Look at pic above to awnser the question

Mathematics

2 answers:

MissTica3 years ago

6 0

Mode- 16 and 5
Median-15
Range- 17
Median-11

alukav5142 [94]3 years ago

4 0

Mode 5 and 16, mean 11, median 14, range 17

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Help 2,4+0,8=.......-0,06=........+1,21=.......+1,78=....-5,14=.

horrorfan [7]

Lets write this out:-

2.4 + 0.8 =______ - 0.06 = _____+ 1.21=_____+1.78=_____- 5.14=___

So to solve d blanks we will do d following:-

2.4 + 0.8 = 3.2

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 - 0.06 = _____+ 1.21=_____+1.78=_____- 5.14=___

Now lets solve again:-

3.2 - 0.06 = 3.14

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21=_____+1.78=_____- 5.14=___

Now lets solve again:-

3.14 + 1.21= 4.25

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78=_____- 5.14=___

Now lets solve again:-

4.25 +1.78= 6.03

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78= 6.03 - 5.14=___

Now lets solve again:-

6.03 - 5.14 = 0.89

Now lets write this out AGAIN.

2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78= 6.03 - 5.14= 0.89

So, 2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78= 6.03 - 5.14= 0.89

Hope I helped ya!! xD

3 0

3 years ago

on january 1,1999,the price of gasoline was$1.39 per gallon .if the price of gasoline increased by 0.5%per month ,what was the w

arlik [135]

$P(t)=1.39.(1.005)^t \ \ \ \ where t \ is \ time \ in \ months\\
\\
P(1)=1.39.(1.005)^1\\
\\
P(1)=1.39*1.005\\
\\
P(1)=\$ \ 1.40$