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il63 [147K]
3 years ago
6

Ana baked 82 muffins for her school's bake sale. She put 4 muffins aside for her friends and placed the remaining muffins on pla

tes to sell. Ana placed 6 muffins on each plate. How many plates did she use?
Mathematics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

Ana used 13 plates

Step-by-step explanation:

Total muffins baked=82

She placed 4 aside for her friends and sell the remaining

Total sellable muffins = 82-4

=78 muffins

Ana placed 6 muffins on each plate

How much plate does she needs?

=Total sellable muffins/ no. per plate

= 78/6

= 13 plates

Ana used 13 plates

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A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro
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Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the  playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

                                                          =2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]    

=$[2(x+y)]  

But the neighbor will play one third of the side x foot.

So the neighbor will play  =\$(\frac13x)

Now homeowner's total cost for the material is

=\$[ 2(x+y)-\frac13x]

=\$[2x+2y-\frac13x]

=\$[2y+x+x-\frac13x]

=\$[2y+x+\frac{3x-x}{3}]

=\$[2y+x+\frac{2}{3}x]

=\$[2y+\frac53x]

\therefore C(x)=2y+\frac53x  

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

                                           =xy square foot

∴xy=200

\Rightarrow y=\frac{200}{x}

Putting the value of y in C(x)

\therefore C(x)=2(\frac{200}{x})+\frac53x

The domain of C is(0,\infty ).

\therefore C(x)=2(\frac{200}{x})+\frac53x

Differentiating with respect to x

C'(x)= - \frac{400}{x^2}+\frac53

Again differentiating with respect to x

C''(x) = \frac{800}{x^3}

To find the critical point set C'(x)=0

\therefore 0= - \frac{400}{x^2}+\frac53

\Rightarrow \frac{400}{x^2}=\frac{5}{3}

\Rightarrow x^2 =\frac{400\times 3}{5}

\Rightarrow x=\sqrt{240}

\Rightarrow x=15.49 \approx15.5

Therefore

\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in y=\frac{200}{x} we get

\therefore y=\frac{200}{15.5}

    =12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

                                             

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A path is 0.75m wide mrkessel extends the width by placing blocks 20cm wide on each side of the path what is the new width of th
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The five-number summary for a data set is given below. 1.3 3.8 4.1 4.4 12.4 Using the 1.5 x IQR Rule, any data value _____ than
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Answer:

Any data value <u>less</u> than 2.9 should be flagged as a suspected outlier.

Step-by-step explanation:

We have been given the five-number summary for a data set is given below. 1.3  3.8  4.1  4.4  12.4. Using the 1.5 x IQR Rule, any data value _____ than 2.9 should be flagged as a suspected outlier. We are asked to fill in the given blank.

Our given points (1.3  3.8  4.1  4.4  12.4) represent minimum point, 1st quartile, median, 3rd quartile and maximum point.

First of all, we will find interquartile range by subtracting 1st quartile from 3rd quartile.

IQR=4.4-3.8=0.6

Using 1.5 x IQR Rule, we will get our outliers as:

3.8-(1.5\times IQR)\Rightarrow3.8-(1.5\times 0.6)=3.8-(0.9)=2.9    

12.4+(1.5\times IQR)\Rightarrow12.4+(1.5\times 0.6)=12.4+(0.9)=13.3

Therefore, any data value less than 2.9 and greater than 13.3 should be flagged as a suspected outlier.

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