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vovangra [49]
4 years ago
13

Write the expression for the nth term for m9, -27, 81

Mathematics
1 answer:
Varvara68 [4.7K]4 years ago
6 0
I don't think that that "m" belongs in this sequence.  Focus on 9, -27, 81, ...

Multiplying 9 by what number produces the product -27?

Mult. -27 by what number produces 81?

This number is called the "common ratio" of this "geometric sequence."

Finish this recursive formula:

a_(n+1) = (a_n) * (         )

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To build a fence is perimeter so 45.5 x 2= 91
155 - 91= 64
Since that is still two sides you divide 64 by 2 and get 32.
Then area is l x w so 32 x 45.5= 1456
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Please help me with 1-8
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Will no explain too much in details.

The bottom rectangle has sizes 21cm*9cm, so its area is 189cm^2.

The remaining rectangle has sizes 6cm*7cm(16-9=7cm),so its area is 42cm^2.

The area of the object is 231 cm^2.

The perimeter is 21+9+16+6+(16-9)+(21-6) = 74 cm.

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B) 5:2
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Compare the process of solving<br> |x - 1 + 1 &lt; 15 to that of solving<br> 1x – 1| + 1 &gt; 15.
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x-1+1<15; this gonna be x<15 and 1x-1+1>15 this is gonna be the same

Step-by-step explanation:

8 0
4 years ago
A research team conducted a study showing that approximately 10% of all businessmen who wear ties wear them so tightly that they
PIT_PIT [208]

Answer:

a) 0.794 = 79.4% probability that at least one tie is too tight.

b) 0.184 = 18.4% probability that more than two ties are too tight.

c) 0.206 = 20.6% probability that no tie is too tight.

d) 0.816 = 81.6% probability that at least 13 ties are not too tight.

Step-by-step explanation:

For each tie, there are only two possible outcomes. Either they are too tight, or they are not too tight. The probability of a businessmen wearing a tie too tight is independent of any other businessmen, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

10% of all businessmen who wear ties wear them so tightly

This means that p = 0.1

15 businessmen

This means that n = 15

a. at least one tie is too tight

This is:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.206

Then

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.206 = 0.794

0.794 = 79.4% probability that at least one tie is too tight.

b. more than two ties are too tight 0.913

This is:

P(X > 2) = 1 - P(X \leq 2)

In which:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.206

P(X = 1) = C_{15,1}.(0.1)^{1}.(0.9)^{14} = 0.343

P(X = 2) = C_{15,2}.(0.1)^{2}.(0.9)^{13} = 0.267

Then

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.206 + 0.343 + 0.267 = 0.816

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.816 = 0.184

0.184 = 18.4% probability that more than two ties are too tight.

c. no tie is too tight

This is P(X = 0). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.206

0.206 = 20.6% probability that no tie is too tight.

d. at least 13 ties are not too tight 0.816

At most 2 are too tight, which is P(X \leq 2) = 0.816, found in option b.

0.816 = 81.6% probability that at least 13 ties are not too tight.

6 0
3 years ago
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