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3 years ago

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1.06g - 3 = 0.71 I need this answer

2 answers:

Arte-miy333 [17]3 years ago

7 0

<span>Simplifying 1.06g + -3 = 0.71 Reorder the terms: -3 + 1.06g = 0.71 Solving -3 + 1.06g = 0.71 Solving for variable 'g'. Move all terms containing g to the left, all other terms to the right. Add '3' to each side of the equation. -3 + 3 + 1.06g = 0.71 + 3 Combine like terms: -3 + 3 = 0 0 + 1.06g = 0.71 + 3 1.06g = 0.71 + 3 Combine like terms: 0.71 + 3 = 3.71 1.06g = 3.71 Divide each side by '1.06'. g = 3.5 Simplifying g = 3.5</span>

kkurt [141]3 years ago

7 0

1.06g -3=0.71 1.06g =0.71+3 1.06g=0.74 both side by 1.06 g =0.74/1.06

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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:

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

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

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

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

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

3 0

3 years ago

Welp !!!!!!!!!!!!!!!!!!

iren [92.7K]

It is <span>B</span>. All u have to do is -2a/b which is 2 then plug 2 in and you get (2,1)

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