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3 years ago

How do i write, Two increased by three equals the quotient of ten and two. Use x to represent any unknown number?

Mathematics

1 answer:

inn [45]3 years ago

7 0

Answer:

$2+3=\frac{10}{2}$

Step-by-step explanation:

we know that

The phrase " Two increased by three equals the quotient of ten and two" is equal to the algebraic expression

$2+3=\frac{10}{2}$

$5=5$ -----> is true

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Suppose we select, without looking, one marbles from a bag containing 4 red marbles and 10 green marbles.what is the probability

OlgaM077 [116]

Since 10 out of 14 marbles are green, the probability of selecting a green marble is 10/14=0.714.

Similarly, the probability of selecting a red marble would be 4/14.

Answer: 0.714

5 0

3 years ago

What is the difference between the mean and the median of the data set?<br> (22, 8, 10, 18, 12, 20)

IgorC [24]

Let D = difference between mean and median of data.

Mean = (22 + 8 + 10 + 18 + 12 + 20)/6

Mean = 90/6

Mean = 15

Let m = median

m = 8, 10, 12, 18, 20, 22

m = (12 + 18)/2

m = 30/2

m = 15

D = M - m

D = 15n- 15

D = 0

7 0

3 years ago

1 If the population of a town increases by 280 people per year, then the growth is LINEAR (or constant). If the population in 20

Stella [2.4K]

The population in the year 2023 is 120,780.

The growth is linear hence, the equation of the growth is
y=ax+b where b represents the initial population and a is the number of people growing yearly.

So, according to the question statement, we have that

y=280x+b where x is the number of years past and y is the population in the xth year.

According to the given problem population in 2022 will be 120,500.

Suppose that the initial population is taken at 2000.

Hence, 120,500=280×(2022-2000)+b

⇒b=120,500-280×22

⇒b=120,500-6,160

=114,340

So, the Linear equation becomes

y=280x+114,340

⇒y=280×(2023-2000)+114,340

⇒y=120,780

Hence, the population in the year 2023 is 120,780.

Learn more about linear equations here-

brainly.com/question/11897796

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7 0

2 years ago

What is the value of the expression?

user100 [1]

3a= 3(5) = 15
15 - - 3 = 15+3 = 18
18/6 = 3
b + a = 5 + -3 = -3 + 5 = 2
3 x 2 = 6

4 0

3 years ago

4. Parking fees at IIUM are RM 5.00 for IIUM students and RM 7.50 for non-IIUM students. At the

professor190 [17]

Answer:

(a) 780 students and 960 non-students

(b) No. The maximum revenue is RM9000 from 1200 non-students.

(c). Revenue is maximum of RM9000 at 1200 non-students, decreasing by RM2.50 per student to a minimum of RM6000 at 1200 students

Step-by-step explanation:

Let x = IIUM students and

and y = non-IIUM students

You have two conditions

(a) x + y = total vehicles parked

(b) 5.00x + 7.50y = total gross receipts

(a) Wednesday

From your table,

x + y = 1740

5.00x + 7.70y = RM11 100

Solve the simultaneous equations

$\begin{array}{rrcrl}(1) & x + y & = &1740&\\(2) & 5.00x + 7.50y & = & 11 100\\(3)& 5.00x + 5.00y & = & 8700 & \text{Multiplied (1) by 5}\\&2.50 y & = &2400 &\text{Subtracted (3) from (2)}\\(4)&y& = &\mathbf{960} &\text{Divided each side by 2.50}\\& x +960& = &1740& \text{Substituted (4) into (1)}\\& x& = &\mathbf{780}& \\\end{array}\\\text{There are $\large \boxed{\textbf{780 students and 960 non-students}}$}$

(b) Can 1200 vehicles bring in RM10000?

No. Even if all the cars were from non-students, the most you could get is

1200 × 7.50 = RM9000

(c) Possible combinations for 1200 vehicles

Revenue = 5.00x + 7.50y = 5.00x + 7.50(1200 -x) = 5.00x + 9000 - 7.50x =

Revenue = 9000 - 2.50x

The maximum revenue of RM9000 occurs when there are no student cars and 1200 non-student cars.

For each student car that enters and displaces a non-student, the revenue drops by RM2.50.

Finally. when there are 1200 student cars and no non-students, the revenue has dropped to a minimum of RM6000.

3 0

4 years ago