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Vladimir [108]
3 years ago
6

Sound Intensity

Mathematics
1 answer:
Tasya [4]3 years ago
4 0

Answer:

\frac{d\beta}{dI}=\frac{10^5}{ln(10)}

Step-by-step explanation:

We are given that a relation between the number of decibles B and the intensity of a sound I

\beta=10log_{10}(\frac{I}{10^{-16}})

\beta=10(log_{10}I-(-16)log_{10}10

By using property

log(\frac{A}{B})=log A-log B

logA^b=blog A

Log_{10}10=1

\beta=10(log_{10}I+16)

Log 10=1

We have to find the rate of change in the number of decibles when the intensity I=10^{-4}watt/cm^2

Differentiate w.r.t I

\frac{d\beta}{dI}=10(\frac{1}{Iln(10)}

\frac{d\beta}{dI}=\frac{10}{Iln(10)}

Substitute I=10^{-4}

\frac{d\beta}{dI}=\frac{10}{10^{-4}ln(10)}=\frac{10^{1+4}}{ln(10)}=\frac{10^5}{ln(10)}

By using property\frac{a^x}{a^y}=a^{x-y}

\frac{d\beta}{dI}=\frac{10^5}{ln(10)}

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vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

For the second condition, we need to find the derivative y' first. In this case, we have:

y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.

Therefore:

y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.

This means that we must solve the following system of equations:

\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.

If we add the equations above, we get:

\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e}  \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.

If we now substitute c_1 = 0 into either of the equations in the system, we get:

c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}

This means that the solution obeying the initial conditions is:

\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

Indeed, we can see that:

y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3

y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,

which do correspond to the desired initial conditions.

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3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Some parts of California are particularly earthquake- prone. Suppose that in one metropolitan area, 25% of all homeowners are in
aleksandrvk [35]

Answer:

a. Binomial random variable (n=4, p=0.25)

b. Attached.

c. X=1

Step-by-step explanation:

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a. The probability that X=k homeowners, from the sample of 4, have eartquake insurance is:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{4}{k} 0.25^{0}\cdot0.75^{4}

The sample space for X is {0,1,2,3,4}

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b. The histogram is attached.

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Answer:

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