During a rock concert, the noise level (in decibels) in front row seats has a mean of 95 dB with a standard deviation of 8 dB. Without assuming a normal distribution, find the minimum percentage of noise level readings within 3 standard deviations of the mean. (Round your answer to 2 decimal places.)Minimum percentage%

Minimum percentage = 88.89%

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 95 dB

Standard Deviation, σ = 8 dB

Chebyshev's rule:

We have to find the minimum percent of data lying within 3 standard deviations of the mean.

Putting values, we get,

Thus, minimum 88.895 of data lies within 3 standard deviations of the mean.

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1