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2 years ago

Find the next two terms. 2, 10, 50, ?, ?

Mathematics

2 answers:

Setler79 [48]2 years ago

3 0

Hi there!

The next 2 terms are 250 and 1,250.

We have a geometric sequence, which means we use multiplication or division to get to the next term.

Here, we multiply by 5 to get to the next term.

So we multiply 50 by 5:

50*5=250

We multiply 250 by 5 to get to the next term:

250*5=1,250

And we're done!

<h2>Therefore, the next two terms are: 250 and 1,250.</h2>

Hope this helps!

~Just a felicitous girlie

#HaveASplendidDay

$SilentNature :)$

emmasim [6.3K]2 years ago

3 0

Answer:

250, 1250

Step-by-step explanation:

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by 5 gives the next term.

Hope that helps!

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Laguna del carbon, a salt lake located in. Argentina, lies 344 feet below sea level. El alto, Bolivia, is located approximately

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Answer:

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Given that Laguna del carbon, a salt lake located in Argentina, lies 344 feet below sea level, while El Alto, Bolivia, is located approximately 13,000 feet above sea level, to determine what is the difference in elevation between Laguna del Carbon and El Alto, the following calculation must be performed:

13,000 - 344 = X

12.656 = X

Thus, the difference in elevation between Laguna del Carbon and El Alto is 12,656 feet.

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2 years ago

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2 years ago

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3 years ago

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3 years ago

Read 2 more answers

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

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3 years ago