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mote1985 [20]
3 years ago
10

A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150

​days? after 300 ​days?
Mathematics
1 answer:
shtirl [24]3 years ago
6 0

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

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Lorico [155]
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3 years ago
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A lottery game has balls numbered 1 through 21. What is the probability of selecting an even numbered ball or an 8? Round to nea
Viktor [21]

Answer: 0.476

Step-by-step explanation:

Let A = Event of choosing an even number ball.

B = Event of choosing an 8 .

Given, A lottery game has balls numbered 1 through 21.

Sample space: S= {1,2,3,4,5,6,7,8,...., 21}

n(S) = 21

Then, A= {2,4,6,8, 10,...(20)}

i.e. n(A)= 10

B= {8}

n(B) = 1

A∪B = {2,4,6,8, 10,...(20)} = A

n(A∪B)=10

Now, the probability of selecting an even numbered ball or an 8 is

P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}

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Hence, the required probability =0.476

7 0
3 years ago
Find equations of the spheres with center(1, −1, 6)that touch the following planes.a) xy-planeb) yz-planec) xz-plane
Misha Larkins [42]

Given :

Center of sphere , C( 1 , -1 , 6 ) .

To Find :

Find equations of the spheres with center (1, −1, 6) that touch the following planes.a) xy-plane b) yz-plane c) xz-plane .

Solution :

a)

Distance of the point from xy-plane is :

d = 6 units .

So , equation of circle with center C and radius 6 units is :

(x-1)^2+(y-(-1))^2+(z-6)^2=6^2\\\\(x-1)^2+(y+1)^2+(z-6)^2=36

b)

Distance of point from yz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1

c)

Distance of point from xz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1

Hence , this is the required solution .

4 0
3 years ago
ANSWER FAST PLEASEEE
KATRIN_1 [288]

Answer:

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To find the y intercept, x=0.

It should be year 0. 44000x2003=88132000+790000=88922000

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6 0
2 years ago
What is lim x→-3 sqrt x^2-8
vitfil [10]

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If the -8 is not under the square root, then...

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Either way, we replace x with -3 and simplify.

For more information, refer to the direct substitution rule for limits.

4 0
1 year ago
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