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.
We are going to rewrite both numbers:
(4.2 × 10 ^ 6) = 4200000
(2.25 × 10 ^ 5) = 225000
Adding we have:
4200000 + 225000 = 4425000
Rewriting in exponential notation we have:
4425000 = 4,425 * 10 ^ 6
Answer:
(4.2 × 10 ^ 6) + (2.25 × 10 ^ 5) is equal to:
4,425 * 10 ^ 6
Answer:
A capacidade dessa piscina é de 150.000 litros de água.
Step-by-step explanation:
A capacidade da piscina é de x.
Para encher 2/5 de uma piscina são necessários 60.000 litros de água.
Isto implica que:
A capacidade dessa piscina é de 150.000 litros de água.
Answer:
2/3 , 13/18, 7/9 , 5/6 (least to greatest order)
Step-by-step explanation:
1. You need to convert all the fractions into decimal form .
2. 7/9.00 = 0.78 (rounded)
3. 13/18.00 = 0.72
4. 5/6.00 = 0.83
5 . 2/3.00= 0.67 (rounded)
