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
Answer:
-11
Step-by-step explanation:
You need to take the absolute value of 12-15 which would be 3. Then take -8-3 which would be -11
Answer:
563673773837737477377374
Step-by-step explanation:
JK
Answer:
18, 19, 20, 21
Step-by-step explanation:
Just like any of these problems, we should start by forming an equation for us to get a reference and plug in. We'll be using x as our variables.
As we have four consecutive integers (and not multiples) we can assume that the integers will be x, x+1, x+2, and x+3.
The sum of the two largest integers we have equals: n+2 + n+3 = 2n+5
and three times the sum of the two smallest = 3(n + n+1) = 6n+3
and the sum of t he two largest subtracted from three times the sum of the two smallest = (6n+3) - (2n+5) = 4n-2
4n-2=70
4n=68
n = 18.
n = 18, n+1 = 19, n+2 = 20, n+3 = 21
Answer:
I dont know im really sorry
