Before the assembly began, only 35 percent of the students were seated. If 91 students were not seated, how many students were there in all?

Answers

Answer 1

Answer: total=100%
seated==35%
100%-35%=65%=not seated

91=not seated=65% of of all
find 100%

91=65% of all
91=0.65 times all
divide both sides by 0.65
140=all

140 students

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The sum of 2 numbers is 76. 1/3 of the first number plus 1/2 of the second number is 34. What are the two numbers? What is the difference between the second number and first number?

Answers

Ok so we have 2 equations here...

x + y = 76
1/3x + 1/2y = 34

First solve for x in the 1st equation...

x + y = 76
x = 76 - y

Now substitute 76-y for x in the 2nd equation...

1/3x + 1/2y = 34
1/3(76-y) + 1/2y = 34
76/3 - 1/3y + 1/2y = 34
-1/3y + 1/2y = 34 - 76/3
-2/6y + 3/6y = 34 - 25 1/3
1/6y = 8 2/3
y = 8 2/3 ÷ 1/6
y = 8 2/3 × 6
y = 52

Now substitute 52 into the 1st equation for y

x + y = 76
x + 52 = 76
x = 76 - 52
x = 24

The 2 numbers are 24 and 52.

The difference between the numbers is
52 - 24 = 28

28 difference and the numbers are 13 and 12 121 21 212 12 55 65 545 46 54 6 5 6 6666666 777.0-0000000

Find two consecutive integers whose sum is 123 please show work

Answers

The required two consecutive integers are 61 and 62, whose sum is 123.

What is the number system?

A number system is defined as a way to represent numbers on the number line using a set of symbols and approaches. These symbols, which are known as digits, are numbered 0 through 9.

To find two consecutive integers whose sum is 123, we can start by letting the smaller integer be "x." The larger integer will be "x+1," since they are consecutive.

We can then set up the equation "x + (x+1) = 123" to represent the sum of the two integers.

Solving for x, we get:

2x + 1 = 123

2x = 122

x = 61

Therefore, the two consecutive integers are 61 and 62.

Learn more about the number system here:

brainly.com/question/21751836

#SPJ2

The equation is as follows:

n + (n + 1) = 123

This can be simplified to:

2n + 1 = 123

Now we just have to solve.

Subtract 1 from both sides.

2n = 122

Divide.

$(2n)/(2) = (122)/(2)$

n = 61

That's the smallest of our first integer. We just have to add the next number. (or 123 - 61)

61 + 62 = 123

Let me know if you find any errors.

What is the median for the numbers 3.60, 3.80, 5.44, 5.52, 5.72, 7.22?I know it's between 5.44 and 5.52

Answers

okay so first of all we need to know the definition of our median first*
median: Given any group of numbers we put the numbers in order from least to greatest then find the number that's right in the middle.
Here ill show you how we do it:)
3.60, 3.80, 5.44, 5.52, 5.72, 7.22?
its already in order for us:)
but when we find the middle number it turns out that we have two middle numbers:/
simple:) all we do is take the two numbers add them together and divide by two
since our middle numbers are 5.44 and 5.52 we are going to use these:)
first add!
5.44 + 5.52 = 10.96
now divide that by 2
10.96 ÷ 2 = 5.48
so our median for this set of numbers is
5.48!
hope this helps;)

The answer is 5.48 because the middle numbers are 5.44 and 5.52 so what you do is find the mean or average of the two numbers by adding them together then dividing them by two, so: (5.44 + 5.52) / 2 = 5.48

a committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions.

Answers

Answer:

$(1)/(990)$

Step-by-step explanation:

The full question:

"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current positions?"

The permutation of choosing 3 members from a group of 11 would be:

P(n,r) = $(n!)/((n-r)!)$