Let us say that the procedure of completely filling up the
pool with water is called “1 job”. Therefore the rates of the equipment in
doing the job would be:
<span>Rate of handheld hose = 1 job / 8 minutes ---> 1</span>
<span>Rate of lawn sprinkler = r --->
2</span>
<span>Rate of the two combines = 1 job/ 5 minutes ---> 3</span>
So we are to find the equation to use, in this case, we
simply have to add equations 1 and 2 to get 3:
1 / 8 + r = 1 / 5
Multiplying both sides by 5:
<span>(5 / 8) + 5 r = 1 --->
ANSWER</span>
That's a lot of money and words. I don't appreciate this. I would probably just break down and cry tbh
Answer:
Step-by-step explanation:
If DE = 4x+10,EF =2X -1, and DF= 9x - 15 find DF
(de + ef ) - df = 2e || 2e/2 = e || ed - e = d || ef - e = f || f + d = df
de + ef = 6x + 11 || (6x +11) - df = -3x -6 || 2e/2 = -3x/2 - 3 || (-3x/2 - 3) - de = x/2 + 7|| (- 3x/2 - 3) - ef = -x/2 - 1|| d + f = 6
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.
