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lakkis [162]
3 years ago
5

a group of 25 sixth graders, 40% of them said they had never visited another state. How many students said they had never visite

d another state?
Mathematics
2 answers:
diamong [38]3 years ago
6 0
Correct Answer: 10 students have never visited another state
40/100 × 25 =
40 is out of 100 because percentages always is represented by 100 (cent)

cross out the zero's equally top and bottom 
so it would become: 4/100 × 25 = 4 ×25 = 100
 then add the denominator
100/10 is our final resolution 

cross out the zeros evenly and answer will give you 10.
Neporo4naja [7]3 years ago
5 0
25 students* (40/100)= 10 students.

10 students said they had never visited another state.
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NNADVOKAT [17]

Answer:

72.7

Step-by-step explanation:

First multiply 72 by 10

because 72 is 72 10s, making it a whole number.

then add 720 (72 times 10) to 7.

then all of that over 10.

then just divide.

727/10

72.7

this is what you do for all mixed numbers like that one.

7 0
3 years ago
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Convenient store sells prepaid mobile phones a purchase of them for $12 each and uses a markup rate of 250% what is the markup r
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250%=2.5

Explanation: it just does
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3 years ago
Write 3x^2-18x-6 in vertex form
Vedmedyk [2.9K]
The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

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3 years ago
PLEASE HELP<br> What is the value of f(2)?
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Step-by-step explanation:

I think your answer is 3

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7 0
2 years ago
Which equation has the solutions x=1+or-square root of 5?
stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> x^{2}+2x+4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=-4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=-4+1

x^{2}+2x+1=-3

Rewrite as perfect squares

(x+1)^{2}=-3

(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i

therefore

case a) is not the solution of the problem

<u>case b)</u> x^{2}-2x+4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=-4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=-4+1

x^{2}-2x+1=-3

Rewrite as perfect squares

(x-1)^{2}=-3

(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i

therefore

case b) is not the solution of the problem

<u>case c)</u> x^{2}+2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=4+1

x^{2}+2x+1=5

Rewrite as perfect squares

(x+1)^{2}=5

(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}

therefore

case c) is not the solution of the problem

<u>case d)</u> x^{2}-2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=4+1

x^{2}-2x+1=5

Rewrite as perfect squares

(x-1)^{2}=5

(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

x^{2}-2x-4=0

5 0
3 years ago
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