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podryga [215]
2 years ago
12

Can someone help? If b/a=c/b=d/c=2/3, find a:b:c:d.

Mathematics
1 answer:
blagie [28]2 years ago
8 0

Answer: a:b:c:d = 16:24:30:40

You can also divide the ratio by 2 because all terms have 2 in common. Which would be  a:b:c:d = 8:12:15:20

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Answer:

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Step-by-step explanation:

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3 years ago
Which expression is the factored form of -4.5n + 3​
IgorLugansk [536]

Answer:

-1.5 (3n-2)

Step-by-step explanation:

Factor -1.5 out of -4.5n+3

6 0
3 years ago
In a geometric sequence, the ratio between consecutive terms is...
Leokris [45]
<h3>Answer: The same or equal</h3>

========================================================

Explanation:

Consider an example like 3, 6, 12, 24, 48, ...

The ratio between consecutive terms is

  • 6/3 = 2
  • 12/6 = 2
  • 24/12 = 2
  • 48/2 = 2

Each time we divide any given term over its previous one, we get the same ratio 2. We call this the common ratio. In terms of notation, the variable r is used for the common ratio, so r = 2 in this case.

As another example, the geometric sequence 5, 50, 500, 5000, ... has r = 10 as the common ratio because we multiply each term by 10 to get the next one. Moving forward has us multiply by r, moving backward and we divide by r. The value of r cannot be zero, but it can be negative.

An example with r being negative would be something like

1, -1, 1, -1, 1, -1, ....

we just bounce back and forth between those two values. In this case, r = -1.

6 0
3 years ago
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victus00 [196]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
In a horse race with 5 horses, you make a bet by predicting the ranking of all 5 horses. Suppose you place your bet at random. W
lapo4ka [179]

Answer:

P( top two horses are predicted incorrectly in incorrect order)

= \frac{1}{2}

Step-by-step explanation:

In the horse race the outcome can be predicted in 5! = 120 ways.

Now suppose the top two horses were predicted incorrectly in incorrect order. Now, the  top horse can be predicted incorrectly in 4 ways.

Suppose the top horse was predicted to be in k-th position where k = 2, 3 ,4,5

so the second horse can be predicted to be in place from 1 to (k - 1)

So, the top two horses can be predicted  incorrectly in incorrect order

in \sum_{k =2}^{5}(k - 1) = 10 ways  

and for each prediction of the two the remaining horses may be predicted in 3! = 6 ways.

Hence ,

P( top two horses are predicted incorrectly in incorrect order)

= \frac{6 \times 10}{120}

=\frac{1}{2}

 

8 0
3 years ago
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