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A low P-Value in ANOVA table mean that the population mean of each treatment effects are different. the population mean of at le

Verizon [17]

Answer:

The population mean of at least one treatment effect are different.

Step-by-step explanation:

An analysis of variance (ANOVA) is conducted in order to determine if there are significant differences between the values of the population mean with respect to the response variable for the domains that under the effects of different treatments. A low p-value leads to reject the null hypothesis of the following hypothesis system:

$H_0: \mu_1=\mu_2=...=\mu_k\\\\\\H_a: \mu_i \neq\mu_j, for $ for some i, j between 1 and k.$$

Rejecting H0 means that this hypothesis is false and, in turn, allows us to conclude that the population mean of one of the domains is different from the others.

4 0

4 years ago

The graphs of the functions f(x)=|x-3| + 1 and G(x)=2x + 1 are drawn. which statement about these functions is true?

Roman55 [17]

The only true statement that compares the two functions is:

B: The solution to f(x)=g(x) is 1

<h3>Which statements are correct?</h3>

Here we have the functions:

f(x) = |x - 3| + 1

g(x) = 2x + 1

First, let's find the solution for:

f(x) = g(x)

|x - 3| + 1 = 2x + 1

|x - 3| = 2x

Notice that we have two options, x = 3:

|3 - 3| = 2*3

0 = 6 (x = 3 is not a solution)

And x = 1:

|1 - 3| = 2*1

2 = 2 (x = 1 is a solution).

Now to get the y-value where the graphs intersect, we just evaluate one of the functions in the solution we found above;

f(1) = |1 - 3| + 1 = 2 + 1 = 3

The graphs intersect when y = 3.

Then we conclude that the only true statement is:

B: The solution to f(x)=g(x) is 1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

8 0

3 years ago

Solve for y.

DochEvi [55]

The answer is b y=6 Use photonath if any more equations like this one

5 0

3 years ago

Read 2 more answers

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

6 0

3 years ago

Read 2 more answers

If x2 − 8xy + 12y2 = 8 and x − 6y = 3, then x − 2y =<br> A) 18<br> B) 24<br> C) 36 <br> D) 8 3

IgorLugansk [536]

It is the last answer letter D 8/3

3 0

3 years ago