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Firlakuza [10]
3 years ago
12

PLEASE HELP FAST NO LINKS!! 3/2 (x-6)= -3

Mathematics
2 answers:
cricket20 [7]3 years ago
6 0

Answer: x=4

Step-by-step explanation:

−3×2 =−6 then − 6/3=−2 after that −2+6 which equals x=4

HOPED THIS HELPED :))

 

​  

 

e-lub [12.9K]3 years ago
4 0

Answer:

x= 4

hope this helps!

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schepotkina [342]
Evaluate A² for A = -3.

(-3)² = (-3) * (-3) = 9

Your answer is 9.

If the question is, however, evaluate A2, which is 2A, for A = -3, then the answer is:

2A = 2(-3) = -6.
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3 years ago
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Use strong mathematical induction to prove the existence part of the unique factorization of integers theorem (Theorem 4.4.5). I
valentina_108 [34]

Answer:

Lets say that P(n) is true if n is a prime or a product of prime numbers. We want to show that P(n) is true for all n > 1.

The base case is n=2. P(2) is true because 2 is prime.

Now lets use the inductive hypothesis. Lets take a number n > 2, and we will assume that P(k) is true for any integer k such that 1 < k < n. We want to show that P(n) is true. We may assume that n is not prime, otherwise, P(n) would be trivially true. Since n is not prime, there exist positive integers a,b greater than 1 such that a*b = n. Note that 1 < a < n and 1 < b < n, thus P(a) and P(b) are true. Therefore there exists primes p1, ...., pj and pj+1, ..., pl such that

p1*p2*...*pj = a

pj+1*pj+2*...*pl = b

As a result

n = a*b = (p1*......*pj)*(pj+1*....*pl) = p1*....*pj*....pl

Since we could write n as a product of primes, then P(n) is also true. For strong induction, we conclude than P(n) is true for all integers greater than 1.

5 0
3 years ago
Current rules for telephone area codes allow the use of digits​ 2-9 for the first​ digit, and​ 0-9 for the second and third​ dig
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Using the fundamental counting theorem, we have that:

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The fundamental counting principle states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are ways to do both things.

For the area code:

  • 8 options for the first digit.
  • 9 options for the second and third.

Thus:

8 \times 9 \times 9 = 648

648 different area codes are possible with this rule.

For the number of 10-digit phone numbers:

  • 7 digits, each with 10 options.
  • 648 different area codes.

Then

648 \times 10^7 = 6,480,000,000&#10;

There are 6,480,000,000 possible 10-digit phone numbers.

The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

A similar problem is given at brainly.com/question/24067651

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3 years ago
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shusha [124]

Answer:

You must survey 784 air passengers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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M = z\sqrt{\frac{\pi(1-\pi)}{n}}

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Assume that nothing is known about the percentage of passengers who prefer aisle seats.

This means that \pi = 0.5, which is when the largest sample size will be needed.

Within 3.5 percentage points of the true population percentage.

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0.035\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.035}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.035})^2

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3 years ago
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