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DanielleElmas [232]

3 years ago

12

Pan Kowalski wpłacił na lokatę terminową –na 2 lata , kwotę 6000zł, na oprocentowanie 6%. Podatek od zysku z oszczędności wynosi

19%. Ile wyniesie rzeczywisty zysk pana Nowaka z tej lokaty terminowej , jeżeli:
a) kapitalizacja odsetek będzie roczna?
b) kapitalizacja odsetek będzie kwartalna
b) bez kapitalizacji odsetek.

1 answer:

vlabodo [156]3 years ago

3 0

Answer:

Hesdfhjwebsdfjwefb

Step-by-step explanation:

sdfjhwefsjhbsdvdbhje

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(i)Find the area of the parallelogram with base 14 cm and height 15 cm.

gulaghasi [49]

Answer:

The area should be "210 cm to the second power"

7 0

4 years ago

Read 2 more answers

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

Can someone plz help would be nice

Tems11 [23]

DONT GO TO THAT LINK ITS A VIRUS AND ITS A BOT

6 0

3 years ago

A spinner is divided into 10 congruent sections.Each section is labeled with a number as shown.What is the probability that the

Rufina [12.5K]

Answer:

50%

Step-by-step explanation:

The only numbers that are multiples of 2 are all even numbers. From 1 to 10 there are a total of 5 even numbers, meaning that only 5 sections are multiples of 2. Therefore, the probability of the arrow landing in a section that is labeled with a number that is a multiple of 2 would be

5/10 if we simplify this it would be 1/2 or 50%

Finally, we can say that the probability of the arrow landing on one of these sections is 50%

8 0

3 years ago

Which postulate or theorem can be used to prove that AABC = ADCB?<br> SSS<br> ASA<br> SAS<br> AAS

koban [17]

<h3>Answer : </h3>

SSS congruency criteria can be used here to prove ∆ABC ≅ ∆DCB

<h3>Solution : </h3>

In ∆ABC and ∆DCB

AB = CD (given)

AC = BD (given)

BC = BC (common)

So, by SSS congruency criteria,

∆ABC ≅ ∆DCB

6 0

3 years ago