Answer:
a. Yes
b. Yes
c. Yes
d. Degree 8
Step-by-step explanation:
a. Yes, n(x) is a polynomial of one single term (also called monomial) because it contains variables raised to positive integers.
b. Yes, m(x) is a polynomial of also one single term (also called monomial) because it contains variables raised to positive integers.
c. The quotient of n(x) / m(x) can be reduced to a polynomial of one single term as follows:
which as can be seen, also contains variables raised to positive integers.
d. The degree of the polynomial resultant is the addition of the powers of all variables present (x and y) which results in: 2 + 6 = 8
Therefore the degree of this polynomial is 8.
Hello!
First you have to list the data in both classes
Class A
41, 42, 45, 46, 47, 48, 52, 53, 54, 59, 61, 61, 64, 68, 71, 82, 85, 90
Class B
41, 42, 59, 62, 64, 69, 71, 75, 77, 78, 78, 80, 83, 84, 84, 86, 86, 87, 92, 92, 95
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First we are going to find the mode and mean of class A
The mode is the number that appears the most
The number that appears the most is 61
The mode is 61
To find the mean you add all the numbers together and divide the sum by the amount of numbers added
41 + 42 + 45 + 46 + 47 + 48 + 52 + 53 + 54 + 59 + 61 + 61 + 64 + 68 + 71 + 82 + 85 + 90 = 1069
Divide this by the amount of numbers added
1069 / 18 = 59.3888...
The mean is 59.3888
The mode is 61 and the mean is 59.39 for class A
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We are going to find the range and median for class B
To find the range you subtract the smallest number from the largest on
The smallest number is 41
The largest number is 95
Subtract these
95 - 41 = 54
The range is 54
To find the median you list the numbers from least to greatest and look for the number in the middle
41, 42, 59, 62, 64, 69, 71, 75, 77, 78, 78, 80, 83, 84, 84, 86, 86, 87, 92, 92, 95
The number in the middle is 78
The range is 54 and the median is 78
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Hope this helps!
Let the total toys be x.
The first has (1/10)x = x/10
The third child has one more toys than the first = x/10 + 1
The fourth has double of the third: = 2* (x/10 + 1) = 2x/10 + 2
Since we know that the second child has 12 more toys than the first
Therefore the Second minus First = 12.
First let us find the second.
1st + 2nd + 3rd + 4th = x
x/10 + 2nd + (x/10 + 1) + (2x/10 + 2) = x
2nd + x/10 + x/10 + 1 + 2x/10 + 2 = x
2nd + x/10 + x/10 + 2x/10 + 1 + 2 = x
2nd + 4x/10 + 3 = x
2nd = x - 4x/10 - 3
2nd = 6x/10 - 3
But recall
2nd - 1st = 12
6x/10 - 3 - x/10 = 12
6x/10 - x/10 = 12 + 3
5x/10 = 15
5x = 15 * 10
x = 15*10 /5
x = 30
So there were 30 toys in all.
