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3 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]3 years ago

7 0

Answer:

They would expect 75 chairs to be understuffed.

Step-by-step explanation:

fenix001 [56]3 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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Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

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The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

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