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Mila [183]
3 years ago
12

The half-life of a radioactive isotope is the time it takes for a quantity of the Isotope to be reduced to half its initial mass

. Starting with 210 grams of a
radioactive isotope, how much will be left after 6 half-lives?
Round your answer to the nearest gram
Mathematics
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:

3 grams will be left after 6 half-lives

Step-by-step explanation:

Half-live:

Time it takes for the substance to be reduced by hall.

After n half lives:

The amount remaing is:

A(n) = A(0)(0.5)^{n}

In which A(0) is the initial amount and n is the number of half-lives.

Starting with 210 grams of a radioactive isotope, how much will be left after 6 half-lives?

This is A(6) when A(0) = 210. So

A(n) = A(0)(0.5)^{n}

A(6) = 210(0.5)^{6} = 3.3

Rounding to the nearest gram

3 grams will be left after 6 half-lives

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Answer the question below
lys-0071 [83]

Answer:

A) -1

Step-by-step explanation:

\frac{3 + 2\sqrt{5} }{2\sqrt{5}-3 } \\

We can rewrite 2√5 - 3 as -(3 + 2√5)

Therefore,

\frac{3+ 2\sqrt{5} }{-(3 + 2\sqrt{5} )} \\

We can cancel out 3 + 2√5 both in the numerator and the denominator. So we get

= 1/-1

= -1

Answer : A) -1

Thank you.

6 0
3 years ago
A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard
bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

In point a)  

\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

  = \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2

value= (1- \frac{1}{k^2}) \times 100 \% =75\%

In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

8 0
2 years ago
In a circle with a radius of 7 feet, the radian measure of the central angle subtended by an arc with a length of 4 feet is . Th
ivann1987 [24]
We know that
length of a sector=[∅]*r--------> when ∅ is in radians
so
∅=length of a sector/r
for r=7 ft
length of a sector=4 ft
∅=4/7-----> 0.57 radians

the answer part 1) is 0.57 radians

part 2) 
area of a sector=(∅/2)*r²--------> when ∅<span> is in radians
</span>area of a sector=(4/7/2)*7²-----> (4/14)*49----> 14 ft²

the answer Part 2) is 14 ft²
3 0
3 years ago
Please someone help me on this &lt;33
dangina [55]

Answer:

B

Step-by-step explanation:

The initial value for a function is the value of y when x=0. For function A, we can see from the table that when x = 0, we have y = 4. For function B, we plug in 0 in our equation and solve: y = 3(0) + 5, or y = 5. So, because function B has a larger value of y when x = 0, function B has the greater initial value because the initial value for Function A is 4 and the initial value for function B is 5.

7 0
2 years ago
True or False?
oee [108]

Answer:

False

Step-by-step explanation:

y = 2x + 1

Substitute the point (4,3) into the equation and see if it is true

3 = 2(4) +1

3 =8+1

3 =9

This is false, so the point is not a solution

4 0
3 years ago
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