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2 years ago

Helium decays to form lithium. 6 2 he → 6 3 li 0 -1 e this type of nuclear decay is called .

1 answer:

8 0

To understand this question, we need some knowledge on radioactive decay. This type of nuclear decay is called a beta decay

<h3>Beta Decay</h3>

This is the process where a radioactive material is fired or naturally emits a beta particle which is equivalent to the lose of an electron and spontaneously becomes an entirely different element.

$^6_2He \to ^6_3 Li + ^0_-_1e$

The participating actors in this nuclear reaction are

helium
lithium
electron

This is the decay process of helium particle in which it became a lithium particle and emitted an electron in the process.

Learn more on beta decay here;

brainly.com/question/9615302

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You might be interested in

Evaluate 4-0.25g+0.5h when g=10 and h=5

skad [1K]

Answer:

$4$

Step-by-step explanation:

$4 - 0.25g + 0.5h = 4 - 0.25(10) + 0.5(5) = 4 - 2.5 + 2.5 = 1.5 + 2.5 = 4$

8 0

4 years ago

Read 2 more answers

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.08

kvv77 [185]

Answer:

a) 44.93% probability that there are no surface flaws in an auto's interior

b) 0.03% probability that none of the 10 cars has any surface flaws

c) 0.44% probability that at most 1 car has any surface flaws

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the binomial probability distributions.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So $\mu = 10*0.08 = 0.8$

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a $p = 0.4493$ probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003$

0.03% probability that none of the 10 cars has any surface flaws

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

At least 9 cars without surface flaws. So

$P(X \geq 9) = P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041$

$P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044$

0.44% probability that at most 1 car has any surface flaws

5 0

3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

4 years ago

In order to inscribe a circle in a triangle, which line must you construct?

Neporo4naja [7]

A single line?? well i believe if you draw the 3 angle bisectors for the 3 angles of the triangle you can then inscribe your circle. But I am not sure if you mean only a single line and thats it.

4 0

3 years ago

Read 2 more answers

a company starts out with 800 units if a particular item in inventory.Assume that their cost is $10 ach. Over the next three mon

Nikitich [7]

Last in First out - The last item/inventory purchased are sold first. Hence your answer is a.$10.

5 0

3 years ago