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3 years ago

How many permutations exist of the letters a, b, c, d taken three at a time? 24 12 4

Mathematics

2 answers:

sweet [91]3 years ago

6 0

Answer:

Option A. 24

Step-by-step explanation:

There are four letters a, b, c, d and we have to tell the permutations made when taken 3 at a time.

This phenomenon exist = $^{4}P_{3}$ = $\frac{4!}{(4-3)!}$

= $\frac{4!}{1!}$ = 4! = 4 × 3 × 2× 1 = 24

Therefore, option (A) 24 are the permutations exist.

Travka [436]3 years ago

4 0

The correct answer is 24

You have to apply factorials here and formulas. You have 4 letters which means that your n=4, and you have 3 objects at a time which, means that you have 4 different groups of combinations with 6 permutations in each due to the 3! factorial. 6 x 4 is 24.

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Rewrite the form In exponential form: Log100 = x

andrey2020 [161]

Answer:

$10^x=100$

Step-by-step explanation:

You know how subtraction is the opposite of addition and division is the opposite of multiplication? A logarithm is the opposite of an exponent. You know how you can rewrite the equation 3 + 2 = 5 as 5 - 3 = 2, or the equation 3 × 2 = 6 as 6 ÷ 3 = 2? This is really useful when one of those numbers on the left is unknown. 3 + _ = 8 can be rewritten as 8 - 3 = _, 4 × _ = 12 can be rewritten as 12 ÷ 4 = _. We get all our knowns on one side and our unknown by itself on the other, and the rest is computation.

We know that $3^2=9$ ; as a logarithm, the exponent gets moved to its own side of the equation, and we write the equation like this: $\log_3{9}=2$ , which you read as "the logarithm base 3 of 9 is 2." You could also read it as "the power you need to raise 3 to to get 9 is 2."

One historical quirk: because we use the decimal system, it's assumed that an expression like $\log1000$ uses base 10, and you'd interpret it as "What power do I raise 10 to to get 1000?"

The expression $\log100=x$ means "the power you need to raise 10 to to get 100 is x," or, rearranging: "10 to the x is equal to 100," which in symbols is $10^x=100$ .

(If we wanted to, we could also solve this: $10^2=100$ , so $\log100=2$ )

6 0

2 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$