The third tree diagram is correct.
The probabilities for the first draw are 3/7 for green and 4/7 for red.
If you draw a green pear first, then your probabilities for the second pear would be 2/6 for green and 4/6 for red (since it is done without replacement, the number of pears to choose from changes).
If you draw a red pear first, then your probabilities for the second pear would be 3/6 for green and 3/6 for red.
Answer:
yes
Step-by-step explanation:
Answer:
correct answer to the question is option B
Step-by-step explanation:
as 3 raise to the power 4 equal to 3 raise to power 3 and 3 get cancel and the answer is
Answer:
x = 109
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!