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3 years ago

H(n) = -10 + 12n Complete the recursive formula of h(n). h(1) = h(n) = h(n-1)+

Mathematics

1 answer:

OlgaM077 [116]3 years ago

5 0

Answer:

h(n) = h(n - 1) + 12

Step-by-step explanation:

Generate the first few terms using the explicit formula

h(1) = - 10 + 12 = 2

h(2) = - 10 + 24 = 14

h(3) = - 10 + 36 = 26

h(4) = - 10 + 48 = 38

This is an arithmetic sequence. To find a term in the sequence add 12 to the previous term, thus

h(n) = h(n - 1) + 12 with h(1) = 2

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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

Solve for g. Show your work! g+6-3=31

Zielflug [23.3K]

Answer:

g=28, that may not be how the teacher would want you to show your work, but that's how I find the answer!

Step-by-step explanation:

6-3=3

31-3=28

28=g

So therefore 28+6-3=31

6 0

3 years ago

Graph the circle (x+6)^2 + (y+5)^2 =9

Andrei [34K]

Answer:

This a circle centered at the point $(-6,-5)$ , and of radius "3" as it is shown in the attached image.

Step-by-step explanation:

Recall that the standard formula for a circle of radius "R", and centered at the point $(x_0,y_0)$ is given by:

$(x-x_0)^2+(y-y_0)^2=R^2$

Therefore, in our case, by looking at the standard equation they give us, we extract the following info:

1) $R^2 = 9\,\,\,then\,\,\,R=3$ since the radius must be a positive number and ( $R=-3$ ) is not a viable answer.

2) $x_0=-6$ for ( $x-x_0$ ) to equal $(x+6)$

3) $y_0=-5$ for ( $y-y_0$ ) to equal $(y+5)$

Therefore, we are in the presence of a circle centered at the point $(-6,-5)$ , and of radius "3". That is what we draw as seen in the attached image.

5 0

3 years ago

Which symbol replaces the box to make the statement true? 2⋅18−18□2⋅[28−3⋅(2+4)] <br>A. <br>C. =

Misha Larkins [42]

Answer:

2⋅18−18 is less than 2⋅[28−3⋅(2+4)]

Step-by-step explanation:

7 0

3 years ago

What is the approximate area of the shaded sector in the circle shown below?

stich3 [128]

Answer: 26.5cm2

Step-by-step explanation:

4 0

3 years ago