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3 years ago

Two unique letters are chosen at random from the alphabet.

2 answers:

7 0

A - choosing letter A first

$|\Omega|=P(26,2)=\dfrac{26!}{24!}=25\cdot26\\
|A|=1\cdot25=25\\\\
P(A)=\dfrac{25}{25\cdot26}=\dfrac{1}{26}\approx0.0385$

Tamiku [17]3 years ago

4 0

Answer:

A 0.0385

Step-by-step explanation:

There are in total 26 letters in English alphabet

Two unique letters are chosen at random.

I letter can be taken from any one of the 26 hence

probability for I letter to be a out of 26 letters = 1/26

We can convert this into decimal as

Probability that the first letter chosen is A =0.0385

Option A is right answer.

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2 years ago

Please Help! My question is in the picture.

Varvara68 [4.7K]

First we need to determine the type of progression in the question.
That's geometric progression. Because the pattern from one sequence to the others are about multiplying.

Second, determine the ratio of the progression
r = a₂/a₁
r = a₂ ÷ a₁
r = 1/2 ÷ 2
r = 1/2 × 1/2
r = 1/4

Third, determine the formula to know the recursive rule
a₂ = a × 1/4
a₂ = 1/4 × a
$a_{n} = \frac{1}{4} a_{(n-1)}$

Fourth, determine a₁. a₁ is the first term of the progression
a₁ = 2

Final answer:
Recursive rule
$a_{n} = \frac{1}{4} a_{(n-1)}$
a₁ = 2

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3 years ago