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3 years ago

5)

Mathematics

1 answer:

stealth61 [152]3 years ago

7 0

Answer:

the correct answer is option D. median is 7 and mean is 5.93

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3x = 0.5.2 x = -6 or x = 0 x = -4 or x = 3 x = -2 or x = 1.5 x = 0 or x = 6

tiny-mole [99]

that is unsolvable coz no double decimal can be used

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3 years ago

Simplify each of the following expressions.

Veronika [31]

Here you go :) have a good day

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3 years ago

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How many cards must be drawn (without replacement) from a standard deck of 52 to guarantee that ten of the cards will be of the

marta [7]

37 cards. 9 cards of every suit + 1 more card to guarantee the 10 cards by pigeonhole principle.

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4 years ago

Mr. Ramadhan wants to save money to buy a house so he puts 21% of his earnings into his savings account. How much money does he

Trava [24]

Answer:

79%

Step-by-step explanation:

He has 100% to start off. If he puts 21% into savings you subtract that from the starting amount. 100 - 21 = 79. Therefore the answer is 79%.

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3 years ago

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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