11 is the expected value of the escaped time.
Bv (Bt - Bo0 + (Bt - Bs0
= 4 - 0 + 4+3 = 11
Escape time refers to the time available (and necessary) for a person to escape from a particularly dangerous situation.
Escape time refers to the time available (and required) for a person to escape from a particularly hazardous situation. Understanding escape times is an important part of the hazard assessment process and is a particularly important component of any hazard assessment involving fire, explosion, or potential exposure to toxic substances.
The escape time concept is not simply a function of the time it takes for workers to escape the hazard. It also takes into account and constrains the amount of time workers have to escape before being overwhelmed by the danger. This means that two of her workers trying to escape the same fire may have different escape times if one is wearing a fire suit and the other is not.
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Answer:
Divide both sides by 3
∣x−1∣≥12/3
Simplify 12/3 to 4
∣x−1∣≥4
Break down problem into 2 equations
x−1≥4
(x−1)≥4
Then solve...
ANSWERS:
x≥5
x≤-3
Just put it in a calculator!
276/11
25.09090909....
The answer is 25.09 repeating
Hope this helps!
Answer:
its correct
Step-by-step explanation:
Answer:
23 years
Step-by-step explanation:
Step 1: Calculate the rate constant (k) for the radioactive decay
A radioactive substance with initial concentration [A]₀ decays to 97% of its initial amount, that is, [A] = 0.97 [A]₀, after t = 1 year. Considering first-order kinetics, we can calculate the rate constant using the following expression.
ln [A]/[A]₀ = - k.t
k = ln [A]/[A]₀ / -t
k = ln 0.97 [A]₀/[A]₀ / -1 year
k = 0.03 year⁻¹
Step 2: Calculate the half-life of the substance
We will use the following expression.
= ln2/ k = ln 2 / 0.03 year⁻¹ = 23 years