Answer:
Step-by-step explanation:
Hello!
Suppose that the objective of the experiment is to test if a certain treatment modifies the mean of the population of interest.
If for example, the treatment is "new fertilizer" and the population of interest is "yield of wheat crops"
Then you'd expect that using the new fertilizer will at least modify the average yield of the wheat crops.
The hypotheses will be then
H₀: μ = μ₀
H₁: μ ≠ μ₀
Where μ₀ represents the known average yield of wheat crops. (is a value, for this exercise purpose there is no need to know it)
We know that the treatment modifies the population mean, i.e. the null hypothesis is false.
The sample we took to test whether or nor the new fertilizer works conducts us to believe, it does not affect, in other words, we fail to reject the null hypothesis.
Then we are in a situation where we failed to reject a false null hypothesis, this situation is known as <em><u>Type II error</u></em>.
I hope this helps!
Answer:
A
Step-by-step explanation:
I think that is A.......
Answer:
33.3333%
Step-by-step explanation:
each side would be 50% and if one side is split between three it would equal if it was 16.6667 each
since tehre are two red sides multiply 16.6667 by 2 and u get 33.3333%
Step-by-step explanation:
Number of boxes that will fit =2340
Step-by-step explanation:
The shipping box volume can be calculated by the formula V = Lxwxh ,where L ,w ,h are length ,width and height of the box.
The length of the shipping box =3\frac{3}{4} =\frac{15}{4} ft.
Width = 3 ft.
Height =3\frac{1}{4} =\frac{13}{4} ft.
Volume of shipping box =\frac{15}{4} .3 .\frac{13}{4} =\frac{585}{16} = 36.6 ft.cubic ft.
Volume of shipping box can also be calculated by multiplying base area with the height of box.
Base area of box = \frac{15}{4} .3=\frac{45}{4} square ft.
Volume of box = base area x height =\frac{45}{4} .\frac{13}{4} = 36.6 cubic ft.
Volume of packing cubes =\frac{1}{4} .\frac{1}{4} .\frac{1}{4} =\frac{1}{64} cubic ft.
Number of fish boxes that will fit in the shipping box = volume of shipping box ÷ volume of fish food box = 36.6 ÷ \frac{1}{64} =2340 .
