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3 years ago

If m PQS=16, M SQR=(9x+17),and m PQR=(12x-6) find m PQR Please answer with work

Mathematics

1 answer:

docker41 [41]3 years ago

4 0

Answer:

$\angle PQR=30^{\circ}$

Step-by-step explanation:

It is given that,

$\angle PQS=16^{\circ}\\\\\angle SQR=(9x+17)^{\circ}\\\\\angle PQR=(12x-6)^{\circ}$

We need to find the measure of $\angle PQR$ .

From the given figure, $\angle PQR=\angle PQS+\angle SQR$

Putting all the values, we get :

$12x-6=16+9x+17\\\\12x-9x=16+17+6\\\\3x=39\\\\x=13$

$\angle PQR=12x-6\\\\=12(3)-6\\\\=30^{\circ}$

So, the measure of $\angle PQR$ is 30 degrees.

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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.

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

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3 years ago

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Answer:

Step-by-step explanation:

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slavikrds [6]

Answer:

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4 years ago

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