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4 years ago

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Before a chair manufacturer sells its beanbag chairs, they spot check a random sample of chairs on the production line. The tabl

e below shows the number of common problems found during one such spot check. Common Problems Frequency Open seam 4 Cuts in upholstery 14 Understuffed 15 None 267 Total 300 If the manufacturer makes 1500 beanbag chairs per day, how many of those chairs would they expect to be understuffed?

2 answers:

Colt1911 [192]4 years ago

7 0

Answer:

They would expect 75 chairs to be understuffed.

Step-by-step explanation:

fenix001 [56]4 years ago

6 0

Answer:

75

Step-by-step explanation:

Given : In beanbags

4 Cuts in upholstery

15 are Understuffed

None 267

Total 300

To Find : If the manufacturer makes 1500 beanbag chairs per day, how many of those chairs would they expect to be understuffed?

Solution :

We are given that out of total of 300 bags there are 15 understuffed

Let x be the no. of understuffed bean bags out of total of 1500.

⇒ $\frac{15}{300} =\frac{x}{1500}$

⇒ $\frac{15*1500}{300} =x$

⇒ $75=x$

Thus If the manufacturer makes 1500 beanbag chairs per day, 75 of those chairs would they expect to be understuffed .

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The point (0, 0) is a solution to which of these inequalities? A. y – 4 < 3x – 1 B. y – 1 < 3x – 4 C. y + 4 < 3x – 1 D.

miskamm [114]

Answer:

A

Step-by-step explanation:

Replace the point (0,0) in each inequality

A. y - 4 < 3x - 1

B. y - 1 < 3x - 4

C. y + 4 < 3x - 1

D. y + 4 < 3x + 1

A. 0 - 4 < 3(0) - 1

- 4 < - 1

True

B. 0 - 1 < 3(0) - 4

- 1 < - 4

False

C. y + 4 < 3x - 1

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False

D. y + 4 < 3x + 1

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4 years ago

Suppose sarah is trying to design a metal box (a rectangular prism), which will hold 450 cubic centimeters of liquid. the length

schepotkina [342]

<span>So, the volume of a container can also be thought of as the amount of liquid that the container is capable of holding. It follows that 450=(length)(width)(height). The constraints of the problem say that the length of the box must be twice as large as the width, so it follows that j(height)=(2w)(w)(height)-450. This then implies that j(height)=(2w^2)(height)-450. In simpler terms, the height is defined as 450/(2w^2), or j(w)=450/(2w^2).</span>

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3 years ago

Use the grouping method to factor x³ + x² + 3x+3.

Annette [7]

Answer:

D. $(x + 1)(x^2 + 3)$

Step-by-step explanation:

Hello!

We can group the first two terms and the last two terms.

<h3>Factor by Grouping</h3>

$x^2 + x^2 + 3x + 3$
$x^2(x + 1) + 3(x + 1)$
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Factoring by grouping is the process of breaking down larger polynomials to smaller ones to factor. We can then combine like factors.

In the second step, we can see that we can rewrite $x^3 + x^2$ as $x^2(x + 1)$ , as both the two terms share a common factor of $x^2$ . We can pull out $x^2$ from that expression. Similarly, $3x$ and $3$ share a common factor of $3$ , so we can pull that out.

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2 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

——————————

Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

3 years ago

Jayden just graduated from college and owes $19,400 on his student loans. The

ELEN [110]

Answer:

M=19400(1.5)/1-(1+1.5)-(15x12)

M=28917.50

Then you would round up so it wo9uld be $28,918.

6 0

3 years ago