Answer:
critical value = 5.29
Step-by-step explanation:
Given that they are divided into 4 groups and a sample of 5 test was selected
N = 5 * 4 = 20
k = 4
∝ = 0.01
Df for numerator ( SS group )= k - 1 = 3
Df for denominator ( SSE group ) = N - k = 20 - 4 = 16
DF ( degree of freedom )
Next we will use the F table to determine the critical value
Critical value = 
= 5.29
H1 (t) = 196 - 16 t-squared. / / / H2 (t) = 271-16t-squared. / / / In each function, 't' is the number of seconds after that ball is dropped. / / / Each function is only true until the first time that H=0, that is, until the first bounce. Each function becomes very complicated after that, and we would need more information in order to write it.
The answer is y=2x+5
To get it use point slope by taking two points and solving .
Slope formula
M= y2-y1/x2-x1
With two points
(0,5)(-5,-5)
M= -5-5/-5-0
M= -10/-5
M=2
2 is slope
Now get one of the points
(0,5) And slope to create equation y=mx+b . Now find b
5=2(0)+b
5=b
So now you can put it all together
Y= 2x+5
f(5) means to replace x in the equation with 5 then solve.
f(x) = x^2 +2x
f(5) = 5^2 + 2(5)
f(5) = 25 +10
f(5) = 35
The answer is D)35
