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PolarNik [594]
3 years ago
12

The product of a particular number and a second value, which is given by the sum of 8 and the original number, is 65. What is th

e number?
Mathematics
1 answer:
stiv31 [10]3 years ago
5 0

Answer:

there are infinite values of sum X + Y = (Y + 65/Y) since there are infinite values of Y.

Step-by-step explanation:

XY = 65. Given that question is asking you find X + Y, it kind of indicates that there could be very limited possible values of X and Y which you can work out. Again it is an indication of such silly questions. It need not be true that you have limited solutions.

One such solution is represented here. I have assumed X and Y are positive integers only.

Factorize 65 = 65 x 1 or 13 x 5

Therefore (X,Y) can be (65,1) or (13,5) .

So X + Y = 66 or 18.

If we include negative integers as well ,

then XY = 65 => Factors are 65 x 1, 13 x 5 or (-13) x (-5) or (-65) x (-1)

So X + Y = -66 or -18 or 18 or 66.

In general,

XY = 65 => X = 65/Y (assuming Y is not 0 else XY would have been 0 too).

Therefore X + Y = Y + 65/Y.

If X, Y belong to real or complex numbers except 0,

there are infinite values of sum X + Y = (Y + 65/Y) since there are infinite values of Y.

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Problem written out
8x = 24
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x = 3
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A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where p_\alpha is the probability of picking the letter \alpha, and \ell_\alpha is the length of code needed to encode \alpha. p_\alpha is given to us, and we have

\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

bits to the code per encoded letter. For a string of length n, we would then expect E[L]=1.375n.

By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

so that the variance for the length such a string is

\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

"squared" bits per encoded letter. For a string of length n, we would get \mathrm{Var}[L]=1.859n.

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3 years ago
I like turtles so how fast can a turtle cross the street if it was going 2 mils per hour and 2,934,000 hours past?
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Answer:

5,868,000

Step-by-step explanation:

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3 0
3 years ago
Dave is comparing the circumferences of several trees in his yard. The oak tree is 0.539 meters in circumference, the ash tree h
stepan [7]

Answer:   c. elm

Step-by-step explanation:

Given : The circumference of the oak tree= 0.539 meters

=0.539\times100=53.9\text{ cm}   [∵ 1 m = 100 cm]             (1)

The circumference of the ash tree= 0.509 yards

=0.509\times3\text{ feet}   [∵ 1 yard = 3 feet]

=0.509\times3\times 30.48\text{ cm}  [∵ 1 foot = 30.48 cm]

=46.54296\approx46.54\text{ cm}        (2)

The circumference of the elm tree = 6281.70 millimeters

=\dfrac{6281.70}{10}=628.17\text{ cm}          (3)

The circumference of the poplar tree = 0.000385 miles

=0.000385 \times5280\text{ feet}   [∵ 1 mile = 5280 feet]

=0.000385 \times5280\times30.48\text{ feet}  [∵ 1 foot = 30.48 cm]

=61.959744\approx61.96\ \text{ cm}              (4)

From (1) , (2) , (3 ) and (4) it is clear that

46.54< 53.9 < 61.96 < 628.17

Hence, the elm tree has the greatest circumference.

5 0
3 years ago
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