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3 years ago

(( 20 pts. please please help with this, and explain if you can!! ))

Mathematics

1 answer:

liubo4ka [24]3 years ago

8 0

Answer:

-8,-40,-200,-1000,-5000,-25000

Step-by-step explanation:

a1 = -8

Let n = 2

a2 = 5 a1 = 5*(-8) = -40

Let n =3

a3 = 5 * a2 = 5 * -40 = -200

Let n =4

a4 = 5 * a3 = 5 * -200 = -1000

Let n =5

a5 = 5 * a4 = 5 * -1000 = -5000

Let n =6

a6 = 5 * a5 = 5 * -5000 = -25000

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Divide 54 into two parts so that four times the greater equals five time the less

LUCKY_DIMON [66]

Let the both the parts be 'x' and 'y'
x>y
We know that 4x = 5y
so x = 5y/4
x + y = 54
Replacing x we get
(5y/4) + y = 54
(5y +4y)/4 = 54
9y = 216
y = 24
Replacing y in ' x+y = 54'
we get x = 30

Hence the larger part is 30 whereas the smaller part is 24

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3 years ago

Round 25,678 to the nearest ten thousand

Art [367]

25,678 rounded to the nearest 10,000 would be 30,000

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4 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

ILL GIVE BRAINLIEST PLS HELP

yan [13]

Answer:

1/5

Step-by-step explanation:

The stick has a length of 5 units

The stick is broken at two points chosen at random

First break: the probability that you get a piece that is 1 unit or longer than 1 units= 1/5.

Second break, the probability that you get a piece that is 1 unit or longer than 1 units is 1/5.

Therefore,

The total probability =probability of first break * probability of second break * original stick unit

=1/5 * 1/5 * 5

= 1/25 *5

=5/25

=1/5

4 0

3 years ago

If angle 1 is not congruent to angle 2, which conclusion must be true?

netineya [11]

That they are not congruent

3 0

3 years ago