The best and most correct answer providedfrom your question about the entrance examination paper is the second option which is 2,250. The problem can be solved by:
3-1-1 : 5c3*5c1*5c1 *3!/2! = 750
<span>2-2-1 : 5c2*5c2*5c1 *3!/2! = 1500
</span>
Adding the two answers:
750 + 1500 = 2250
I hope it has come to your help.
The weight of the new student is 27 kg.
Average weight
= total weight ÷total number of students
<h3>
1) Define variables</h3>
Let the total weight of the 35 students be y kg and the weight of the new student be x kg.
<h3>2) Find the total weight of the 35 students</h3>
<u>
</u>
y= 35(45)
y= 1575 kg
<h3>3) Write an expression for average weight of students after the addition of the new student</h3>
New total number of students
= 35 +1
= 36
Total weight
= total weight of 35 students +weight of new students
= y +x
<h3>4) Substitute the value of y</h3>
<h3>5) Solve for x</h3>
36(44.5)= 1575 +x
1602= x +1575
<em>Subtract 1575 from both sides:</em>
x= 1602 -1575
x= 27
Thus, the weight of the new student is 27 kg.
Answer:
AREA =18
Step-by-step explanation
think of a rectangle, 12 by 3 which is also 12*3
so you do 12*3=36
but since it is a triangle you have to divide it by 2
36/2=18
18
They are all averages
mean=average 4+6=10, 10 divded by 2= 5, the mean is 5
median= middle number, 5 6 7, 6 is the median
mode= most repeated number, so 4,4,4,4,4,4,4,5.5,6 4 is the mode
2 tables. Table 1 is titled Population data with entries 2, 4, 8, 5, 1, 1, 10, 5, 4, 2, 3, 3. Table 2 is titled Sample data with
alekssr [168]
Answer: //The mean of the population is 4
//The mean of the population is 3.25
//The difference between the mean of the sample and the mean of the population is 0.75
Step-by-step explanation: EGDEUNITY
