Percentage of part time employees = (part-time employees/total employees) * 100 = (6/60) * 100 = 1/10 * 100 = 10%
X = $20 bills
6x = $5 bills
x(20) + 6x(5) = 1450
20x + 30x = 1450
50x = 1450
x = 29
6(29)= 174
First we would change the fractions into decimals and compare them. That way it would be easy to solve when needed.
1/2=0.50
5/8=0.625
1/8=0.125
Now we can see that 1/8 (0.125) is the smallest, followed by 1/2 (0.50), and 5/8 the largest. Now we can match the sizes with the insects.
Since the tiger beetle is the largest we will place that with the length of 5/8 since that is the largest.
The carpenter ant is second largest so it would be matched with 1/2
The aphid is the smallest so that would be with 1/8 as the smallest
*Hope that helped :D*
1. 11.32
2. 69.12
Hope it helps
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
