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3 years ago

ASAP !!

Mathematics

1 answer:

brilliants [131]3 years ago

8 0

Answer:

Step-by-step explanation:

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Compute the mean, median, and mode for the following three sets of scores saved.

igor_vitrenko [27]

Answer/Step-by-step explanation:

Score 1:

3, 7, 5, 4, 5, 6, 7, 8, 6, 5

Mean = $\frac{3 + 7 + 5 + 4 + 5 + 6 + 7 + 8 + 6 + 5}{10} = \frac{56}{10} = 5.6$

Median: order the data from least to the greatest.

3, 4, 5, 5, 5, 6, 6, 7, 7, 8

The median is average of the fifth and sixth data value in the data set

Median = (5 + 6)/2 = 11/2 = 5.5

Mode = value with the highest frequency = 5

Score 2:

34, 54, 17, 26, 34, 25, 14, 24, 25, 23

Mean = sum of all values/10

Mean = 276/10 = 27.6

Median: order the data set from min to max.

14, 17, 23, 24, 25, 25, 26, 34, 34, 54

The median is average of the fifth and sixth data value in the data set

Median = (25 + 25)/2 = 50/2 = 25

Mode = 25 and 34 (both have frequencies of 2)

Score 3:

154, 167, 132, 145, 154, 145, 113, 156, 154, 123

Mean = sum of all values/10

Mean = 1443/10 = 144.3

Median: order the data set from min to max.

113, 123, 132, 145, 145, 154, 154, 154, 156, 167

The median is average of the fifth and sixth data value in the data set

Median = (145 + 154)/2 = 299/2 = 149.5

Mode = 154

8 0

3 years ago

True or False? The sum of the differences must be zero for any distribution consisting of n observations.

MakcuM [25]

Its TRUE!!! :) Hope this helps.

3 0

3 years ago

Read 2 more answers

*Find the formula of the series 1+2²+3²+4²+....+n²*<br>

Sophie [7]

The formula is $\frac{n(n+1)(2n+1)}{6}$

What are series?

In mathematics, we can describe a series as adding infinitely many numbers or quantities to a given starting number or amount.

We will find the formula as shown as below:

Let $S=1+2^2+3^2+4^2+................+n^2$

We know $(n+1)^3=n^3+3n^2+3n+1$

$(1+1)^3=1^3+3(1)^2+3(1)+1$

$(2+1)^3=2^3+3(2)^2+3(2)+1$

$(3+1)^3=3^3+3(3)^2+3(3)+1$

$(n+1)^3=n^3+3(n)^2+3(n)+1$

On adding

$2^3+3^3+4^3......(n+1)^3=(1^3+2^3+3^3+.....+n^3)+3(1^2+2^2+.....+n^2)+3(1+2+3....n)+(1+1+1+....+1)$

$2^3+3^3+4^3......(n+1)^3-(1^3+2^3+3^3+.....+n^3)=3S+\frac{3n(n+1)}{2}+(1+1+1+....+1)$

$(n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n$

$n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n$

$2n^3+6n^2+6n=6S+3n^2+3n+2n$

$6S=2n^3+3n^2+n$

$6S=2n^2(n+1)+n(n+1)$

$6S=(n+1)(2n^2+n)$

$6S=n(n+1)(2n+1)$

$S=\frac{n(n+1)(2n+1)}{6}$

Hence, the formula is $\frac{n(n+1)(2n+1)}{6}$

Learn more about Series here:

brainly.com/question/24643676

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3 0

2 years ago

The sum of two numbers is 73. Twice the smaller number is one less than the larger number. The smaller number is

Ilya [14]

Answer:

The big number is 41 and the small number is 32.

Step-by-step explanation:

b + s = 73

2b - s = 50 Add these equations together

3b = 123 Divide both sides by 3

b = 41

Now plug this into the first equation to find s

41 + s = 73

-41 - 41

s = 32

6 0

3 years ago

Need HELP<br> Evaluate the expression <br> 14.3-2x5²÷5

Likurg_2 [28]

I got 4.3. . . . . . . . . .

5 0

3 years ago

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