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3 years ago

I have a math problem.

Mathematics

2 answers:

Mademuasel [1]3 years ago

7 0

•ω• Hewo fren!

☆☆●◉✿Answer:✿◉●☆☆

20740

☆☆●◉✿Step-by-step explanation:✿◉●☆☆

Since the oranges cost 2.50 cents each then the ratio would be 2.50 cents to 1 orange, 2.5:1. James is buying 8296 oranges so we would have to multiply the ratio by 8296 since the oranges increased. 1 × 8296 = 8296.

2.5 × 8296 = 20740. James paid 20740 dollars for 8296 oranges. (20740:8296 would be the ratio.)

HOPE I HELPED! ∧∧

→⇒brainliest please? ∑(OΔO )♥♥︎

{WHY SO MANY ORANGES ono}

Naddik [55]3 years ago

3 0

Answer:

Step-by-step explanation:

20740

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Solve for z w=x+y/z please help

Nadusha1986 [10]

Answer:

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8 0

3 years ago

Use the model below to estimate the average annual growth rate of a certain country's population for 1950, 1988, and 2010, where

Morgarella [4.7K]

Answer:

The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.

Step-by-step explanation:

The given equation is

$Y=-0.0000084x^3+0.00211x^2-0.205x+8.423$

Where Y is the annual growth rate of a certain country's population and x is the number of years after 1900.

Difference between 1950 and 1900 is 50.

Put x=50 in the given equation.

$Y=-0.0000084(50)^3+0.00211(50)^2-0.205(50)+8.423$

$Y=2.398$

Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.

Difference between 1988 and 1900 is 88.

Put x=88 in the given equation.

$Y=-0.0000084(88)^3+0.00211(88)^2-0.205(88)+8.423$

$Y=0.9984752\approx 0.9985$

Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.

Difference between 2010 and 1900 is 110.

Put x=110 in the given equation.

$Y=-0.0000084(110)^3+0.00211(110)^2-0.205(110)+8.423$

$Y=0.2236$

Therefore the estimated average annual growth rate of the country's population for 2010 is 0.2236.

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3 years ago

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I think it's 9 but I could be wrong

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hoa [83]

$n^{-x}=\frac1{n^x}$

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3 years ago