Complete question is:
Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 29.7 grams per pound. Find the probability of observing a sample mean of 29.7 grams of fat per pound or less in a random sample of 34 farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Answer:
Probability = 0.0277
Step-by-step explanation:
We are given;
Mean: μ = 32
Standard deviation;σ = 7
Random sample number; n = 34
To solve this question, we would use the equation z = (x - μ)/(σ/√n) to find the z value that corresponds to 29.7 grams of fat.
Thus;
z = (29.7 - 32)/(7/√34)
Thus, z = -2.3/1.200490096
z = -1.9159
From the standard z table and confirming with z-calculator, the probability is 0.0277
Thus, the probability to select 34 fish whose average grams of fat per pound is less than 29.7 = 0.0277
<span>(1/2) [SIN(X-Y)-SIN(X+Y)]= COS(X)SIN(Y)</span>
8 Students are not enrolled. Reason being 9 are taking German and 9 are taking Spanish so 9+9= 18 but because 3 are taking both you subtract 3 so that would give you 15 and 23-15=8
Answer: 8 Students
Answer:
Commutative Property of Multiplication.
Answer:
The answer to your question is: y = -1/2 x + 1
Step-by-step explanation:
Data
A (-4 , 3)
B (6, -2)
Formula
(y - y1) = m(x - x1)
Process
(y - 3) = -1/2 (x + 4)
2(y - 3) = -1(x + 4)
2y - 6 = -x - 4
2y = - x - 4 + 6
2y = - x +2
y = -1/2 x + 1
