(a) The probability of a computer failure from Company A is 0.001; from Company B is 0.002; and from Company C is 0.005.
Therefore, the probability that a computer will experience a hard drive failure within one year is:(0.50 x 0.001) + (0.30 x 0.002) + (0.20 x 0.005)= 0.0012. The probability of a randomly selected computer experiencing a hard drive failure within one year is 0.0012 or 0.12%.
(b) Bayes' theorem will be used to calculate this probability:Let A be the event that the computer's hard drive was manufactured by Company C. Let B be the event that the computer experienced a hard drive failure. P(A|B) is the probability that the hard drive was manufactured by Company C given that a hard drive failure was experienced.
P(A|B) = P(B|A) P(A) / P(B) Where: P(B|A) = 0.005 (the probability of failure if the hard drive was manufactured by Company C)P(A) = 0.20 (the proportion of hard drives that the computer manufacturer gets from Company C)P(B) = (0.50 x 0.001) + (0.30 x 0.002) + (0.20 x 0.005) = 0.0012 (as in part a)
Therefore: P(A|B) = (0.005 x 0.20) / 0.0012 = 0.0833 or 8.33%.
(c)Let A be the event that both hard drives were manufactured by Company A; B be the event that both hard drives were manufactured by Company B; and C be the event that both hard drives were manufactured by Company C. Then we need to find the probability of event A or B or C, given that a hard drive failure was experienced:P(A U B U C|F) = P(F|A U B U C) P(A U B U C) / P(F)where F is the event that the hard drive in the replacement computer fails.P(F|A U B U C) = P(F) = (0.50 x 0.001) + (0.30 x 0.002) + (0.20 x 0.005) = 0.0012P(A U B U C) = (0.50)^2 + (0.30)^2 + (0.20)^2 = 0.46P(F) = P(A U B U C) P(F|A U B U C) + P(A' n B n C) P(F|A' n B n C)= 0.46 x 0.0012 + 0.04 x 0.3 = 0.000552P(A U B U C|F) = P(F|A U B U C) P(A U B U C) / P(F)= (0.0012 x 0.46) / 0.000552 = 1.00or 100%. Therefore, the probability that the original and replacement computers were produced by the same company is 100%.
(d) Bayes' theorem will be used to calculate this probability:Let A be the event that the hard drive was manufactured by Company C. Let B be the event that the computer did not experience a hard drive failure. P(A|B) is the probability that the hard drive was manufactured by Company C given that no hard drive failure was experienced.P(A|B) = P(B|A) P(A) / P(B)Where:P(B|A) = 1 - 0.005 = 0.995 (the probability that the hard drive did not fail if it was manufactured by Company C)P(A) = 0.20 (as in part b)P(B) = 1 - (0.50 x 0.001) - (0.30 x 0.002) - (0.20 x 0.005) = 0.9988
Therefore:P(A|B) = (0.995 x 0.20) / 0.9988 = 0.1989 or 19.89%. Therefore, the probability that the hard drive was manufactured by Company C given that it did not fail is 19.89%.
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The probability that the hard drive was manufactured by company C given that a failure was not experienced by the computer within one year is approximately 0.256.
(a)Probability that a randomly chosen computer purchased from this manufacturer will experience a hard drive failure within one year = 0.5 x 0.001 + 0.3 x 0.002 + 0.2 x 0.005 = 0.0016
(b)Let's denote the event that a computer failure is experienced within one year by F and the event that the hard drive is made by company C by C.
Then we are required to calculate P(C | F), which is the probability that the hard drive was manufactured by company C given that a failure was experienced by the computer within one year. This can be found by using the Bayes' rule as follows:
[tex]$$P(C|F) = \frac{P(F|C)P(C)}{P(F|A)P(A) + P(F|B)P(B) + P(F|C)P(C)}$$[/tex]
where P(C) = 0.2, P(A) = 0.5 and P(B) = 0.3.$$P(F|A) = 0.001, P(F|B) = 0.002, P(F|C) = 0.005$$
Thus, we have:[tex]$$P(C|F) = \frac{0.005 \times 0.2}{0.001 \times 0.5 + 0.002 \times 0.3 + 0.005 \times 0.2} \approx 0.476$$[/tex]
Therefore, the probability that the hard drive was manufactured by company C given that a failure was experienced by the computer within one year is approximately 0.476.
