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3 years ago

Write the first five terms of the geometric sequence in which a1 = 27 and the common ratio is 4/3

Mathematics

1 answer:

tangare [24]3 years ago

5 0

Answer:

c) 27, 36, 48, 64, 256/3

Step-by-step explanation:

We are given a geometric sequence with;

First term, a = 27
Common ratio, r = 4/3

We are supposed to write the sequence;

We know that a given term in a GP is given by;

$T_n=ar^n^-^1$ , where a is the first term, n is the nth term

Therefore;

$First term, a = 27 \\Second term = ar^1= 27(4/3) = 36\\Third term = ar^2=27(4/3)^2=48\\Fourth term =ar^3=27(4/3)^3=64\\Fifth term = ar^4 = 27(4/3)^4 =\frac{256}{3} \\And so forth$

Therefore;

The geometric sequence is;

$27,36,48,64,\frac{256}{3}$

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In 1980, the median age of the U.S. population was 30.0; in 2000, the median age was 35.3.

miss Akunina [59]

Answer:

y = 0.265x - 494.7

Step-by-step explanation:

Let median age be represent by 'a' and time be represent by 't'

In 1980, median age is given 30

which means that

a₁ = 30

t₁ = 1980

In 2000, the median age is given 35.3

which means that.

a₂ = 35.3

t₂ = 2000

The slope 'm' of the linear equation can be found by:

m = (a₂ - a₁) /(t₂ - t₁)

m = (35.3 - 30)/(2000-1980)

m = 0.265

General form of linear equation is given by:

y = mx + c

y = 0.265x +c

Substitute point (1980,30) in the equation.

30 = 0.265(1980) + c

c = -494.7

Hence the the linear equation can be written as:

y = mx + c

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3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

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