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Afina-wow [57]
2 years ago
12

The half-life of cesium-137 is 30 years. Suppose we have a 170 mg sample.(a) Find the mass that remains after t years.(b) How mu

ch of the sample remains after 60 years? (Round your answer to two decimal places.)(c) After how long will only 1 mg remain? (Round your answer to one decimal place.)
Mathematics
1 answer:
saul85 [17]2 years ago
5 0

Answer:

212.9 years

Step-by-step explanation:

Given that the half-life of cesium-137 is 30 years. Suppose we have a 170 mg sample

P0 = 175

P(30) = 87.5

So we can write equation as

P(t) = 170(\frac{1}{2} )^{\frac{t}{30} }

b) After 60 years t = 30

In 30 years it becomes half and hence in 60 years it would become 1/4

i.e. P(60) = 137(1/2^4) = 34.25 mg

c) If P(t) =1, let us find t

137(1/2)^{\frac{t}{30} } =1\\\\n = 7.098 years\\n = 7.1*30 = 212.9 years

212.9

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Thomas wants to invite Madeline to a party. He has an 80% chance of bumping into her at school. Otherwise, he’ll call her on the
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Answer:

84%

Step-by-step explanation:

The probability of Thomas bumping into her at school is 80%, so the probability of not bumping into her is 100% - 80% = 20%.

If he doesn't bump into her (20% chance), he will call her, and the probability of asking her in this case is 60%, so the final probability of asking her in this case is:

P_1 = 20\% * 60\% = 12\%

If he bumps into her (80% chance), the probability of asking her is 90%, so the final probability of asking her in this case is:

P_2 = 80\% * 90\% = 72\%

To find the probability of Thomas inviting Madeline to the party, we just have to sum the probabilities we found above:

P = P_1 + P_2

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5 0
2 years ago
Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $
nlexa [21]

Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

For delivery within 2\frac{1}2 mi - 2\frac{3}4 mi, charges = $3.50

For delivery within 2\frac{3}4 mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3\frac{1}4 mi, charges = $4.40

So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

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2 years ago
Betty took a doll she inherited from her grandmother to the antique store. The dolls original price tag says $6.80. The antique
mariarad [96]
There was a 12.35% increase in the price.
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2 years ago
3 kg of rump steak costs £42 Adel buys 4 kg of rump steak Work out how much Adel Pays
OLEGan [10]

Answer:

£56

Step-by-step explanation:

you first have to find out how much one kg is so you do 42/3=14

then times is by 4, 14x4=56

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2 years ago
If the cost of 1 kg of oranges is Rs. 42.50, find the cost of 3.5 kg of oranges.
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Answer:

1kg = Rs. 42.50

3.5kg = 3.5*42.50

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I hope you know this answer

4 0
2 years ago
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