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3 years ago

14

Lee and barry play a trivia game in which questions are worth different numbers of points. If a question is answered correctly,

a player earns point. If a question is answered incorrectly, the player loses points. Lee currently has -350 points.
a. Before the game ends, lee answers a 275 point question correctly, a 70 point question correctly, and a 50 point question incorrectly. write and find the value of an expression to find lee's final score

2 answers:

Genrish500 [490]3 years ago

7 0

1st) 275+70-50 Lastly: Subtract -350 - 295 = - 55

275+70 = 345 The answer is - 55

345 - 50 = 295

dem82 [27]3 years ago

5 0

Answer: The required expression,

$-350+275+70-50$

Lee's final score is -55.

Step-by-step explanation:

Given,

The current score of Lee = - 350 points,

Now, he answered 275 point question correctly,

∴ Score he gained = + 275 points,

Again he answered 70 point question correctly,

∴ score he gained = + 70

Finally, he answer 50 point question incorrectly,

∴ Score he gained = - 50 points,

So, the total gaining in score = 275 + 70 - 50

Therefore,

Final score = Current score + Gaining scores = -350+275+70-50 = -55

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Substitute the value of the variable into the equation and simplify.

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Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

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2 years ago