Using the normal distribution, it is found that a production worker has to make $542.64 a week to be in the top 30% of wage earners.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean 
and standard deviation 
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of
.
- The standard deviation is of
.
- The lower bound of the top 30% is the 70th percentile, which is X when Z has a p-value of 0.7, so <u>X when Z = 0.84.</u>
A production worker has to make $542.64 a week to be in the top 30% of wage earners.
You can learn more about the normal distribution at brainly.com/question/24663213
Answer:
A. 
= 
Step-by-step explanation:
of the students wore shorts.
of the students wore jeans.
Converting the fractions of the students who wore shorts and jeans respectively;
= 0.25
= 0.25
This means that the ratio of students who wore shorts is the same as that of students who wore jeans.
i.e =
Answer:
0%
Step-by-step explanation:
Don't quote me on that but I'm pretty sure
Nah fam math hard math can solve its own problems 82773919
Answer:
a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Step-by-step explanation:
for the equation
(1 + x⁴) dy + x*(1 + 4y²) dx = 0
(1 + x⁴) dy = - x*(1 + 4y²) dx
[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx
∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx
now to solve each integral
I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁
I₂= ∫[-x/(1 + x⁴)] dx
for u= x² → du=x*dx
I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂
then
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C
for y(x=1) = 0
1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C
since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be
1/2 * 0 = - π/4 + C
C= π/4
therefore
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
