If you want to know the concept of square roots and cube roots ?
the square root of a number is that number which when multiplied by itself two times gives us that number
e.g.
√64 = x , so that (x)(x) = 64
and from our multiplication table we know that
(8)(8) = 64 , so that
√64 = 8
the cube root of a number is that number which when multiplied by itself three times gives us that number
e.g.
∛64 = x , so that (x)(x)(x) = 64 , and thus x = 4 because 4x4x4 = 64
so ∛64 = 4
By the inequality for x over on the right side if x is equal to or greater than 2 you use the bottom equation.
G(2) means x is 2.
Using the bottom equation replace the x’s with 2 and solve.
X^3 -9x^2 +27x-25
2^3 -9(2)^2+27(2)-25
Simplify:
8 -36 + 54-25 =1
The answer is A. 1
Answer:
2 1/3 planes
Step-by-step explanation:
3.5 lego planes / 60 minutes
So fraction =
3.5/60
In 40 minutes he will build x planes
So the fraction equation is:
3.5/60 = x/40
Solve:
Cross multiply =
140 = 60x
x = 2 1/3
2 1/3 planes
If my answer is incorrect, pls correct me!
If you like my answer and explanation, mark me as brainliest!
-Chetan K
Answer:
Null hypothesis: 
Alternative hypothesis: 
The alternative hypothesis for this case is that at least one mean is different from the others.
And the best method for this case is an ANOVA test.
Step-by-step explanation:
For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:
Null hypothesis: 
Alternative hypothesis: 
The alternative hypothesis for this case is that at least one mean is different from the others.
And the best method for this case is an ANOVA test.