(c)Let's denote the event that the original hard drive is manufactured by company A, B and C by A, B, and C respectively.
Similarly, let's denote the event that the replacement hard drive is manufactured by company A, B, and C by A', B', and C' respectively.
We are required to calculate P(A = A', B = B', C = C' | F), which is the probability that the hard drives in the original and replacement computers were manufactured by the same company given that a failure was experienced by both computers within one year.
This can be found by using the Bayes' rule as follows:
[tex]$$P(A = A', B = B', C = C'|F) = \frac{P(F|A = A', B = B', C = C')P(A = A')P(B = B')P(C = C')}{P(F)}$$[/tex]
where: [tex]$$P(F) = P(F|A = A', B = B', C = C')P(A = A')P(B = B')P(C = C') + P(F|A \ne A', B \ne B', C \ne C')P(A \ne A')P(B \ne B')P(C \ne C')$$[/tex]
Here, we are assuming that the probabilities of computer failure are independent of each other and the company that manufactured the hard drives of the two computers are independent of each other. Therefore, we have:
[tex]$$P(F|A = A', B = B', C = C') = P(F|A)P(F|B)P(F|C) = 0.001 \times 0.002 \times 0.005$$[/tex]
[tex]$$P(F|A \ne A', B \ne B', C \ne C') = 0$$[/tex]
Also, we have:$$P(A = A') = P(B = B') = P(C = C') = \frac{1}{3}$$
[tex]$$P(A \ne A', B \ne B', C \ne C') = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$$[/tex]
Thus, we have:$$P(A = A', B = B', C = C'|F) = \frac{0.001 \times 0.002 \times 0.005 \times (\frac{1}{3})^3}{P(F)}$$
[tex]$$P(A \ne A', B \ne B', C \ne C'|F) = \frac{P(F) - 0.001 \times 0.002 \times 0.005 \times (\frac{1}{3})^3}{\frac{8}{27}}$$[/tex]
Now, we need to find P(F). This can be done as follows:
[tex]$$P(F) = P(F|A = A', B = B', C = C')P(A = A')P(B = B')P(C = C') + P(F|A \ne A', B \ne B', C \ne C')P(A \ne A')P(B \ne B')P(C \ne C')$$$$= 0.001 \times 0.002 \times 0.005 \times (\frac{1}{3})^3 + 0 = 4.6296 \times 10^{-8}$$Thus, we have:$$P(A = A', B = B', C = C'|F) = 0.0296$$[/tex]
[tex]$$P(A \ne A', B \ne B', C \ne C'|F) = 0.9704$$[/tex]
Therefore, the probability that the hard drives in the original and replacement computers were manufactured by the same company given that a failure was experienced by both computers within one year is 0.0296.(d)Let's denote the event that the hard drive is made by company C by C and the event that a computer failure is not experienced within one year by F'. We are required to calculate P(C | F'), which is the probability that the hard drive was manufactured by company C given that a failure was not experienced by the computer within one year. This can be found by using the Bayes' rule as follows:
[tex]$$P(C|F') = \frac{P(F'|C)P(C)}{P(F'|A)P(A) + P(F'|B)P(B) + P(F'|C)P(C)}$$[/tex]
where P(C) = 0.2, P(A) = 0.5 and P(B) = 0.3.$$P(F'|A) = 0.999, P(F'|B) = 0.998, P(F'|C) = 0.995$$
Thus, we have: [tex]$$P(C|F') = \frac{0.995 \times 0.2}{0.999 \times 0.5 + 0.998 \times 0.3 + 0.995 \times 0.2} \approx 0.256$$[/tex]
Therefore, the probability that the hard drive was manufactured by company C given that a failure was not experienced by the computer within one year is approximately 0.256.
