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alexgriva [62]
3 years ago
8

In a college philosophy class during the fall semester, there were 323 students,42% of whom were male and the rest of whom were

female. In the springsemester, the same number of females was in the class as in the fall, but there were only 298 total students in the class. In the spring, what percent of students were female?
A. 46%B. 58%C. 63%D. 92%
Mathematics
1 answer:
11111nata11111 [884]3 years ago
7 0
C i think...........
You might be interested in
Linear Algebra question! Please help!
kozerog [31]

Answers:

  1. false
  2. false
  3. true
  4. false
  5. True

==================================================

Explanation:

Problem 1

This is false because the A and B should swap places. It should be (AB)^{-1} = B^{-1}A^{-1}.

The short proof is to multiply AB with its inverse (AB)^{-1}  and we get: (AB)*(AB)^{-1} = (AB)*(B^{-1}A^{-1}) = A(B*B^{-1})*A^{-1} = A*A^{-1} = I

The fact we get the identity matrix proves that we have the proper order at this point. The swap happens so that B matches up its corresponding inverse B^{-1} and the two cancel each other out.

Keep in mind matrix multiplication is <u>not</u> commutative. So AB is not the same as BA.

-------------------------

Problem 2

This statement is true if and only if AB = BA

(A+B)^2 = (A+B)(A+B)

(A+B)^2 = A(A+B) + B(A+B)

(A+B)^2 = A^2 + AB + BA + B^2

(A+B)^2 = A^2 + 2AB + B^2 ... only works if AB = BA

However, in most general settings, matrix multiplication is <u>not</u> commutative. The order is important when multiplying most two matrices. Only for special circumstances is when AB = BA going to happen. In general,  AB = BA is false which is why statement two breaks down and is false in general.

-------------------------

Problem 3

This statement is true.

If A and B are invertible, then so is AB.

This is because both A^{-1} and B^{-1} are known to exist (otherwise A and B wouldn't be invertible) and we can use the rule mentioned in problem 1. Make sure to swap the terms of course.

Or you can use a determinant argument to prove the claim

det(A*B) = det(A)*det(B)

Since A and B are invertible, their determinants det(A) and det(B) are nonzero which makes the right hand side nonzero. Therefore det(A*B) is nonzero and AB has an inverse.

So if we have two invertible matrices, then their product is also invertible. This idea can be scaled up to include things like A^4*B^3 being also invertible.

If you wanted, you can carefully go through it like this:

  1. If A and B are invertible, then so is AB
  2. If A and AB are invertible, then so is A*AB = A^2B
  3. If A and A^2B are invertible, then so is A*A^2B = A^3B

and so on until you build up to A^4*B^3. Therefore, we can conclude that A^m*B^n is also invertible. Be careful about the order of multiplying the matrices. Something like A*AB is different from AB*A, the first of which is useful while the second is not.

So this is why statement 3 is true.

-------------------------

Problem 4

This is false. Possibly a quick counter-example is to consider these two matrices

A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}

both of which are invertible since their determinant is nonzero (recall the determinant of a diagonal matrix is simply the product along the diagonal entries). So it's not too hard to show that the determinant of each is 1, and each matrix shown is invertible.

However, adding those two mentioned matrices gets us the 2x2 zero matrix, which is a matrix of nothing but zeros. Clearly the zero matrix has determinant zero and is therefore not invertible.

There are some cases when A+B may be invertible, but it's not true in general.

-------------------------

Problem 5

This is true because each A pairs up with an A^{-1} to cancel out (similar what happened with problem 1). For more info, check out the concept of diagonalization.

5 0
2 years ago
Plssss help 20 points plssa<br> If K Mis the perpendicular bisector of JL, find x.<br> Plsss
Viefleur [7K]

Answer:

x= 7

Step-by-step explanation:

4x+9 =11x-61

9 = 7x - 61

70 = 7x

x = 7

8 0
2 years ago
Seven years ago, lan purchased a $265,000 home with a 30-year mortgage at 3.5%. Having recently lost his job, he can no longer a
MariettaO [177]

Answer:

980.82

Step-by-step explanation:

5 0
3 years ago
A field is a rectangle with a perimeter of 1240 feet. The length is 400 feet more than the width. Find the width and length of t
Wewaii [24]

Answer:  The length is 510 and the width is 110.

Step-by-step explanation:

To find the area of a rectangle, you will have to add the 2 times the length plus 2 times the width because  a rectangle have 4 sides. Two widths and two  lengths.

You can now use the formula  P= 2l + 2w  

were P is the perimeter , l is the length, and w is the width.

the length is 400 more than the width, so we can represent that by the equation,  l = w + 400  

And now we know that the width is w.

So now we will input the perimeter, length, and into the formula to solve for w.

1240 = 2(w + 400) + 2w

1240 = 2w + 800 + 2w

1240 = 4w + 800

-800            -800

   440 = 4w

    w = 110

L= 110 + 400

L = 510  

Check :

1240 = 2(510) + 2(110)

1240 = 1020 + 220

1240 = 1240

5 0
2 years ago
Determine whether the given value is a solution of the equation: x + 15 = 10; x=5 *
kobusy [5.1K]

Answer:

It is not correct, x = -5

Step-by-step explanation:

5 0
2 years ago
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