Answer:
what how do you even do this
2.6/1.3 * 10^(9-2) = 2*10^7
By using and X,Y grid starting at zero and then making its way to 1,4 for example showing that for every 1 X there are 4 Y's
1. Getting either all heads or all tails is a 1/1024 chance
2. Rolling a pair of dice and getting a sum of one has a chance of 0
3. Rolling a sum of 5 has a 2 in 21 chance
4. Rolling a sum of 12 has a 1 in 21 chance
Answer:
a) possible progressions are 5
b) the smallest and largest possible values of the first term are 16 and 82
Step-by-step explanation:
<u>Sum of terms:</u>
- Sₙ = n/2(a₁ + aₙ) = n/2(2a₁ + (n-1)d)
- S₂₀ = 20/2(2a₁ + 19d) = 10(2a₁ + 19d)
- 2020 = 10(2a₁ + 19d)
- 202 = 2a₁ + 19d
<u>In order a₁ to be an integer, d must be even number, so d = 2k</u>
- 202 = 2a₁ + 38k
- 101 = a₁ + 19k
<u>Possible values of k= 1,2,3,4,5</u>
- k = 1 ⇒ a₁ = 101 - 19 = 82
- k = 2 ⇒ a₁ = 101 - 38 = 63
- k = 3 ⇒ a₁ = 101 - 57 = 44
- k = 4 ⇒ a₁ = 101 - 76 = 25
- k = 5 ⇒ a₁ = 101 - 95 = 16
<u>As per above, </u>
- a) possible progressions are 5
- b) the smallest and largest possible values of the first term are 16 and 82