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Can someone pleaseeee helppp i dont know how to do this
Answer:
C=3e
Step-by-step explanation:
Slope=
You surveyed the number of tree species along the American River watershed, and obtain the following data set. Please respond to the following questions. Species a b Forest A Number 10 8 3 Forest B Number 5 6 0 7 10 Forest C Number 8 8 5 2 NINO d 1 e 1 2 Which forest has the lowest species richness, A, B, or c?
After considering the given data and analysing the information carefully we conclude that the lowest species richness observed is Forest A with only 18 species.
Let us get into the explanation part by first keeping in mind that to determine this, we need to evaluate the total number of species in each forest.
From the given data set, we clearly see that Forest A has 18 species, Forest B has 28 species, and Forest C has 25 species.
Hence, Forest A has the lowest species richness with only 18 species then Forest A has the lowest species richness among the three forests.
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4th grade math helplolll
Answer:
175 dollars
Step-by-step explanation:
from that word problem the equation that i was able to get out of it was
1225/ 7
^ the travel allowance and how much days they are going to be there
i dont know if you are allowed to use a calculator in your class but puting 1225/ 7 does give you a answer of a positive number of 175 dollars
meaning that they can only use 175 dollars a day so they dont go over the allowance
i hope this helps you! please stay safe and have a good day :)
Answer:
Um, the answer for that is 175.
Step-by-step explanation:
all you need to do is divide $1,225 by 7 which is 175.
Problem 8-28 The exponential distribution applies to lifetimes of a certain component. Its failure rate is unknown. Find the probability that the component will survive past 5 years assuming:
(a) lambda=.5
Pr=
(b) lambda=0.9
Pr=
(c) lambda=1.1
Pr=
The exponential distribution applies to the lifetimes of a certain component. Its failure rate is unknown. The probability that the component will survive the past 5 years assumes:
(a) lambda=.5
Pr= 0.082
(b) lambda=0.9
Pr= 0.082
(c) lambda=1.1
Pr= 0.036
In the exponential distribution, the failure rate is a degree of the way fast the factor is expected to fail. It is regularly denoted through the parameter lambda (λ).
The opportunity that a thing will continue to exist beyond a positive time may be calculated using the exponential survival function, which is given by:
[tex]Pr(X > t) = e^(-λt)[/tex]
where X represents the random variable denoting the life of the thing, t is the specific time, and e is the bottom of the herbal logarithm.
Now let's calculate the possibilities for each case:
(a) lambda = 0.5, t = 5
Pr(X > 5) = [tex]e^(-0.5 * 5)[/tex] ≈ 0.082
In this example, with a lambda of 0.5, the element has a notably low failure price. The opportunity of the thing surviving beyond 5 years is about 0.082, or 8.2%.
(b) lambda = 0.9, t =5
Pr(X > 5) = [tex]e^(-0.9 * 5)[/tex] ≈ 0.082
With a lambda of 0.9, the issue has a slightly higher failure rate as compared to the previous case. The probability of the aspect surviving beyond 5 years stays at about 0.082, or 8.2%.
(c) lambda = 1.1, t = 5
Pr(X > five) = [tex]e^(-1.1 * 5)[/tex]≈ 0.036
In this situation, with a lambda of one.1, the factor has a better failure fee. The possibility of the element surviving beyond 5 years decreases to approximately 0.036, or 3.6%.
In precis, the possibility of a component surviving the past five years in an exponential distribution relies upon the failure price parameter lambda.
A lower failure price ends in a higher chance of survival, at the same time as a higher failure price decreases the opportunity of survival. It is essential to don't forget these chances when assessing the reliability and toughness of additives in diverse packages.
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tricia runs 520 meters each day for 3 days. how many total kilometers does tricia run in these 3 days?
Answer: 1.56 kilometers
Step-by-step explanation:
you multiply 520 by 3 because she ran for three days
so she ran 1560 meters
1000 meters= 1 kilometer
that makes it 1.56 kilometers
Help ASAP! Find The Area Of A Circle With R =20.5
Answer:
Step-by-step explanation:
Pi*r^2 = Area
20.5^2 * Pi = 1320.25
At the 90% Confidence Interval, what are the (lower bound; upper bound)?
The lower bound of the interval is given as follows:
28.1.
