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.
<span>1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?
25% chance
</span><span>2. Describe how to use two coins to simulate which two pets the family has.
</span>
You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).
<span>3. Flip both coins 50 times and record your data in a table like the one below.
</span><span>Based on your data, what is the experimental probability that the family has two dogs or two cats?
</span>
Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance
<span>4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
1/8 chance (accidentally messed up there) or 12.5%
</span><span>5. How could you change the simulation to generate data for three pets?
</span><span>
To flip 3 coins and add more spots on the chart.
I hope that this helps because it took a while to write out. If it does, please rate as Brainliest
</span>
Hello there! Thank you for asking your question here at Brainly. I will be assisting you today with how to handle this problem, and will teach you how to handle it on your own in the future.
First, let's evaluate the question.
"The circumference of a circle is 6.28. What is the area of a circle?"
Now, let's remember the different formulas for area and circumference.
The circumference is "2•3.14•r", while the area is "3.14•r•r".
We have our circumference, 6.28.
However, we are looking for the area. Since we have the circumference, we need to narrow down to the radius (so we can solve for the area).
Let's set this up as an equation;
C = 2 • 3.14 • r
Plug in the value for our circumference.
6.28 = 2 • 3.14 • r
Multiply 2 by 3.14 and r to simplify the right side of the equation.
2 • 3.14 • r = 6.28 • r = 6.28r
We're now left with:
6.28 = 6.28r
Divide both sides by 6.28 to solve for r.
6.28 / 6.28 = 1
6.28r / 6.28 = r
We are now left with the radius:
R = 1.
Now, we can solve for the area.
Remember our formula for the area.
A = r • r • 3.14.
Plug in 1 for r.
A = 1 • 1 • 3.14
A = 3.14.
Your area is 3.14 units^2.
I hope this helps, and has prepared you for your future problems in relation to this topic!
Answer:
As x ⇒-∞, P(x) ⇒ -∞
As x ⇒ ∞, P(x) ⇒ ∞
Step-by-step explanation:
To find left hand end behavior, plug in negative infinity into the function and evaluate...
P(x) = 3(-∞) = -3(∞) = -∞
The 'y' values of the function decrease towards negative infinity as the 'x' values approach negative infinity
P(x) = 3(∞) = ∞
The 'y' values of the function increase towards positive infinity as the 'x' values approach positive infinity
