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Mnenie [13.5K]
3 years ago
9

Which of the following lists has a mode of 213?

Mathematics
2 answers:
quester [9]3 years ago
4 0

Answer:

213, 278, 108, 213, 157

Step-by-step explanation:

The mode is the number with the highest frequency in a given list of data. Simply, it is the number with most occurrence.

Now, let us chart the frequency of each numbers in the list provided:

1........ 111, 108, 213, 198, 205

Arrange in ascending order

108, 111, 198, 205 and 213

Numbers Frequency

108 1

111 1

198 1

205 1

213 1

No modal value

2..... 212, 215, 213, 211, 220

Arranging in ascending order:

211, 212, 213, 215 and 220

Numbers Frequency

211 1

212 1

213 1

215 1

220 1

No modal value

3......213, 278, 108, 213, 157

Arranging in ascending order:

108, 157, 213 and 278

Numbers Frequency

108 1

157 1

213 2

278 1

The mode here is 213. It has a frequency of 2 compared to other numbers whose mode is 1

4........ 210, 200, 213, 221, 221

Arranging in ascending order:

200, 210, 213, 221

Numbers Frequency

200 1

210 1

213 1

221 2

The mode here is 221

bixtya [17]3 years ago
3 0

Answer is: 213, 278 , 108, 213, 157

213 is repeated 2 times; more than any other number.

A number that appears most often is the mode.

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The cast of the play makes 275 programs for 5 shows. How many programs will be needed for 13 shows?
Advocard [28]
715 because 275/5= 55 then 55x13 will give you 715
8 0
3 years ago
A dog kennel owner has 121 ft of fencing to enclose a rectangular dog run. She wants it to
Snezhnost [94]

Answer:

605/12 ft by 121/12 ft

Step-by-step explanation:

So, the dimensions of the dog run would be 5x as the length and x as the width.

The perimeter of this dog run would therefore be :

5x + 5x + x + x = 12x

From here, we know that 12x = 121 ft, so x = 121/12 ft

so the final dimensions of the dog run would be 605/12 ft by 121/12 ft

5 0
2 years ago
Answer all questions please
Marat540 [252]

Answer:

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4 0
3 years ago
9x2+4y2 = 36
valentinak56 [21]

Answer:

<em>The foci of the given ellipse  = (0 ,-√5 ) and ( 0, √5 )</em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given Ellipse   9 x² + 4 y² = 36

                     \frac{9x^{2} }{36} + \frac{4y^{2} }{36} =1

               ⇒     \frac{x^{2} }{4} + \frac{y^{2} }{9} = 1

<u><em>Step(ii)</em></u>:-

  The Major axis lies on y-axis

<em>    Foci of the Formula C² = a² -b² = 9-4 =5</em>

<em>                                  C = √5 </em>

The Foci is always lie on the major axis(longest) axis, spaced equally each side of the centre

<em>Given ellipse of the foci is lie on Y- axis</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The foci of the given ellipse  = (0 ,-√5 ) and ( 0, √5 )</em>

8 0
3 years ago
a pound of popcorn is popped for a class party. The popped corn is put into small popcorn boxes that each hold 130 popped kernel
Nostrana [21]

Answer:

12 boxes are needed

Last box contain only 20 kernels

Step-by-step explanation:

It is given that,

a pound of popcorn is popped for a class party.

The popped corn is put into small popcorn boxes that each hold 130 popped kernels.

There are 1450 kernels in a pound of unpopped popcorn.

If all the boxes are filled except for the last box,

<u>To find the number of boxes</u>

number of boxes  = Total number of kernels/Number of kernel in each box

= 1450/130 = 11 2/13

Therefore total number of boxes = 11 + 1 =12 boxes

number of kernels in 11 boxes = 11x 130 =1430 kernels

So last box contain = 1450 - 1430 = 20

Last box have only 20 kernels

3 0
3 years ago
Read 2 more answers
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