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Charra [1.4K]
3 years ago
14

In a geometric sequence, the ratio between consecutive terms is _____ .

Mathematics
2 answers:
Rama09 [41]3 years ago
8 0

Answer:

In a geometric sequence, the ratio between consecutive terms is _____ .

Step-by-step explanation:

A geometric progression<em> is a sequence of real numbers in which the next element is obtained by multiplying the previous element by a constant called</em> the reason or factor of the progression.

5, 15, 45, <em>The reason would be to</em> multiply by 3 <em>the previous element.</em>

<u><em>The answer is</em></u>: <u>The reason or factor of the progression.</u>

juin [17]3 years ago
7 0
Tn = ar^(n-1) n = 1,2,3,4 a = 1st term r = common ratio, T2/T1 = T3/T2 = r
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Which of the following situations calls for a hypothesis test about a population mean?
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Answer:

Step-by-step explanation:

Hypothesis test about a population mean is done to evaluate two exclusive statements about a population in order to determine which is best supported by the sample data.

From the situations given, the correct options are

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2 years ago
Jade buys a blouse and a skirt for 3/6 of their original price. Jade pays a total of $31 for the two items. If the original pric
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Answer:

3/4 times 18= 13.5

31.5-13.5=18  

It cost 18 dollars for the skirt with a discount.

It cost 13.5 for the blouse as opposed to the original price of 18 dollars.

18(skirt)+13.5(blouse)=31.5

If you set up a proportion to find the original price of the skirt you will get the answer.  

18/x=75/100

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Step-by-step explanation:

4 0
3 years ago
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

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5x-3x+2x=4x=13

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\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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