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4 years ago

908,145,623,777 ten-millions place of this number.

1 answer:

7 0

Start from the right-most 7, and go one digit to the left each time:
7 - units place
7 - tens place
7 - hundreds place
3 - thousands place
2 - ten-thousands place
6 - hundred-thousands place
5 - millions place
4 - ten-millions place <----- answer to this question
1 - hundred-millions place
8 - billions place
0 - ten-billions place
9 - hundred billions place

The ten-millions place is the 4.

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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

el producto de tres números es 1000 y su suma es 39. Si se sabe que el primero es al segundo como el segundo es al tercero, hall

pentagon [3]

Answer: The difference between the larger and smaller number is 21.

Step-by-step explanation:

This can be translated to:

"the multiplication between 3 numbers is 1000 and the addition is 39.

If the first number is at the second like the second at the third, find the difference between the smaller and the larger number"

If A is the first number, B the second and C the third we have that:

A*B*C = 1000

A + C + B = 39

A/B = B/C

From the third equation we have that:

A*C = B*B

we can replace it in the first equation and get:

B*(A*C) = 1000

B*B*B = B^3 = 1000

$B = \sqrt[3]{1000} = 10$

So now we have the value of B.

Now we have the two equations:

A + 10 + C = 39

A*C = B*B = 10*10 = 100

We can isolate A in the first equation and get:

A = 39 - 10 - C = 29 - C

now we can replace it in the second equation:

(29 - C)*C = 100

29C - C^2 - 100 = 0

We need to find the roots of that equation, we can use the Bhaskara's function:

$C = \frac{-29 +- \sqrt{29^2 - 4*(-1)*(-100)} }{2*(-1)} = \frac{-29 +- 21}{-2}$

Where we have two possible values for C

C = (-29 - 21)/(-2) = 25

C = (-29 + 21)/(-2) = 4

Let's chose C = 4, because C is the smaller value. (A will take the other value)

Then we have:

A + B + C = 39

A + 10 + 4 = 39

A = 39 - 14 = 25

Then the difference between the larger and smaller number is:

A - C = 25 - 4 = 21

4 0

4 years ago

A quadrilateral has vertices A(4, 5), B(7, 7), C(6, 8), and D(4, 7). The image of the quadrilateral has vertices A"(–6, 0), B"(–

professor190 [17]

Answer:

Step-by-step explanation:

6 0

3 years ago

Which sampling method is generally the weakest

Rudiy27

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4 years ago

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nataly862011 [7]

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8 0

3 years ago

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