Class 12th

The following bar chart gives the increase in internet user percentage of villages, towns and cities in the four states for the period between 2018-19 and 2020-21. The table provides information about the number of households who uses internet, average expenditure by a household and growth on expenditure on internet by a household.

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

The average income for the Tamil Nadu in 2020-21 was

0 Follower 2 Views

Contributor-Level 10

Average expenditure on internet for the Tamil Nadu

= $\frac{(400 * 30 + 100 * 50 + 50 * 75)}{400 + 100 + 50}$ = 37727.

0 0

New answer posted

5 months ago

Class 12th Maths

If the value of the integral $\int_{0}^{5} \frac{x + [x]}{e^{x - [x]}} d x = α e^{- 1} + β,$ where $α, β ε R, 5 α + 6 β = 0,$ and [x] denotes the greatest integer less than or equal to x; then the value of ${(α + β)}^{2}$ is equal to

0 Follower 11 Views

Contributor-Level 10

$\int_{0}^{5} \frac{x + [x]}{e^{x - [x]}} d x$

= \sum_{r = 1}^{5} \int_{r - 1}^{r} \frac{(x + 1 - r)}{e^{x + 1 - r}} d x

Put x + 1 – r = t ->dx = dt

$= 5 {[- t e^{- t} - e^{- t}]}_{0}^{1}$

$= 5 [- 1 e^{- 1} - e^{- 1} + e^{0}] = 5 (1 - \frac{2}{e})$

$\Rightarrow α = - 10 β = 5 \Rightarrow {(α + β)}^{2} = 25$

0 0

New answer posted

5 months ago

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

The total expenditure on the internet in the cities than the town increased by how much percentage in year 2018-19?

0 Follower 3 Views

Contributor-Level 10

Total expenditure on internet by cities users = 7500 * 50000 = 375000000

Total expenditure on internet by town users = 5000 * 100000 = 50000000

= $\frac{125000000}{500000000} * 100$ = 25%

0 0

New answer posted

5 months ago

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

What is the percentage increase in total numbers of households that uses internet in towns of Tamil Nadu's over the given periods ?

0 Follower 3 Views

Contributor-Level 10

The percentage increase in total number of households that uses internet in Tamil Nadu is 200%.

0 0

New answer posted

5 months ago

Directions: Answer the questions based on the following Information.

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

What was the total households expenditure on internet in towns of Tamil Nadu's town who uses internet?

0 Follower 1 View

Class 12th Inverse Trigonometric Functions

Contributor-Level 10

Statewise breakup is not available. Hence, question cannot be answered.

0 0

New answer posted

5 months ago

The domain of the function cosec^-1 $(\frac{1 + x}{x}) i s :$

0 Follower 2 Views

Contributor-Level 10

$\Rightarrow \frac{1 + x}{x} \geq 1 a n d \frac{1 + x}{x} \leq - 1$

$\Rightarrow \frac{1}{x} \geq 0 \frac{1 + 2 x}{x} \leq 0$

Domain $[- \frac{1}{2}, 0) \cup (0, \infty)$

$= [- \frac{1}{2}, \infty) - {0}$

0 0

New answer posted

5 months ago

Class 12th Maths

Let [t] denote the greatest integer less than or equal to t.

Let f(x) = $x - [x], g (x) = 1 - x + [x], a n d h (x) = m i n {f (x), g (x)}, x \in [- 2, 2]$

Then h is:

0 Follower 5 Views

Contributor-Level 10

$f (x) = {x}, g (x) = 1 ? {x}$

0 0

New answer posted

5 months ago

The first table gives the percentage of employees in skyroot, who work in the department of finance marketing and software. The second table gives the average starting salaries of the employees per month (Rs. in thousands) in these. The line graph gives the no. of employees who are working for Skyroot each year.

First Table
	Finance	Marketing	Software	Others
2016	22	36	19	23
2017	17	48	23	12
2018	23	42	21	13
2019	19	37	16	28
2020	32	32	20	16

	Second Table
		Finance	Marketing	Software
	2016	54500	51700	52900
	2017	63800	63900	64400
	2018	75500	76300	70500
	2019	89200	89600	77600
	2020	92100	102200	86400

In 2018 employees working in finance department earned more than those working for software (per annum)?

0 Follower 2 Views

Contributor-Level 10

Earning of employee in finance = $\frac{23}{100} * 550 * 75500$

= 9550750

Earning of employee working in software department = $\frac{21}{100} * 550 * 75500 = 8720250$

Difference in the amount earned = 9550750 – 8720250 = 830500

0 0

New answer posted

5 months ago

Class 12th Maths

The local maximum value of the function

$f (x) = {(\frac{2}{x})}^{x^{2}}, x > 0,$ is:

0 Follower 3 Views

Contributor-Level 10

$f (x) = e^{x^{2} l n (\frac{2}{x})} = e^{x^{2} (l n 2 - l n x)}$

$f' (x) = e^{x^{2} (l n 2 - l n x)} [2 x (l n 2 - l n x) - x]$

$= {(\frac{2}{x})}^{x^{2}} x [2 (l n 2 - l n x) - 1]$

$f' (x) = 0 \Rightarrow 2 (l n 2 - l n x) - 1 = 0$

2ln x = $l n \frac{4}{e} \Rightarrow x^{2} = \frac{4}{e}$

0 0

New answer posted

5 months ago

First Table
	Finance	Marketing	Software	Others
2016	22	36	19	23
2017	17	48	23	12
2018	23	42	21	13
2019	19	37	16	28
2020	32	32	20	16

	Second Table
		Finance	Marketing	Software
	2016	54500	51700	52900
	2017	63800	63900	64400
	2018	75500	76300	70500
	2019	89200	89600	77600
	2020	92100	102200	86400

What is the percentage increase in the average salary of finance employee from 2016 to 2020 ?

0 Follower 6 Views