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
80,000.
To round up to the nearest ten thousand, we would see if the next lower place value has a amount either 5 or higher or 4 or lower.
If it’s five or higher, we round up.
If it’s four or lower, we round down.
We can see that the number in the ten thousands place is 8. The number in the place value after 8 is 2.
2<5 so we round down.
The answer is 80,000.
Hope this helps!
Answer:
Step-by-step explanation:
21. (24)(2)/6+4=48
=48/6+4
=8+4
=12
Answer:
The answer is Sunna and Qur'an on edgeunity
Step-by-step explanation:
