Answer:
a
Step-by-step explanation:
-2 over 3 is it cuz -4 over 6 simplifies to that and all the others do too
The answer is a.0.32 km.
The speed that a tsunami can travel is modeled by the equation is s = 356√d.
It is given:
s = 200 km/h
d = ?
Now, let's substitute s in the equation and find d:
s = 356√d
200 = 356√d
√d = 200 ÷ 356
√d = 0.562
Now, let's square both sides of the equation:
(√d)² = (0.562)²
d = (0.562)² = 0.316 ≈ 0.32
Therefore, <span> the approximate depth (d) of water for a tsunami traveling at 200 kilometers per hour is 0.32 km.</span>
Answer:
a) 151lb.
b) 6.25 lb
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, a large sample size can be approximated to a normal distribution with mean and standard deviation 
.
In this problem, we have that:
So
a) The expected value of the sample mean of the weights is 151 lb.
(b) What is the standard deviation of the sampling distribution of the sample mean weight?
This is 
Comparing map distance to real distance we get 2cm/4km. That means 1cm = 2km.
So the map distance is half the real distance (well, technically not as one is in cm and the other in km but it’s enough to think this way) and a real distance of 10km must mean a map distance of half that (again ignoring the units) so we get 5cm.
