What’s the the math problem? I don’t see it..
Answer:
Step-by-step explanation:
a. The probability of selecting a 6 from the first draw and a 7 on the second draw when two balls are selected without replacement from a container with 10 balls numbered 1 to 10
Not independent because without replacement
Prob for both = 
b. The probability of selecting a 6 on the first draw and a 7 on the second draw when two balls are selected with replacement from a container with 10 balls numbered 1 to 10
Here independent because with replacement makes probability independent.
Prob for both = P(A) *P(B) =
d
c. The probability that two people selected at random in a shopping mall on a very busy Saturday both have a birthday in the month of June. Assume that all 365 birthdays are equally
likely, and ignore the possibility of a February 29 leap-year birthday.
Here independent because one person birthday will not affect the other person birthday
Prob for both = 
d. The probability that two socks selected at random from a drawer containing 10 black socks and 6 white socks will both be black
Prob for I sock black = 10/16 and II sock black if first sock is black = 9/15
Hence not independent
Prob for both = 
The total amount of money that Jenny and Susan spent is the sum of the money spent for the games and pizza. That is,
T = $93 + $6 = $99
We let x be the amount that Susan has. With this representation, the amount of money that Jenny has is x + 17. The equation that would represent this situation is,
x + 17 + x = 99
The value of x from the equation is 42.
ANSWER: $42.
Answer:

Step-by-step explanation:
<u><em>The correct question is</em></u>
If Triangle ABC is congruent with triangle DEF, DE=17, EF =13, DF =9, and BC = 2x-5, then which of the following is the correct value of x?
we know that
If two triangles are congruent, then their corresponding sides and their corresponding angles, are congruent
In this problem
The corresponding sides are
AB and DE
AC and DF
BC and EF
The corresponding angles are
∠A and ∠D
∠B and ∠E
∠C and ∠F
so
AB≅DE
AC≅DF
BC≅EF
and
∠A≅∠D
∠B≅∠E
∠C≅∠F
<em>Find the value of x</em>
BC≅EF
substitute the given values

Solve for x



<h3>
Answer: D) infinitely many solutions</h3>
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Explanation:
Let's solve the first equation for y
4x - 2y = 6
4x-6 = 2y
2y = 4x-6
y = (4x-6)/2
y = (4x/2) - (6/2)
y = 2x - 3
After doing so, we see that 4x-2y = 6 is equivalent to y = 2x-3
Therefore, the original system of equations is effectively listing the same equation twice (one has a different form compared to the other).
Both equations in this system produce the same graph, which leads to infinitely many solutions. All solutions are on the line y = 2x-3.
You can say that all solutions are in the form (x, 2x-3) where x is any real number you want.
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Here's another approach using substitution
4x - 2y = 6 ... start with the first equation
4x - 2( y ) = 6
4x - 2( 2x-3 ) = 6 .... replace y with 2x-3; ie plug in y = 2x-3
4x - 2(2x) - 2(-3) = 6
4x - 4x + 6 = 6
0x + 6 = 6
0 + 6 = 6
6 = 6
We get a true statement. The last equation is always true regardless of what we plug in for x, so this is another way to see how we get to infinitely many solutions.
Side note: the system is considered dependent since one equation depends on the other. The system is also consistent since it has at least one solution.