0.08
0.2
+
=
0.28 that’s how u work it out
Answer:
b. 2.28%.
Step-by-step explanation:
Mean temperatue (μ) = 1000°F
Standard Deviation (σ) = 50
°F
For any temperature value, X, the z-score is given by:
For X= 900°F

A z-score of -2.0 corresponds to the 2.28-th percentile of a normal distribution. Therefore, the probability that X<900 is:

(1/8) / 6 =
1/8 * 1/6 =
1/48 of a lb per fish tank <==
The middle right is 145-90, so 55°. I'm not sure about the others
The answer to question 1 is B
the answer to question 2 is B