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timama [110]
4 years ago
9

at a factory that produces pistons for cars, machine 1 produced 700 satisfactory pistons and 300 unsatisfactory pistons today. m

achine 2 produced 707 satisfactory pistons and 303 unsatisfactory pistons today. suppose that one piston from machine 1 and one piston from machine 2 are chosen at random from today's batch. what is the probability that the piston chosen from machine 1 is unsatisfactory and the piston chosen from machine 2 is satisfactory?
Mathematics
1 answer:
viva [34]4 years ago
5 0
<span>the proccess of choosing is independent so probability of choosing from machin1 is:300/1000 =0.3
choosing from machin2 is:707/1010=0.7
the total result is 0.3*0.7=0.21 ;</span>
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Before being simplified, the instructions for computing income tax in Country R were to add 2 percent of one’s annual income to
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Answer:

Option C)

50 + \dfrac{I}{40}          

Step-by-step explanation:

We are given the following in the question:

I is the annual income of a person in a country R

2 percent of one’s annual income =

\dfrac{2}{100}\times I = 0.02I

1 percent of one’s annual income =

\dfrac{1}{100}\times I = 0.01I

Average of  100 units of Country R’s currency and 1 percent of one’s annual income.

=\dfrac{0.01I + 100}{2}

Income tax =

2 percent of one’s annual income + Average of  100 units of Country R’s currency and 1 percent of one’s annual income.

= 0.02I + \dfrac{100+0.01I}{2}\\\\=50 + \dfrac{0.04I + 0.01I}{2}\\\\=50 + \dfrac{0.05I}{2}\\\\= 50 + \dfrac{I}{40}

Thus, income tax is given by

Option C)

50 + \dfrac{I}{40}

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3 years ago
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Answer:

a) X \sim Binom(n=16, p=0.2)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

The expected value is given by this formula:

E(X) = np=16*0.2=3.2

b) P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Part a

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=16, p=0.2)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

The expected value is given by this formula:

E(X) = np=16*0.2=3.2

Part b

For this case we want this probability:

P(X=0)

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

And using this function we got:

P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815

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defon
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The answer is -1/16 or -0.0625, do you need to see work?
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