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3 years ago

7

After the first term in a sequence, the ratio of each term to the preceding term is r. If the first term is a, what is the third

term?

1 answer:

irinina [24]3 years ago

7 0

Answer:

The 3rd term is

$ar^2$

Step-by-step explanation:

From the given information, the first term of the sequence is

$a_1=a$

The common ratio of the sequence is

$r$

The explicit formula for a geometric sequence is :

$a_n=a_1(r^{n-1})$

To find the third term, we put n=3 to get:

$a_n=a(r^{3-1})$

This simplifies to:

$a_3=ar^2$

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A baseball coach is creating a nine-player batting order by selecting from a team of 15 players. How many different batting orde

zzz [600]

Answer:

1,816,214,400 batting orders are possible.

Step-by-step explanation:

The order is important.

Suppose we had a two player batting order.

A batting order of Jonathan Schoop and Manny Machado is a different order than Manny Machado and Jonathan Schoop. So we use the permutations formula.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!)}$

A baseball coach is creating a nine-player batting order by selecting from a team of 15 players. How many different batting orders are possible?

Selection of 9 players from a set of 15 players. So

$P_{15,9} = \frac{15!}{(15-9)!} = \frac{15!}{6!} = 1816214400$

1,816,214,400 batting orders are possible.

4 0

2 years ago

A circle with radius of 1 cm sits inside a 3 cm x 3 cm rectangle.

alina1380 [7]

Answer:

6.14

Step-by-step explanation:

Find area of shaded region:

regtangle area - circle area = shaded area

Regtangle area:

3x3=9

Circle area:

Pi (3.14) x radius x radius= 3.14

area all together if shaded region

9-3.14=6.14

4 0

2 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

i.

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

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3 years ago

Mr. Kaplan bought 11 tickets to the circus and spent $50. He bought child tickets for $4 each and bought adult tickets for $7 ea

NNADVOKAT [17]

Adult - 2 tickets . child- 9 tickets.

7 0

2 years ago

What is the range of possible credit scores?

gogolik [260]

What class requires this tho? But its 300-850

8 0

2 years ago