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valentina_108 [34]
2 years ago
9

A factory produces 90 packages of mixed nuts in 1 min. A quality control manager selected 10 of these 90 packages at random and

then counted the number of nuts in each package. The manager recorded the results.
30, 31, 36, 32, 35, 35, 35, 33, 31, 36

What is the estimated median number of nuts in all 90 packages?

Enter your answer in the box.

nuts =
Mathematics
1 answer:
valentina_108 [34]2 years ago
7 0
Answer:

34

explanation:

First of all, put the numbers in order

30, 31, 31, 32, 33, 35, 35, 35, 36, 36

Then, find the middle number.
In this case, there is an even amount of numbers so, we have to pick the 2 middle numbers which is 33 and 35.

Now all you have to do is add these two numbers together then divide by 2 which will give you 34.

or, in simpler questions like this one, you can just say 34 as you know it is between 33 and 35.


In other questions, it might have and odd amount of numbers, for example:

3, 3, 5, 8, 10

so all you would do here is pick the middle number which would be 5. (it has 2 numbers on each side of it)
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Andre45 [30]

Answer:

a_n=-3(3)^{n-1} ; {-3,-9, -27,- 81, -243, ...}

a_n=-3(-3)^{n-1} ; {-3, 9,-27, 81, -243, ...}

a_n=3(\frac{1}{2})^{n-1} ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

a_n=243(\frac{1}{3})^{n-1} ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

a_n=-3(3)^{n-1}

At n=1,

a_1=-3(3)^{1-1}=-3

At n=2,

a_2=-3(3)^{2-1}=-9

At n=3,

a_3=-3(3)^{3-1}=-27

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

a_n=-3(-3)^{n-1}

At n=1,

a_1=-3(-3)^{1-1}=-3

At n=2,

a_2=-3(-3)^{2-1}=9

At n=3,

a_3=-3(-3)^{3-1}=-27

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

a_n=3(\frac{1}{2})^{n-1}

At n=1,

a_1=3(\frac{1}{2})^{1-1}=3

At n=2,

a_2=3(\frac{1}{2})^{2-1}=1.5

At n=3,

a_3=3(\frac{1}{2})^{3-1}=0.75

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

a_n=243(\frac{1}{3})^{n-1}

At n=1,

a_1=243(\frac{1}{3})^{1-1}=243

At n=2,

a_2=243(\frac{1}{3})^{2-1}=81

At n=3,

a_3=243(\frac{1}{3})^{3-1}=27

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

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Ray EF is the bisector of angle AET. Find the measure of angle FEA.* 70° E A Your answer​
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