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To succeed in Disaster Management, you will need a blend of technical and soft skills which are mentioned below:    

• Crisis Management
• Quick Decision-Making
• Team Coordination
• Public Speaking & Awareness Building
• Geographic Information Systems (GIS)
• Data Collection & Report Writing
• Emotional Resilience & Stress Handling
• Basic Health & Safety Knowledge
• Cultural Sensitivity

It's a career that's challenging, fast paced, and deeply rewarding.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhatn=18,{\displaystyle \sum _{}^{}\left(x-5\right)=3,}{\displaystyle \sum _{}^{}{\left(x-5\right)}^2=43}\\ \therefore Mean=A+\frac{{\displaystyle \sum _{}^{}\left(x-5\right)}}{n}=5+\frac{3}{18}=\frac{93}{18}=5.166=5.17\\ andS.D.\left(\sigma \right)=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{\left(x-5\right)}^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}\left(x-5\right)}}{n}\right)}^2}=\sqrt[]{\frac{43}{18}-{\left(\frac{3}{18}\right)}^2}\\ =\sqrt[]{2.39-{\left(0.166\right)}^2}=\sqrt[]{2.39-0.027}=1.54\\ Hence,therequiredmeanis5.17andSD=1.54\end{array}$

To apply for the BTech course at UPES, students should appear for JEE Main or CUET-UG exam. After results announcement, fill the application form on the official website of UPES. The application process includes paying a non-refundable fee of INR 2,000 and sspecialisation selection during the online counselling round.

The ICFAI University Mizoram BCom admission process begins with the application process. The steps to apply are as follows:

1. Go to the official website of the university.

2. Click on the 'Admissions' tab.

3. Fill out the available application form.

4. Upload the scanned documents.

5. Pay the application fee, i.e., INR 300.

6. Click on the submit button to apply.

$1+\left(1-2^2.1\right)+\left(1-4^2.3\right)+\dots \dots .+\left(1-{20}^2.19\right)$

$=\alpha -220\beta$

$=11-\left(2^2\cdot 1+4^2\cdot 3+\dots ..+{20}^2\cdot 19\right)$

$=11-2^2\cdot {\sum }_{r=1}^{10}?r^2(2r-1)=11-4\left(\frac{{110}^2}{2}-35\times 11\right)$

$=11-220\left(103\right)$

$\Rightarrow \alpha =11,\beta =103$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat\overline{x}=50,n=100andSD\left(\sigma \right)=4\\ \overline{x}=\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\Rightarrow 50=\frac{{\displaystyle \sum _{}^{}{x}_i}}{100}\Rightarrow {\displaystyle \sum _{}^{}{x}_i}=5000\\ andvariance{\sigma }^2=\frac{{\displaystyle \sum _{}^{}{f}_i{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{f}_i{x}_i}}{N}\right)}^2\\ {\left(4\right)}^2=\frac{{\displaystyle \sum _{}^{}{f}_i{x}_i^2}}{100}-{\left(50\right)}^2\Rightarrow 16=\frac{{\displaystyle \sum _{}^{}{f}_i{x}_i^2}}{100}-2500\\ \therefore {\displaystyle \sum _{}^{}{f}_i{x}_i^2}=\left(2500+16\right)\times 100\\ \Rightarrow {\displaystyle \sum _{}^{}{f}_i{x}_i^2}=2516\times 100=251600\\ Hence,therequiredsumare5000and251600.\end{array}$

Yes, a candidate can apply for admission at Mangalmay Institute of Management and Technology if class 12 marks are pending. The institute may offer them provisional admission in this case. However, the candidate must submit the marksheet and certificate for their class 12 once the results are declared. Only then the admission will be confirmed.

The Indira Group of Institutes, Pune has had a strong record when it comes to placements provided. Refer to the table below for more placement details:

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{n}_1=60,\overline{x_1}=650and{s}_1=8\\ and{n}_2=80,\overline{x_2}=660and{s}_2=7\\ Weknowthatforthecombinedtwoseriesthat\\ \therefore \sigma =\sqrt[]{\frac{n_1{s}_1^2+n_2{s}_2^2}{n_1+n_2}+\frac{n_1{n}_2{\left(\overline{x_1}-\overline{x_2}\right)}^2}{{\left(n_1+n_2\right)}^2}}=\sqrt[]{\frac{60\times {\left(8\right)}^2+80\times {\left(7\right)}^2}{60+80}+\frac{60\times 80{\left(650-660\right)}^2}{{\left(60+80\right)}^2}}\\ =\sqrt[]{\frac{6\times 64+8\times 49}{14}+\frac{60\times 80\times 100}{140\times 140}}=\sqrt[]{\frac{192+196}{7}+\frac{1200}{49}}\\ =\sqrt[]{\frac{388}{7}+\frac{1200}{49}}=\sqrt[]{\frac{2716+1200}{49}}=\sqrt[]{\frac{3916}{49}}=\frac{62.58}{7}=8.9\\ Hence,therequiredSD=8.9\end{array}$

Purdue University placement rate stands over 90% meaning that within six-months of graduation, students have a high chance of securing employment, especially backed up by the university's efforts such as Fast Track Partners that help their students and alumni attain employment opportunities faster. Purdue placement rate for MSBAIM stands at a perfect rate of 100%, whereas Purdue placement rate for MSGSCM is 82%.

