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2 years ago

Here's another question, same one (kind of) 5x - 8 = 3x + 4

Mathematics

1 answer:

cluponka [151]2 years ago

6 0

Answer:

x=6

Step-by-step explanation:

5x - 8 = 3x + 4

+8 +8

5x=3x+12

-3x -3x

2x=+12

/2 /2

x=6

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Help me with problems 39 a-d

BabaBlast [244]

A) -4
d) -3.1
for these questions you have to subtract the two numbers since they have different signs and the biggest sign is the ending sign

8 0

3 years ago

Read 2 more answers

In triangle $ABC,$ point $D$ is on $\overline{AC}$ such that $AD = 3CD = 12$. If $\angle ABC = \angle BDA = 90^\circ$, then what

Neporo4naja [7]

Answer:

$BD=4\sqrt{3}\ units$

Step-by-step explanation:

we know that

$AD=12\ units$

$3CD=12$ ----> $CD=4\ units$

see the attached figure to better understand the problem

Triangles ABD and BCD are similar by AA Similarity Theorem

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional

$\frac{BD}{AD}=\frac{CD}{BD}$

substitute the given values

$\frac{BD}{12}=\frac{4}{BD}$

$BD^2=48$

$BD=\sqrt{48}\ units$

simplify

$BD=4\sqrt{3}\ units$

8 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

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

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

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

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

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