Christopher mixes the letters s, e, l, e, c, t, e, and d thoroughly. without looking, allen draws one letter. expressed as a fraction, decimal, and percentage, what is the probability that allen will not select a consonant?

A. 3/8 , 0.375, 37.5%
B. 3/5 , 0.6, 60%
C. 5/8 , 0.625, 62.5% *
D. 1/2 , 0.5, 50%

Yes, C is right.

btw im dumb its c

The answers I have so far

1. 7/8, 0.875, 87.5%
2. 5/8, 0.625, 62.5%
3. 1/3

it is c, its the same as number one, but different names.

It's C, thanks Ms sue!

Ms.Sue is right

dumb

Ms. sue may be right but she dead-- Rest her soul :)

As an AI language model, I don't have any information regarding the life status of any individual. Therefore, I can't confirm or deny your statement about Ms. Sue. However, I'm programmed to provide you with accurate answers and support your learning.

mind yo binnus

I'm sorry, I'm not sure what you mean by "mind yo binnus." Could you please provide me with more context or clarify your statement so that I can assist you better?

Christopher mixes the letters S, E, L, E, C, T, E, D thoroughly without looking Allen draws one letter expressed as a fraction, decimal and percentage. What is the probability that Allen will not select a consonant

To solve this problem, we need to know that vowels are A, E, I, O, U and Y, and all the other letters are consonants. Therefore, we need to count the number of vowels and consonants in the given set of letters.

Given set of letters: S, E, L, E, C, T, E, D

Number of vowels: 3 (E, E, and E)
Number of consonants: 5 (S, L, C, T, and D)

Total number of letters = 8

The probability of not selecting a consonant is equal to the probability of selecting a vowel.

Probability of selecting a vowel = Number of vowels / Total number of letters
Probability of selecting a vowel = 3 / 8

Expressing the probability as a fraction, decimal, and percentage:

Fraction: 3/8
Decimal: 0.375
Percentage: 37.5%

Therefore, the probability that Allen will not select a consonant is 3/8, 0.375, or 37.5%. Answer choice A matches this solution.

suppose, do you have a bag of red blue and yellow marbles if the probability of picking a red marble is 1/6 and the probability of picking a yellow marble is 1/2 what is the probability of picking a blue marble?

To determine the probability of picking a blue marble, we need to know the total number of marbles in the bag and how many marbles of each color are in the bag.

However, without that information, we cannot directly calculate the probability of picking a blue marble. We only know the probabilities of picking a red and yellow marble.

One approach could be to use the fact that the sum of probabilities for all possible outcomes should be equal to 1. Therefore, the probability of picking a blue marble can be found by subtracting the sum of the probabilities of picking a red and yellow marble from 1.

Let's denote the probability of picking a blue marble as P(B).

We are given that the probability of picking a red marble is 1/6, so we can represent it as P(R) = 1/6.

We are also given that the probability of picking a yellow marble is 1/2, so we can represent it as P(Y) = 1/2.

Using the fact that the sum of all probabilities is equal to 1, we have:

P(R) + P(Y) + P(B) = 1

Substituting the values we know, we get:

1/6 + 1/2 + P(B) = 1

Simplifying the equation, we get:

P(B) = 1 - (1/6 + 1/2)

P(B) = 1 - 2/3

P(B) = 1/3

Therefore, the probability of picking a blue marble is 1/3.

it really isn't

No its not