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sertanlavr [38]
3 years ago
8

The product of three consecutive odd numbers is 6,783. What are the numbers?

Mathematics
1 answer:
Katarina [22]3 years ago
7 0
<span>The numbers are 2260, 2261, and 2262. Here’s how i figured it out:
Let x = The smallest numberThe second number is the the first number with one added to it, so it can be represented as x+1.The third number is the first number added 2 to it, so we can say it's x+2.Put this in a equation. The three numbers add together to become 6783, so it will be the sum.6783 = x + x + 1 + x + 2I just replaced the second and third numbers with what I explained above.We also can simplify the expression by combining the like terms to become 6783 = 3x + 3.Subtract 3 on both sides.6780 = 3xDivide by three on both sides to isolate the variable.2260 = xBut, we’re not done yet. Since x is only the starting number, we have to add 1 and 2 to it to get the consecutive  numbers. I think you can add, so the three consecutive numbers will be 2260, 2261, and 2262. If you’re not sure then you can add the three numbers together and you will get 6783.<span>I hope this answer helped and if you have any other questions, don’t hesitate to contact me!</span></span>
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Ilia_Sergeevich [38]

Answer:

Sample space= 1, 2, 3, 4, 5, 6

Outcomes= 2, 4, 6

Probability= Number of outcomes/ total number of outcomes

= 3/6 times 2( as 2 dices are rolled)

=6/12

=1/2

Therefore, the probability of getting 2 even number is 1/2

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Ab with slope = 5 and ST with slope = 5
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4 birlik + 9 binlik + 1 yüz binlik + 9 yüzlük ten oluşan sayı?​
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2 years ago
John's current salary is 40000 per year.His annual pay raise is always a percent of his salary.What would his salary be if he re
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Answer:

The salary after four consecutive increase = $ 46794.34

Step-by-step explanation:

According to question,

salary of MR. John = $ 40000 per year

percentage py increase = 4%

Let the salary after four consecutive pay increase = x

So, X = 40000 (1 +\frac{4}{100})^4

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2 years ago
he data represents the daily rainfall​ (in inches) for one month. Construct a frequency distribution beginning with a lower clas
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Answer:

It is not normally distributed as it has it main concentration in only one side.

Step-by-step explanation:

So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).

So, let us begin the groupings into their different classes, shall we?

Data given:

0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.

(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.

(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31

(3). 0.4 - 0.6 : there are 2 values that falls into this category.

(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.

Class interval            frequency.

0.00 - 0.20.                   15.

0.20 - 0.40.                    2.

0.4 - 0.6.                        2.

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