How many distinct products can be formed using two different integers from the given set: {–6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4
zhannawk [14.2K]
Number of distinct products that can be formed is 144
<h3>Permutation</h3>
Since we need to multiply two different integers to be selected from the set which contains a total of 12 integers. This is a permutation problem since we require distinct integers.
Now, for the first integer to be selected for the product, since we have 12 integers, it is to be arranged in 1 way. So, the permutation is ¹²P₁ = 12
For the second integer, we also have 12 integers to choose from to be arranged in 1 way. So, the permutation is ¹²P₁ = 12.
<h3>
Number of distinct products</h3>
So, the number of distinct products that can be formed from these two integers are ¹²P₁ × ¹²P₁ = 12 × 12 = 144
So, the number of distinct products that can be formed is 144
Learn more about permutation here:
brainly.com/question/25925367
1 ft = 12 in
192/12 = 16
16 ft = 192 in
Answer:
16 days
Step-by-step explanation:
Given that:
Number of paramecium Aurelia doubles every 1.25days
5 days ago ; number of paramecium Aurelia = 1
How many are there now?
Using the relation :
Multiplier^(total number of days / time it takes to multiply)
Multiplier = doubles = 2
Total number rof days =. 5
Time it takes to. Multiply = 1.25
The formula becomes :
Number(n) of ptotozoan in 5days is given by 2^(5/1.25)
n = 2^(5/1.25)
n = 2^4
n = 2 * 2 * 2 * 2 =. 16
Answer:
Number of each ticket is;
$10 tickets = 1115
$20 tickets = 1251
$30 tickets = 934
Step-by-step explanation:
Let x,y and z represent the number of $10,$20 and $30 tickets sold.
Given;
Total number of tickets n = 3300
x+y+z = 3300 .....1
Total sales = $64,190
10x + 20y + 30z = 64,190 .....2
It has sold 136 more $20 tickets than $10 tickets
y = x +136 ........3
Substituting equation 3 into equation 1 and 2;
For 1;
x+y+z = 3300
x+(x+136)+z = 3300
2x + z = 330-136
2x + z = 3164 ........4
For 2;
10x + 20y + 30z = 64,190
10x + 20(x+136) + 30z = 64,190
10x + 20x + 2720 + 30z = 64190
30x + 30z = 64190-2720
30x+30z = 61470
divide through by 30
x+z = 2049 ......5
Subtract equation 5 from 4
2x-x +z-z = 3164-2049
x = 1115
From equation 3
y = x + 136 = 1115+136
y = 1251
From equation 1;
z = 3300 - (x+y)
z = 3300- (1115 + 1251)
z = 934
Number of each ticket is;
$10 tickets = 1115
$20 tickets = 1251
$30 tickets = 934
The answer would be 52w -39
