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2 years ago

15

Cathy did a survey in four bird parks and wrote the following observations: park p number of green birds 18 total number of bird

s 82. Park q 39 green birds 88 total birds. Park r 38 green birds 64 total birds. Park service 22 green birds 50 total birds. Based on the survey, which park had the greatest percentage of green birds.

2 answers:

Fofino [41]2 years ago

6 0

park with greatest percentage of green birds is park "R"

Nutka1998 [239]2 years ago

3 0

Park p :
18/82 = 0.219 = 21.9% green birds

park q :
39/88 = 0.443 = 44.3% green birds

park r :
38/64 = 0.593 = 59.3% green birds

park service :
22/50 = 0.44 = 44%

park with greatest percentage of green birds is park r <==

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tankabanditka [31]

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tan(∠CAE) = $\frac{CE}{AE}$

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2 years ago

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3 years ago

Read 2 more answers

I need help please.

makkiz [27]

Answer:

see explanation

Step-by-step explanation:

Using the rules of radicals/ exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$

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Simplifying each term

7 $\sqrt{x^{3} }$ = 7 $x^{\frac{3}{2} }$

x $\sqrt{9x}$

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= 3 × $x^{(1+\frac{1}{2}) }$

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Subtracting the 2 simplified like terms, that is

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2 years ago

What is the value of y if 13/9 = y/18

Maslowich

Answer:

$y=26$

Step-by-step explanation:

1. Swap sides

$\frac{13}{9}=\frac{y}{18}$

Swap sides:

$\frac{y}{18}=\frac{13}{9}$

2. Isolate the y

$\frac{y}{18}=\frac{13}{9}$

Multiply to both sides by 18:

$\frac{y}{18}\cdot 18=\frac{13}{9}\cdot 18$

Group like terms:

$\frac{1}{18}\cdot 18y=\frac{13}{9}\cdot 18$

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Multiply the fractions:

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Why learn this:

Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends?
Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives.

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Terms and topics

Linear equations with one unknown

The main application of linear equations is solving problems in which an unknown variable, usually (but not always) x, is dependent on a known constant.

We solve linear equations by isolating the unknown variable on one side of the equation and simplifying the rest of the equation. When simplifying, anything that is done to one side of the equation must also be done to the other.

An equation of:

$ax+b=0$

in which $a$ and $b$ are the constants and $x$ is the unknown variable, is a typical linear equation with one unknown. To solve for $x$ in this example, we would first isolate it by subtracting $b$ from both sides of the equation. We would then divide both sides of the equation by $a$ resulting in an answer of:

$x = -b/a$

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