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nydimaria [60]
3 years ago
14

The half-life of chromium−51 is 27.8 days. How much of a 7.6−oz sample of chromium−51 will remain after 139 days?

Mathematics
1 answer:
In-s [12.5K]3 years ago
6 0
Ending Amount = Beginning Amount / 2^n
where "n" is the number of half-lives
n = 139 / 27.8 = 5 half-lives
Ending Amount = 7.6 oz / 2^5
Ending Amount = 7.6 oz / 32
Ending Amount = .2375 ounces

Source:
http://www.1728.org/halflife.htm


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Bella is buying balloons for a back to school celebration. She sees there are 4
loris [4]

Answer:

The first deal is better because it has more balloons for not too much of a price (individually)

Step-by-step explanation:

15*4= 60

60/18= 3.33

Deal 1= 3.33 for each balloon

10*5= 50

50/16=3.125

3 0
3 years ago
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Order these numbers from least to greatest:<br> |-17|<br> |-18|<br> -18<br> |19|<br> 20
victus00 [196]

Answer:

-18, |-17|, |-18|, |19|, 20

Step-by-step explanation:

the absolute value of -17 is 17, the absolute value of -18 is 18, and the absolute value of 19 is just 19. Absolute values are never negative.

Hope this helps!

5 0
3 years ago
Which of the following correctly uses absolute value to show the distance between -60 and 13? (5 points)
Sauron [17]

Answer:

l-60 – 13| = |-73| = 73 units

Step-by-step explanation:

Required

Distance between -60 and 13

Using absolute values, distance is calculated as:

Distance = |Start - End|

Where

Start = -60

End = 13

So, we have:

Distance = |-60 - 13|

Distance = |-73|

Take absolute value of -73

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4 0
3 years ago
If (NN)^3 = 8518N, what digit does N represent?
sveticcg [70]

Answer:

(C) 4

Step-by-step explanation:

I am going to solve this by trial and error:

(A) 1

Is: (11)^{3} = 85181?

(11)^3 = 1331

So 1 is not the digit of N.

(B) 2

Is: (22)^{3} = 85182?

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(C) 4

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44^3 = 85184

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7 0
3 years ago
If 2160= 2^a3^b5^c the solution set of a,b,c is<br>{4,3,0}<br>{1,0,3}<br>{4,3,1}<br>{2,3,4}​
Readme [11.4K]

Answer:

4,3,1

Step-by-step explanation:

2^a3^b5^c=2160

2^a3^b5^c=2^4 3^3 5^1

a=4,b=3 and c=1

8 0
3 years ago
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