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Vlad1618 [11]
3 years ago
12

What is the answer for number 3

Mathematics
1 answer:
kozerog [31]3 years ago
8 0
6 tenths I thought I already answered this guess not lol
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What is f(5)?<br><br> –8<br> –1<br> 1<br> 8
ValentinkaMS [17]
The function of 5 (or y when x is 5) can not be determined with this little information.
7 0
3 years ago
Sarah Sandoval is a coffee farmer trying to decide how many tons of coffee to produce. She can sell each ton of coffee for $2,40
Alexandra [31]

Answer:

Sarah should produce four tons of coffee and total cost for four ton is $6,000.

Step-by-step explanation:

According to the scenario, calculation of the given data are as follows,

Selling price per ton = $2,400

Cost of production is increasing by $600 for each ton.

So, Cost of production first ton = $600

Cost of production for second ton = $600 + $600 = $1,200

Similarly, Cost of production for third ton = $1,200 + $600 = 1,800

Cost of production of fourth ton = $1,800 + $600 = $2,400

Here, Sarah can produce coffee till selling price = cost of production.

As, cost of production of four ton = selling price, then Sarah can produce only four tons of coffee.

Total cost for four ton coffee = $600 + $1,200 + $1,800 + $2,400

= $6,000

8 0
3 years ago
Yoshi is 5% taller today than she was one year ago. Her current height is 168 cm. How tall was she one year ago
NemiM [27]
155.6 CM
I divided 168 by 100 to find 1% then i got 1.68cm then I times that by 5 to find 5% which was 8.4cm. Finally I subtracted 168 and 8.4 to get 155.6 CM
I hope that helped :)
5 0
3 years ago
Read 2 more answers
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
Which statements describe the function f(x) = 2x-5/2(x^2-1) Check all that apply.
Paul [167]

Answer:

1, 4, 6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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