To obtain the probability of obtaining heads at least 3 times out of 5 times we recall that in her simulation she assigned odd digits to represent heads and even digits to represent tails. Thus, we count the simulated numbers to see in how many numbers do we have 3 or more odd digits.

Below, the simulated numbers with 3 or more odd digits are bolded.
 32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 65311

We have 6 simulated numbers having 3 or more odd digits.
Therefore, P("heads" at least 3 out of 5 times) = 6 / 16 = 3 / 8.

5 0

4 years ago

Find Four Numbers that form a geometric progression such that the second term is less than the first by 35, and the third term i

Olegator [25]

Answer:

-35/3, -140/3, - 560/3, -2240/3
7, -28, 112, - 448

Step-by-step explanation:

The terms are:

a₁, a₂, a₃, a₄

Given:

a₁ = a₂ + 35
a₃ = a₄ + 560

Use the nth term formula:

a₂ = a₁r
a₃ = a₁r²
a₄ = a₁r³

Substitute:

a₁ = a₁r + 35 ⇒ a₁(1 - r) = 35
a₁r² = a₁r³ + 560 ⇒ a₁(1 - r)r² = 560

Divide the second equation by the first:

r² = 560/35
r² = 16
r = √16
r = ± 4

Use the first equation to find the first term:

a₁( 1 ± 4) = 35
1. a₁ = 35/-3 = -35/3
2. a₁ = 35/5 = 7

We have two sequences:

r = 4

-35/3, -140/3, - 560/3, -2240/3

r = -4

7, -28, 112, - 448

8 0

3 years ago