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kobusy [5.1K]
3 years ago
13

Please help

Mathematics
1 answer:
Alex17521 [72]3 years ago
8 0
B is the answer llllllllllllllllll
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Agri-small business limited expenses on petrol for their fleet is r 4 850.00 at the end of september and they have a balance of
kotykmax [81]

The total income and expenditure of Agri-small business limited in September is R 299596 and expenditure is R 193236.

Given that the expenses on petrol be R4850.00 at the end of September, balance is R 106360.00, usage on income of salaries be 25%, on electricity,rates and taxes be 11%, 42% of the remaining on insurance and investments.

We are required to find the total income and expenditure in September.

Generally the equation for insurance is mathematically given as :

Insurance=42% of the balance

Insurance=0.42*0.64A

Insurance=0.2688A.

Generally the equation for the total after all the other expenditures is mathematically given as:

Total expenditure after all other expenditures=0.64A-0.2688A

Total expenditures after all other expenditures=0.3712A.

Total expenditures after all other expenditures is as 111210 (106360+4850)

Equating both the equation and value:

111210=3712A

A=299596

The total income is 299596.

Therefore the calculation of expense is given as:

Expense=0.25A+0.11A+0.2688A+4850

Expense=0.6288A+4850

=0.6288*299596+4850

=193236

The expense is 193236.

Hence the total income and expenditure in September are "Income is R 299596 and expenditure is R 193236.

Learn more about income at brainly.com/question/25745683

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Is the question even complete ?

Step-by-step explanation:

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Jackson earned 22 out of 25 points on a class project. Determine the percentage grade that he earned by setting up an equivalent
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it would be a 85 percent

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A relation is what in math
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A relation is a relationship between sets of values. In math, the relation is between the x-values and y-values of ordered pairs. The set of all x-values is called the domain, and the set of all y-values is called the range.
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5/14 in in simple form
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Answer:

Step-by-step explanation:

1. Find the GCD ( or HCF) of numerator and denominator

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2. Divide both the numerator and denominator by the GCD

5 divide 1 and 14 divide 1

3 reduced fraction

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