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2 years ago

Ben can type 2/5 of a page of an essay in 1/2 minute. How much time would

Mathematics

2 answers:

KiRa [710]2 years ago

7 0

Answer:

ok so first i want to figure out what 2/5 converted to percent is......

you can find it out on a calculator or on paper by dividing 2 / 5 = 0.4

so this is 40% and if it takes him half a minute to do this, it takes him about 30 seconds to finish 40% of the essay and there is a full 60% still waiting to be done, so lets see,

40% + 40% = 80%

so that is about 1 full minute to write 80% we are still left with 20% so let's see how much time does it take to write 20% of the essay, in 40% it takes 30 seconds and in 80% it takes 60 seconds. So since 20% is half of 40% then we must divide the time in half so 30 / 2 = 15

So if we add all these numbers up it equals 1:15 minutes to write an essay

I hope this helps and please mark brainliest if i got it right!

DiKsa [7]2 years ago

4 0

Answer:

1.25 minutes, or 1 minute 15 seconds

Step-by-step explanation:

We know that is speed is 0.4 pages per 0.5 minutes. This is 0.8 pages per 1 minute. If we divide 1/0.8, we get 1.25 or 1 minute 15 seconds.

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Translate: 18 less then a number c

Sergio [31]

Answer:

18 less than a number “c” is 18 - c.

7 0

2 years ago

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

2 years ago

Im new to this, and feeling hopeless '-'

Ede4ka [16]

$\left[\begin{array}{ccc}22&18\end{array}\right]\times\left[\begin{array}{cccc}5&18&32&40\\25&40&38&12\end{array}\right]\\\\=\left[\begin{array}{cccc}22\cdot5+18\cdot25&22\cdot18+18\cdot40&22\cdot32+18\cdot38&22\cdot40+18\cdot12\end{array}\right]\\\\=\left[\begin{array}{cccc}560&1116&1388&1096\end{array}\right]$

second question:

January: 32 · 22 + 38 · 18 = 704 + 684 = 1388

December: 18 · 22 + 40 · 18 = 396 + 720 = 1116

1388 - 1116 = 272

Answer: $272.

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2 years ago

How does a hexagon look

hoa [83]

A hexagon is an 6 sided shape. It has 6 angles therefore. here is a picture of a hexagon:

3 0

3 years ago

Read 2 more answers

One angle of a triangle measures 36.5 degrees. What is the measure of the smaller angle?

Sunny_sXe [5.5K]

The final answer would depend in the type of triangle we are analyzing, however here are the possible outcomes:

1.) If it was a right triangle, 36.5 would be the smaller angle.

2.) It cannot be an equilateral triangle since all angles would be 60°.

3.) In a isosceles triangle, 36.5° would be the smaller, since the others would be 72°.

4.) In an scalene triangle it cannot be determined unless we had 2 angles since in that kind of triangle all angles can be different.

5.) In an acute triangle, 36.5° would be the smaller angle.

6.) In an obtuse triangle it cannot be determined unless we had 2 angles, since it can have highly acute angles.

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3 years ago

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