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3 years ago

How many ounces of

1 answer:

7 0

Find how many oz she ate: 1/2 of 3 is 1.5. 3/4 of 3 is 2.25. 1/2 of 2 is 1. Add all of the oz: 1.5+2.25+1=4.75. She ate 4.75 oz :)

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A tow truck company in Houston charges a $40 service call fee and an additional fee of $15 per mile that the vehicle is towed. W

Juliette [100K]

Answer:

Step-by-step explanation:

Step one:

given data

charged = $40 call fee

additional charges= $15 per mile

let the number of miles be m

and let the total charges be c

Required, the linear function which represents the total charges

Step two:

the equation for the charges in slope-intercept form is

c=15m+40

7 0

3 years ago

How many 3/4-foot lengths of pipe can be cut from a 6 1/3-foot pipe?

Ymorist [56]

8, 3/4-foot lengths of pipe can be cut from a 6 1/3-foot pipe.
Hope this helps you :)

5 0

3 years ago

A model of a 39 foot tall building was made by using the scale 1 in : 4.5 feet. What is the height of the model? Round your answ

Gnom [1K]

1 / 4.5 = x / 39....1 inch to 4.5 ft = x inches to 39 ft
cross multiply
(4.5)(x) = (1)(39)
4.5x = 39
x = 39 / 4.5
x = 8.7 inches <==

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

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3 years ago

Madelynn has a budget of $1,484 to spend this month. if she wants to split her money evenly over 4 weeks,how much can she spend

marshall27 [118]

She can spend 371 each week (1484 divided by 4)

5 0

3 years ago

Read 2 more answers