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ra1l [238]
3 years ago
5

The heights of 500 students at a local school were recorded and found to be approximated by the normal curve below. Which

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

option C.

Step-by-step explanation:

The given graph is the bell shaped graph.

From the graph we can observe that the peak is at 69.

So, the mean of the data is = 69 .

Standard deviation is the deviation of the graph from the mean value.

 =  72-69 = 3

Hence, the standard deviation is equal to 3.

So, the correct answer is option C.

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1. 8y^2-4y+1 divided by 2y-1
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Q1. The answer is 4y+ \frac{1}{2y-1}

\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}


Q2. The answer is 2a+3+ \frac{6}{a-1}

\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}=  \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\  \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\  \\ =2a+3+ \frac{6}{a-1}
<span>

Q3. The answer is </span>2 x^{2} +5x+2<span>

</span>\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4  }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\  \\ =  \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\  \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\  \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2}  + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2
<span>

Q4. The answer is 2x + 7

</span>\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} =  \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\  \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7
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Q5. The answer is </span>x+1- \frac{3}{x-1}<span>
      
</span>\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} =  \\  \\ \frac{(x+1)(x-1)-3}{x-1} =  \frac{(x+1)(x-1)}{x-1}  -\frac{3}{x-1} =x+1- \frac{3}{x-1}


Q6. The answer is y^{2} -2y+3

\frac{ y^{3}-4 y^{2}+7y-6  }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\  \\ = \frac{y^{2}(y-2)}{y-2}+   \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} =  \\  \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\  \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3
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Q7. The answer is </span>x^{2} +xy+ y^{2}}{x-y}<span>

</span>\frac{ x^{3} - \frac{x}{y}  y^{3} }{x-y} =  \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}
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Q8. The answer is </span>(a^{2} +2ab+2b^{2})<span>

</span>\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})<span>


Q9. The answer is a^{n-8} - a^{-14}

</span>\frac{ (a^{2}) ^{n} - a^{n-6}  }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}    \\  \\   (x^{y}) ^{z}= x^{y*z}   \\  \\  \frac{ x^{y} }{ x^{z} } = x^{y-z}  \\  \\ &#10;\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}  =\frac{a^{2n}  }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}}  =  a^{2n-(n+8)} - a^{n-6-(n+8)} = \\  \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}
5 0
3 years ago
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3 years ago
Natalie's group brought in pizza but is buying drinks at the concession stand medium sodas are a dollar and $.25 how many medium
Phantasy [73]

Keywords:

<em>Medium sodas, buy, dollars, divide </em>

For this case we must find the amount of medium sodas that Natalie's group can buy, taking into account that they have 20 dollars and that each medium soda costs 1.25 dollars. To solve, we must divide:

Let "x" be the number of medium sodas you can buy, then:

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So, Natalie's group can buy 16 medium sodas with 20 dollars

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