After 6 hours the radioactive compound with mass 470 grams becomes 391.49 grams and the equation will be C is equal to 470(1-0.03)^6.
<h3>What is exponential decay?</h3>
During exponential decay, a quantity falls slowly at first before rapidly decreasing. The exponential decay formula is used to calculate population decline and can also be used to calculate half-life.
We know the decay can be as:

We have:
a = 470 grams
r = 3% = 0.03
t = 6 hours

C = 391.49 grams
Thus, after 6 hours the radioactive compound with mass 470 grams becomes 391.49 grams and the equation will be C is equal to 470(1-0.03)^6.
Learn more about the exponential decay here:
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Answer:

Step-by-step explanation:
This is hard to explain, but basically you use exponent rules to switch it to an exponent
![(\sqrt[4]{x^{2} +4} )^{3}](https://tex.z-dn.net/?f=%28%5Csqrt%5B4%5D%7Bx%5E%7B2%7D%20%2B4%7D%20%29%5E%7B3%7D)
It's easiest if you calculate
first
4³ = 4 x 4 x 4 = 48
Then convert the root to a fraction exponent
![\sqrt[4]{x^{2}} = x^{\frac{2}{4}} = x^{\frac{1}{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B2%7D%7D%20%3D%20x%5E%7B%5Cfrac%7B2%7D%7B4%7D%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)
Combine them using addition, since addition was in the problem

I hope this helps!
Answer:
1.15:)
Step-by-step explanation:
Answer:
35
Step-by-step explanation:
4/14 divided by 2/2 is 2/7 and then 2/7 times 5/5 is 10/35 so its 35
<h2>
Answer:</h2>

<h2>
Step-by-step explanation:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is
</u></em>