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3 years ago

If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15,

Mathematics

1 answer:

never [62]3 years ago

6 0

Answer: Smart.

Average is like 80

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I got 1792 do I have to divide this number by 32?

bulgar [2K]

Yeah, you'd divide 1792 by 32 to get 56.

8 0

3 years ago

How long is a line of 1 million centimeter cubes be?

fenix001 [56]

Answer:

1m cm

10m mm

Step-by-step explanation:

4 0

2 years ago

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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

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

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2 years ago

What is the partial product for 1,600 and 280

Likurg_2 [28]

Answer:

Hope this helps!

Step-by-step explanation:

4 0

2 years ago

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What is common tangent?

rusak2 [61]

A common tangent is a line that happens to be tangent ("touching but not intersecting a curve or curved surface") to two different coplanar ("in the same plane) circles. I do hope that helped you.

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2 years ago