Please assist me with this problem
1 answer:
Answer:
The answer to your question is 90 dB
Step-by-step explanation:
Data
I = 10⁻³
I⁰ = 10⁻¹²
Formula
Loudness = 10log ()
Process
1.- To solve this problem, just substitute the values in the equation and do the operations.
2.- Substitution
Loudness = 10 log
3.- Simplify
Loudness = 10log (1 x 10⁹)
Loudness = 10(9)
Loudness = 90
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Answer: a) 79.10 % b) 7% c)19.85 %
Step-by-step explanation:
a) Z = ( × - μ ) ÷ σ Where μ mean of population σ standard deviation Z is the abscissa to give the area or probability we are looking for associated to the value 10 ounces ( × )
So: Z = ( 10 - 9 ) ÷ 1.2 ⇒ Z = 1/1.2 ⇒ Z = 0.83 it has to be below this value
From Z tables we get P [ Z ≤ 0.82 ] = 0.7910 0r 79.10 %
b) Following the same procedure: We look for
P [ Z > ( x - μ ) ÷ σ ] ⇒ Z = ( 12 -10 ) ÷ 1.2 = 2.5
From Z table we get the area under the curve from the left tail up to the point Z < 2.5 ( 2.5 not included) but we were asked for the area out of that previous so 1- 0.9930 = 0.007 is the area we are looking for
So P (b) = 0.007 or 7 %
Finally between the two points above mentioned ( 10 ≤ Z ≤ 12 ) we use the previous values (taking in consideration the limits, according to the problem statement )
Z ≤ 10 Z ⇒( 10-9 ) ÷ 1.2 Z = 0.7967
Z ≥ 12 Z ⇒ ( 12 - 9 ) ÷ 1.2 Z = 0.9952
The interval is between these two points
0.9952 - 0.7967 = 0.1985 ⇒ or 19.85 %
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