Please Solve Now!!! Will Give Brainliest!!!

Mathematics

1 answer:

zhuklara [117]3 years ago

5 0

Answer:

Step-by-step explanation:

1, sec theta * sin theta = 1/cos * sin = sin/cos =tan theta

2. cosec theta * cos theta = 1/ sin * cos = cos/sin = cot theta

3, tan theta * cosec theta =sin/cos * 1/sin = 1/cos

4, cot theta * sec theta = cos /sin * 1/cos= 1/sin

5, 1 - cos^2 theta = sin ^2 theta

6, sec^2 theta - 1 =tan ^2 theta

7 , cosec^2 theta - cot^2 theta = 1

8, 1 - sin^2 theta = cos^2

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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)

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