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2 years ago

What is the common difference between ratio between successive terms in the sequence 1.5, 1.2, 0.96, 0.768

Mathematics

2 answers:

nikklg [1K]2 years ago

6 0

First, we should see if there are any common differences. We do this by subtracting each term from its previous:

1.5 - 1.2 = 0.3

1.2 - 0.96 = 0.24

Already, we can tell that there is no common difference.

Let's look at common ratio:

1.2/1.5 = 0.8

0.96/1.2 = 0.8

0.768/0.96 = 0.8

Therefore, there is a common ratio and it is 0.8.

sladkih [1.3K]2 years ago

4 0

Answer:

Answer is D: 0.8

Step-by-step explanation:

r = 1.2 / 1.5 = 0.8

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A 90% confidence interval for the mean height of students

Westkost [7]

Answer:

$ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635$

And the best answer on this case would be:

b) m = 4.635

Step-by-step explanation:

Let X the random variable of interest and we know that the confidence interval for the population mean $\mu$ is given by this formula:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

The confidence level on this case is 0.9 and the significance $\alpha=1-0.9=0.1$

The confidence interval calculated on this case is $60.128 \leq \mu \leq 69.397$

The margin of error for this confidence interval is given by:

$ME =t_{\alpha/2} \frac{s}{\sqrt{n}}$

Since the confidence interval is symmetrical we can estimate the margin of error with the following formula:

$ME = \frac{Upper -Lower}{2}$

Where Upper and Lower represent the bounds for the confidence interval calculated and replacing we got:

$ME= \frac{69.397-60.128}{2}= 4.6345 \approx 4.635$

And the best answer on this case would be:

b) m = 4.635

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3 years ago

Suppose that farmers grew and unexpectedly large number of tomatoes this year .how would this increase in production affect the

zloy xaker [14]

Answer:

The price would go down as people would be more willing to buy cheaper potatoes and the producers would want to sell them for a lower price to get rid of them easy and fast.

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Aleksandr-060686 [28]

$D:x>0\\
\log x+\log x^2+\log x^3=-6\\
\log (x\cdot x^2\cdot x^3)=\log 10^{-6}\\
x^6=10^{-6}\\
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