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lapo4ka [179]
3 years ago
5

HELP PLEASE:

Mathematics
1 answer:
saw5 [17]3 years ago
4 0
For this case we have the following scenario:

 Carlos enrolled in a gym to play sports. Carlos pays a fee for the registration and must pay 1 dollar for each day he trains in the gym. Write an equation that models the problem.

 The equation that models the problem is:
 y = x + 1.
 The slope of the line is 1 and represents the payment of 1 dollar for each day trained.
 The intersection with the y axis is 1 and represents the payment of 1 dollar for the initial inscription.
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Tony bought x cans of tennis balls for $3 each. He bought the same number of water bottles for $5. He also
NeX [460]
What are you asking exactly?
7 0
2 years ago
Can someone help me with this problem
Ede4ka [16]
-13/24 because u have to multiply to get the same denominator which would end up being -15/24 and 2/24 add that to get -13/24
7 0
2 years ago
Steve and Marcia bought soda, chips, and gum. • They bought twice as many chips as pieces of gum. • They bought 3 fewer sodas th
serg [7]

Answer:

ok so S=soda C=chips and G=gum

Step-by-step explanation:

From the information,

s+3=c

(1/2)c=g

from this P=s+c+g

P=s+(s+3)+((s+3)/2)

P=5/2s+9/2

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.

5 0
2 years ago
Carly is traveling to visit family. She drives x miles on the first day and y miles on the second day for a total of 800 miles.
Dmitrij [34]

Answer:

The domain and range of this relationship is (-∞,∞) and (-∞,∞) respectively.

Step-by-step explanation:

We are given that Carly is traveling to visit family.

She drive distance on first day = x

She drive distance on second day = y

We are also given that Carly is travelling a total of 800 miles.

So,x+y=800

We are supposed to find domain and range of this relationship.

y=800-x

Domain : (-∞,∞)

Range:(-∞,∞)

Hence the domain and range of this relationship is (-∞,∞) and (-∞,∞) respectively.

5 0
3 years ago
Read 2 more answers
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