Hello!
You can only add variables with the same base and exponent
8u^3 + 4u^3 = 12u^3
8u^2 + 0 = 8u^2
0 + -6u = -6u
6 + 3 = 9
Put the sums together
12u^3 + 8u^2 - 6u + 9
The answer is
Hope this helps!
Answer:
m = -6
Step-by-step explanation:
3m= 5(m + 3)-3
Distribute
3m = 5m +15 -3
Combine like terms
3m = 5m +12
Subtract 5m
3m-5m = 5m+12-5m
-2m = 12
Divide by -2
-2m/-2 = 12/-2
m = -6
Answer:
one and fifty six hundred thousandths
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
