R - 4 x 8 is the expanded expression
Answer:
H0: μ = 5 versus Ha: μ < 5.
Step-by-step explanation:
Given:
μ = true average radioactivity level(picocuries per liter)
5 pCi/L = dividing line between safe and unsafe water
The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.
A type I error, is an error where the null hypothesis, H0 is rejected when it is true.
We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.
Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.
You could do the question the way it is written, but it is far easier to bring the negative power up to the numerator.
y= x^2 - 3. The derivative of that is
dy/dx = 2x The three is a constant and is always dropped when a derivative is taken
d(-3)/dx = 0
If you are a purist and want to solve the question the way it is written, you could do it this way.
dy/dx = d(1)/dx x^-2 - d(x^-2)/dx * 1
======================
(x^-2)^2
dy/dx = - (-2 x^ - 3) / x^-4
dy/dx = 2 x^-3 * x^4
dy/dx = 2 x^(-3 + 4)
dy/dx = 2x ^ 1
dy/dx = 2x <<<<< answer
Answer: <u>Archimedes</u>
Step-by-step explanation:
Answer:
The expectation of the policy until the person reaches 61 is of -$4.
Step-by-step explanation:
We have these following probabilities:
0.954 probability of a loss of $50.
1 - 0.954 = 0.046 probability of "earning" 1000 - 50 = $950.
Find the expectation of the policy until the person reaches 61.
Each outcome multiplied by it's probability, so:
The expectation of the policy until the person reaches 61 is of -$4.
