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2 years ago

Find the 7th term of the geometric sequence with the given terms:

1 answer:

3 0

For a geometric progression, a(n) = a(1)*r^(n-1)
where a(n) is the nth term
r is the common ratio
n is the number of terms

First we need to obtain the common ratio by using a(2) and a(4). The number of terms (n) is 3.

Therefore

48 = 768* (r)^2

thus, r = 0.25

solving for a(7)

a(7) = 768 * (0.25)^(5)
a(7) = 0.75

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The erea of a square is given by x2, where x is the length of one side. Mary's original garden was in the shape of a square. She

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Answer: The expression that represents the area of Mary's new garden =2x^2

Step-by-step explanation:

The area of a square is given by x^2, where x is the length of one side. The length of the other side is also x. In a square, all sides are equal. each of the four sides equal x

Area = x × x

Area = x^2

She decided to double the area of her garden

New area = 2(x × x)

New area = 2x^2

The expression that represents the area of Mary's new garden =2x^2

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2 years ago

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Leya [2.2K]

The answer is 6.4

It is positive because the 11.4 is larger than the negatvie 5.

I hope this helps!

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Make it into 3 shapes

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2 years ago

Heidi grew her hair out for many years. Her hair was 1\dfrac{1}{3}1 3 1 1, start fraction, 1, divided by, 3, end fraction of a

RSB [31]

Answer:

Heidi hass really kong hair and should cut it

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2 years ago

A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander aroun

Brut [27]

Answer:

The expected number of minutes the rat will be trapped in the maze is 21 minutes.

Step-by-step explanation:

The rat has two directions to leave the maze.

The probability of selecting any of the two directions is, $\frac{1}{2}$ .

If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability $\frac{1}{3}$ after 2 minutes. And with probability $\frac{2}{3}$ the rat will return to the starting point after 5 minutes of wandering.

Let <em>X</em> = number of minutes the rat will be trapped in the maze.

Compute the expected value of <em>X</em> as follows:

$E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21$

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.

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2 years ago