The upper bound of the interval is given as follows:
29.9.
What is a t-distribution confidence interval?The t-distribution is used when the standard deviation for the population is not known, and the bounds of the confidence interval are given according to the equation presented as follows:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
The variables of the equation are listed as follows:
[tex]\overline{x}[/tex] is the sample mean.t is the critical value.n is the sample size.s is the standard deviation for the sample.The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 30 - 1 = 29 df, is t = 1.6991.
The lower bound of the interval is given as follows:
[tex]39 - 1.6991 \times \frac{3}{\sqrt{30}} = 38.1[/tex]
The upper bound is given as follows:
[tex]39 + 1.6991 \times \frac{3}{\sqrt{30}} = 39.9[/tex]
Missing InformationThe complete problem is:
"If n=30, (x-bar)=39, and s=3, at the 90% Confidence Interval, what are the (lower bound; upper bound)".
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If $120.99 is charged for 654 units of electricity used,find the cost of one unit of electricity
Answer:
$5.41
Step-by-step explanation:
654 divided by 120.99=5.405... therefore the answer is $5.41
(50 POINTS) Write out each sum.
Step-by-step explanation:
12. n^2+2n
if you insert 1 for k and then work up by inserting 2 for k and adding those together and stoping at n.
13. 8-2(2^n)
if you insert 3 for k and then work up by inserting 4 for k and adding those together and keep on going but stopping at n.
Hope that helps :)
Gary is 4 years less than three times his brothers age, \displaystyle bb. The sum of Gary and his brothers' age is 52. Write an equation to represent this relaitonship.
(PLS HELP ME WITH THAT QUESTION! IF YOU HELP ME I GIVE YOU BRAINLIEST I SWEAR)
For each of the following questions, draw the phase portrait as function of the control parameter μ. classify the bifurcations that occur as μ varies, and find all the bifurcation values of μ .
1. θ = μ sin θ - sin 2θ
2. θ = sin θ/ μ+cos θ
3. θ = sin θ / μ + sin θ
4. θ = μ + cos θ + cos 2 θ
5. θ = μ sin θ + cos 2θ
6. θ = sin 2θ/ 1 + μ sin θ
Phase portrait as a function of the control parameter μ and the classification of bifurcations that occur as μ varies in the following questions are:
1. θ = μ sin θ - sin 2θA) μ<0, stable equilibrium at θ = nπ, where n is an odd integerB) μ>0, stable equilibrium at θ = 0, unstable equilibrium at θ = nπ, where n is a non-zero even integer. Hence, we have homoclinic bifurcation at μ = 0.
2. θ = sin θ/ μ+cos θA) μ<1, stable equilibrium at θ = nπ, where n is an integerB) μ>1, stable equilibrium at θ = sin−1 (μ) + nπ, where n is an integer. Hence, we have a pitchfork bifurcation at μ = 1.
3. θ = sin θ / μ + sin θA) μ<−1, stable equilibrium at θ = nπ, where n is an integerB) μ>−1, stable equilibrium at θ = 0, unstable equilibrium at θ = nπ, where n is a non-zero integer. Hence, we have homoclinic bifurcation at μ = −1.
4. θ = μ + cos θ + cos 2θA) μ>−1, stable equilibrium at θ = nπ, where n is an even integerB) μ<−1, no equilibrium point exists. Hence, we have fold bifurcation at μ = −1.
5. θ = μ sin θ + cos 2θA) μ>0, stable equilibrium at θ = sin−1 (−μ) + 2nπ, where n is an integerB) μ<0, stable equilibrium at θ = sin−1 (−μ) + (2n+1)π, where n is an integer. Hence, we have pitchfork bifurcation at μ = 0.
6. θ = sin 2θ/ 1 + μ sin θA) μ<−1, unstable equilibrium at θ = nπ/2, where n is an odd integerB) μ>−1, unstable equilibrium at θ = 0, stable equilibrium at θ = π. Hence, we have pitchfork bifurcation at μ = −1.
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help me bros, this question is a big part of my grade!!
Answer:
6.
m1 : 61
m2 : 61
m3 : 29
7.
m1 : 87
m2 : 45
m3 : 45
m4 : 52
What is the simple interest on $4,000 for 2 and a half years at 4 percent a year?
