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pickupchik [31]
2 years ago
11

A survey determined that 7/18 of the seventh grade students at Mills Middle School preferred round chocolate candies to hard can

dy. When Jennifer duplicated the survey, 1/9 of the students changed their preference from chocolate candy to hard candy.
What fraction of the students in the second survey preferred round chocolate candy?

A. 8/27
B. 7/162
C. 5/ 18
D. 1/2
Mathematics
1 answer:
andre [41]2 years ago
8 0

Answer:

The correct answer is C) 5/18

Step-by-step explanation:

In order to find this answer, you need to take the fraction that answered it on the first try and subtract the number that changed their mind.

7/18 - 1/9

In order to complete this operation, you need to give them common denominators.

7/18 - 2/18 = 5/18


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Both answers plssss ?????!?
Troyanec [42]

Answer:

4) C

5) C

Step-by-step explanation:

7 0
3 years ago
Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the
Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right], to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B=\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]

the system Bx=0 can be represented in matrix form as:

\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right] ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have B_{0}:      

B_{0}=\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]. The rank of B_{0} can be found by using the second column and third column pair as follows:

|B_{0}|=(3*0)-(0*2)=0 i.e, B_{0} is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is \neq0.

Comparing the rank of both B and B_{0}, it is obvious that

Rank of B\neqRank of B_{0} since (-2)<1.

Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution.     </em>

(2) If B=\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right] is the transpose of matrix A=\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right], then

Then the equation Bx=0 is represented as:

\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right]..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0  i.e B has a rank of order 1.

B_{0}=\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right],

|B_{0}|=(5*0)-(0*10)=0-0=0   i.e B_{0} has a rank of order 1.

we can therefor conclude that since

rank B=rank B_{0}=1,  equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (X_{1} and X_{2}).

<u>Summary:</u>

  • Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
  • Determine the rank of both the coefficients matrix B and B_{0} which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
  • If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.
5 0
3 years ago
What is the range of possible sizes for side x? _ &lt; x &lt; _
qwelly [4]

Answer:

2.8 < x < 5.8

Step-by-step explanation:

We must apply the Triangle Inequality Theorem which states that for any triangle with sides a, b, and c:

a + b > c

b + c > a

c + a > b

Here, let's arbitrarily denote a as 4.1, b as 1.3, and c as x. So, let's plug these values into the 3 inequalities listed above:

a + b > c  ⇒  4.1 + 1.3 > x  ⇒  5.8 > x

b + c > a  ⇒  1.3 + x > 4.1  ⇒  x > 2.8

c + a > b  ⇒  x + 4.1 > 1.3  ⇒  x > -2.8

Look at the last two: clearly if x is greater than 2.8 (from the second one), then it will definitely be greater than -2.8 (from the third), so we can just disregard the last inequality.

Thus, the range of possible sizes for x are:

2.8 < x < 5.8

<em>~ an aesthetics lover</em>

8 0
2 years ago
Read 2 more answers
Y varies inversely with x k = 4.5 What is the value of x when y is 9?
masya89 [10]
C because i just took de test
7 0
3 years ago
on a mathematics test, Mary scored 27 points out of a possible 60 points. what was her percentage mark?
Oksana_A [137]

Answer:

45%

Step-by-step explanation:

\frac{27}{60}  \times 100 = 45\%

4 0
2 years ago
Read 2 more answers
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