The Karnataka PGCET score refers to the actual marks a candidate secures in the PGCET exam. KEA calculates the Karnataka PGCET marks based on correct and incorrect answers marked by a candidate. On the other hand, Karnataka PGCET rank represents the candidate's relative position among all test-takers, determined using the scores of everyone who appeared. Karnataka PGCET scores indicates individual performance of a candidate, whereas the PGCET rank indicates how a candidate performed in comparison to others and decides eligibility and priority during counselling.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{n}_1=20,\overline{x_1}=17and{\sigma }_1=5\\ and{n}_2=20,\overline{x_2}=22and{\sigma }_2=5\\ Weknowthatforthecombinedtwoseriesthat\\ \therefore \sigma =\sqrt[]{\frac{n_1{s}_1^2+n_2{s}_2^2}{n_1+n_2}+\frac{n_1{n}_2{\left(\overline{x_1}-\overline{x_2}\right)}^2}{{\left(n_1+n_2\right)}^2}}=\sqrt[]{\frac{20\times {\left(5\right)}^2+20\times {\left(5\right)}^2}{20+20}+\frac{20\times 20{\left(17-22\right)}^2}{{\left(20+20\right)}^2}}\\ =\sqrt[]{\frac{1000}{40}+\frac{400\times 25}{1600}}=\sqrt[]{25+\frac{25}{4}}=\sqrt[]{\frac{125}{4}}=\sqrt[]{31.25}=5.59\\ Hence,therequiredSD=5.59\end{array}$

No, qualification alone does not guarantee admission in the courses. The final admission depends on availability of seats, choice of course, counseling process, and document verification.

Candidates are selected for DPharma course at Mangalmay Institute of Management and Technology on the basis of marks in the JEECUP Entrance Exam.  Candidates who qualify State Entrance Examination JEECUP Lucknow are eligible to apply and are admitted through counselling. Candidate must have passed Class 12 examination or its equivalent with Physics and Chemistry as compulsory subjects along with Mathematics/Biology subject (PCB/PCM) for being eligible to take admission in the course. 

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Let{x}_i,i=1,2,3,4,\dots n_1\\ and{y}_j,j=1,2,3,4,\dots n_2\\ \therefore \overline{x_1}=\frac{1}{n_1}{\displaystyle \sum _{i=1}^n{x}_i}and\overline{x_2}=\frac{1}{n_2}{\displaystyle \sum _{i=1}^n{y}_j}\\ \Rightarrow {\sigma }_1^2=\frac{1}{n_1}{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_1}\right)}^2}and{\sigma }_2^2=\frac{1}{n_2}{\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x_2}\right)}^2}\\ Nowmeanofthecombinedseriesisgivenby\\ \overline{x}=\frac{1}{n_1+n_2}\left[{\displaystyle \sum _{i=1}^n{x}_i}+{\displaystyle \sum _{i=1}^n{y}_j}\right]=\frac{n_1\overline{x_1}+n_2\overline{x_2}}{n_1+n_2}\\ Therefore,{\sigma }^{2}ofthecombinedseriesis\\ {\sigma }^2=\frac{1}{n_1+n_2}\left[{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x}\right)}^2}\right]\\ Now,{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}={\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_j}+\overline{x_j}-\overline{x}\right)}^2}\\ ={\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_j}\right)}^2}+n_1{\left(\overline{x_j}-\overline{x}\right)}^2+2\left(\overline{x_j}-\overline{x}\right){\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_j}\right)}^2}\\ But{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x_i}\right)=0}\left[\begin{array}{l}?Thealgebraicsumofthedeviationofvaluesoffirst\\ seriesfromtheirmeaniszero.\end{array}\right]\\ Also{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}=n_1{s}_1^2+n_1{\left(\overline{x_1}-\overline{x}\right)}^2\\ =n_1{s}_1^2+n_1{d}_1^2\\ where{d}_1=\left(\overline{x_1}-\overline{x}\right)\\ Similarly,wehave\\ {\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x}\right)}^2}={\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x_i}+\overline{x_i}-\overline{x}\right)}^2}=n_2{s}_2^2+n_2{d}_2^{2}where{d}_2=\left(\overline{x_2}-\overline{x}\right)\\ NowcombinedStandardDeviation\left(SD\right)\\ \sigma =\sqrt[]{\frac{n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)}{n_1+n_2}}\\ where{d}_1=\overline{x_1}-\overline{x}=\overline{x_1}-\left(\frac{n_1\overline{x_1}+n_2\overline{x_2}}{n_1+n_2}\right)=\frac{n_2\left(\overline{x_1}-\overline{x_2}\right)}{n_1+n_2}\\ d_2=\overline{x_2}-\overline{x}=\overline{x_2}-(\frac{n_1\overline{x_1}+}{}\end{array}$