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4 years ago

Find the mean, median, and mode of the date. 4, 8, 11, 6, 4, 5, 9, 10, 10,4

Mathematics

1 answer:

exis [7]4 years ago

7 0

Answer:

Mean: 7.1, median: 7, mode: 4

Step-by-step explanation:

The mean, also known as the average, is equal to the sum of all the numbers divided by how many numbers there are.

4 + 8 + 11 + 6 + 4 + 5 + 9 + 10 + 10 + 4 = 71

10 numbers

71/10 = 7.1

The median is the number in the middle of the set.

4, 4, 4, 5, 6, 8, 9, 10 ,10 ,11

6 and 8 are in the middle, and the number between 6 and 8 is 7.

The mode of a set of numbers is the most common number. In this case, the number is 4 as it appears 3 times.

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3 years ago

A line has slope -3/4 and passes through the point (8,3). Write an equation of the line

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3 years ago

Write an equation for the description.A number x increased by 13 is 116

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Step-by-step explanation:

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3 years ago

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A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

3 years ago

Find the mean, median, and mode of the date. 4, 8, 11, 6, 4, 5, 9, 10, 10,4​

Find the mean, median, and mode of the date. 4, 8, 11, 6, 4, 5, 9, 10, 10,4