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3 years ago

Henry believes that if you square the expression x^4 The result will be x^16. Is Henry correct? (Help please)

Mathematics

1 answer:

ella [17]3 years ago

8 0

He is incorrect because using the exponent properties, you’re supposed to be multiplying 4*2 instead of 4*4.

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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i

vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable X be defined as the number of students at a particular college who are smokers

The random variable X follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

$np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10$

Thus, a Normal approximation to binomial can be applied.

So,

$X\sim N(\mu=600, \sigma=\sqrt{120})$

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

= P (X > 280.50)

$=P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

8 0

3 years ago

The function f(x) is shown on the graph.

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I think C is the answer

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