Nth term of sequence 1.5 3 4.5 6 7.5

1. Pick a sport and name two categorical variables and two numerical variables

Varvara68 [4.7K]

Answer:

Football

Categorical variables:

Type of game= Ball game

Governing body= Federation Internacionale de Football Association (FIFA)

Numerical Variables

Number of players in each team= 11

Number of teams involved= 2

Step-by-step explanation:

A categorical variable is data about a subject that tells its category, type or description. It answers the what, when, why, which, and where questions. The game of football is categorized as a ball game and the governing body that supervises it is FIFA. This answers the question of 'Which governing body supervises the game of football?

Numerical variables give details of a subject in terms of figures, or measurements. To answer the question of how many people play the game of football in each team, we state that there are 11 players and the number of teams that can play at the same time is 2.

6 0

2 years ago

Please help me out !!!!

Mkey [24]

Answer:

ok

Step-by-step explanation:

7 0

2 years ago

In arithmetic sequence tn find: S10, if t2=5 and t8=15

makkiz [27]

Given:

$\bold{t_2=5}\\\\\bold{t_8=15}\\\\$

To find:

$\bold{T_n=?}\\\\\bold{S_{10}=?}$

Solution:

$\to \bold{t_2=5}\\\\ \bold{a+d=5} \\\\ \bold{a=5-d} ................(i)\\\\ \to \bold{t_8=15}\\\\ \bold{a+ 7d=15}.................(ii)\\\\$

Putting the equation (i) value into equation (ii):

$\to \bold{5-d+7d=15}\\\\\to \bold{5+6d=15}\\\\\to \bold{6d=15-5}\\\\\to \bold{6d=10}\\\\\to \bold{d=\frac{10}{6}}\\\\\to \bold{d=\frac{5}{3}}\\\\$

Putting the value of (d) into equation (i):

$\to \bold{a=5-\frac{5}{3}}\\\\\to \bold{a=\frac{15-5}{3}}\\\\\to \bold{a=\frac{10}{3}}\\\\$

Calculating the $\bold{t_n :}$

$\to \bold{t_n=a+(n-1)d}\\\\\to \bold{t_n=\frac{10}{3}+(n-1)\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}+\frac{5}{3}n-\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}-\frac{5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{10-5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{5}{3} +\frac{5}{3}n}\\\\ \to \bold{t_n=\frac{5}{3}(1+n)}\\\\$

Formula:

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

Calculating the $\bold{S_{10}} :$

$\to \bold{S_{10} = \frac{10}{2}[2\frac{10}{3} + (10- 1)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + (9)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 9 \times \frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 3\times 5 ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 15 ] }\\\\\to \bold{S_{10} = 5[\frac{20+ 45}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{65}{3} ] }\\\\\to \bold{S_{10} = 5 \times \frac{65}{3} }\\\\\to \bold{S_{10} = \frac{325}{3} }\\\\\to \bold{S_{10} =108.33 }$