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3 years ago

For each sequence write an explicit formula 96,48,24,12,6

1 answer:

8 0

$a_1=96;\ a_2=48;\ a_3=24;\ a_4=12;\ a_5=6\\\\a_1=a_2:2\to96:2=48\\a_2=a_3:2\to48:2=24\\a_3=a_2:2\to24:2=12\\a_4=a_3:2\to12:2=6\\\\therefore\\\\a_n=96:2^{n-1}=\frac {96}{2^{n-1}}\\\\check:\\a_1=\frac{96}{2^{1-1}}=\frac{96}{1^0}=96\\a_2=\frac{96}{2^{2-1}}=\frac{96}{2^1}=\frac{96}{2}=48\\a_3=\frac{96}{2^{3-1}}=\frac{96}{2^2}=\frac{96}{4}=24\\\vdots$

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What is 72% of 50?....

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72%*50=36
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3 years ago

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A coin is to be tossed as many times as necessary to turn up one head. Thus the elements c of the sample space C are H, TH, TTH,

slavikrds [6]

Answer:

Step-by-step explanation:

As stated in the question, the probability to toss a coin and turn up heads in the first try is $\frac{1}{2}$ , in the second is $\frac{1}{4}$ , in the third is $\frac{1}{8}$ and so on. Then, P(C) is given by the next sum:

$P(C)=\sum^{\infty}_{n=1}(\frac{1}{2} )^{n}=1$

This is a geometric series with factor $\frac{1}{2}$ . Then is convergent to $\frac{1}{1-\frac{1}{2}}-1=1.$ . With this we have proved that P(C)=1.

Now, observe that

$P(H)=\frac{1}{2}, P(TH)=\frac{1}{4},P(TTH)=\frac{1}{8},P(TTTH)=\frac{1}{16},P(TTTTH)=\frac{1}{32},P(TTTTTH)=\frac{1}{64}.$

Then

$P(C1)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} =\frac{31}{32}$

$P(C2)=P(TTTTH)+P(TTTTTH)=\frac{1}{32}+\frac{1}{64} =\frac{3}{64}$

$P(C1\cap C2)=P(TTTTH)=\frac{1}{32}$

and

$P(C1\cup C2)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)+P(TTTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} +\frac{1}{64}=\frac{63}{64}$

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3 years ago

- Lulah is buying beach balls for her beach

Elan Coil [88]

The set of the ordered pairs is a {(1, 6), (2, 12), (3, 18), (4, 24)}.

According to the statement

We have to find that the set of the ordered pairs.

So, For this purpose, we know that the

A function is a set of ordered pairs in which no two different ordered pairs have the same x -coordinate.

From the given information:

The package contains 6 beach balls And

the number of packages x for a total of 1, 2, 3, and 4 packages.

Then

(packages, balls) = (1, 6)

Multiplying by 2, 3, and 4, we find the set of ordered pairs to be ...

{(1, 6), (2, 12), (3, 18), (4, 24)}

So, The set of the ordered pairs is a {(1, 6), (2, 12), (3, 18), (4, 24)}.

Learn more about ordered pairs here

brainly.com/question/1528681

#SPJ9

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2 years ago

Please answer all :3

My name is Ann [436]

Answer:

1 - First question = 6.00 | Second Question = 0.06

2 - Didnt know

3 - 1360, from 1.36kg

4 - 0.3175

5 - 20.5

6 - 18

7 - 965606.4

Step-by-step explanation:

sorry I couldn't give you number 2, but here are the rest of the answers

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3 years ago

In a Match 6 Lotto, winning the jackpot requires that you select six different numbers from 1 to 49, and that the same six numbe

zimovet [89]

Answer:

$P=\frac{1}{13,983,816}=7.2\ x\ 10^{-8}$

Step-by-step explanation:

Consider the following characteristics of the problem:

The numbers that are selected are different between 1 and 49. This means that the same number is not repeated twice.

The order in which the selected numbers appear does not matter:

This means that (123) = (312)

With this in mind, we know that it is a problem of combinations without repetition. It is not calculated using permutations because in the permutations the order of selection is important, for example: (123) is not equal to (312)

The formula for calculating combinations without repetition is:

$nCr=\frac{n!}{r!(n-r)!}$

Where n is the number of "elements" you can choose and choose r from them

In this case:

n=49

r=6

So:

$49C6=\frac{49!}{6!(49-6)!}$

$49C6=13,983,816\ outcomes$

There are 13,983,816 possible results

This is the best method to calculate the number of possible outcomes.

<em><u>"Besides your method, is there another method to determine the number of outcomes?"</u></em>

Sure, make a list of the 13,983,816 different sets of 6 numbers.

To win it is necessary to obtain the 6 winning numbers in any order. The number of ways this can occur is calculated by combining 6 in 6

$6C6=\frac{6!}{6!(6-6)!}=1$

Finally the probability of winning is:

$P=\frac{1}{13,983,816}=7.2\ x\ 10^{-8}$

Note that the probability of winning is very close to 0. It is practically impossible to win the lottery, you will probably never win anything. Therefore it is better not to invest money in this

6 0

3 years ago