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g100num [7]
3 years ago
6

What is the median value of the data set shown on the line plot?

Mathematics
2 answers:
MrRa [10]3 years ago
3 0

Answer:

6

Step-by-step explanation:

To find the median, order the numbers from smallest to largest. Since this is already done, count how many x's there are. Then find the very center number or, if there is an even amount, the center two numbers.

There are 22 x's or pieces of data. Since 22 is an even number we will have to find the center two numbers which will be the 11th and 12th x. Both of these are the number 6 so your median will be 6.

[In some cases, the two center numbers won't always be the same number. If that happens, just add the two numbers together and divide by 2. You will have your median.

Ex: Say the center two numbers are 7 and 5.

You will add them together: 7 + 5 = 12

And then divide by 2: 12/2 = 6

Your median would be 6.]

i hope this helps!

Soloha48 [4]3 years ago
3 0
Your answer would be six!
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Step-by-step explanation:

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Calculus3 - Infinite sequences and series ( URGENT!!)​
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Answer:

Limit=0

Converges

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Step-by-step explanation:

If a_n=\frac{2^n n!}{(3n+4)!}

then a_{n+1}=\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}.

Let's rewrite a__{n+1} a little.

I'm going to hone in on (3(n+1)+4)! for a bit.

Distribute: (3n+3+4)!

Combine like terms (3n+7)!

I know when I have to find the limit of that ratio I'm going to have to rewrite this a little more so I'm going to do that here. Notice the factor (3n+4)! in a_n. Some of the factors of this factor will cancel with some if the factors of (3n+7)!

(3n+7)! can be rewritten as (3n+7)×(3n+6)×(3n+5)×(3n+4)!

Let's go ahead and put our ratio together.

a_{n+1}×\frac{1}{a_n}

The second factor in this just means reciprocal of {a_n}.

Insert substitutions:

\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}×\frac{(3n+4)!}{2^nn!}

Use the rewrite for (3(n+1)+4)!:

\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)(3n+4)!}×\frac{(3n+4)!}{2^nn!}

Let's go ahead and cancel the (3n+4)!:

\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}

Use 2^(n+1)=2^n × 2 with goal to cancel the 2^n factor on top and bottom:

\frac{2^{n}2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}

\frac{2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}

Use (n+1)!=(n+1)×n! with goal to cancel the n! factor on top and bottom:

\frac{2(n+1)×n!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}

\frac{2(n+1)}{(3n+7)(3n+6)(3n+5)}×\frac{1}{1}

Now since n approaches infinity and the degree of top=1 and the degree of bottom is 3 and 1<3, the limit approaches 0.

This means it absolutely converges and therefore converges.

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Step-by-step explanation:

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