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2 years ago

Which geometric series represents 0.4444... as a fraction?

Mathematics

2 answers:

Alenkasestr [34]2 years ago

8 0

The answer, for future reference, is C. 4/10 + 4/100 + 4/1,000 + 4/10,000

Aliun [14]2 years ago

6 0

Here the first term of this series is 0.4. The next is 0.1(0.4), or 0.04. The next is (0.1)(0.1)(0.4), or 0.004. Thus, the common ratio, r, is 0.1.

The sum of this series is a/(1-r), where a is the first term and r is the common ratio.

Subst. the known values of a and r, the sum of this series is 0.4/(1-0.1), or

0.4 4
------ = ---
0.9 9

Divide 4 by 9 on your calculator. You'll find that the result is the repeating fraction 0.44444....

Thus, 4/9 is equivalent to 0.4444.....

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Collete mapped her vegetable garden on the graph below. Each unit represents 1 foot.

ipn [44]

Answer:HMMMMM I don’t really know

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

11) The Cost of maintaing a

dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

Equation will be,

15705 = 50a + b -------(1)

Cost of 40 pupils = $13305

Equation will be,

13305 = 40a + b --------(2)

By subtracting equation (2) from equation (1),

15705 - 13305 = (50a + b) - (40a + b)

2400 = 10a

a = 240

From equation (1),

b = 3705

Equation representing the total cost will be,

C = 240x + 3705

If x = 44

C = 240(44) + 3705

C = $14265

b). If the fee per pupil 'a' = $360

Let the number of pupils = p

Total fee of 'p' pupils = $360p

Total cost to run the school will be = 3705 + 240p

For the school not to be in the loss,

360p ≥ 3705 + 240p

360p - 240p ≥ 3705

120p ≥ 3705

p ≥ $\frac{3705}{120}$

p ≥ 30.875

Therefore, to run the school without loss, number pupils should be at least 31.

3 0

2 years ago

a(0.3−y)+1.1+2.4x (y−1.2) =0 =−1.2(x−0.5) Consider the system of equations above, where aaa is a constant. For which value o

Veseljchak [2.6K]

Answer:

For a = 1.22 there is one solution where y = 1.3

Step-by-step explanation:

Hi there!

Let´s write the system of equations:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

-1.2(x-0.5) = 0

Let´s solve the second equation for x:

-1.2(x-0.5) = 0

x- 0.5 = 0

x = 0.5

Now let´s repalce x = 0.5 and y = 1.3 in the first equation and solve it for a:

a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0

a(0.3 - 1.3) + 1.1 + 2.4(0.5)(1.3 -1.2) = 0

a(-1) + 1.1 + 1.2(0.1) = 0

-a + 1.22 = 0

-a = -1.22

a = 1.22

Let´s check the solution and solve the system of equations with a = 1.22. Let´s solve the first equation for y:

1.22(0.3 - y) + 1.1 +2.4(0.5)(y-1.2) = 0

0.366 - 1.22y + 1.1 + 1.2 y - 1.44 = 0

-0.02y +0.026 = 0

-0.02y = -0.026

y = -0.026 / -0.02

y = 1.3

Then, the answer is correct.

Have a nice day!

4 0

2 years ago

HELP PLEASE I NEEED HELP!!!!

Nina [5.8K]

It is d just trust me

4 0

2 years ago

a person on a tour has rupees 12000 for his daily expenses. In order to extend his journey for 2 more days he had to cut down hi

myrzilka [38]

Answer:

Duration of the tour he planned first is 8 days.

Step-by-step explanation:

Given that a person has 12000 rupees for his daily expenses.

Let x be the number of days.

Then daily expenses per day $=\frac{12000}{x}$

Given that number of day is increased by 2 more days, that is number of days is x+2.

New daily expense per day $\frac{12000}{x+2}$

Given that this new daily expenses are 300 less than original.

That is $\frac{12000}{x}-\frac{12000}{x+2}=300$

$\frac{12000(x+2)-12000x}{x(x+2)}=300$

$\frac{24000}{x(x+2)}=300$

$x(x+2)=\frac{24000}{300}$

$x^{2}+2x-80=0$

(x-8)(x+10)=0

x=8 or -10.

Since number of days cannot be negative, duration of the tour planned=8.

7 0

2 years ago