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earnstyle [38]
3 years ago
10

What is the common ratio for the geometric sequence? 18, 12, 8, 16/3,…

Mathematics
2 answers:
alex41 [277]3 years ago
6 0

Answer:

2/3

Step-by-step explanation:

Given : A geometric sequence : 18, 12, 8, 16/3,…

Solution :

Each term is found by multiplying the previous term by a constant in a Geometric Sequence .

In a Geometric Sequence  common ratio is denoted by r


r = \frac{a_{n} }{a_{n-1} }


for n = 2


r = \frac{a_{2} }{a_{1} }


⇒r = \frac{12 }{18 }


⇒r = \frac{2 }{3 }


for n = 3


r = \frac{a_{3} }{a_{2} }


⇒r = \frac{8 }{12 }


⇒r = \frac{4 }{6 }


⇒r = \frac{2}{3 }


Thus the common ratio 'r' = 2/3

ivolga24 [154]3 years ago
6 0
Basically, what should we multiply each term by to get next term

18 times what=12
what=12/18=2/3

12 times what=8
what=8/12=2/3

common ratio is 2/3
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Lydia drives from city a to city b to transport goods. her return speed is 3 times her departure speed and she takes 40 minutes
gizmo_the_mogwai [7]

Answer:

1 hour

Step-by-step explanation:

Hello, let's say that her departure trip takes t in minutes, as her return speed is 3 times her departure speed, she took t/3 for the return and we know that this 40 minutes less, so we can write.

t/3=t-40

We can multiply by 3

t = 3t -40*3 = 3t - 120

This is equivalent to

3t -120 = t

We subtract t

2t-120 = 0

2t = 120

We divide by 2

t = 120/2 = 60

So this is 60 minutes = 1 hour.

Thank you.

4 0
2 years ago
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Nataly_w [17]
108y-52y= y 56 Years Old
4 0
3 years ago
Admission tickets are $80 for adults and $50 for kids. The ration of kids to adults is 5:2. If they paid $10,000, find the numbe
Elan Coil [88]

Answer:

  80 adults

Step-by-step explanation:

The number of kids is $10,000/$50 = 200. The number of adults is 2/5 of that, or 80.

There are 80 adults in the group.

4 0
3 years ago
Please help with this I still do not know what I am doing
amm1812

Answer:

See Below

Step-by-step explanation:

150 = p + 78    ------ you want to get the p to one side and the other number the the other side of the =

(78 - 150) = p (78 - 78)   ---- what ever you do to one side you do to the other

72 = p

s - 51 = 12 (same thing as above)

s (-51 + 51)  = 12 + 51

s = 63

65 = blank + 7 ( you can put a letter like x or y because they number is not determined yet)

65 = x + 7 (same as above)

( 7 - 65 ) = x ( 7 - 7 )

58 = x

6 0
3 years ago
Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre
Whitepunk [10]

Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

Answer:

P(three quarters given two are quarters) = 1/7

PROBLEM 2:

Again, this is conditional probability. To help count the ways, let's instead count the ways to *not* draw any dimes. That means you have 2 choices for the first coin, 2 choices for the second coin and 2 choices for the third coin.

So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

3 0
3 years ago
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