Answer:
Step-by-step explanation:
You can't receive money if you withdraw in the midst of a year.
So 4000 * 1/25 * 2 = $4320
Find the volume of this triangular pyramid.
Answer:
v = 340 cm³
Step-by-step explanation:
base area = 12 x 10 x 0.5 = 60 cm²
v = 60 x 17 x 1/3 = 340 cm³
Find the volume of a cylinder with a diameter 4 mm and a height 8 mm.
Answer:100.53mm
Step-by-step explanation:
Solve the word problem using the plotted points on the graph.
Answer:
In the graph, we can see that the relation between length and weight is given by the adjusted line, which passes through the points (24, 16) and (28, 25)
Remember that a linear relation can be written as:
y = a*x + b
Where a is the slope and b is the y-intercept.
If this line passes through the points (x₁, y₁) and (x₂, y₂) the slope can be calculated as:
a = (y₂ - y₁)/(x₂ - x₁)
Then in our case, the slope will be:
a = (25 - 16)/(28 - 24) = 9/4
y = (9/4)*x + b
Knowing that this line passes through (24, 16), we know that when x = 24, y must be equal to 16.
If we replace these in the equation, we can find the value of b.
16 = (9/4)*24 + b
16 = 54 + b
16 - 54 = b - 38
Then the equation is:
y = (9/4)*x - 38
Now that we know the equation, we can simply replace y by 34 pounds to find the value of x.
34 = (9/4)*x - 38
34 + 38 = (9/4)*x
72*(4/9) = x = 32
So we can estimate that the length of a fish that weighs 34 pounds is 32 (I do not know the unit of length, I can't see the horizontal axis on the image)
The star running back on our football team got most of his total yardage running. The rest was catching passes. He caught passes for 60 yards. His total yardage was 150 yards. The running back for the other team got 200 yards. How many yards did the star running back on our football team get running?
Answer: The other team is extra information. 150 – 60 = 90
He got 90 yards running.
Step-by-step explanation:
Over which interval does f(t) have positive average rate of change?
A) -8,-2
B) -5,-1
C) -9,-8
D) 2,4
Answer:
D) 2,4
Step-by-step explanation:
The only answer with positive numbers
Answer:
D) 2,4
Step-by-step explanation:
A new experimental tank is in the shape of a cone, cylinder and sphere. All of the tanks have a volume of 10,000 cm3 . One condition to this tank is that the Radius should be 10 cm. Follow up the questions below based on this scenario.
Find the height of cylinder . Keep the answer in terms of π
Answer:
100/π cm
Step-by-step explanation:
Volume of a cylinder = πr²h
Volume = 10,000cm³
Radius = 10cm
The formula for the height of a cylinder is obtained as:
V = πr²h
h = V/ πr²
h = 10000 /π × 10²
h = 10000 /π × 100
h = 100/π cm
The height of the cylinder in terms of π = 100/π cm
Event A: You roll a double. Event B: The sum of the two scores is even. Event C: The score on the blue die is greater than the score on the red die. Event D: You get a 6 on the red die. Complete the following as a group. Discuss each question together and enter your answers. When you are done, be sure to finish the last few steps of the meeting agenda ("reflect" and "share recording"). = 1. Circle the pairs of events for which PIX and Y) = P(X) x P(Y) A&B A&C A&D B&C B&D C&D
Among the pairs of events A&B, A&C, A&D, B&C, B&D, and C&D, the pairs A&C and B&D satisfy the condition P(A∩B) = P(A) × P(B), indicating that they are independent events.
To determine if two events are independent, we compare the product of their individual probabilities to the probability of their intersection. If the product of the individual probabilities is equal to the probability of the intersection, then the events are independent.
Let's examine each pair of events:
A&B: Rolling a double and getting a sum of even scores are not independent events. The occurrence of one event does not guarantee the occurrence of the other.
A&C: Rolling a double and having the score on the blue die greater than the score on the red die are independent events. The probability of rolling a double is solely dependent on the outcome of the dice roll, while the probability of the blue die having a greater score than the red die is independent of the outcome of rolling a double.
