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4 years ago

6

Sven has 11 more than twice as many customers as when he started selling newspapers.He now has 73 customers.How many did he have

when he started

2 answers:

Daniel [21]4 years ago

7 0

Sven had 31 customers when he first started selling newspapers.

sladkih [1.3K]4 years ago

5 0

the correct answer to this and im not 100% sure but I think its 51.

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Find the answer to each of the fot

stich3 [128]

Answers:
1 is -8
2 is 29
3 is -15
4 is -1

Hope this answer helps and have a nice day

3 0

3 years ago

An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is

vladimir1956 [14]

Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

b) $P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

c) $P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

Part b

We want this probability:

$P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

We find the individual probabilities like this:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

Part c

For this case we want this probability:

$P(X \geq 5)$

And we can use the complement rule like this:

$P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

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Define the term isomer

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3 years ago

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Solve the following system. x-y=2. 3x+z=11. y-2z=-3

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3 years ago