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Goryan [66]
3 years ago
13

The meadian of 26 29 31 33 36

Mathematics
2 answers:
mojhsa [17]3 years ago
5 0

..................................................................

kramer3 years ago
3 0
The answer is 31 the median is the number in the middle and if you cancel out the 2 numbers on the left and right 31 is the middle number
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Help me please thank you!!
Alexxandr [17]

The answer to your problem is A because you must add them altogether but i might be wrong im just trying to help you out

5 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
HELP PLEASE WILL GIVE BRAINLIEST BE QUICK
slamgirl [31]

Answer:

(2,0) and (4,0)

Step-by-step explanation:

The roots of the parabola are the points where the value of the function is zero, i.e., where the graph crosses the x-axis.

(2,0) and (4,0)

7 0
3 years ago
Ella has 10 more CDs than her friend Krystal has. Iris has 3 times as many CDs as Ella has. Which of the following can be used t
pav-90 [236]

Answer:

3(10+k)=total

Step-by-step explanation:

5 0
2 years ago
Express the following numbers as the sum of two consecutive integers of 3 square and 5 square.first correct answer with explanat
Nostrana [21]
3^2 = 9 = 4+5
5^2 = 25 = 12+13


work:
If you want to write an odd number as the sum of two consecutive integers, just subtract 1 then divide by 2 and you'll get the first integer.

Like with 25: first subtract 1 and get 24. Then divide by 2 and get 12. So the integers are 12 and 13.
8 0
3 years ago
Read 2 more answers
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