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2 years ago

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The fish population is growing 3 % each year in a local lake. The function to model this is A = A (1 + r)' where A is the ending

population, A, is the starting population, r is the annual growth rate, and 1 is the number of years. If the fish population was 1, 300 in 2010, how many fish can be expected in the lake 5 years later? Round to the nearest whole number. 1

1 answer:

Schach [20]2 years ago

7 0

Answer:

that all persons held as slaves" within the rebellious states "are, and henceforward shall be free."

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"If the probability of an event is observed repeatedly during independent repetitions than the ratio of the observed to the tota

PIT_PIT [208]

Answer:

This is the Law of <u>large</u> Numbers.

Step-by-step explanation:

The Law of Large numbers states that as the sample size is increased the sample mean value gets closer to the population mean.

That is, as <em>n</em> → ∞ the value of $\bar x$ → μ.

For instance, the theoretical probability of getting a Heads in a toss of a coin is, 0.50.

To check this perform an experiment whether the coin is tossed 20 times.

The probability of heads will be somewhere near 0.50.

Now toss the coins 50 times.

The probability of heads will be now closer to 0.50.

Now toss the coins 100 times.

The probability of heads will be approximately equal to 0.50.

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This is the Law of <u>large</u> Numbers.

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2 years ago

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Answer:

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Step-by-step explanation:

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How many gal is in 32 qt?

spin [16.1K]

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3 years ago

greg wants to divide a 2 and 1/2 pound block of cheese into four equal pieces. How much will each piece weigh?

Alex73 [517]

let's firstly convert the mixed fraction to improper fraction and then divide.

$\bf \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{5}{2}\div 4\implies \cfrac{5}{2}\div \cfrac{4}{1}\implies \cfrac{5}{2}\cdot \cfrac{1}{4}\implies \cfrac{5}{8}$

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3 years ago

Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) ∈ R if and only if ad = bc. Ar

Veseljchak [2.6K]

Answer:

The given relation R is equivalence relation.

Step-by-step explanation:

Given that:

$((a, b), (c, d))\in R$

Where $R$ is the relation on the set of ordered pairs of positive integers.

To prove, a relation R to be equivalence relation we need to prove that the relation is reflexive, symmetric and transitive.

1. First of all, let us check reflexive property:

Reflexive property means:

$\forall a \in A \Rightarrow (a,a) \in R$

Here we need to prove:

$\forall (a, b) \in A \Rightarrow ((a,b), (a,b)) \in R$

As per the given relation:

$((a,b), (a,b) ) \Rightarrow ab =ab$ which is true.

$\therefore$ R is reflexive.

2. Now, let us check symmetric property:

Symmetric property means:

$\forall \{a,b\} \in A\ if\ (a,b) \in R \Rightarrow (b,a) \in R$

Here we need to prove:

$\forall {(a, b),(c,d)} \in A \ if\ ((a,b),(c,d)) \in R \Rightarrow ((c,d),(a,b)) \in R$

As per the given relation:

$((a,b),(c,d)) \in R$ means $ad = bc$

$((c,d),(a,b)) \in R$ means $cb = da\ or\ ad =bc$

Hence true.

$\therefore$ R is symmetric.

3. R to be transitive, we need to prove:

$if ((a,b),(c,d)),((c,d),(e,f)) \in R \Rightarrow ((a,b),(e,f)) \in R$

$((a,b),(c,d)) \in R$ means $ad = cb$ .... (1)

$((c,d), (e,f)) \in R$ means $fc = ed$ ...... (2)

To prove:

To be $((a,b), (e,f)) \in R$ we need to prove: $fa = be$

Multiply (1) with (2):

$adcf = bcde\\\Rightarrow fa = be$

So, R is transitive as well.

Hence proved that R is an equivalence relation.

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