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solong [7]
3 years ago
5

Can a graph be linear with a outlier

Mathematics
1 answer:
Makovka662 [10]3 years ago
7 0

Answer:

yes it can be.

Step-by-step explanation:

jrjrjekekekdkdkdkdmdmdmrmmedmdmdmdmdmdm

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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
Recycling Center A processes aluminum at a rate of 100 pounds per hour, while Recycling Center B processes aluminum at a rate of
maria [59]
I believe it is Recycling center A
6 0
2 years ago
Read 2 more answers
In the circle below, suppose m FEH=272º and m EFG=116º. Find the following.​
adoni [48]

Answer:

m∠FEH = 44°

m∠EHG =  64°

Step-by-step explanation:

1) The given information are;

The angle of arc m∠FEH = 272°, the measured angle of ∠EFG = 116°

Given that m∠FEH = 272°, therefore, arc ∠HGF = 360 - 272 = 88°

Therefore, angle subtended by arc ∠HGF at the center = 88°

The angle subtended by arc ∠HGF at the circumference = m∠FEH

∴ m∠FEH = 88°/2 = 44° (Angle subtended at the center = 2×angle subtended at the circumference)

m∠FEH = 44°

2) Similarly, m∠HGF is subtended by arc m FEH, therefore, m∠HGF = (arc m FEH)/2 = 272°/2 = 136°

The sum of angles in a quadrilateral = 360°

Therefore;

m∠FEH + m∠HGF + m∠EFG + m∠EHG = 360° (The sum of angles in a quadrilateral EFGH)

m∠EHG = 360° - (m∠FEH + m∠HGF + m∠EFG) = 360 - (44 + 136 + 116) = 64°

m∠EHG =  64°.

5 0
3 years ago
20 POINTS!<br> HELP ME HOW TO GET THIS. I REALLY NEED ASAP. EXPLAIN HOW. THANK YOUUUUUU
dimaraw [331]
Left side = 170-58 =121'

Bottom side = 96-4 =92'

Now let's calculate the upper oblique (slantwise) side by Pythagoras

oblique² = 92²+28² = 9248 & oblique =√9248 = 96.167'

The perimeter of the backyards = 96.167+93+92+121 = 402.167 ft

7 0
2 years ago
Answer I need help this question because I don’t know what it is and I need help
Lesechka [4]

Answer:

17

Step-by-step explanation:

QS and RS equal the same

5x - 8 = 3x + 2

-3x  +8    -3x +8

2x = 10

x = 5 you would then plug x into QS

5(5) - 8

25 - 8

17

5 0
3 years ago
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