A&D: Rolling a double and getting a 6 on the red die are not independent events. The occurrence of rolling a double does not affect the probability of getting a 6 on the red die.
B&C: Getting a sum of even scores and having the score on the blue die greater than the score on the red die are not independent events. The occurrence of one event does not guarantee the occurrence of the other.
B&D: Getting a sum of even scores and getting a 6 on the red die are independent events. The probability of getting a sum of even scores is solely dependent on the outcome of the dice roll, while the probability of getting a 6 on the red die is independent of the sum of the scores.
C&D: Having the score on the blue die greater than the score on the red die and getting a 6 on the red die are not independent events. The occurrence of one event does not guarantee the occurrence of the other.
In summary, the pairs of events A&C and B&D are the only pairs that satisfy the condition P(A∩B) = P(A) × P(B), indicating that they are independent events.
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can someone pls simplify this
[tex] \frac{105}{12} [/tex]
A rectangular hall is 55 feet long and 48 feet wide. How long is a walkway along the diagonal?
Answer:
[tex]73[/tex]
Step-by-step explanation:
[tex]73=\sqrt{55^{2} +48^{2} }[/tex]
pls helpppppp !!
tysmmmm <33
Answer:
180
Step-by-step explanation:
From basic rules of a triangle we know that the interior angles of a triangle have to add up to equal 180°
But here is a formula for future references for finding the sum of the interior angles
interior angle sum = (n-2)180
where n = number of sides
and here is an example:
a triangle has 3 sides so we plug in 3 for n
(3-2)180
3-2=1
1*180 = 180
so the interior angles add up to equal 180
Students plant 148 flowers at a community park. Seventy-eight percent of the flowers are pansies. Use
rounding to estimate how many flowers are pansies
Solve the system using substitution: x = -4y and x + 5y = 2
Please and thank you.
Answer:
x = - 8, y = 2
Step-by-step explanation:
[tex]x = - 4y......(1) \\ x + 5y = 2....(2) \\ plug \: x = - 4y \: in \: equation \: (2) \\ - 4y + 5y = 2 \\ y = 2 \\ plug \: y = 2 \: in \: equation \: (1) \\ x = - 4(2) \\ x = - 8\\[/tex]
How to divide 49 yd in the ratio 1:6?
Answer:
7:42
Step-by-step explanation:
First off you add the ratio together-
6+1=7
Then you divide-
49÷7= 7
7 is equal to 1 in this ratio.
To write out the ratio you need to multiplicate-
7×1=7
and
6×7= 42
Leaving the as-
7:42
Answer:
7:42
Step-by-step explanation:
First, add up the two numbers in the ratio to get 49.
Next, divide the total amount by 49, i.e. divide £16 by 8 to get £5. £5 is the amount of each 'unit' in the ratio.
Then you need to divide the total amount using that number i.e. 49/16 = 7/42.
To work out how much each person gets, you then multiply their share by the ratios. Therefore, the answer is 7 yd and 42 yds.
Consider the following scenario. Suppose that
1000 people are involved in a conspiracy.
Suppose that every single individual involved in
the conspiracy can be trusted 99.9%, that is,
there is a 0.01% probability they reveal the
secrets of the conspiracy to the media within 50
years. This probability is the same for everyone
involved, and never changes over the course of
the 50 years. Moreover, suppose that the event
where anyone goes to the media (or not) is
independent if anyone else has (or hasn't)
already.
What is the probability that at least one
person involved in the conspiracy reveals their
involvement to the media within 50 years?
The probability that at least one person involved in the conspiracy reveals their involvement to the media within 50 years is approximately 0.994.
Consider the following scenario: There are 1000 individuals involved in a conspiracy.
Every person involved in the conspiracy can be trusted 99.9%, which means that there is a 0.01% probability that they will reveal the secrets of the conspiracy to the media within 50 years.
The probability of revealing the conspiracy is the same for everyone involved and remains constant throughout the 50 years.
Additionally, the probability of someone revealing the conspiracy is independent of whether anyone else has already done so or not.
