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3 years ago

Radium-226 has a half life of 1622 years. If you start off with a 200

Mathematics

1 answer:

sweet [91]3 years ago

3 0

Answer:

≈23.6 kg left.

Step-by-step explanation:

For determining this, we can simply use an exponential decay equation representing this scenario:

$y= a(0.5)^{\frac{t}{h} }$

a: the initial amount

t: the amount of time

h: the half life

Now, substitute the numbers into the equation:

$y=200(0.5)^{\frac{5000}{1622} }$

Now, solve this equation:

$y=220(0.5)^{3.08}$

Use a calculator to solve for y:

y= 200(0.118)

y≈23.6 kg left.

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2 years ago

Combined, there are Asians, Africans, Europeans, and Americans in a village. The number of Asians exceeds the number of Afric

yanalaym [24]

There are 123 Asians, 25 Africans, 21 Europeans and 8 Americans in the village.

To determine the number of Asians, Africans, Europeans, and Americans in the village, the following calculation must be performed:

A.S + A.F + E.U + A.M = 177
A.S = A.F + E.U + 69
E.U - A.M = 13
E.U = A.M + 13
2A.F = E.U + A.M + 29
A.F + E.U + 69 + A.F + E.U + A.M = 177
A.F + A.M + 13 + 69 + A.F + A.M + 13 + A.M = 177
2A.F + 3A.M = 177 - 13 - 13 - 69
E.U + A.M + 29 + 3A.M = 82
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5A.M = 82 - 13 - 29
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2 years ago

A coin is tossed and a single 6 sided number cube is rolled. Find the probability of landing on the heads side of the coin and r

Delicious77 [7]

Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

Now, for tossing a coin, the total possible outcomes are head and tail. So, there are 2 possible outcomes.

So, probability of event 'H' is given as:

$P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5$

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, $n(S)=6$

Now, probability of rolling a 3 is given as:

$P(R3)=\frac{n(R3)}{n(S)}\\\\P(R3)=\frac{1}{6}$

Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

$P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083$

Note: Independent events are those events whose intersection is an empty set of events or the outcome of one event doesn't affect the outcome of the other event.

Therefore, the probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

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3 years ago

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Temka [501]

Answer:

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