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lana [24]
2 years ago
9

Kendra is also buying souvenirs for the family reunion. She wants to spend under $2.25 for each item.

Mathematics
1 answer:
PIT_PIT [208]2 years ago
5 0

Answer:

1st,2nd,4th,

Step-by-step explanation:

You have to divide the price by the numbers of items!

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A transcriptionist types 62 words per minute. Which statement is correct? As the number of words increases, the number of minute
Monica [59]

Answer:

The number of words, w, is the dependent variable.

Step-by-step explanation:

8 0
3 years ago
Help with #2 it has to do with number 1 that is the polygon you need to find the area of it.
Katena32 [7]
Okay, it's the equation to how find area any other way. A=L×W

This means *picture above*

So then you count over each line meaning 1.
This means L=9 and a 1/2. Since its in between the 2 an 3 on the x-axis.
Now for W or width. It equals 3 since you can count 2 lines and one either ends of the polygon is 1/2 each this creates one when added together.

All you do from there is multiply 9 and 3. This should give you A=area.

8 0
3 years ago
Find a line that is parallel to y=-2x+7 and passes through a point of (1,-1)
navik [9.2K]

Answer:

y=-2x +1

Step-by-step explanation:

hope this helps

6 0
3 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.

5 0
3 years ago
What is a good estimate for 1,710 2,740 4.322 5,700 7,810 6,395
timama [110]

Answer:

What do you mean by a "Good estimate"? There's no picture or questions.

Step-by-step explanation:

3 0
2 years ago
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