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olga55 [171]
2 years ago
8

You are analyzing the survey results tabulated below. The first survey question (yes/no) is whether the respondent is over 40, a

nd the second survey question (low-cal, regular) is whether the respondent prefers your low-cal butter or the regular.
Mathematics
2 answers:
ankoles [38]2 years ago
7 0
2 because the 2 is the number on the left and can be bad and good and uhh yea
ivolga24 [154]2 years ago
7 0
It’s 2…………………………………..
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wolverine [178]

B.)between 3 and 4 tons.

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3 years ago
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20% of Ms. Martin's class did not complete their homework. If there are 30 students in her class, how many did not complete home
padilas [110]

Answer:

Step-by-step explanation:

20% of 30 is 6

Words: 6 Students in Ms. Martin's class did not do their homework.

hope this helps :)

7 0
2 years ago
The length of the hypotenuse of a right triangle is 157 units. The length of one leg of the triangle is 132. Lara wrote the foll
ASHA 777 [7]

the correct answer for this is the last choice:

It is incorrect because the length of the unknown side is the square root of 7,225.

4 0
4 years ago
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Danny bought 5 Pes to $0.30 each six pencils at $0.25 since in The Notebook for $3 and the tax is $8 how much did he pay will ma
Paha777 [63]
The answer is 14. This is because 0.30 multiplied by 5 is 1.5. Then 6 multiplied by 0.25 is 1.5 as well. Adding these two together is 3. 3 plus 3 for the notebook is 6. Than 6 plus 8 dollar tax is 14 dollars.
6 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 \\\\

Volume of the box at 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 the box at x= 0.53:

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

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

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0
3 years ago
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