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Zepler [3.9K]
3 years ago
10

Assuming that Switzerland's population is growing exponentially at a continuous rate of 0.24 percent a year and that the 1988 po

pulation was 6.7 million, write an expression for the population as a function of time in years. (Let t=0 in 1988.)
Mathematics
1 answer:
Sergio [31]3 years ago
7 0

Answer:St = 6700000(1 - 1.04^t ) / 0.76

Step-by-step explanation:

Since the population is growing exponentially at a continuous rate of 0.24 percent a year, then a geometric progression is defined. The formula for the sum of n terms, Sn of a geometric progression is expressed as

Sn = a(r^n - 1)/r - 1

Where

a is the first term of the sequence

n is the number of terms

r is the common ratio

From the given information,

a = 6.7 million

r = 0.24

n = t

St = the population after t years

The expression for the population as a function of time in years will be

St = 6700000(1.04^t - 1)/(0.24-1)

Since 0.24 is lesser than 1, it can be rewritten as

St = 6700000(1 - 1.04^t ) / (1 - 0.24)

St = 6700000(1 - 1.04^t ) / 0.76

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Step-by-step explanation:

A 5 pound bag of apples costs 4.50.

The unit price will be 4.50 divided by 5 which will be:

4.50 ÷ 5 = 0.9

The unit price is 0.9 per pound

A 8 pound bag of the same type of apples costs 7.52.

The unit price will be 7.52 divided by 8 which will give:

7.52 ÷ 8 = 0.94

The unit price is 0.94 per pound

The difference in the unit prices will be the subtraction of 0.90 from 0.94.

0.94 - 0.90 = 0.04

The difference in the unit prices is 0.04 per pound.

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<h2>The answer is 1000100</h2>

Step-by-step explanation:

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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(
Umnica [9.8K]

Answer:

(a) A_1 and A_2 are indeed mutually-exclusive.

(b) \displaystyle P(A_1\; \cap \; B) = \frac{1}{20}, whereas \displaystyle P(A_2\; \cap \; B) = \frac{1}{25}.

(c) \displaystyle P(B) = \frac{9}{100}.

(d) \displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}, whereas P(A_1 \; |\; B) = \displaystyle \frac{4}{9}

Step-by-step explanation:

<h3>(a)</h3>

P(A_1 \; \cap \; A_2) = 0 means that it is impossible for events A_1 and A_2 to happen at the same time. Therefore, event A_1 and A_2 are mutually-exclusive.

<h3>(b)</h3>

By the definition of conditional probability:

\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}.

Rearrange to obtain:

\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot  P(A_1) = 0.25 \times 0.20 = \frac{1}{20}.

Similarly:

\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot  P(A_2) = 0.80 \times 0.05 = \frac{1}{25}.

<h3>(c)</h3>

Note that:

\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}.

In other words, A_1 and A_2 are collectively-exhaustive. Since A_1 and A_2 are collectively-exhaustive and mutually-exclusive at the same time:

\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}.

<h3>(d)</h3>

By Bayes' Theorem:

\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}.

Similarly:

\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}.

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