3 years ago

120, 60, 30, 15, ... 8th term: 10th term:

Mathematics

1 answer:

Kryger [21]3 years ago

8 0

Answer:

8th term would be .9375

10th term would be .234375

Step-by-step explanation:

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myrzilka [38]

Answer:

$2\pi \: r$

<h3>4th answer is correct</h3>

Step-by-step explanation:

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3 years ago

R/19 > -1<br> please help solve for R

pantera1 [17]

Answer:

Step-by-step explanation:

$\frac{r}{19}>-1$

Multiply both sides by 19

$\frac{r}{19}*19>-1*19 \\\\r > -19$

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3 years ago

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Elena L [17]

Answer:

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Step-by-step explanation:

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3 years ago

The number of typing errors made by a typist has a Poisson distribution with an average of two errors per page. If more than two

wlad13 [49]

Answer: 0.6767

Step-by-step explanation:

Given : Mean = $\lambda=2$ errors per page

Let X be the number of errors in a particular page.

The formula to calculate the Poisson distribution is given by :_

$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$

Now, the probability that a randomly selected page does not need to be retyped is given by :-

$P(X\leq2)=P(0)+P(1)+P(2)\\\\=(\dfrac{e^{-2}2^0}{0!}+\dfrac{e^{-2}2^1}{1!}+\dfrac{e^{-2}2^2}{2!})\\\\=0.135335283237+0.270670566473+0.270670566473\\\\=0.676676416183\approx0.6767$

Hence, the required probability :- 0.6767

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3 years ago

-6+(-1)=<br> ..................

Mice21 [21]

Answer:

It's -7

Step-by-step explanation:

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3 years ago

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