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Mademuasel [1]
3 years ago
11

When a product of integers has an odd number of negative factors, then sign of the product is?

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0

Answer:

<em>The product of an even number of negatives give us a positive number. An odd number of - 1's gives us a negative one, -1. This can be applied to any set of products. The product of an odd number of negatives give us a negative number. </em>

Step-by-step explanation:

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A motorboat takes 3 hours to travel 108 going upstream. The return trip takes 2 hours going downstream. What is the rate of the
Natalija [7]
Let s and c be the rate of the boat in still water and the rate of the current, respectively. Then:
108/s-c=3
108/s+c=2
3s-3c=108
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4 0
2 years ago
what is the probability that a positive integer less than 100 picked at random has all distinct digits
bazaltina [42]

the probability that a positive integer less than 100 picked at random has all distinct digits is 90.

He has 99 positive integers less than 100.

The integers that do not have all distinct digits have the same digit twice.

11, 22, 33, 44, 55, 66, 77, 88, 99

There are 9 integers that do not have distinct digits, therefore there are 99-9 integers that do.

Probability is without a doubt how probable something is to show up. whenever we are uncertain about the final results of an occasion, we can talk about the possibilities of sure outcomes and how probable they are. The analysis of events governed by opportunity is referred to as records.

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Learn more about probability here brainly.com/question/24756209

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8 0
2 years ago
Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
Oliga [24]

The base case is the claim that

\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}

which reduces to

\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}

By the induction hypothesis,

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}

Now compare this to the upper bound we seek:

\dfrac{2k+1}{k+1}  > \dfrac{2k+2}{k+2}

because

(2k+1)(k+2) > (2k+2)(k+1)

in turn because

2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0

6 0
2 years ago
Read 2 more answers
ILL GIVE YOU BRAINLIST !!! SHOW WORK
alexgriva [62]

Answer:

-0.004 repeating

Step-by-step explanation:

5 times negative three is -15 and that to the negative second power is -0.004 with the 4 repeating.

6 0
2 years ago
How to find the original price of a discounted item?
k0ka [10]
You can use the percentage such as 20% off, make it 0.20 times the discounted price. Then add that into the discounted price.
7 0
3 years ago
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