All categories

3 years ago

2. The average age of 25 students is 10years

Mathematics

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

12.7 years to the nearest tenth.

Step-by-step explanation:

The total of the ages for the 25 students = 10*25 = 250.

For the 30 students it is 15*30 = 450.

The average for the 2 classes = (250 + 450) / (25 + 30)

= 700 / 55

= 12.7 years to the nearest tenth.

You might be interested in

Jasmine has taken an online boating safety course and is now completing her end-of-course exam. As she answers each question, th

dsp73

Answer:

There are 60 total questions on the course exam.

Step-by-step explanation:

Given that:

Number of questions finished by Jasmine = 12

Percentage shown on the bar = 20%

The total number of questions will be 100%, therefore, 20% of the questions are equal to 12.

Let,

x be the total number of questions on the course exam

20% of x = 12

$\frac{20}{100}x=12\\0.2x=12$

Dividing both sides by 0.2

$\frac{0.2x}{0.2}=\frac{12}{0.2}\\x=60$

Hence,

There are 60 total questions on the course exam.

8 0

2 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers

9 times the quantity of 12 less than a number x is 24. Write in an algebraic expression.

kiruha [24]

Answer:

9(x-12)=24

Step-by-step explanation:

5 0

2 years ago

What do I have to do

hichkok12 [17]

Answer:

+15

-721

+462

-1200

The positives are because it said 'above sea level', the negatives because it said 'below sea level' and zero because it said 'sea level'.

3 0

3 years ago

WILL MARK BRAINLIEST!!!!!!! REALLY NEED HELP! The graph of the following compound inequality shows all the numbers between 4 and

DIA [1.3K]

True, because it is x is greater than OR EQUAL TO 4 so that means 4 is included. Then x is less than 8 so it’s not including 8. x could be 4,5,6, or 7

4 0

3 years ago