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2 years ago

Match each fraction with its equivalent decimal number.

1 answer:

6 0

Changing a fraction to its decimal form can be done by long division method. Dividing the denominator by its numerator will change the fraction into decimal form. Alternatively, you can use a calculator to do that for you. Below are the fractions with their decimal forms listed:

1.) 2/5 = 0.4 (D)
2.) 7/20 = 0.35 (A)
3.) 3/25 = 0.12 (C)
4.) 7/10 = 0.7 (B)

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What is 78.000.000.000 in scientific notation?

borishaifa [10]

I think it is 78x10^9

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2 years ago

Another fair dice is rolled 300 times. How many times would you expect it to land with a multiple of three on top? I need help t

ira [324]

Answer:

100 times

Step-by-step explanation:

multiples of 3 on a number cube are 3 and 6

probability of rolling a '3' or '6' is 1/6 + 1/6 = 2/6 or 1/3

1/3 x 300 = 100

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2 years ago

Angles: can somebody tell me the angles of X, Y, Z please p.s 25 points if answered correctly

antoniya [11.8K]

Answer:

X=30; Z=80; Y=70

Step-by-step explanation:

The straight lines equal 180 and you subtract 180 to 150 and 180 to 100 to get 30 and 80 and the inside of a triangle is 180 and you subtract 30 and 80 to 180 to get 70.

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2 years ago

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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2 years ago

Please help me on this algebra 1 question, i dont understand it

Y_Kistochka [10]

I believe the answer they're looking for is 3, but it's hard to tell, because there's a negative sign before the box, and a polynomial cannot contain a negative exponent.

Ignoring that sign though, the answer should be 3 because Polynomials typically have a term with each degree up to the maximum of the polynomial.

In this polynomial, there is a term with degree 4, degree 2, and degree 0, but none for degree 3 or 1. Because of this, the only acceptable answer is 3.

3 0

3 years ago