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matrenka [14]
3 years ago
12

HELPPPPPPPPPPP!!!!!!!!!!!!!

Mathematics
1 answer:
Molodets [167]3 years ago
3 0
Choice A is false. There are 15% of freshman in her BIO class, but there are 35% of her total students in it.
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For every team competing in a robotics competition, there are 4 students. How many students are on 3 teams?
givi [52]

Answer:

<em>There are 12 students.</em>

Step-by-step explanation:

This is because there are 4 students per team, and there is 3 teams.

So...

3 * 4 = 12 students

8 0
3 years ago
Y=x+1<br> y+x=5<br> Cant figure out what i am suppose to do with this problem
exis [7]

Answer:can you give me more detail

Step-by-step explanation:

8 0
3 years ago
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Find the area of the circle below. Use the formula <br> Use 3.14 for pi.
kirill [66]

Answer:

314

Step-by-step explanation:

Area is pie(r^2

10x10=100

100x3.14 is 314

6 0
2 years ago
Read 2 more answers
Select all that apply. (PLEASE HELP!!)
AlexFokin [52]
A and B are the answers.

If you need proof, just ask in the comments, and I’ll get back to you.

I hope this helps!
5 0
3 years ago
The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in gr
raketka [301]

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass <em>M</em>₀, the mass after 12 hours is half that:

1/2 <em>M</em>₀ = <em>M</em>₀ exp(12<em>k</em>)

for some decay constant <em>k</em>. Solve for this <em>k</em> :

1/2 = exp(12<em>k</em>)

ln(1/2) = 12<em>k</em>

<em>k</em> = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass <em>M</em>₀, the mass <em>M</em> remaining after time <em>t</em> is given by

<em>M</em> = <em>M</em>₀ exp(<em>kt </em>)

So if <em>M</em>₀ = 590 g and <em>t</em> = 36 h, plugging these into the equation with the previously determined value of <em>k</em> gives

<em>M</em> = 590 exp(36<em>k</em>) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74

6 0
3 years ago
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