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:
2/15
Step-by-step explanation:
The answer choices to this question do not make sense since probability is a measure that is always <= 1.0
Based on the number of times a particular colored tile is drawn we can assume that at the minimum there are 6 green, 20 red, 7 blue, 8 purple and 4 black tiles
Total number of tiles = 6 + 20 + 7 + 8 + 4 = 45
Probability of drawing a green tile = 6/45 = 2/15
Answer: 64
Step-by-step explanation:
Answer:
It is the 4th one- 4x^5
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
.4 as we know can also be handled as .40, as the zero helps us determine the difference in value. 50%, though it is a percentage, can be put into decimal form. That would be .5, or again, we could add a zero to help us out, which would make it .50. As of right now, .4 is less than 50%. 3 and 13 are whole numbers, so they do not really need deep observation, except that 3 is less than 13. That would make the order of number from least to greatest .4, 50%, 3, and 13.
