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3 years ago

Calculate the chance of dice rolling the same time every roll

Mathematics

1 answer:

lilavasa [31]3 years ago

4 0

I think it would be a one i eight chance of that happening 1/8

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The ac is a leg of the right triangle True False

Vinil7 [7]

Answer:

False

Step-by-step:

The hypotenuse of the triangle is the side opposite to the right angle. The legs are the other two sides of the triangle. Since AC is on the opposite side of the right angle, it is the hypotenuse and therefore not a leg. Hope this helps!

3 0

3 years ago

Aliun [14]

The answer:

according to the image, the main theorem concerning right triangle similarity is as follow:
the altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and each other,

in our case, KL is the altitude, and by applying theorem, we get three triangles that are similar:
therefore:
△JKL ~ △JKM
△JKM ~ △JKM
△JMK ~ △KML

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3 years ago

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adell [148]

Answer:

1/2?

Step-by-step explanation:

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3 years ago

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Please help a girl out:

12345 [234]

355 cm because 3.55x100= 355

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3 years ago

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An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

5 0

3 years ago