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Gre4nikov [31]
2 years ago
10

The mean length of 5 childrens' index finger is 5.6cm.

Mathematics
1 answer:
rjkz [21]2 years ago
3 0

The mean length (rounded to 2 DP) of these 8 people's index finger is 5.83

<h3>What is the mean length (rounded to 2 DP) of these 8 people's index finger?</h3>

The given parameters are:

<u>Children</u>

Mean = 5.6 cm

Frequency = 5

<u>Adult</u>

Mean = 6.2 cm

Frequency = 3

The mean length of these 8 people's index finger is calculated as:

\bar x = \frac{\sum fx}{\sum f}

So, we have

Mean = (5.6 * 5 + 6.2 * 3)/(5 + 3)

Evaluate

Mean = 5.83

Hence, the mean length (rounded to 2 DP) of these 8 people's index finger is 5.83

Read more about mean at

brainly.com/question/20118982

#SPJ1

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Answer:

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See attachment for plots of 2x + y > 10 and x + 2y > 8

From the attached plot, the point that satisfy x,y>0 is:

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2 years ago
PLS PLS HELP, I DONT WANT TO FAIL!!
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<u>Given quadratic functions</u>:

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To find the time, in seconds, that the balloons collided at the highest point, <u>substitute</u> one equation into the other equation and rearrange to <u>equal zero</u>:

\begin{aligned}-6x^2+23x+5 & = -7x^2+26x+3\\-6x^2+23x+5+7x^2 & = -7x^2+26x+3+7x^2\\x^2+23x+5 & = 26x+3\\x^2+23x+5-26x & = 26x+3-26x\\x^2-3x+5& = 3\\x^2-3x+5-3& = 3-3\\x^2-3x+2& = 0\end{aligned}

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\begin{aligned}x^2-3x+2 & = 0\\x^2-2x-x+2 & = 0\\x(x-2)-1(x-2) & = 0\\(x-1)(x-2) & = 0\\\end{aligned}

Apply the <u>zero-product property</u> to solve for x:

\implies (x-1)=0 \implies x=1

\implies (x-2)=0 \implies x=2

Therefore, the balloons collided at 1 second and 2 seconds.  

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f(2)=-6(2)^2 +23(2)+5=27

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Learn more about quadratic systems of equations here:

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