If m<BEG = (19x + 3)° and m<EGC = (m<GCB + 4x)°, which of the following statements is true about quadrilateral BEGC? Se

Answer: 3hk - 12.4h

Step-by-step explanation:

h(3k-12.4)

h multiplied by 3k results in 3hk, and h multiplied by 12.4 gives -12.4h.

So, the outcome is 3hk - 12.4h

An even function can be reflected over the y-axis and still remain unchanged.
Example: y=x^2
On the other hand, an odd function can be reflected around the origin and also remains unchanged.
Example: y=x^3


A straightforward method to determine this is:

if f(x) is even, then f(-x)=f(x)
if f(x) is odd, then f(-x)=-f(x)


Hence, for an even function
substitute -x in for each and check for equivalence
make sure to fully expand the expressions
g(x)=(x-1)^2+1=x^2-2x+1+1=x^2-2x+2 is the original expression
g(x)=(x-1)^2+1
g(-x)=(-x-1)^2+1
g(-x)=(1)(x+1)^2+1
g(-x)=x^2+2x+1+1
g(-x)=x^2+2x+2
Not the same, as the original contains -2x
Therefore, it is not even
g(x)=2x^2+1
g(-x)=2(-x)^2+1
g(-x)=2x^2+1
It matches, hence it is even
g(x)=4x+2
g(-x)=4(-x)+2
g(-x)=-4x+2
Not equivalent, thus not even
g(x)=2x
g(-x)=2(-x)
g(-x)=-2x
Not equal, therefore not even



g(x)=2x²+1 is the confirmed even function.

Y = 6 Here’s a step-by-step explanation: On the initial day of ticket sales, the school sold 3 adult tickets and 8 student tickets for a total of $72. On the following day, the school collected $152 from 7 adult tickets and 16 student tickets. What is the price of a student ticket? Day 1: 3x + 8y = 72 Day 2: 7x + 16y = 152 This leads to a system of equations. By manipulating these equations, we find: Ultimately, y = 6. If you wish to continue from this point, substituting y into the equation will give the value for x.

Response:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) signifies the likelihood that a student with a visa card also possesses a MasterCard.

p(A/B) indicates the probability that a student with a MasterCard also has a visa card.

e. 0.35

f. 0.31

Detailed explanation:

a. p(AUBUC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

        = 0.6 + 0.4 + 0.2 - 0.3 - 0.11 - 0.1 + 0.07 = 0.76

b. P(AnBnC') = P(AnB) - P(AnBnC)

        = 0.3 - 0.07 = 0.23

c. P(B/A) = P(AnB)/P(A)

        = 0.3/0.6 = 0.5

e. P((AnB)/C) = P((AnB)nC)/P(C)

        = P(AnBnC)/P(C)

        = 0.07/0.2 = 0.35

f. P((AUB)/C) = P((AUB)nC)/P(C)

        = (P(AnC) U P(BnC))/P(C)

        = (0.11 + 0.1)/0.2

        = 0.21/0.2 = 0.31

The probability mass function of X equals 0.03. To clarify:
Assuming the requirement for winning is one side as heads and the opposing side as tails, the likelihood of both outcomes is 1/2 or 0.5. Thus, we can construct a graph to calculate all probabilities related to achieving heads. In this context, X indicates the dollar amounts won during the coin flips, while the probability of heads reflects the likelihood of each outcome and the potential winnings. The chances of winning decrease as the winning amount rises.