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3 years ago

Help if you see a troll report them

Mathematics

1 answer:

12345 [234]3 years ago

7 0

Answer:

I've seen multiples of trolling going on in brainly, and i could barely report them all.

Step-by-step explanation:

You might be interested in

What is 1,562 divided by 34

marishachu [46]

This would be 45.9
Get a calculator or try long division :-)

3 0

2 years ago

Read 2 more answers

In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that

attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let X = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252$

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0

2 years ago

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

If ED = 10 and DB = 3x-8, what is the length of EB 1. 32 2. 22 3. 10 4. 20

Debora [2.8K]

Based on the picture, we know that ED is equal to DB. Therefore:

2 times ED=EB
2(10)=20

EB is 20

4 0

3 years ago

Read 2 more answers

Please help me out!!! :)

alexgriva [62]

Answer:

x = 26°

Step-by-step explanation:

The measure of the secant- secant angle x is one- half the difference of the measures of the intercepted arcs , larger subtract smaller

x = $\frac{1}{2}$ (66° - 14°) = 0.5 × 52° = 26°

8 0

3 years ago