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Sever21 [200]
3 years ago
8

There are 58 girls in a junior class of 146 students. find the ratio of girls

Mathematics
1 answer:
Alexandra [31]3 years ago
7 0
For ever 146 Students, there are 58 females.

73:28 is the simplified ratio, 28 being the female students!

Hope this helps!
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2 years ago
We are driving to Las Vegas. The sign says that it is one hundred forty-five miles to Las Vegas.
nata0808 [166]

Answer:

233.35488 km (Rounded to 233 km)

Step-by-step explanation:

If you convert miles to kilometers then you would use this conversion factor:

1 mile = 1.609344 km (or 1.61 km)

so in this case,

145 mile = 1.609344 km x 145

145 mile = 233.35488 km

I could also round it to 233 km.

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10 months ago
4x(5x_)=(_x_)x3 associative property
11111nata11111 [884]

4x(5x3)=(4x5)x3 the associative property basically just moves the parentheses.

6 0
3 years ago
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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.
3 0
3 years ago
Evaluate. 3−1⋅(4⋅6)⋅2−3
SSSSS [86.1K]
-48
put it in a calculator, it'll spit the answer out
8 0
2 years ago
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