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3 years ago

The grades on a geometry midterm at Springer are roughly symmetric with u = 73 and 0 = 3.0.

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

z = - 0.33

Step-by-step explanation:

Given that:

Mean (μ) = 73, standard deviation (σ) = 3.0

Z score is a measure in statistics to determine the number of standard deviation a value is above or below the mean of what is measured. If the value is above the mean it is positive but if it is below the mean it is negative. It is given by the equation:

$z=\frac{x-\mu}{\sigma}$

To calculate z-score for Umaima's exam grade, we need to calculate the z score of 72

For x = 72, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{72-73}{3} =-0.33$

z = - 0.33

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Hello there!

otez555 [7]

Answer:

24 Domain: s>=2 or s<=-2

25. 3x^2 +14x +10

26. x^2 -2x+5

Step-by-step explanation:

24. Domain is the input or s values

square roots must be greater than or equal to zero

s^2-4 >=0

Add 4 to each side

s^2 >=4

Take the square root

s>=2 or s<=-2

25. f(g(x)) stick g(x) into f(u) every place you see a u

f(u) = 3u^2 +2u-6

g(x) = x+2

f(g(x) = 3(x+2)^2 +2(x+2) -6

Foil the squared term

= 3(x^2 +4x+4) +2x+4-6

Distribute

= 3x^2 +12x+12 +2x+4-6

Combine like terms

=3x^2 +14x +10

26 f(g(x)) stick g(x) into f(u) every place you see a u

f(u) = u^2+4

g(x) = x-1

f(g(x) = (x-1)^2 +4

Foil the squared term

= (x^2 -2x+1) +4

= x^2 -2x+5

7 0

3 years ago

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

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

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

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

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

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

So, the volume of the box becomes,

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

In order to maximize the volume enclosed by the box, 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 are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

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

Where

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

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