Answer:
1. Proved down
2. proved down
3. f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5
Step-by-step explanation:
Let us explain how to solve the question
∵ f(0) = -20, f(n) = f(n - 1) - 5 for n > 1
→ That means we have an arithmetic sequence with constant
difference -5 and first term -20
1. → f(1) means we need to find the second term, which equal the
term - 5
∵ f(1) means n = 1
∴ f(1) = f(1 - 1) - 5
∴ f(1) = f(0) - 5
∵ f(0) = -20
∴ f(1) = -20 - 5 → Proved
2. → f(3) means we need to find the third term, which equal the
second term - 5
∵ f(3) means n = 3
∴ f(3) = f(3 - 1) - 5
∴ f(3) = f(2) - 5
→ f(2) = f(1) - 5
∵ f(1) = -20 - 5
∴ f(2) = [-20 - 5] - 5 = -20 - 5 - 5
∴ f(3) = [-20 - 5 - 5] - 5
∴ f(3) = -20 - 5 - 5 - 5 → Proved
3. → From 1 and 2 we notice that the number of -5 is equal to n,
at n = 1 there is one (-5), when n= 3 there are three (-5)
∵ n = 10
∴ There are ten (-5)
∴ f(10) = -20 - 5(10)
∴ f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 → Proved
P(2 or H) = P(2) + P(H) - P(2 and H)
What is the probability of getting a 2 P(2)? = 1/6
What is the probability of getting heads P(H)? = 1/2
P(2 and H) is the product of those two events since the events are independent. = 1/6 * 1/2 = 1/12
P(2 or H) = P(2) + P(H) - P(2 and H)
P(2 or H) = 1/6 + 1/2 - 1/12 = 7/12
The given equations are incomprehensible, I'm afraid...
You're given that osmium-183 has a half-life of 12 hours, so for some initial mass <em>M</em>₀, the mass after 12 hours is half that:
1/2 <em>M</em>₀ = <em>M</em>₀ exp(12<em>k</em>)
for some decay constant <em>k</em>. Solve for this <em>k</em> :
1/2 = exp(12<em>k</em>)
ln(1/2) = 12<em>k</em>
<em>k</em> = 1/12 ln(1/2) = - ln(2)/12
Now for some starting mass <em>M</em>₀, the mass <em>M</em> remaining after time <em>t</em> is given by
<em>M</em> = <em>M</em>₀ exp(<em>kt </em>)
So if <em>M</em>₀ = 590 g and <em>t</em> = 36 h, plugging these into the equation with the previously determined value of <em>k</em> gives
<em>M</em> = 590 exp(36<em>k</em>) = 73.75
so 73.75 ≈ 74 g of Os-183 are left.
Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:
590/8 = 73.75 ≈ 74
Answer:
(1, 1)
Step-by-step explanation:
1-5/8=3/8
If he had 5/8 leftover, then he used 3/8.
