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vampirchik [111]
3 years ago
11

18x (14 - 9) ÷[12 - (19 - 13)] = what is the answer​

Mathematics
1 answer:
marysya [2.9K]3 years ago
5 0

Answer:

15

Step-by-step explanation:

18 . ( 14 - 9 ) ÷ [ 12 - ( 19 - 13 ) ]

18 . ( 5 ) ÷ [ 12 - 6 ]

90 ÷ 6

15

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27% of the students in a school are in grade 6. This is 540 students. How many students are in the school?
Anettt [7]

There are a total of 2,000 students in the school.

1.) There are 540 students in the 6th grade. 540 is only 27% of students compared to the total. To find the total students in the whole school, we divide 540 by 27%.

2.) We first change 27% into a decimal by moving the point two times to the left. (Rule to change the percent into a decimal)

27% = .27

3.) Before we divide, we remove the decimal by moving the decimal two times to the right and so do we to 540.

.27 = 27

540. = 54000.

4.) Now we have 54000 divided by 27 which give us the answer of 2,000.

Hope this Helps!!!

Please Mark as Brainliest!!!

6 0
3 years ago
Please help me. I need help ASAP. ​
lara [203]

Answer:

a) 72

b) $259.2

Step-by-step explanation:

A- The bus travels 40 miles on 8 gallons of gasoline. The bus is traveling 360 miles in total.

360/40= 9

9*8= 72

B- $3.60 per gallon and 72 gallons in total.

3.60*72=259.2

$259.2

5 0
3 years ago
Convert into 9 kilogram into metre​....
IrinaVladis [17]

Answer:

0.009000000000000001

5 0
3 years ago
Read 2 more answers
A town has a population of 32,000 in the year 2002; 35,200 in the year 2003; 38,720 in the year 2004; and 42,592 in the year 200
Romashka-Z-Leto [24]
The answer is 102,442
3 0
3 years ago
Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

3 0
3 years ago
Read 2 more answers
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