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3 years ago

A randomly generated password has four characters. Each character is either A, B, C, D, E or F or a number from 0 -

Mathematics

1 answer:

krek1111 [17]3 years ago

5 0

Sample space is 36C4

Now, we want to know all of the combinations that have 1 digit in it.

So, we can have one here:

1XXX

X1XX

XX1X

XXX1

But we have 10 different digits to choose from. So, we need to introduce the combination term, nCr, where n is a list of all digits and r is how many we want.

Since we only want one, we will need 10C1 for the number of digits. But we need to choose three lowercases, so it becomes 10C1 × 26C3

Since it's a probability question, we need to divide that by our sample space, 36C4, and our percentage becomes 44%

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What's the table plz help

Veronika [31]

If you mean what goes in the blanks, it is 5 for the first empty box and 10 for the second empty box.

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3 years ago

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In a college parking lot, the number of ordinary cars is larger than the number of sport utility vehicles by 42.9%. the differen

fenix001 [56]

Let
x = number of common cars
y = number of suvs
we can write the following system of equations
x-y = 12
x = 1.429y
Solving the system we have
x-y = 12
1.429y-y = 12
and (1,429-1) = 12
y = (12) / (1,429-1) = 27.97202797
then,
x = 1429 * (27.97202797)
x = 39.97202797
answer
The number of suvs in the parking lot is 28

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3 years ago

1) Diego measured the number of cups of water in 15 bottles of various sizes. He

Ivan

Answer:

$8\frac{3}{4}$ cups.

Step-by-step explanation:

From the line plot made by Diego which represents bottle sizes in cups, we can see that the most common size is the bottle having 1¾ cups (there are 5 of this sizes).

If he poured all 5 of this bottle sizes in 1 container, he would have the following:

$1\frac{3}{4} * 5 = \frac{7}{4}*5 = \frac{35}{4}$

$= 8\frac{3}{4}$ cups.

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3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

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3 years ago

If y=22 when x=8,find y when x=-16

inysia [295]

Y = 44

8 x 2 = 16
22 x 2 = 44

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3 years ago

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