Answer:
−1(5−2)(5+2)
Step-by-step explanation:
Eliminate redundant parentheses
(−11²+6)−1(14² +2)
Distribute
−11² + 6 -- 1(14² +2)
−11² + 6 - 14² − 2
Answer:
megl;r g;rmbr
Step-by-step explanation:
Answer:

Step-by-step explanation:

Answer:
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 95
Given that the standard deviation of the Population = 5
Let 'X' be the random variable in a normal distribution
Let X⁻ = 96.3
Given that the size 'n' = 84 monitors
<u><em>Step(ii):-</em></u>
<u><em>The Empirical rule</em></u>


Z = 2.383
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = P(Z≥2.383)
= 1- P( Z<2.383)
= 1-( 0.5 -+A(2.38))
= 0.5 - A(2.38)
= 0.5 -0.4913
= 0.0087
<u><em>Final answer:-</em></u>
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
If
is the amount of strontium-90 present in the area in year
, and it decays at a rate of 2.5% per year, then

Let
be the starting amount immediately after the nuclear reactor explodes. Then

or simply

So that after 50 years, the amount of strontium-90 that remains is approximately

or about 28% of the original amount.
We can confirm this another way; recall the exponential decay formula,

where
is measured in years. We're told that 2.5% of the starting amount
decays after 1 year, so that

Then after 50 years, we have
