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3 years ago

A recipe requires 1/3 cup of milk for each 1/4 cup of water. How many cups of milk per cups of water are there?

Mathematics

1 answer:

Dafna1 [17]3 years ago

3 0

1/3 / 1/4 = 1/3 x 4/1 = 4/3

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If the sales tax in a town is 5% and Amanda pays $10 in tax on her purchase what was the cost of the item?

Vikentia [17]

Answer:

$200

Step-by-step explanation:

If the tax is 5%, then she pays $5 for every $100. If she paid $10 in tax, her purchase was $200.

5 0

2 years ago

Please help I’m not good at this

tia_tia [17]

23/4

Step by step:
5/1 + 3/4
5/1 • 4 = 20/4
20+3/4
23/4

8 0

3 years ago

Read 2 more answers

If one factor of x2 + 2x - 24 is (x-6)

kolbaska11 [484]

Answer:

for this equation one factor is (x+6) and the other is (x-4)

5 0

3 years ago

How do I solve this equation: (3+2i)+(2+bi)=5-4i

oksian1 [2.3K]

$(3+2i)+(2+bi)=5-4i$

6 0

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 $\mathrm{Var}[L]=1.859n$ .

5 0

2 years ago