All categories

3 years ago

Can someone plz help me solved this problem I need help ASAP plz help me! Will mark you as brainiest!

Mathematics

1 answer:

miv72 [106K]3 years ago

3 0

Answer:

-1, 1

13, 15

Step-by-step explanation:

x and x+2 are the integers

x*(x+2)= 7(x+x+2) -1
x²+2x= 14x+14-1
x² - 12x -13= 0

Roots of the quadratic equation are: -1 and 13.

So the integers pairs are: -1, 1 and 13, 15

You might be interested in

A sequence of transformations maps ∆ABC to ∆A′B′C′. The sequence of transformations that maps ∆ABC to ∆A′B′C′ is a reflection ac

7nadin3 [17]

Answer:

the answe is d

Step-by-step explanation:

3 0

3 years ago

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

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

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

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

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

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

$P( X \le 4 ) = 0.0054$

4 0

3 years ago

|x|, if x=10; .3; 0; −2.7; −9

alekssr [168]

Answer:

For $\left | x \right |$ we will have positive values only.

$\left | -2.7 |\right =2.7$

$\left | -9 |\right =9$

Rest are positive so will give positive values out of the modulus function.

Step-by-step explanation:

The absolute value of any number is its positive value only.

As for example $(-2.7)$ and $(2.7)$ both are having equal distance from $0$ in a number line, one on the left other on the right side of the number line but are equidistant from $0$ .