on June 9 of an year is independent of rain on June 9 on any other year, which means that the binomial probability distribution is used to solve this question.

---------------------------------------

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

---------------------------------------

0.86 probability of rain means that $p = 0.86$
A decade means 10 years, so $n = 10$

---------------------------------------

What is the probability that it rains there in June at least 9 times in a decade?

This is:

$P(X \geq 9) = P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 9) = C_(10,9).(0.86)^(9).(0.14)^(1) = 0.3603$

$P(X = 10) = C_(10,10).(0.86)^(10).(0.14)^(0) = 0.2213$

Thus

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.3603 + 0.2213 = 0.5816$

0.5816 = 58.16% probability that it rains there in June at least 9 times in a decade.

A similar question is found at brainly.com/question/9631195

R - it is rain (probablity: 0.86)
N - it is not rain (probablity: 1-0.86 = 0.14)
You've got these cases:

R R R R R R R R R N
R R R R R R R R N R
R R R R R R R N R R
. . .
N R R R R R R R R R - it is 10 cases. Probablity each case is
(0.86)^9 * 0.14 - it is approximately 0.036 (9R and 1N - you multiply everything)

It is 10 cases so you have to multiply by 10 and you get 0.36

There is also eleventh case:

R R R R R R R R R R

Probablity is (0.86)^10 approximately 0.22

Add these cases and you'll get your probablity: 0.36 + 0.22 = 0.58

The sum of two numbers is 21. The first number is 2/5 of the second number. What are the numbers?

Answers

Let's call the bigger number n
So the smaller number is 2/5n
So n + 2/5n = 21
Which is equivalent of n + 2n/5 = 21
Then you need to get rid of the fraction by multiplying everything by 5
5n + 2n = 105
Or simply
7n = 105
Divide both sides by 7 to isolate n
n = 15
So the larger number is 15
Now we need to work out 2/5 of 15 to work out the smaller number
(15/5 [the denominator])*2 [the numerator] = 6
So the two numbers are 15 and 6

Final answer:

The two numbers are 6 and 15; since 6 is 2/5 of 15 and their sum is 21.

Explanation:

In this problem, we want to find two numbers, which we'll call x and y, such that their sum is 21. That is, x + y = 21. Additionally, we know that the first number is 2/5 of the second number, or x = 2/5 * y.

First, solve for x in the equation x = 2/5 * y, we get x = 2y/5.

Next, substitute x in the equation x + y = 21 by 2y/5 to get 2y/5 + y = 21. When you solve that, you'll find y = 15. Then substitute y = 15 in the equation x = 2/5 * y, you'll get x = 6.

So, the two numbers are 6 and 15.

Learn more about Solving Equations here:

brainly.com/question/18322830

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