Answer:
Randomized block design
Step-by-step explanation:
From the question, we can see the following:
- There are 30 plants of each variety. This means that they are divided into variety subgroups which we will call blocks.
- Now, we are told each plant in each block all are potted in the same amount and type of soil, given the same amount of water, and exposed to the same amount of light. This means that each plant in each block is assigned a treatment condition.
- The procedure is repeated by subjecting each plant one after the other in teach Block to different treatments and this will reduce variability.
Looking at all the statements above, it is clear that this is a randomized block design because a randomized block design is when the experimenter/researcher divides members/participants into subgroups called blocks in a manner that the variability within the blocks is less than the variability between the blocks. Thereafter, the participants within each block will now be randomly assigned to treatment conditions.
Answer:
-5a-15/4
______
6
Step-by-step explanation:
Answer:
Where
and ![\sigma=13.1](https://tex.z-dn.net/?f=%5Csigma%3D13.1)
We are interested on this probability
![P(X>140)](https://tex.z-dn.net/?f=P%28X%3E140%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule:
![P(z>1.924)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E1.924%29%3D1-P%28z%3C1.924%29%3D0.0272)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and ![\sigma=13.1](https://tex.z-dn.net/?f=%5Csigma%3D13.1)
We are interested on this probability
![P(X>140)](https://tex.z-dn.net/?f=P%28X%3E140%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule:
![P(z>1.924)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E1.924%29%3D1-P%28z%3C1.924%29%3D0.0272)
F(x) = 16ˣ
A. g(x) = 8(2ˣ)
g(x) = (2³)(2ˣ)
g(x) = 2ˣ⁺³
The answer is not A.
B. g(x) = 4096(16ˣ⁻³)
g(x) = (16³)(16ˣ⁻³)
g(x) = 16ˣ
The answer is B.
C. g(x) = 4(4ˣ)
g(x) = 4ˣ⁺¹
The answer is not C.
D. g(x) = 0.0625(16ˣ⁺¹)
g(x) = (16⁻¹)(16ˣ⁺¹)
g(x) = 16ˣ
The answer is D.
E. g(x) = 32(16ˣ⁻²)
g(x) = (2⁵)(2⁴ˣ⁻⁸)
g(x) = 2(⁴ˣ⁻³)
The answer is not E.
F. g(x) = 2(8ˣ)
g(x) = 2(2³ˣ)
g(x) = 2³ˣ⁺¹
The answer is not F.
The answer is B and D.