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2 years ago

I need the answer this question hard

Mathematics

1 answer:

Genrish500 [490]2 years ago

6 0

Answer:

option B is the correct answer of this question.

plz mark my answer as brainlist if you find it useful plz .hope this will be helpful to you ☺️.

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Pls if anyone can solve NOW

Yuki888 [10]

Answer:

Step-by-step explanation:

For the graphing, we can graph both equations, and where they intercept is the answer. Okay, so we can give x random inputs and get y as an output which is one pair of coordinates. I usually like using 0 as x. y-3(0)=12. y-0=12. y=12. So for our first equation, on of the coordinates is (0,12). Now we can insert another input for x! ( I chose 1.) y-3(1)=12. y-3=12. y=15. So our other pair of coordinates for the first equation is (1,15). We can do the same with the second equation. 2y+8(0)= -4. 2y+0= -4. y=-2. The first pair of coordinates for the second equation is (0,-2). Another input we can put in is 1, again. 2y+8(1)=-4. 2y+8=-4. 2y= -12. y= -6. So our second pair of coordinates for our second equation is (1,-6). We can graph this with a graphing calculator, or mark these points and draw a straight line through them. When we draw a line through them, the part where the two lines intersect is the answer.

When we do substitution, we need to solve for x or y in the bottom equation. I want to solve for x. ( NOTE: IF YOU SOLVE FOR y, YOU STILL GET THE SAME ANSWER) x=-56-3y. Then we replace the x on the top equation with 56-3y. And we get: 2(56-3y)-y=0. We can use the distributive property. The answer I have is 112-6y-y=0. -6y-y is -7y. 112-7y=0. We can add 7y to both sides so they seperate the variables and the numbers. 112=7y. Lastly, divide by 7. For y, we get 16. To get x, we insert y, AKA 16 into x+3y= -56. x+3(16)=-56. x+48= -56. Our last step to get x is to subtract 48 from both sides leaving us with: x= -100. Our final answer is y= 16 and x= -100.

3 0

2 years ago

use a graphing calculator or other technology to answer the question which quadratic regression equation best fits the data set

Greeley [361]

Answer:

Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

Step-by-step explanation:

x y

3 470

4 416

5 403

Analyzing Option A:

Considering the equation

$y=32.86\:\left(x\right)^2+379.14\left(x\right)-1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2+379.14\left(3\right)-1369.14\:$

$y=64.02$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2+379.14\left(4\right)-1369.14\:\:$

$y=673.18$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2+379.14\left(5\right)-1369.14\:$

$y=1348.06$

Analyzing Option B:

$y=32.86\:\left(x\right)^2-379.14\left(x\right)$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)$

$y=-841.68$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)$

$\:y=-990.8$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)$

$\:y=-1074.2$

Analyzing Option C:

Considering the equation

$y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)+1369.14\:$

$y=527.46$

So, the approximately result is (3, 527)

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)+1369.14\:$

$y=378.34$

So, the approximately result is (4, 378)

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)+1369.14\:\:\:$

$y=294.94$

So, the approximately result is (5, 295)

Analyzing Option D:

Considering the equation

$y=-1369.14\:\left(x\right)^2-379.14\left(x\right)+32.86$

From (3, 470), putting x = 3

$y=-1369.14\:\left(3\right)^2-379.14\left(3\right)+32.86\:\:$

$y=-13426.82$

From (4, 470), putting x = 4

$y=-1369.14\:\left(4\right)^2-379.14\left(4\right)+32.86$

$y=-23389.94$

From (5, 403), putting x = 5

$y=-1369.14\:\left(5\right)^2-379.14\left(5\right)+32.86\:\:$

$y=-36091.34$

Therefore, from the above calculations and analysis, we conclude that Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

3 0

3 years ago

Solve for x leave answer as decimal 6/5=x/3

777dan777 [17]

Answer:

3 3/5 =x

3 .6 =x

Step-by-step explanation:

6/5 = x/3

Use cross products

6*3 = 5*x

18 = 5x

Divide each side by 5

18/5 = 5x/5

18/5 =x

3 3/5 =x

8 0

3 years ago

Read 2 more answers

Draw a picture of the standard normal curve and shade the area that corresponds to the requested probabilities. Then use the sta

elena-14-01-66 [18.8K]

Answer:

a)P [ z > 1,38 ] = 0,08379

b) P [ 1,233 < z < 2,43 ] = 0,1012

c) P [ z > -2,43 ] = 0,99245

Step-by-step explanation:

a) P [ z > 1,38 ] = 1 - P [ z < 1,38 ]

From z-table P [ z < 1,38 ] = 0,91621

P [ z > 1,38 ] = 1 - 0,91621

P [ z > 1,38 ] = 0,08379

b) P [ 1,233 - 2,43 ] must be P [ 1,233 < z < 2,43 ]

P [ 1,233 < z < 2,43 ] = P [ z < 2,43 ] - P [ z > 1,233 ]

P [ z < 2,43 ] = 0,99245

P [ z > 1,233 ] = 0,89125 ( approximated value without interpolation)

Then

P [ 1,233 < z < 2,43 ] = 0,99245 - 0,89125

P [ 1,233 < z < 2,43 ] = 0,1012

c) P [ z > -2,43 ]

Fom z-table

P [ z > -2,43 ] = 1 - P [ z < -2,43 ]

P [ z > -2,43 ] = 1 - 0,00755

P [ z > -2,43 ] = 0,99245

8 0

3 years ago

What is the difference? −43−(−18) Enter your answer in the box.

gulaghasi [49]

Hello!

Answer:

-25

Step-by-step explanation:

−43+18

−25

Hope this helps!

5 0

2 years ago

Read 2 more answers