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3 years ago

Happy Birthday Scorpios!!

Mathematics

2 answers:

Sonja [21]3 years ago

8 0

Answer:

HAPPY B-DAY :))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

Step-by-step explanation:

juin [17]3 years ago

6 0

Answer:

Happy Birthday Scorpios!!

Step-by-step explanation:

You might be interested in

What is the expectation for a binomial distribution with p = 0.5 and n = 8?

Natasha2012 [34]

Answer: 4

Step-by-step explanation:

Given the following :

P = probability of success = 0.5

n = number of trials = 8

The expected value of a binomial distribution with probability of success P and number of trials n is defined by:

E(n, p) = n * p

Therefore, expected value when P = 0.5 and n = 8

E(8, 0.5) = 8 × 0.5

= 4

The expected value of the binomial distribution is 4

8 0

3 years ago

(5x4 + 5x3 + 4x - 9) + (2x4 + 7x2 + 2x + 14)

ladessa [460]

Answer:

7x⁴ + 5x³ + 7x² + 6x + 5

Step-by-step explanation:

The given expression is

(5x4 + 5x3 + 4x - 9) + (2x4 + 7x2 + 2x + 14)

The first step is to open the brackets by multiplying each term inside each bracket by the term outside each bracket. Since the term outside each bracket is 1, the expression becomes

5x⁴ + 5x³ + 4x - 9 + 2x⁴ + 7x² + 2x + 14

We would collect like terms by combining each term with the same exponent or raised to the same power. The term would be arranged in decreasing order of the exponents. It becomes

5x⁴ + 2x⁴ + 5x³ + 7x² + 4x + 2x - 9 + 14

7x⁴ + 5x³ + 7x² + 6x + 5

7 0

3 years ago

QUADRATIC FORMULA SIMPLIFIED PLZ HELP

olga2289 [7]

Answer:

3x² + 5x + 1 = 0 simplified

x = (-5 ± $\sqrt{13}$ )/6 roots

Step-by-step explanation:

Given quadratic formula

3x² + 5x - 5 = - 6

Standard form

ax² + bx + c = 0

Simplifying

3x² + 5x - 5 + 6 = 0
3x² + 5x + 1 = 0

Solving

x = (-5 ± $\sqrt{5^2 -4*3*1}$ )/(2*3)
x = (-5 ± $\sqrt{13}$ )/6

6 0

3 years ago

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.

8 0

2 years ago

7. Joe spent of his money and Cara

goldenfox [79]

Hi how are you doing today Jasmine

4 0

3 years ago