What is the likelihood that at least one person will reveal their involvement to the media within 50 years? We can use the binomial distribution to determine the probability that at least one person in the conspiracy will reveal their involvement in the media in 50 years.
We'll use the following formula for the binomial distribution: P(X ≥ 1) = 1 - P(X = 0) where X is the number of individuals that reveal the secrets of the conspiracy.
In this situation, we know that there are 1000 people in the conspiracy, and the probability that each person will reveal the conspiracy is 0.01%.
We can therefore use the formula for the binomial distribution to solve for the probability of at least one person revealing the conspiracy:
P(X ≥ 1) = 1 - P(X = 0) P(X ≥ 1) = 1 - [tex](0.999)^{1000}[/tex] P(X ≥ 1) ≈ 0.994
Therefore, The probability that at least one person involved in the conspiracy reveals their involvement to the media within 50 years is approximately 0.994.
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A spinner has five equal sections labeled 1-5. In 60 spins, how often can you expect to spin a 3?
ok so lets say that you have a pizza cutted out in 5 sections and want to pick up only a certain piece, let's say pepperoni. If you took a piece at random, the chances of getting that certain piece of pepperoni in 60 spins is: 12 / 60
Find a solution, an, for the recurrence relation given below where ao = 7, a1 = 8 and for n > 2 an = -20 x an-1-90 x an 2
The solution an for the recurrence relation is given by the formula,an = (25/19)*(10)^n + (102/19)*(-9)^n for n > 1.
Given the recurrence relation,ao = 7, a1 = 8 and for n > 2 an = -20 x an-1-90 x an 2.
To find the solution, an, of the recurrence relation we need to follow the below steps.
Step 1:Find the general formula for the recurrence relation. We have an = -20 x an-1-90 x an 2. This is a second-order recurrence relation.
To solve a recurrence relation of this order, we assume the solution of the form an = r^n.Then substituting this value of an in the given relation we have r^n = -20r^(n-1) - 90r^(n-2).
Dividing both sides by r^(n-2), we have the characteristic equation r^2 = -20r - 90.On simplifying the above equation we get, r = 10 and r = -9.
Now, the general solution for an is given by, an = c1 * (10)^n + c2 * (-9)^n.
Step 2:Find the value of constants c1 and c2. We have a0 = 7 and a1 = 8.
Substituting n = 0 in the above general formula for an, we get c1 + c2 = 7.
Substituting n = 1 in the above general formula for an, we get 10c1 - 9c2 = 8.
On solving the above two equations we get, c1 = (25)/19 and c2 = (102)/19.
Hence, the solution to the given recurrence relation is,an = (25/19)*(10)^n + (102/19)*(-9)^n.
The solution is valid for n > 1.
Therefore, the solution an for the recurrence relation is given by the formula,an = (25/19)*(10)^n + (102/19)*(-9)^n for n > 1. This is the required answer.
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Eu tenho um ovo de páscoa que custa 24,99 e tem 185 gramas, tbm tenho uma barra de chocolate q custa 5,10 e tem 90 gramad. Qual da mais vantagem para mim?
Answer:
A maior vantagem para mim é a barra de chocolate.
Step-by-step explanation:
Temos que encontrar o custo de 1 grama de ovo de Páscoa e barra de chocolate
Para o ovo de páscoa
Tenho um ovo de Páscoa que custa 24,99 e tem 185 gramas.
185 gramas = 24,99
1 grama = x
Multiplicação cruzada
185 gramas × x = 1 grama × 24,99
x = 1 grama × 24,99 / 185 gramas
x = 0,1612258065
Aproximadamente = 0,16
O custo de 1 grama de ovo de Páscoa = 0,15
Para a barra de chocolate
Também tenho uma barra de chocolate que custa 5,10 e tem 90 gramas.
90 gramas = 5,10
1 grama = x
Multiplicação cruzada
90 gramas × x = 1 grama × 5,10
x = 1 grama × 5,10 / 90 gramas
x = 0,0566666667
Aproximadamente = 0,06
O custo de 1 grama de barra de chocolate = 0,06
A maior vantagem para mim é a barra de chocolate.