The total odd number between 1000 to 9999 without repeating any digit can be such that

Only odd digit can be at one’s position = 5 ways

Thousand’s Position = 8 ways

Hundred's position = 8 ways

Ten’s Position = 7 ways

In the word ‘EQUATION’ letters are A,E,I,N,O,Q,T,U

Before the first word in dictionary starting with ‘E’

A will be first in starting of word.

Then,

In the word REASONING, we treat vowels A,E,O,I as a single letter. So now we have 6 letters i.e., R,S,N,N,G,AEIO.

These six words can be arranged in 6! ways.

But N is repeating 2 times, so the number of arrangements is 6!/2! = 360
The total number of vowels can be arrange = 4! 
now the total number of the arrangement of word REASONING with vowel come
together= ( 6!/2!)(4!)= 8640

Pie diagram is a circle chart which is divded into sectors to illustrate the numerical problem

Therefore these are used to categoral data.

The arc length of sector is propotional to the component value.

An event that is certain to happen has a probability of 1. An event that cannot possibly happen has a probability of zero. If there is a chance that an event will happen, then its probability is between 0 and 1.

A leap year has 366 days

A year has 52 weeks hence there will be 52 Mondays

In a leap year there will be 52 Mondays and 2 days will be left

These days can be:

Sunday , Monday

Monday ,Tuesday

Tuesday ,Wednesday

Wednesday, Thursday

Thursday, Friday

Friday,  Saturday

Saturday Sunday

Out of these total 7 outcomes the favourable outcomes are 2

Hence the probability of getting 53 Mondays in a leap year = 2/7

Given:

Six team can fnish race in 6! ways.

A,B,C can finish first, second, third, then only 3! ways.

x, y, z are in GP,

Then y² = xz

Taking log, 2lny = lnx + lnz,

ln(x), ln(y), ln(z) are in A.P

adding xyz, xyx + ln(x), xyz + ln(y), xyz + ln(z) are in A.P

Taking log, 2ln3y=ln3x+ln3z, So, ln(3x), ln(3y), ln(3z) are in AP.

Given that log₁₀2, log₁₀(2^x – 1), log10(2^x + 3) are in AP, then