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Oksanka [162]
3 years ago
8

Sneakychicken help me ​

Mathematics
1 answer:
12345 [234]3 years ago
5 0

Answer:

x, process, y

-2, -2-1, -3

-1, -1-1, -2

0, 0-1, -1

1, 1-1, 0

2, 2-1, 1

X = -2, -1, 0, 1, 2

Y = -3, -2, -1, 0, 1

Step-by-step explanation:

Coordinates for the graph are ( -2,-3), (-1,-2), (0,-1), (1,0), (2,1)

The graph is the image attached to this.

How to get Y:

-2-1= -3

-1-1= -2

0-1= -1

1-1= 0

2-1= 1

Hope this helps.

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Can anyone help me I did it I'm just checking :)
lukranit [14]

Answer:

<h2>x = 12</h2><h2 />

Step-by-step explanation:

2x + 5 = 3x - 7

5 + 7 = 3x - 2x

12 = x

8 0
2 years ago
1/2a+(−7)+1/2a+ 1/3 <br> what is the value of A?
siniylev [52]
I reduced the equation to a - 6 (2/3)
4 0
3 years ago
Help! Show work please<br>will mark brainliest ​
Arada [10]

Answer:

x = 62.23 ft

y = 47.67 ft    

Step-by-step explanation:

6 0
2 years ago
David bought 3 dvds and 4 books for $40 at a yard sale. Anna bought 1 dvd and 6 books for $18. How much did each dvd and book co
Ber [7]

Answer:

A book is worth $1 and a DVD is worth $12.

Step-by-step explanation:

The equations (2 unknowns and two equations, d is for a DVD and b is for a book):

For David: 3d+4b=40

For Anna: d+6b=18

Now multiply the second equation with -3 and add to the first equation:

3d+4b=40

−3d−18b=−54

Combined equation: −14b=−14 and b=1 (means that each book is worth $1).

 

Now for DVD price, use the second equation:

d=18−6 or d=12 (means that each DVD  is worth $12).

A book is worth $1 and a DVD is worth $12.

8 0
3 years ago
A study indicates that 37% of students have laptops. You randomly sample 30 students. Find the mean and the standard deviation o
Brilliant_brown [7]

Answer:

The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.

Step-by-step explanation:

Let <em>X</em> = number of students who have laptops.

The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.

A random sample of <em>n</em> = 30 students is selected.

The event of a student having a laptop is independent of the other students.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The mean and standard deviation of a binomial random variable <em>X</em> are:

\mu=np\\\sigma=\sqrt{np(1-p)}

Compute the mean of the random variable <em>X</em> as follows:

\mu=np=30\times0.37=1.11

The mean of the random variable <em>X</em> is 1.11.

Compute the standard deviation of the random variable <em>X</em> as follows:

\sigma=\sqrt{np(1-p)}=\sqrt{30\times0.37\times(1-0.37)}=\sqrt{0.6993}=0.836

The standard deviation of the random variable <em>X</em> is 0.836.

5 0
3 years ago
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