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3 years ago

Answer this question, me and my partner have no idea what to do I'm very confused

Mathematics

1 answer:

Ksivusya [100]3 years ago

8 0

The middle 50 percent of data are ages 13-31. Every segment of the plot is 25%, so in this plot the two center big boxes add up to 50%, which span 13-31 years of age.

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Please help me on this will give you brainliest

NNADVOKAT [17]

Answer:

Step-by-step explanation:

8 exponent 9

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3 years ago

5, 18, 6, 18, 13 WHATS THE INTERQUARTILE RANGE??

Harlamova29_29 [7]

Answer: 12.5

Step-by-step explanation:

25th Percentile: 5.5

50th Percentile: 13

75th Percentile: 18

Interquartile Range: 12.5

5 0

3 years ago

OMGGG PLZZZ HELP I'LL GIVR U A MEDAL N FAN U (the answer have to RIGHTT)

Troyanec [42]

The answer is the second one

6 0

3 years ago

Read 2 more answers

Can anyone help me about question 2

madreJ [45]

9514 1404 393

Answer:

3(4/3)^2543

Step-by-step explanation:

Using logarithms base 2, we have ...

$(a_n)^{\log{a_n}}=(a_{n+1})^{\log{a_{n-1}}}\\\\(\log{a_n})(\log{a_n}) = (\log{a_{n-1}})(\log{a_{n+1}}) \\\\ \dfrac{\log{a_{n+1}}}{\log{a_{n}}}=\dfrac{\log{a_n}}{\log{a_{n-1}}}=\dots\dfrac{\log_2{16}}{\log_2{8}}=\dfrac{4}{3}$

That is, the ratio of each term to the previous is a constant equal to 4/3. This is the definition of a geometric sequence. This sequence has first term 3 and common ratio 4/3, so the general term is ...

$\log_2 a_n=3\left(\dfrac{4}{3}\right)^{n-1}$

and the 2544th term is ...

$\boxed{\log_2{a_{2544}}=3\left(\dfrac{4}{3}\right)^{2543}}$

8 0

3 years ago

When you divide the number of permutations of 11 objects taken 3 at a time by 3!, you will get the number of combinations of 11

Sloan [31]

Answer:

Yes

Step-by-step explanation:

Permutation :

nPr = n! ÷ (n - r)!

nCr = n! ÷ (n-r)!r!

11 objects taken 3 at a time :

11P3 = 11! ÷ (11 - 3)!

11P3 = 11! / 8!

11P3 = (11 * 10 * 9 )

11P3 = 990

Dividing by 3!

3! = 3*2*1 = 6

= 990 / 6

= 165

For combination :

11C3 = 11! ÷ 8!3!

11C3 = (11 * 10 * 9) / (3 * 2 * 1)

11C3 = 990 / 6

11C3

4 0

3 years ago