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:
c = 2
Step-by-step explanation:
2c – 4 = 3(4 – 2c)
Expand brackets:
⇒ 2c – 4 = 12 - 6c
Add 6c to both sides:
⇒ 8c – 4 = 12
Add 4 to both sides:
⇒ 8c = 16
Divide both sides by 8:
⇒ c = 2
Scott would make double of the original amount
Answer:

Step-by-step explanation:
For normal distributions only, all data falls within approximately 68% of one standard deviation, 95% of two standard deviations, and close to 100% of three standard deviations. The standard deviation is far too small to represent two or three standard deviations, hence
.
*Important: This problem would be unsolvable if the question did not say her cuts were normally distributed, because the information above is only applicable to normal distributions.
The question has four different answers. The first answer is: 8/12. Second: 4/11. Third: 8/33. Fourth: 24. Hope this helps!