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Nutka1998 [239]
3 years ago
11

Can Someone answer this please

Mathematics
1 answer:
Vsevolod [243]3 years ago
5 0

Answer:

So the scaling means that per every foot in real life, you have 2.4 centimeters, which means that the scale model would be 48 cm tall.

A. To find your answer, you would divide 48 by 1.2, which is 40. So your model would be 40 bottle caps tall

B. Same as A, all you have to do is divide 48 by 15.2, which would be about 3 popsicle sticks tall (rounded)

(Hope this helps)

Step-by-step explanation:

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Answer:

y = -2x + 7 is the equation if that's what you are asking for.

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The heights of the trees for sale at two nurseries are shown below. Heights of trees at Yard Works in feet : 7, 9, 7, 12, 5 Heig
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Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

                                               = 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

                                                             

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f(1) = 8

To find the common ratio, divide any term by the term before it.

We can use any two of the given terms in the sequence EXCEPT for 8 because it is the first term and does not have a term before it.

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