Answer:
c i think sorry if im wrong
Step-by-step explanation:
Answer:
X=7
Step-by-step explanation:
Answer:
= 1.56
Step-by-step explanation:
Hello!
You have two independent samples of fertilizers from distributor A and B, and need to test if the average content of nitrogen from fertilizer distributed by A is greater than the average content of nitrogen from fertilizer distributed by B.
Sample 1
X₁: Content of nitrogen of a fertilizer batch distributed by A
n₁= 4 batches
Sample mean X₁[bar]= 23pound/batch
σ₁= 4 poundes/batch
Sample 2
X₂: Content of nitrogen of a fertilizer batch distributed by B
n₂= 4 batches
Sample mean X₂[bar]= 18pound/batch
σ₂= 5 pounds/batch
Both variables have a normal distribution. Now since you have the information on the variables distribution and the values of the population standard deviations, you could use a pooled Z-test.
Your hypothesis are:
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
The statistic to use is:
Z= <u> X₁[bar] - X₂[bar] - (μ₁ - μ₂) </u>~N(0;1)
√(δ²₁/n₁ + δ²₂/n₂)
=<u> </u>(<u>23 - 18) - 0 </u> = 1.56
√(16/4 + 25/4)
I hope this helps!
Answer:
5,175 fish
Step-by-step explanation:
To figure out the total you have to add all the fish together.
Day 1: 750 + 300 + 250 = 1,300 fish were caught on Day 1
Day 2: 720 + 455 + 400 + 50 = 1,625 were caught on Day 2
Day 3: 2,250 were caught on Day 3
Then, add those three numbers
1,300 + 1,625 + 2,250 = 5,175
<span>-6x - 2 = 40
-6x = 42
x = -7
..................</span>