Answer: 14 years
Step-by-step explanation:
Let the cat's age when it died = x.
Therefore, the child's age is 2 + x because he/she got the cat when he/she was 2 years old.
We can set up an equation.
x + (2+x) = 30
2x + 2 = 30
2x = 28
Therefore, x = 14
Hi,
81 is 37.5% of 216.
"1. We assume, that the number 37.5 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 100% equals 37.5, so we can write it down as 100%=37.5.
4. We know, that x% equals 81 of the output value, so we can write it down as x%=81.
5. Now we have two simple equations:
1) 100%=37.5
2) x%=81
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=37.5/81
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for 81 is what percent of 37.5
100%/x%=37.5/81
(100/x)*x=(37.5/81)*x - we multiply both sides of the equation by x
100=0.46296296296296*x - we divide both sides of the equation by (0.46296296296296) to get x
100/0.46296296296296=x
216=x
x=216 "
Have a great day!
A=P(1+r%)ⁿ , where P is the initial capital, r, the interest rate and n, the # years
A = 2000(1+0.13)³
A = 2000(1.13)³
A = 2,885.79
Answer:
The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.
Step-by-step explanation:
We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.
If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.
The standard deviation of the sample is equal to:

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:
