Anoise level of a lawn tractor is approximately
normallydistributed with a mean of 100 decibels and a
standarddeviation of 3 decibels. what is the probability that a
arandomly slected lawn tractor will generate a noise level of
morethan 105 decibels?

Step-by-step explanation:

Since the noise level of a lawn tractor is approximately

normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = the noise level

u = mean noise level

s = standard deviation

From the information given,

u = 100 decibels

s = 3 decibels.

We want to find the probability that a randomly selected lawn tractor will generate a noise level of

more than 105 decibels. It is expressed as

P(x > 105) = 1 - P(x ≤ 105)

For x = 105,

z = (105 - 100)/3 = 1.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.9525

P(x > 105) = 1 - 0.9525 = 0.0475

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6