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ch4aika [34]
1 year ago
6

The National Science Foundation (NSF) offers grants to students who are interested in pursuing a college education. The student

must be enrolled full time (12 credit hours or more), and hold a grade point average of 3.8. The NSF will pay $9,500 per year towards tuition for 4 years. The initial amount is placed into an account paying 7.8% compounded annually, and the tuition is drawn from the account until the remaining balance is zero dollars. Determine the initial amount invested by the NSF. Round your answer to the nearest dollar.
a.
$30,000
c.
$35,036
b.
$31,606
d.
$38,000
Mathematics
2 answers:
Kruka [31]1 year ago
6 0
PVAO=9500[(1-(1+0.078)^(-4))/0.078]
PVAO=31605.79
So the answer is 31606
Elodia [21]1 year ago
3 0

Answer:

The answer is option b.  $31,606

Step-by-step explanation:

The NSF will pay $9,500 per year towards tuition for 4 years.

So, p = 9500

r = 7.8% or 0.078

t = 4

We can find the initial amount invested by:

p(\frac{1-(1+r)^{-t} }{r})

Putting values in the formula we get

9500(\frac{1-(1.078)^{-4} }{0.078} )

9500(\frac{1-0.74050}{0.078} )

= $31605.769

Rounding this to nearest whole number, we get $31606.

So, option B is the correct answer.

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the area of a trapezium is 594 sq. cm. The lengths of its parallel sides are in the ratio 4 : 5. The height of the trapezium is
Sophie [7]

Answer:

1st side=44 cm, 2nd side= 55cm

Step-by-step explanation:

let a = '4x' , b = '5x'

h= 12cm

Area of trapezium=1/2*(a+b)h

area=594=1/2*9x*12

594*2/12=9x

594/6=9x

99=9x

99/9=11=x

therefore, 4x=4*11=44cm

             &5x=5*11=55cm

5 0
1 year ago
We suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in y
ANTONII [103]

Answer:

D) a chi square test for independence.

Step-by-step explanation:

Given that we  suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in years), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about their ages and insurance premiums.

We are to check whether two variables insurance premiums and driving experience are associated.

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Hence a proper test would be

D) a chi square test for independence.

8 0
1 year ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
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myrzilka [38]
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Part B: f(x)=(5)(4)+20(alligators) and f(x)=(10)(4)+25(crocodiles)
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Part C: None because the rate of the alligators won't catch up with the rate of crocodiles.
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<em>Greetings from Brasil...</em>

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