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Softa [21]
3 years ago
8

At a school, 60% of the sixth-grade students said that pizza is their favorite kind food. If 210 sixth-grade students prefer piz

za, how many sixth-grade students are at the school?
Mathematics
1 answer:
Vika [28.1K]3 years ago
3 0

Answer:

350

Step-by-step explanation:

Let s= total number of sixth-grade students.

It is given that 60% of sixth-grade students prefer pizza which is equivalent to 210 sixth-grade students. With this, the following can be written:

60\%s = 210

0.6s=210 (60\% can be converted into the decimal form 0.6)

s = 210\div 0.6

=350

∴ There is a total of 350 sixth-grade students in the school.

Hope this helps :)

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Answer:

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Step-by-step explanation:

D represents absolute value, and this is what you use to find the answer to this equation.

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I would say that the correct answer is D. because y is always the numerator while x is the denominator for the equation y2 - y1/x2 - x1 which means if there is two y's or two x's on the same line you subtract the second one from the first one. If there is only one y and/or x and the other is 0 on the same line, it stays at y or x without subtracting y2 - y1 or x2 - x1. 

Since b is on the y coordinate and -a is on the x coordinate, you would make it b/a while -a is gonna be a positive since the lines are going up and to the right. Now, since c is the y coordinate and d is the x coordinate, make C the numerator and d the denominator since y is always the numerator and x is the denominator for these parallel line figures on the graph and the equation will be equaled to the fraction to the other fraction for parallel lines. 

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<em>~ ShadowXReaper069</em>
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A gold mine is projected to produce $52,000 during its first year of operation, $50,000 the second year, $48,000 the third year,
marshall27 [118]

Answer:

its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

Step-by-step explanation:

Given that;

gold mine is projected to produce $52,000 during its first year

produce $50,000 the second year

produce $48,000 the third year

the mine is expected to produce for 20yrs; i.e n = 20

annual interest rate = 4% = 0.04%

now let P represent the present worth

we determine the present worth ;

Present worth ⇒ Cashflow(Uniform series present worth) - (2000)(uniform gradient present worth)

⇒Cashflow(P | A,i%,n) - (2000)(P | G,i%,n)

= 52000[ ((1+i)ⁿ - 1) / (i(1+i)ⁿ) ] - (2000)[ {((1+i)ⁿ - 1) / (i²(1+i)ⁿ))} - (n/i(1 + i)ⁿ) ]

= 52000[ ((1+0.04)²⁰ - 1) / (0.04(1+0.04)²⁰) ] - (2000)[ {((1+0.04)20 - 1) / ((0.04)²(1+0.04)²⁰))} - (20/0.04(1 + 0.04)²⁰) ]

= 52000[ ((1.04)²⁰ - 1) / (0.04(1.04)²⁰) ] - (2000)[ {((1.04)²⁰ - 1) / ((0.04)²(1.04)²⁰))} - (20/0.04(1.04)²⁰) ]

= 52000[ ((2.191123 - 1) / (0.04(2.191123) ] - (2000)[ {((2.191123 - 1) / ((0.0016)(2.191123))} - (20/0.04(2.191123) ]

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= 52000(13.59033) - (2000)(111.5647)

= 706695.6 - 223129.4

= 483,566.2 ≈ 483,566

Therefore its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

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