Answer:
c = 8.14 million×(0.9166)^t
4.83 million
Step-by-step explanation:
Data:
t = y - 2007
c₀ = 8.14 million
c₃ = 23 % less than c₁
Part 1. Calculate c₃
c₃ = c₀(1 - 0.23) = 0.77c₀
Part 2. Calculate r
c₃ = c₀r^t
0.77c₀ = c₀r³
0.77 = r³ Divided each side by c₀
r = 0.9166 Took the cube root of each side
The explicit decay model is c = 8.14 million×(0.9166)^t
Part 3. Prediction
t = 2013 - 2007 = 6
c = c₀r^t = 8.14 million×(0.9166)⁶ = 8.14 million × 0.5929 = 4.83 million
The model predicts that there will be 4.83 million cars for sale in 2013.
Answer:
Sports players: For people that need high respiratory performance, rises in the pressure may difficult the things, so they would be affected by high pressure because this makes breathing a little bit harder.
Workers in construction and such: People that work a lot with their muscles, in heat environments (road workers, construction workers) may be affected by the same problem as before, and this makes the workers to exhaust in a shorter time than if they worked in a lower pressure environment.
Scientist: A lot of time, some experiments or investigations need a specific pressure to work, so changes in weather pressure may affect investigations, and because the changes are actually uncontrollable, the scientist are really affected by them.
Answer:
its correct bro ;) click next
Answer:
What is the probability that a randomly selected family owns a cat? 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%
Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:
Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)
So, 30% of the families own a dog = .30*100 = 30
20% of the families that own a dog also own a cat = 0.2*30 = 6
34% of all the families own a cat = 0.34*100 = 34
Dogs and cats: 6
Only dogs: 30 - 6 = 24
Only cats: 34 - 6 = 28
Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42
What is the probability that a randomly selected family owns a cat?
34/100 = 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
A = doesn't own a dog
B = owns a cat
P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%
