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Vlad [161]
3 years ago
14

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

ope B does this in 43 days. What is the approximate difference in the half-lives of the isotopes?
a 3 days
b 10 days
c 13 days
d 33 days

Mathematics
2 answers:
Luden [163]3 years ago
8 0

Answer:

3 days

Step-by-step explanation:

took the test

Molodets [167]3 years ago
4 0
It should be c cause I just took the test
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Solve the following system of equations using the substitution method.
Novay_Z [31]
X=2y
2x+5y=9

substitute the x=2y into the second equation
2(2y)+5y=9
multiple the 2y by 2. 4y+5y=9
combine like terms. 9y=9
divide the 9 out from both sides. y=1
plug the y back into the first equation
x=2(1)
multiply. x=2

your answer is: x=2
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7 0
3 years ago
If X denotes the number of heads in n tosses of a coin, what is the standard deviation of the random variable X? Does this stand
vfiekz [6]

Answer:

\sigma=\sqrt{np(1-p)}=\sqrt{n*0.5(1-0.5)}=\frac{\sqrt{n}}{2}

As we can see the deviation is proportional to the value of n and if n increase then the deviation increases too. So then the deviation would be larger when n gets larger.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

We can define the following random variable X ="Number of heads in n tosses of a coin".

We assume that the coin is fair and then P(H)= 0.5 for any trial so then we can model X with the following distribution:

X \sim Bin (n, p=0.5)

For this distribution the mean and variance are given by:

E(X)=np=n*0.5=\frac{n}{2}

Var(X)= np(1-p) = n*0.5*(1-0.5) = 0.25 n = \frac{n}{4}

And the deviation would be just the square root of the variance and we got:

\sigma=\sqrt{np(1-p)}=\sqrt{n*0.5(1-0.5)}=\frac{\sqrt{n}}{2}

Does this standard deviation get larger or smaller when n gets larger?

As we can see the deviation is proportional to the value of n and if n increase then the deviation increases too. So then the deviation would be larger when n gets larger.

3 0
3 years ago
3<br>If x=<br>)<br>X Х<br>find the value of x-2<br>3<br>2.​
Slav-nsk [51]

Step-by-step explanation:

Answer

Open in answr app

Correct option is

B

±2

x=3+22   x1=3+221×3−223−22

=3−22

(x−21)2=x+x1−2

=3+22+3−22−2=4

So, x−x1=4=±2 Ans 

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Answer:

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