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2 years ago

Please help to solve these 3 questions

Mathematics

1 answer:

mixas84 [53]2 years ago

7 0

Based on the number of goals scored in the football matches, the mode is 4 goals. The median is 3 goals and the mean goals scored is 2.6 goals.

<h3>What are the statistics from the football matches?</h3>

The mode is the number of goals that was scored in the most matches. This number is 4 goals as this was scored in 6 matches.

To find the median, order the numbers:

0, 0, 0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 6

The median is:

= (3 + 3) / 2

= 3

The mean is:

= (0 + 0 + 0 + 0 + 1 + 1 + 2 +2 +2 + 3 + 3 + 3 + 4 + 4+ 4 + 4 +4 + 4 + 5 + 6) / 20

= 2.6 goals

Find out more on mean at brainly.com/question/1136789

#SPJ1

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Find the 75th term of the arithmetic sequence -17, -13, -9....

Sphinxa [80]

Answer:

The 75th term of the arithmetic sequence -17, -13, -9.... is:

$a_{75}=279$

Step-by-step explanation:

Given the sequence

$-17, -13, -9....$

An arithmetic sequence has a constant difference 'd' and is defined by

$a_n=a_1+\left(n-1\right)d$

computing the differences of all the adjacent terms

$-13-\left(-17\right)=4,\:\quad \:-9-\left(-13\right)=4$

The difference between all the adjacent terms is the same and equal to

$d=4$

The first element of the sequence is:

$a_1=-17$

now substitute $d=4$ and $a_1=-17$ in the nth term of the sequence

$a_n=a_1+\left(n-1\right)d$

$a_n=4\left(n-1\right)-17$

$a_n=4n-21$

Now, substitute n = 75 in the $a_n=4n-21$ sequence to determine the 75th sequence

$a_n=4n-21$

$a_{75}=4\left(75\right)-21$

$a_{75}=300-21$

$a_{75}=279$

Therefore, the 75th term of the arithmetic sequence -17, -13, -9.... is:

$a_{75}=279$

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