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ra1l [238]
3 years ago
15

I need help solving these for class and it’s too difficult

Mathematics
1 answer:
bezimeni [28]3 years ago
7 0
Idek tbh I have the same one
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a teacher promised a movie day to the class that did better, on average, on their test. The box plot shows the results of the te
natka813 [3]

The question is incomplete. The complete question is

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.

Which class should get the reward, and why?

- The 2nd-period class should get the reward. They have the highest score, a perfect 100.

- The 2nd-period class should get the reward. They have a higher median.

- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.

The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is

\frac{77+78+89+92+100}{5}=87.2

In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.

Hence, the mean of the results of the 4th period is

\frac{72+83+89+96+98}{5}=87.6

Hence, obviously, the 4th period is having more mean.

Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.

So, just by observing the box plot, we can give the statement that  "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "

Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

Learn more about box plots here-

brainly.com/question/1523909

#SPJ10

6 0
2 years ago
61 ones×<br> ​10<br> ​<br> ​1<br> ​​ = ?
Vitek1552 [10]
The answer is 6161 your welcome
6 0
4 years ago
Read 2 more answers
A Machine can produce six yards of yarn in two minutes. How many yards of yarn can it produce in one hour
grin007 [14]
The answer is 180!!
7 0
3 years ago
Read 2 more answers
Melinda and Marcus are saving money to purchase a present for a friend. Each friend starts with an amount and also saves a speci
Goshia [24]
Let's find the rates at which Melinda and Marcus are saving money and compare them. 
Based on the table, we can see that Melinda originally had $75. Then she got $135 in 5 weeks.  So the rate of saving money is (135–75)/5 = $12 per week.
This rate is unchanged for the next weeks. As we can see, she got $195 in the next 5 weeks. So she saved $60 more those 5 weeks, or the rate is $60/5 = $12 per week again. 

So Melinda saved $12 per week.
As for Marcus, the equation tells us that the rate of saving money is $14 which is the coefficient in front of x.

Hence, t<span>he rate at which Melinda is adding to her savings each week is 2$
less than the rate at which Marcus is adding to his savings each week.</span>
5 0
3 years ago
Read 2 more answers
One Endpoint is 9,18 and the midpoint is 14,16 what the other end point
Valentin [98]
\bf \textit{middle point of 2 points }\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;&({{ 9}}\quad ,&{{ 18}})\quad &#10;%  (c,d)&#10;&({{ x}}\quad ,&{{ y}})&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)

\bf \left( \cfrac{x+9}{2}~,~\cfrac{y+18}{2} \right)=\stackrel{\textit{midpoint}}{(14,16)}\implies &#10;\begin{cases}&#10;\cfrac{x+9}{2}=14\\\\&#10;x+9=28\\&#10;\boxed{x=19}\\&#10;-------\\&#10;\cfrac{y+18}{2}=16\\\\&#10;y+18=32\\&#10;\boxed{y=14}&#10;\end{cases}
7 0
3 years ago
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