Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Answer:
Vc = (π/3)✓17 ≈ 4.12 ft³
Step-by-step explanation:
h = ✓(4²+1²)
h = ✓17
V = πr²h/3
V = π(1)²✓17/3
V = (π/3)✓17 ≈ 4.12
Answer:
The percentage of workers not wearing the helmets is 5.3 %.
Step-by-step explanation:
A safety committee randomly examined 900 construction workers during their work, and found that 48 workers were not wearing helmets. Estimate the percentage of workers who do not wear protective masks during their working time with 98% confidence
total workers = 900
Not wearing helmet = 48
Percentage which are not wearing the helmets
= 
%
<h3>
Answer is the ratio 1000:193:61</h3>
Explanation
1 liter = 1000 mL
For every 1000 mL of water, we need 193 mL of sucrose and 61 mL of saline solution.
The ratio is therefore 1000:193:61
We cannot simplify this ratio any further because the GCF of those three terms is 1.
A quick way to see this is to look at how 1000 = (2*5)^3 has prime factors 2 and 5, but 2 nor 5 are factors of 193 and 61. So only 1 is a factor of all three values 1000, 193, 61
