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RUDIKE [14]
4 years ago
14

Many employees screen gpa in addition to a variety of skills. Suppose a student is in the last semester of college and has a 2.6

8 gpa after 108 credit hours. If he is taking 12 credit hours in his last semester and gets a perfect 4.0 gpa, what will his overall gpa be? Is it possible to get his gpa up to 3.0 for graduation?
Mathematics
1 answer:
wel4 years ago
3 0

Answer:

  • Overall GPA=2.81
  • It is not possible to get his GPA to 3.0 for graduation.

Step-by-step explanation:

The Student already has a GPA of 2.68 after 108 credit hours.

If he is taking 12 credit hours in his last semester and gets a perfect 4.0 GPA

Total Credits Earned Before = 2.68 X 108=2894.4

Projected Credit to be earned = 12 X 4= 48 Credits

Total credit Hour= 108+12=120 Hours

His Cumulative GPA = Total credits earned ÷ Total Credit Hour

=\frac{2894.4+48}{120} =2.81

Since 2.81 is less than 3.00, it is not possible to get his GPA to 3.0 for graduation.

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Find the inverse of each function. Is the inverse a function? y=(x+4)³-1
tatuchka [14]

Inverse of the give function F(x) = ( x + 4 )³ - 1 is F⁻¹(x) = [√( x + 1 ) ] - 4.

Given is a function of x, F(x) = ( x + 4 )³ - 1

These question can be solved by following 4 easy steps

Step 1 : Switch the F(x) with the variable y

This implies, F(x) = ( x + 4 )³ - 1 becomes y = ( x + 4 )³ - 1

Step 2 : Interchange the variable x and y in the above obtained equation

This implies, from the equation y = ( x + 4 )³ - 1, we get

x = ( y + 4 )³ - 1

Step 3 : Solve the new obtained equation for y

This implies, we have to simplify the equation by rearranging the terms to get the equation in terms of the variable x.

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=> x + 1 = ( y + 4 )³

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=> y + 4 = √( x + 1 )

=> y = √( x + 1 ) - 4

Hence we obtain the required equation.

Step 4 : Switch the variable y with F⁻¹(x)

This implies, y = √( x + 1 ) - 4 becomes F⁻¹(x) = √( x + 1 ) - 4

Therefore, we get the inverse of the give function F(x) = ( x + 4 )³ - 1 as F⁻¹(x) = [√( x + 1 ) ] - 4.

Learn more about Inverse of a Function here:

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