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4 years ago

One solution is 3+4i what is the other solution

1 answer:

4 0

So from this that one solution is 3+4i so from this result that the other solution may be 3-4i because these are complex number roots

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A tree cutting business has an insurance policy to cover any damages caused during the tree cutting process. The business pays a

Yakvenalex [24]

Answer:

-$500

Step-by-step explanation:

The expected value of a payment from the policy is ...

$500,000 × 0.002 = $1000

Since the business pays $1500 for the policy, the expected value to the business is ...

$1000 - 1500 = -$500 . . . per year

_____

Of course, the expected value to the insurance company is $500 per year.

_____

We have computed on the basis of 1 claim per year. If we consider the possibility of multiple independent claims, then the expected payment from the insurance goes up by a factor of 1/(1 -0.002) ≈ 1.002004008016.... This has the effect of increasing the expected value by $2.00 per year to -$498.

3 0

4 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.