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BabaBlast [244]
3 years ago
11

out of 30 students surveyed 17 have a dog based on these results predict how many of the 300 students in the school have a dog

Mathematics
2 answers:
Ira Lisetskai [31]3 years ago
7 0
The whole school would have 170 students that have a dog
AfilCa [17]3 years ago
7 0
One hundred seventy is the answer.
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Shaun is the wholesale buyer for a computer store. he purchase laptops wholesale for x dollars and sell them retail at a 10% pro
timama [110]
<span>Shaun is the wholesale buyer for a computer store. he purchase laptops wholesale for x dollars and sell them retail at a 10% profit.what is the retail price for the laptop?

R= retail price
x= original price

R= X(1.10 or 110%)</span>
3 0
3 years ago
Find the equation of the line parallel to y = 2x + 8 and passes through (4, - 6)
uysha [10]

Answer:

y=2x-14

Step-by-step explanation:

y=2x+8

y-y1=m(x-x1)

y-(-6)=2(x-4)

y+6=2x-8

y=2x-8-6

y=2x-14

8 0
4 years ago
Rewrite the equation y=2|x−3|+5 as two linear functions f and g with restricted domains.
Mama L [17]

Answer:

f(x) = 2+5

g(x) = 2(-(x−3))+5

Step-by-step explanation:

If you remember, an absolute value can have two different answer, a positive and negative answer because the absolute value symbol makes all values in it positive. Example: |-2|=2 and |2|=2

So if  y = 2|x−3|+5

then the two possibilities are

y = 2(x−3)+5    and

y = 2(-)+5

Set one of them equal to f(x) and the other one to g(x)

f(x) = 2(x−3)+5

g(x) = 2(-(x−3))+5

You can also write it as a piecewise function.

y(x)=$\begin{array}{cc}  \{ &     \begin{array}{cc}      -(x-3) & x3    \end{array}\end{array}$

5 0
3 years ago
Can someone please help me solve 91
Mashcka [7]
3x^3+7x^2-20x=0\\&#10;x(3x^2+7x-20)=0\\&#10;x(3x^2-5x+12x-20)=0\\&#10;x(x(3x-5)+4(3x-5))=0\\&#10;x(x+4)(3x-5)=0\\&#10;x=0 \vee x=-4 \vee x=\dfrac{5}{3}
5 0
3 years ago
Read 2 more answers
ILL GIVE BRAINIEST TO WHO EVER IS CORRECT!! PLEASE HELP ASAP!!
Ira Lisetskai [31]

The equation of the line is

x

2

−x

1

x−x

1

=

y

2

−y

1

y−y

1

Here (−3,4)≡(x

1

,y

1

),(4,5)≡(x

2

,y

2

)

∴

4+3

x+3

=

5−4

y−4

⇒

7

x+3

=

1

y−4

⇒x+3=7(y−4)

⇒x+3=7y−28

⇒x−7y+3+28=0

⇒x−7y+31=0

please follow me also

6 0
3 years ago
Read 2 more answers
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