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3 years ago

Question 1b.

Mathematics

1 answer:

Ainat [17]3 years ago

8 0

Answer: i) Reasonably symmetric

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Explanation:

The data set is

35,40,45,49,51,54,57,58,58,64,68,68,75,77

Erase the outer pair of values (35 and 77) to end up with this smaller set

40,45,49,51,54,57,58,58,64,68,68,75

Repeat and erase off the outer values (40 and 75) to end up with

45,49,51,54,57,58,58,64,68,68

Keep going until we have either one or two elements left

49,51,54,57,58,58,64,68
51,54,57,58,58,64
54,57,58,58
57,58

At this point, we see that the center is between these two last values. So the median is (57+58)/2 = 57.5

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The center is at 57.5

Let L = {35,40,45,49,51,54,57} consist of the set of numbers lower than the median

Let U = {58,58,64,68,68,75,77} be the upper set of values larger than the median.

The middle of set L is 49 and the middle of set U is 68. So Q1 and Q3 are 49 and 68 respectively.

From that, we see the IQR = Q3 - Q1 = 68-49 = 19

Let A and B be the lower and upper fence

We would then say

A = Q1 - 1.5*IQR = 49-1.5*19 = 20.5

B = Q3 + 1.5*IQR = 68 + 1.5*19 = 96.5

Notice how all of the given data values are between A = 20.5 and B = 96.5, which means we don't have any outliers.

Furthermore, if we plotted a box-and-whisker plot (see below), we can see that the two whiskers are fairly the same length. It's not perfect but they're reasonably symmetric. Similarly, the median is somewhat in the middle of the box itself. So this gives a rough idea that the data set we're dealing with is fairly symmetric.

More confirmation can be in the form of a histogram which is also part of the image below. This histogram is fairly symmetric as the left half is nearly (not perfectly) a mirror image of the right half, and vice versa. No one side pulls too far out to create a highly skewed distribution.

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Answer:

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Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

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Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

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$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

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$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

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$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

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Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

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