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frozen [14]
4 years ago
15

a shop only buys bananas with the lengths between 13 cm and 20 cm. estimate the fraction of the 80 bananas that the shop buys

Mathematics
2 answers:
ipn [44]4 years ago
7 0

Answer:

<h2>42/80</h2>

Step-by-step explanation:

We know that the shop only buys abanans with lengths between 13 cm and 20 cm.

So, the fraction of the 80 bananas is actually a ratio between the y-values of each lenth (13 and 20 cm) and the total 80. First, let's see which y-values belong to each length.

Notice that 13 belons to 20 at Cumulative Frequency. And 20 belongs to 62. So, the pairs are (13,20) and (20,62).

\frac{62-20}{80}=\frac{42}{80}

This fraction means that the shop only buys 42 bananas out of 80, because those are between 13 cm and 20 cm.

ddd [48]4 years ago
4 0

Answer:

44/80

Step-by-step explanation:

draw a line from 13cm upwards to the graph line and then draw a horizontal line from that point to the y axis to find that you get 18 bananas

then draw a line from 20cm upwardsto the graph line and then draw a horizontal line from that point to the y axis to find that you get 62 bananas

62-18=44 you do this because the shop buys banana lengths between 13 and 20

only from 13 to 20 = 44

so out of 80 (the total) it should be 44/80

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Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x        Equation 1

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24        Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

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​    

=24

=24

​    

Equation 2

Substitute 2x for y

​  

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\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}  

y

y

y

​    

=2x

=2(8)

=16

​    

Equation 1

Substitute 8 for x

​  

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}  

y

16

16

​    

=2x

=

?

2(8)

=16

​    

Plug in x = 8 and y = 16

Yes!

​  

Let's check the second equation:

\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}  

x+y

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24

​    

=24

=

?

24

=24

​    

Plug in x = 8 and y = 16

Yes!

​  

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

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