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seraphim [82]
4 years ago
15

Seth wants to find out if students at Hillsboro High School go to the football games. He decides to take a sample rather than as

k every student. He asks the first 15 boys leaving the library. Which statement about the sample is true? Select two answers.
A. This sample includes the entire student population.
B. This sample includes only those students who use the library.
C. This sample is representative of the population.
D. This sample is not a random sample.
E. This sample is a random sample
Mathematics
1 answer:
Sergio [31]4 years ago
3 0

Answer:

c & e

Step-by-step explanation:

not sure tho

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If x = -2 and y = 3, by how much does 2y2 - x exceed 2x2 - y?
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Yes it does, to find this insert the values of x and y into the equation.
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PLEASE!!!!!
DochEvi [55]

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10 meters/sec

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5 0
4 years ago
Please help and please provide the steps to solve this! Thanks
otez555 [7]

slope = m = (y2 - y1)/(x2 - x1)

(y2 - y1)/(x2 - x1) = m

(-1 - 1)/(-1 - x) = 2/7

-2/(-1 - x) = 2/7

2/(x + 1) = 2/7

1/(x + 1) = 1/7

x + 1 = 7

x = 6

Answer: x = 6

8 0
3 years ago
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A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six
Zarrin [17]

Answer:

The volume of the box is maximized when x = 0.53 cm

Therefore,  x = 0.53 cm  and the Maximum volume = 1.75 cm³

Step-by-step explanation:

Please refer to the attached diagram:

The volume of the box is given by

V = Length \times Width \times Height \\\\

Let x denotes the length of the sides of the square as shown in the diagram.

The width of shaded region is given by

Width = 3 - 2x \\\\

The length of shaded region is given by

Length = \frac{1}{2} (5 - 3x) \\\\

So, the volume of the box becomes,

V =  \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V =  \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V =  \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V =  \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\

Take the derivative of volume and set it to zero.

\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15)\\\\18x^2 -38x + 15 = 0\\\\

We may solve the quadratic equation using the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The values of coefficients a, b, c are

a = 18 \\\\b = -38 \\\\c = 15 \\\\

Substituting the values into quadratic formula yields,

x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 +  19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\

Volume of box when x = 1.59:

V =  \frac{1}{2} (5 - 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\

Volume of box when x = 0.53:

V =  \frac{1}{2} (5 - 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3

As you can see, the volume of the box is maximized when x = 0.53 cm

Therefore,  x = 0.53 cm  and the Maximum volume = 1.75 cm³

8 0
3 years ago
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