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Dafna1 [17]
3 years ago
5

Write down five numbers, so that the mean is 6, the median is 5 and the mode is 4.​

Mathematics
1 answer:
Talja [164]3 years ago
3 0

Answer:

10, 4, 5, 7, 4

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Please help me with this question. image attached.
Alona [7]

Answer:

Answer: A as you have indicated.

Step-by-step explanation:

Givens

  • There are two sides that are 3.1 cm in length
  • There are two sides that are 6.2 cm in length

Formula

  • P = 2*(b + a)
  • where b is the base and
  • where a is the side

Solution

  • P = 2*(6.2 + 3.1)       Add what is inside the brackets
  • P = 2*9.3                 Multiply
  • P = 18.6

Answer A


7 0
3 years ago
Fill in the missing number.3 meters=______cm
Brut [27]

Answer:

300

Step-by-step explanation:

there are 100 cm in a meter (I believe)

8 0
3 years ago
Read 2 more answers
How do i find out a
drek231 [11]

Answer:

it must be

\frac{c}{8\pi \: b}

7 0
3 years ago
What is 1/6+7/10 as a equivalent fraction
nekit [7.7K]

Answer: 13/15

Step-by-step explanation:

8 0
3 years ago
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
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