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Vlad [161]
3 years ago
13

What is the equation for the average?

Mathematics
1 answer:
Darina [25.2K]3 years ago
8 0

Answer:

D. Average = (sum of all data values) / (total number of data values)

Step-by-step explanation:

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In a certain​ chemical, the ratio of zinc to copper is 3 to 17. A jar of the chemical contains 799 grams of copper. How many gra
dezoksy [38]

Answer:

let's see what to do buddy...

Step-by-step explanation:

\frac{zinc}{copper} =  \frac{3}{17} \\  \\  \frac{zinc}{799} =  \frac{3}{17}

Multiply the sides of the equation by 799

zinc =  \frac{3}{17} \times 799 \\ \\ zinc =  \frac{3}{17} \times 17 \times 47 \\ \\  zinc = 3 \times 47 = 141

So A jar of the chemical contains 141 grams of <em><u>zinc</u></em>.

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

8 0
3 years ago
Please help with these! or at least some of them.​
Fiesta28 [93]

Answer:

<h2>1) 9/2 = 5/4</h2><h2>5) False</h2><h2>6) $66</h2><h2>7) x=7</h2><h2>8) x=69</h2>

Step-by-step explanation:

5 0
3 years ago
Just fill out the explanation on why it's 6?
g100num [7]

Answer: I subtracted 8 from both sides then divided 3 on both sides

Step-by-step explanation: welcome

8 0
3 years ago
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There are currently 17 frogs in a (large) pond. The frog population grows exponentially, tripling every 6 days. How long will it
Tema [17]

Answer:

t=11,11 days

Step-by-step explanation:

F=frogs poblation, t=time, be the variables dF/dt = KF, dF/F=Kdt, integrating \int\limits^ {} \, dF/F =K\int\limits^ {} \, dt⇒ LnF=Kt+c,F=ce^{Kt}; Knowing t=0, F=17 and t=6 F=51 (tripling every 6 days (17*3)), F=ce^{K0} = F=c=17; F=17e^{6K} =51⇒e^{6K} =51/17; K6=ln\frac{51}{17} ; K=ln\frac{51}{17}/6=0.183, so F=17e^{0,183t}, now if F=130, t=? we have:

130=17e^{0.183t} =e^{0.183t} =130/17; 0.183t=ln(130/17); t=ln(130/17)/0.183 = 11,11

8 0
3 years ago
Ross drives 280 miles in 4 hours. If Ross continues to drive at the same rate, how many miles can he drive in 12 hours?​
const2013 [10]

Answer:

Step-by-step explanation:3,360 is your answer

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3 years ago
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