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3 years ago

Decide whether the sequence is arithmetic geometric or neither 5,-10,20, -40

1 answer:

7 0

Try this option^
note, that all members of this sequence have '-' and '+' and 5*(-2)=-10, -10*(-2)=20; 20*(-2)=-40.
It means, that the suggested sequence is geometric, where b₁=5, q=-2.

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How to solve <img src="https://tex.z-dn.net/?f=log_%7By%7D%285y%29%20%3D%202" id="TexFormula1" title="log_{y}(5y) = 2" alt="log_

Natalija [7]

Answer:

y = 5

Step-by-step explanation:

Expand the logarithm:

$\log_y{(5)}+\log_y{(y)}=2\\\\\dfrac{\log{(5)}}{\log{(y)}}+1=2 \quad\text{change of base formula}\\\\\dfrac{\log{(5)}}{\log{(y)}}=1 \quad\text{subtract 1}\\\\\log{(5)}=\log{(y)} \quad\text{multiply by log(y)}\\\\5=y \quad\text{take the anti-log}$

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You can also take the antilog first:

5y = y²

y(y -5) = 0 . . . . . subtract 5y, factor

y = 0 or 5 . . . . . y=0 is not a viable solution, so y=5.

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How can I estimate 26×22

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3 years ago

Help binomial distributions

Ymorist [56]

We have the formula to compute the probability of having exactly k successed over n trials, given a probability p of success (and implicitly a probability 1-p of failure), which is

$P(\text{k successed on n trials}) = \binom{n}{k}p^k(1-p)^{n-k}$

Now, the probability of at least 3 successes is the union of the following event: exactly three successes,exactly four successes and exactly five successes.

We can compute their probability and sum them: $\binom{5}{3}\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^2 + \binom{5}{4}\left(\frac{3}{7}\right)^4\left(\frac{4}{7}\right)^1 + \binom{5}{5}\left(\frac{3}{7}\right)^5 \approx 0.36788$

So, the answer is about 36.79%

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