Answer: About (-2.414, 0) and (0.414, 0)
Step-by-step explanation:
Let us graph the equation given and see what the x-intercepts are. These are points where the line intercepts the x-axis. <em>See attached</em>.
This means our x-intercepts are at about (-2.414, 0) and (0.414, 0).
These are also equal to 
and 
1. x <= 31
2. x < 28
For number 1, x is less than or equal to 31 because the number of days in a month (represented by x) will be less than or equal to 31.
For number 2, x is less than 28 because the number of students (represented by x) in each class will always be less than 28.
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
The valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b
<h3>How to determine the valid exclusion?</h3>
The expression is given as:
8ab^2x/4a^2b - 8ab^2
Set the denominator to 0
4a^2b - 8ab^2 = 0
Divide through by 4ab
a - 2b = 0
Add 2b to both sides
a = 2b
Hence, the valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b
Read more about algebraic fraction at:
brainly.com/question/4344214
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Answer:
28 hours for 7 days
Step-by-step explanation:
8/2 =
4
4 hours for 1 day
1 x 7 =
7
7 x 4
28
28 for 7 days
