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hichkok12 [17]
3 years ago
7

Sorry I don't know what to post this in but I will give BRAINLIEST Max kept 100 grams of radioactive iodine in a container. He o

bserved the amount of iodine left in the container after regular intervals of time and recorded them in the table shown below. Time (days) Amount of iodine in container (in grams) 0 100 8 50 16 25 24 12.5 Based on the observations, which of these is most likely Max's inference?
The half life period of radioactive iodine is 32 days. The half life period of radioactive iodine is 50 days. After 32 days, the amount of iodine left in the container will be 1.25 gram. After 32 days, the amount of iodine left in the container will be 6.25 gram.
Mathematics
1 answer:
jekas [21]3 years ago
8 0

Answer:

the amount of iodine left in the container will be 6.25 gram.

Step-by-step explanation:

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I will give BRAINLIEST to whoever is correct.
maw [93]
Remember, you can do anything to an equation as long as you do it to both sides

we hates fractions
times both sides by 3
(we could have added 1/3b to both sides but I'm lazy)


4b+6=12-b
add b to both sides
5b+6=12
minus 6 both sides
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divide both sides by 5
b=\frac{6}{5}
8 0
4 years ago
12 students share 3 pizzas equally what fraction of the pizza each student gets
alekssr [168]
To find the solution, simply take 3 and divide it by 12. 3/12 as a fraction would be simplified down to 1/4 or 0.25. Therefore, each student would receive one quarter of a pizza.
8 0
3 years ago
Suppose you are managing 16 employees, and you need to form three teams to work on different projects. Assume that all employees
valentina_108 [34]

Answer:

4380 ways

Step-by-step explanation:

We have to form 3 project of 16 employees, they tell us that the first project must have 5 employees, therefore we must find the number of combinations to choose 5 of 16 (16C5)

We have nCr = n! / (R! * (N-r)!)

replacing we have:

1st project:

16C5 = 16! / (5! * (16-5)!) = 4368 combinations

Now in the second project we must choose 1 employee, but not 16 but 11 available, therefore it would be to find the number of combinations to choose 1 of 11 (11C1)

2nd project:

11C1 = 11! / (1! * (11-1)!) = 11 combinations

For the third project we must choose 10 employees, but since we only have 10 available, we can only do a combination of this, since 10C10 = 1, therefore:

3rd project: 1 combination

The total number of combinations fro selecting 16 employees for each project would be:

4368 + 11 + 1 = 4380 combinations, that is, there are 4380 different ways of forming projects with the given conditions.

3 0
3 years ago
Six times the difference of a number and 4 is no more than 10.
klio [65]

Answer:

nnnnnnnnnnnnnnnnnnnnnnnnn

Step-by-step explanation:

7 0
3 years ago
What is 0.04 1/10 of
Marat540 [252]
0.04 is 1/10 of 0.4

0.4 x 0.10 = 0.04

hope this helps
3 0
4 years ago
Read 2 more answers
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