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Degger [83]
3 years ago
6

B) Sally has a GPA of 3.35. What percentage of the students have a GPA above her GPA? Show all work. (Note: GPA is normally dist

ributed).
All GPA's: 2.59
3.49
3.78
3.91
3.47
3.36
3.14
2.80
2.93
3.49
3.32
3.72
3.08
3.90
3.33
1.78
3.22
3.36
3.47
2.89
3.34
2.96
3.10
3.38
3.09
3.59
3.93
3.02
3.45
3.49
3.54
3.49
3.82
1.86
Mathematics
1 answer:
Cerrena [4.2K]3 years ago
3 0
13 gpas are higher than 3.35. There are a total of 34 gpas. 13 divided by 34 is 0.382 (rounded to 3 dp). 0.382 x 100 = 38.2 therefore 38.2% of students have a higher gpa than Sally.
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Answer:

\dfrac{dh}{dt}=21 \text{m/min}

the rate of change of height when the water is 1 meter deep is 21 m/min

Step-by-step explanation:

First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)

V = \dfrac{1}{2}(bh)\times L

here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:

\dfrac{b}{h} = \dfrac{5}{7}

b=\dfrac{5}{7}h this can be substituted back in the volume equation

V = \dfrac{5}{14}h^2L

the rate of the water flowing in is:

\dfrac{dV}{dt} = 6

The question is asking for the rate of change of height (m/min) hence that can be denoted as: \frac{dh}{dt}

Using the chainrule:

\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

V = \dfrac{5}{14}h^2L

\dfrac{dV}{dh} = \dfrac{5}{7}hL

reciprocating

\dfrac{dh}{dV} = \dfrac{7}{5hL}

plugging everything in the chain rule equation:

\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}

\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6

\dfrac{dh}{dt}=\dfrac{42}{5hL}

L = 12, and h = 1 (when the water is 1m deep)

\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}

\dfrac{dh}{dt}=21 \text{m/min}

the rate of change of height when the water is 1 meter deep is 21 m/min

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A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
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Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

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To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

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You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

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Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

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With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

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for xo = a = 0.235 and yo = f(a) = 0.721:

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Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

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