Factorise fully -x-5

Mathematics

2 answers:

PSYCHO15rus [73]2 years ago

8 0

-x: -1 • x
-5: -5 • 1

liraira [26]2 years ago

7 0

The fully factorized version of -x-5 is $-1 \times (x+5)$ .

What is Factorization?

Usually, factorization or factoring means writing a number (or another mathematical object) as a product of minor or more uncomplicated objects of the same kind.

Given -x-5 is a linear expression.

For factorizing, first, make the coefficient of x positive. For this take -1 common.

⇒ -x-5 = $-1\times(x+5)$

Now take the greatest common divisor (gcd) of absolute values of coefficients in x+5.

⇒ gcd(1,5) = 1.

Now take gcd 1 common in the expression.

⇒ $-1\times1\times(x+5)$

⇒ $-1\times(x+5)$ .

∴ The fully factorized version of -x-5 is $-1\times(x+5)$ .

Learn more about factorization here:

brainly.com/question/25829061

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At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$