Given that the number rolled is odd, it can be 1, 3, or 5 = 3 possibilities.
Only one of the possibilities is success . . . rolling a 3 .
So the probability of success is 1/3 = 33 and 1/3 percent .
When it comes to finding points on a graph, think of the saying, "you have to learn how to walk before you climb."
To let the variable alone you can follow these steps: 1) subtract 228 from both sides 2) divide both sides by 4. This is the solution: 1) 4y + 228 - 228 = 352 - 228 => 2) 4y = 124 => 3) (4y)/4 = 124/4 => 4) y = 31. Then, <span>the answer is the option A. y = 31. </span>
1. 9/44
2. 1/3
3. 1/14
4. 4/45
5. 7/12
6. 14/55
7. 5/12
8. 2/9
9. 1/16
10. 7/6
11. 5/12
12. 4/5
13. 6/7
14. 7/12
15. 1
16. 3/2
17. 15/8
18. 4/7
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.
