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3 years ago

Pls can some one help if you like

1 answer:

4 0

Slope is rise/run
gains 4m height for every 3secons
rise=4m
run=3sec
slope=4/3

if takes 54seconds t
54 times 4/3=72 meters

max height is 72 meteres assuming that it starts from the ground level

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Sum to n terms the series 7 +77 +777

julia-pushkina [17]

Answer:

Step-by-step explanation:

=7 (1+11+111+1111......n)

=7/9 (9+99+999+9999....n)

=7/9 ((10-1)+(10^2-1)+(10^3-1)+....n)

=7/9 ((10+10^2+10^3...n)-(1+1+1+1.....n))

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4 0

3 years ago

Please I don't Know Fractions

V125BC [204]

Answer:

11%

Step-by-step explanation:

1/3 (twix)

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5 0

3 years ago

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Shkiper50 [21]

Answer:

a is supplementary with 36 and b is supplementary with 113.

a = 180 - 36 = 144
b = 180 - 113 = 67

Correct choice is D

LM = 1/2(AB + DC) = 1/2(46 + 125) = 85.5

Diagonals of a rectangle are congruent

KM = LN
6x + 16 = 49
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Correct choice is A

8 0

3 years ago

The table shows the total number of pictures Cal took by the end of each week.

CaHeK987 [17]

Answer:

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Step-by-step explanation:

I did this one too but i don't if they changed the answer good luck I would double check

6 0

3 years ago

Read 2 more answers

The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.

nata0808 [166]

Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

2) The slope of regression line b=937.97 represents the rate of change of average annual cost of tuition at 4-year institutions (y) from 2003 to 2010(x). Here,average annual cost of tuition at 4-year institutions is dependent on school years .

Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

and y be the average annual cost of tuition at 4-year institutions from 2003 to 2010.

With reference to table we get

$\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418$

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85$

and

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97$

∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31

4 0

3 years ago