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STatiana [176]
4 years ago
6

Help me solve 6x+5-3x=26

Mathematics
2 answers:
Oksana_A [137]4 years ago
5 0
6x+5-3x=26
 

6x+5-3x=26:x
  |Solve \ for \ x| 


3x+5=26
 
 


3x=26-5
 
  |Subtract \ 5 \ from \ both \ sides| 

3x=21 

x= \dfrac{21}{3}  |Divide \ both \ sides \ by \ 3| 

x=7 

\star \ Hence \ Solved
melamori03 [73]4 years ago
3 0
6x + 5 - 3x = 26
3x + 5 = 26
3x + 5 - 5 = 26 - 5
3x = 21
3x/3 = 21/3
X = 7.

The solution to the equation is X = 7.
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15 volume of the fish tank

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3 years ago
What is the y-intercept and x-intercept of 2x + y = 10x - 1
Sauron [17]

Answer:

Wait I only see the part where it says “what is the y-intercept and x-intercept of” but I don’t see an equation :/

Step-by-step explanation:

6 0
3 years ago
9+x/15=10<br><br> What is happening to the variable? <br><br><br> How do you solve for the variable?
Katarina [22]

Answer:

The variable x is being divided by 15 in the equation and x=15

Step-by-step explanation:

To solve for x we :

9+x/15=10

x/15=1

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4 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
4 years ago
An urn contains eight blue balls and seven yellow balls. if six balls are selected randomly without being​ replaced, what is the
Vlad1618 [11]

to be sure that 2 of them will be blue i have to select 7+2=9 balls
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P=9/15

to be sure that 4 of them will be yellow i have to select 8+4=12 balls
P=12/15


8 0
3 years ago
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