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.
The class width for this Frequency Distribution Table is 5.
<h3>What is the class width?</h3>
The class width is the difference between the upper boundary and the lower boundary of the class.
The class width = upper boundary - lower boundary
5 - 0 = 5
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ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
Simple, .3 centimeters a day.
3/10=.3
Answer:
no solution
Step-by-step explanation:
