Answer:
(a) P(all of the next three vehicles inspected pass) = 0.512 .
(b) P(at least one of the next three inspected fails) = 0.488 .
(c) P(exactly one of the next three inspected passes) = 0.096 .
(d) P(at most one of the next three vehicles inspected passes) = 0.104 .
(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
Step-by-step explanation:
We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.
So, Probability that the next vehicle examined fails the inspection is 20%.
Also, it is given that successive vehicles pass or fail independently of one another.
(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection
= 0.8 * 0.8 * 0.8 = 0.512
(b) P(at least one of the next three inspected fails) =
1 - P(none of the next three inspected fails) = 1 - P(all next three passes)
= 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .
(c) P(exactly one of the next three inspected passes) is given by ;
First vehicle pass the inspection, second and third vehicle doesn't passSecond vehicle pass the inspection, first and third vehicle doesn't passThird vehicle pass the inspection, first and second vehicle doesn't passHence, P(exactly one of the next three inspected passes) = Add all above cases ;
(0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .
(d) P(at most one of the next three vehicles inspected passes) = P(that none
of the next three vehicle passes) + P(Only one of the next three passes)
We have calculated the P(Only one of the next three passes) in the above part of this question;
Hence, P(at most one of the next three vehicles inspected passes) =
(0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)
= 0.008 + 0.096 = 0.104 .
(e) Probability that all three pass given that at least one of the next three vehicles passes inspection is given by;
P(All three passes/At least one of the next three vehicles passes inspection)
= P( All next three passes [tex]\bigcap[/tex] At least one of next three passes) /
P(At least one of next three passes)
= P( All next three passes ) / P(At least one of next three passes)
= P( All next three passes ) / 1 - P(none of the next three passes)
= [tex]\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}[/tex] = 0.516 .
Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
a) P(all of the next three vehicles pass) = 0.512
(b) P(at least one of the next three inspected fails) = 0.488
(c) P(exactly one of the next three inspected passes) = 0.096
(d) P(at most one of the next three vehicles inspected passes) = 0.104
(e) Given that at least one of the next three vehicles passes inspection, the probability that all three pass is approximately 1.049 (rounded to three decimal places).
Let's calculate the probabilities step by step:
(a) To find the probability that all of the next three vehicles pass, we can use the probability of a single vehicle passing (0.80) raised to the power of 3 (because the events are independent).
P(all pass) = (0.80)^3 = 0.512
(b) To find the probability that at least one of the next three vehicles fails, we can use the complement rule. It's easier to find the probability that none of them fail and then subtract that from 1.
P(at least one fails) = 1 - P(all pass) = 1 - 0.512 = 0.488
(c) To find the probability that exactly one of the next three vehicles passes, we need to consider three cases: Pass-Fail-Fail, Fail-Pass-Fail, and Fail-Fail-Pass. Each of these cases has a probability of (0.80) * (0.20) * (0.20).
P(exactly one passes) = 3 * (0.80) * (0.20) * (0.20) = 0.096
(d) To find the probability that at most one of the next three vehicles passes, we sum the probabilities of exactly one passing and none passing.
P(at most one passes) = P(none pass) + P(exactly one passes) = (0.20)^3 + 0.096 = 0.104
(e) To calculate the conditional probability that all three pass given that at least one passes, we use the formula for conditional probability:
P(all three pass | at least one passes) = P(all three pass and at least one passes) / P(at least one passes)
P(all three pass and at least one passes) = P(all pass) = 0.512 (from part a)
So, P(all three pass | at least one passes) = 0.512 / 0.488 ≈ 1.049 (rounded to three decimal places).
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Find the volume of a cube with side length of 7 in.
A.) 147 in3
B.) 343 in3
C.) 49 in3
D.) 215 in3
Answer:
343 in³ - B
Step-by-step explanation:
Volume of cube = l³ = 7 in x 7 in x 7 in = 343 in³
In the 1939 movie The Wizard of Oz, upon being presented with a Th.D. (Doctor of Thinkology), the Scarecrow proudly exclaims, "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side." Did the Scarecrow get the Pythagorean Theorem right? In particular, describe four errors in the Scarecrow's statement.
Answer:
The Scarecrow got it wrong.
Step-by-step explanation:
1. The Pythagoras theorem does not apply to every isosceles triangle. It only applies to a triangle that has one of its sides as a right angle i.e. 90 degrees.
2. The Pythagoras theorem deals with squares of sides not square roots of sides.
3. Pythagoras theorem does not state that the square of any side is equal to the sum of the remaining two. Instead, it states that the square of the hypotenuse of a right angled triangle is equal to the sum of the square of the remaining two sides.
4. The Pythagoras theorem does not mention anything about an isosceles triangle. It deals only with Right angled triangles.
The Scarecrow's statement has errors in terms of the type of triangle it refers to, the mathematical operation used, which sides of the triangle he's considering, and the side with which the sum is equated.
Explanation:The Scarecrow in the 1939 film The Wizard of Oz did not correctly state the Pythagorean Theorem. There are a few errors in his statement:
Firstly, the Pythagorean theorem applies to right triangles, not isosceles triangles.Secondly, the theorem talks about the squares of the sides of the triangle, not the square roots.Thirdly, it's about the sum of the squares of the two shorter sides, not any two sides.Lastly, this sum equals the square of the longest side (hypotenuse), not its square root.The correct statement of the Pythagorean Theorem is: 'In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.'
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Assume the random variable X is normally distributed, with mean 46 and standard deviation 8. Find the 7 th percentile.
Answer:
The 7th percentile is 34.2
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 46, \sigma = 8[/tex]
Find the 7 th percentile.
The value of X when Z has a pvalue of 0.07. So it is X when Z = -1.475.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.475 = \frac{X - 46}{8}[/tex]
[tex]X - 46 = -1.475*8[/tex]
[tex]X = 34.2[/tex]
The 7th percentile is 34.2
A technology committee of wants to perform a test to see if the mean amount of time students are spending in a lab has increased from 54 minutes. Here are the data from a random sample of 12 students.
52 74
57 54
55 137
76 74
63 7
53 63
Are there any? outliers? Select the correct choice below and fill in any answer boxes within your choice.?Yes, 2
b) What do you conclude about the? claim? If there are? outliers, perform the test with and without the outliers present
Write appropriate hypotheses for the test.
H0: u=54
HA: u=>54
The Test Statistic is t= _________
The P- Value is ___________
If there are outliers , calculate the test statistic for the data set without outliers.
The test statistic is t=______________
The P-Value is ____________
To identify outliers from the data, the interquartile range (IQR) is used. Hypotheses are established to test if the mean time spent in the lab has increased. The test statistic and P-value are calculated with and without outliers to reach a conclusion.
Explanation:To determine if there are any outliers in the provided data, we can use the interquartile range (IQR) method. Calculating IQR and then finding the lower and upper bounds will help us identify outliers. An outlier is a value that falls below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR. The hypotheses for our test are as follows:
H0: \\( \\mu = 54 \\) minutesHA: \\( \\mu > 54 \\) minutesThe test statistic and P-value are calculated using the sample mean, standard deviation, and sample size, and comparing them with the presumed population mean of 54 minutes.
If outliers are identified and removed, we would calculate a new test statistic and P-value for the data set without the outliers to see if there is a significant change.
The random variable is the number of nonconforming solder connections on a printed circuit board with 970 connections. Determine the range (possible values) of the random variable.
Answer: X = {0,1,2,3,4,5,....970}
Explanation: Let X → random nonconforming solder connections on a printed circuit board. The number has to be a whole number to indicate the connections. This gives all possible range of X = {0,1,2,3,4,5,....970}
The number of solder connections would always be discrete, hence, a positive integer value. The range of possible values which the variable could have are {1, 2, 3,..., 970}
Number of solder connections would always be a whole number, as it cannot be negative or be a fractional value, hence, the possible range of values will be discrete.
Hence, the possible range can be expressed in the form :
X = {1, 2, 3,..., 970}Learn more : https://brainly.com/question/17225268
You draw one card at random from a standard deck of 52 playing cards. Find the probability that (a) the card is an even-numbered card, (b) the card is a heart or a diamond, and (c) the card is a nine or a face card.?
Answer:
[tex]P(even )=\frac{20}{52} =\frac{5}{13}[/tex]
[tex]P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}[/tex]
[tex]P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}[/tex]
Step-by-step explanation:
You draw one card at random from a standard deck of 52 playing cards
Total cards = 52
5 even numbers in each suit. 4 times 5 = 20 even number cards
[tex]P(even )=\frac{20}{52} =\frac{5}{13}[/tex]
(b) the card is a heart or a diamond
there are 13 heart and 13 diamonds
[tex]P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}[/tex]
(c) the card is a nine or a face card
There are 4 nine and 12 face cards
[tex]P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}[/tex]
A system of identifying call letters uses call letters that begin either with E or W. Some have a total of 2 letters, and others have 3 letters. How many different call letters are possible?
Answer:
If repetition is allowed then the different call letters possible are 1404.
If repetition is allowed then the different call letters possible are 1250.
Step-by-step explanation:
The call letter are of either 2 letter or 3 letters.
And for them to be identified they should start with either E or W.
If Repetition is allowed:Consider the call letters with 2 letter:
Possible number of call letters starting with E : E _
The second place can be filled any of the 26 letters.
Possible number of call letters starting with W : W _
The second place can be filled any of the 26 letters.
Consider the call letters with 3 letter:
Possible number of call letters starting with E : E _ _
The second place and third place can be filled with any of the 26 letters.
Possible number of call letters starting with W : W _ _
The second place and third place can be filled with any of the 26 letters.
Total number of possible letters: 26 + 26 + (26)² + (26)² = 1404
Thus, if repetition is allowed then the different call letters possible are 1404.
If Repetition is not allowed:Consider the call letters with 2 letter:
Possible number of call letters starting with E : E _
The second place can be filled any of the 25 letters.
Possible number of call letters starting with W : W _
The second place can be filled any of the 25 letters.
Consider the call letters with 3 letter:
Possible number of call letters starting with E : E _ _
The second place can be filled with any of the remaining 25 letters and the third can be filled with the remaining 24 letters.
Possible number of call letters starting with W : W _ _
The second place can be filled with any of the remaining 25 letters and the third can be filled with the remaining 24 letters.
Total number of possible letters: 25 + 25+ (25 × 24) + (26 × 24) = 1250
Thus, if repetition is allowed then the different call letters possible are 1250.
Complete an amortization schedule for a $27,000 loan to be repaid in equal installments at the end of each of the next three years. The interest rate is 12% compounded annually. If an amount is zero, enter "0". Do not round intermediate calculations. Round your answers to the nearest cent.
Answer:
Amount paid annually = $11241.42
Step-by-step explanation:
The steps are as shown in the attached file
"Suppose the scenario that the standard devation of semiannual changes in the price of wheat is $0.79, the standdard devation of changes in the futures contract on wheat over the same tim eperiod is $0.93, and the correlation coeffieient relating the assest and futures contract is $0.86. What is the optimal headge ratio for the six month contract on wheat
Answer:
The optimal Hedge Ratio is 0.7305.
Step-by-step explanation:
Optimal Hedge ratio is given as
[tex]HR_{optimal}=\epsilon_{correlation} \times \frac{\sigma_{current}}{\sigma_{future}}[/tex]
Here
HR_optimal is the Hedge Ratio for the next 6 months which is to be calculated.ε_correlation is the correlation coefficient relating the assets and futures contract whose value is give as $0.86.σ_current is the standard deviation of the semiannual changes of the wheat which is given as $0.79σ_future is the standard deviation of the changes in the future over the same time period which is given as $0.93So the Hedge Ratio is given as
[tex]HR_{optimal}=\epsilon_{correlation} \times \frac{\sigma_{current}}{\sigma_{future}}\\HR_{optimal}=0.86 \times \frac{0.79}{0.93}\\HR_{optimal}= \$0.7305[/tex]
So the optimal Hedge Ratio is 0.7305.
Find the equation for the circle with center (4,3) and passing through (2,-4)
Answer:
[tex](x - 4)^{2} + (y - 3)^{2} = 53[/tex]
Step-by-step explanation:
The general equation of a circle is as follows:
[tex](x - x_{c})^{2} + (y - y_{c})^{2} = r^{2}[/tex]
In which the center is [tex](x_{c}, y_{c})[/tex], and r is the radius.
In this problem, we have that:
[tex]x_{c} = 4, y_{c} = 3[/tex]
So
[tex](x - 4)^{2} + (y - 3)^{2} = r^{2}[/tex]
Passing through (2,-4)
We replace into the equation to find the radius.
[tex](2 - 4)^{2} + (-4 - 3)^{2} = r^{2}[/tex]
[tex]4 + 49 = r^{2}[/tex]
[tex]r^{2} = 53[/tex]
The equation of the circle is:
[tex](x - 4)^{2} + (y - 3)^{2} = 53[/tex]
A boxer takes 3 drinks of water between each round for the first four rounds of a championship fight. After the fourth round he starts to take his three drinks plus one additional drink between each of the remaining rounds. If he continues to increase his drinks by 1 after each round, how many drinks will he take between the 14th and 15th round
Answer:
56 drinks
Step-by-step explanation:
Given
n = 15-4 = 11 rounds (with one additional drink)
a₀ = 3*4 = 12 drinks (between the first and the fourth round)
r = 4 (number of drinks per round)
We can use the formula
aₙ = a₀ + n*r
⇒ a₁₁ = 12 + 11*4 = 12 + 44
⇒ a₁₁ = 56 drinks
A statistician selected a sample of 16 accounts receivable and determined the mean of the sample to be $5,000 with a standard deviation of $400. He reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. He neglected to report what confidence level he had used. Based on the above information, determine the confidence level that was used. Assume the population has a normal distribution.
Answer:
The confidence level that was used is 0.25% .
Step-by-step explanation:
We are given that mean of the selected sample of 16 accounts is $5,000 and a standard deviation of $400.
It has also been reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. This represents the Confidence Interval for population mean.
But we have to find that at what confidence level this information about range of population men has been stated.
Since we know that Confidence Interval for population mean is given by :
C.I. for population mean = Sample mean(xbar) [tex]\pm[/tex] z value * [tex]\frac{Standard deviation}{\sqrt{n} }[/tex]
i.e.,if we have 95% C.I. = xbar [tex]\pm[/tex] 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] .
So, our Confidence Interval for population is written as :
[$4,739.80 , $5,260.20] = $5000 [tex]\pm[/tex] z value * [tex]\frac{400}{\sqrt{16} }[/tex]
$5000 - z value * 100 = $4739.80 { Solving these we get Z value = 2.602}
$5000 + z value * 100 = $5260.20
In z table we find that at value of 2.60 the probability is 0.99534 so subtracting this from 1 we get confidence level for one tail i.e.0.5%(approx).
Therefore, for two tail Confidence level will be 0.5%/2 = 0.25% .
Using the t-distribution, it is found that a confidence level of 98% was used.
We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.
The information given is:
Sample mean of [tex]\overline{x} = 5000[/tex]. Sample standard deviation of [tex]s = 400[/tex]. Sample size of [tex]n = 16[/tex].The margin of error is of:
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
In which t is the critical value.
For this problem, the margin of error is:
[tex]M = \frac{5260.2 - 4739.8}{2} = 260.2[/tex]
Hence, the critical value is found solving the equation of the margin of error for t.
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
[tex]260.2 = t\frac{400}{\sqrt{16}}[/tex]
[tex]100t = 260.2[/tex]
[tex]t = \frac{260.2}{100}[/tex]
[tex]t = 2.602[/tex]
Looking at the t-table, with 16 - 1 = 15 df, t = 2.602 is associated with a confidence level of 98%.
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what are the coordinates of point A
Answer:
-5,7 those are the coordinates on the graph
Answer:
-5,7 this should be the answer
Sammy read the line plot and said that there were 8 little league players who are 3 years old.
(View the picture.)
Is Sammy correct? If not, construct a viable argument and critique his response.
Answer:
where is the age three she is incorrect
Step-by-step explanation:
Answer:
she is correct its 8 people who are 3 years old and it says it
Step-by-step explanation:
A baker blends 600 raisins and 400 chocolate chips into a dough mix and, from this, makes 500 cookies. (a) Find the probability that a randomly picked cookie will have no raisins. (b) Find the probability that a randomly picked cookie will have exactly two chocolate chips. (c) Find the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
Answer:
a) P(X = 0) for raisins = 0.33287
b) P(X = 2) for chocolate chips = 0.14379
c) P(X ≥ 2) for both bits = 0.59399
Step-by-step explanation:
The average amount of raisin per cookie is 600/500 = 1.2
The average amount of chocolate chips per cookie = 400/500 = 0.8
a) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For raisin, Mean = λ = 1.1
x = 0
P(X = 0) = (e⁻¹•¹)(1.1⁰)/(0!) = 0.33287
b) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For chocolate chips, Mean = λ = 0.8
x = 2
P(X = 2) = (e⁻⁰•⁸)(0.8²)/(2!) = 0.14379
c) the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
For this, the average number of bit in a raisin = (600+400)/500 = 2 bits per raisin.
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
For P(X = 0)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 0
P(X = 0) = (e⁻²)(2⁰)/(0!) = 0.13534
For P(X = 1)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 1
P(X = 1) = (e⁻²)(2¹)/(1!) = 0.27067
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
P(X ≥ 2) = 1 - (0.13534 + 0.27067) = 1 - 0.40601 = 0.59399
Using the Poisson distribution, it is found that there is a:
a) 0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.
b) 0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.
c) 0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
x is the number of successes e = 2.71828 is the Euler number [tex]\mu[/tex] is the mean in the given interval.Item a:
600 raisins in 500 cookies, hence, the mean is:
[tex]\mu = \frac{600}{500} = 1.2[/tex]
The probability is P(X = 0), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.2}1.2^{0}}{(0)!} = 0.3012[/tex]
0.3012 = 30.12% probability that a randomly picked cookie will have exactly two chocolate chips.
Item b:
400 chips in 500 cookies, hence, the mean is:
[tex]\mu = \frac{400}{500} = 0.8[/tex]
The probability is P(X = 2), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-0.8}0.8^{2}}{(2)!} = 0.1438[/tex]
0.1438 = 14.38% probability that a randomly picked cookie will have exactly two chocolate chips.
Item c:
1000 bits, as 600 + 400 = 1000, in 500 cookies, hence, the mean is:
[tex]\mu = \frac{1000}{500} = 2[/tex]
The probability is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}2^{0}}{(0)!} = 0.1353[/tex]
[tex]P(X = 1) = \frac{e^{-2}2^{1}}{(1)!} = 0.2707[/tex]
Hence:
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1353 + 0.2707 = 0.4060[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.406 = 0.594[/tex]
0.594 = 59.4% probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
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A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms. Find the current i(t) if i(0) = 0. i(t) = Determine the current as t → [infinity]. lim t→[infinity] i(t) =
Answer:
attached below
Step-by-step explanation:
The current as t → [infinity] is 2/7 A if A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms
What is ohms law?It is defined as the relationship between current and voltage according to the ohms law the voltage is directly proportional to the current.
It is given that:
A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms.
Using the equation:
LdI/dt + RI = v(t)
I(0) = 0
LdI/dt = v(t)
V = 20 volts
After integration:
I = 2/7 + c
c = e⁻⁷⁰⁰ⁿ (n = t) = 0
t → ∞
I = 2/7 A
Thus, the current as t → [infinity] is 2/7 A if A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 ohms
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What is the probability that more than twelve loads occur during a 4-year period? (Round your answer to three decimal places.)
Answer:
Given that an article suggests
that a Poisson process can be used to represent the occurrence of
structural loads over time. Suppose the mean time between occurrences of
loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a
4-year period? c). How long must a time period be so that the probability of no loads
occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a
4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads
occurring during that period is at most 0.3 is given by: 3.3 years
Step-by-step explanation:
Suppose you find $20. If you choose to use the $20 to go to the football game, your opportunity cost of going to the game is a. nothing, because you found the money. b. $20 (because you could have used the $20 to buy other things). c. $20 (because you could have used the $20 to buy other things) plus the value of your time spent at the game. d. $20 (because you could have used the $20 to buy other things) plus the value of your time spent at the game, plus the cost of the dinner you purchased at the game.
the property of society getting the most from its scarce recources ... Fairness might require that everyone get an equal share because they were ... If you choose to use the $20 to go to the football game, your opportunity cost of going to the game is. $20 because you could have used that money on other things plus the value
The opportunity cost of going to the football game is $20 because you could have used that money to buy other things.
Explanation:The correct answer to this question is b. $20 (because you could have used the $20 to buy other things).
Opportunity cost is the value of the next best alternative that you give up when making a decision. In this case, if you choose to use the $20 to go to the football game, you are giving up the opportunity to buy other things with that money.
There are no additional costs like the value of your time spent at the game or the cost of dinner because they are not explicitly mentioned in the question.
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A large tank holds 400 gallons of salt water. A salt water solution with concentration 2 lb/gal is being pumped into the tank at a rate of 3 gal/min. The tank is simultaneously being emptied at a rate of 1 gal/min. If 100 pounds of salt was dissolved in the tank initially, how much salt is in the tank after 250 minutes?
Final answer:
The question involves calculating how much salt remains in a tank after 250 minutes when saltwater of a certain concentration is added and the solution is simultaneously drained at a different rate. A basic calculation suggests an addition of 1500 pounds to the initial 100 pounds, but the real answer must consider dilution and the tank's capacity, which complicates the calculation.
Explanation:
The question involves calculating the amount of salt in a tank after 250 minutes when saltwater is being pumped in and out at different rates. Initially, the tank has 100 pounds of salt and is then filled with a solution at a rate of 3 gallons per minute, with a concentration of 2 pounds of salt per gallon. Meanwhile, the tank is emptied at a rate of 1 gallon per minute.
To find the amount of salt in the tank after 250 minutes, we need to consider both the inflow and outflow of the tank. The net flow rate is 2 gallons per minute (3 gallons in - 1 gallon out). This leads to an increase in the volume of the tank by 2 gallons for every minute.
Every minute, 6 pounds of salt (3 gallons x 2 pounds/gallon) are added to the tank. However, since the tank's volume increases by 2 gallons per minute but 1 gallon is removed, we should adjust for the concentrate reduction due to dilution. Over 250 minutes, the total added salt is 250 minutes x 6 pounds/minute = 1500 pounds. However, the calculation must be adjusted to consider the dilution effect which is beyond the scope of this immediate calculation.
Initially, the tank contains 100 pounds of salt. Without accounting for dilution or maximum capacity effects, a simple addition suggests that after 250 minutes, 1600 pounds of salt would be in the tank. Yet, this simplistic approach does not account for dilution effect properly or the fact the tank has a maximum capacity, indicating that a more complex calculation is required for precise results.
Kareem bought 5 feet of fabric. how much is this in yards?
Answer:
1.66667
Step-by-step explanation:
You would do 5 divided by 3.
5 divided by 3 = 1.66667
You can verify this answer by multiplying 1.66667 * 3
1.66667 * 3 = 5
Hope this helps!!!
A closed box with a square bottom is three times high as it is wide. a) Express the surface area of the box in terms of its width. b) Express the volume of the box in terms of its width. c) Express the surface area in terms of the volume. d) If the box has a volume of 24 m³, what is its surface area?
Answer:
a) [tex]S(s) = 14s^2[/tex]
b) [tex]V(s) = 3s^3[/tex]
c) [tex]S(s) = \dfrac{14V(s)}{3s}[/tex]
d) 56 square meter
Step-by-step explanation:
We are given the following in the question:
A closed box with a square bottom is three times high as it is wide.
Let s be the side of square and h be the height.
[tex]h = 3s[/tex]
a) Surface area of box
[tex]2(lb + bh + hl)[/tex]
where l is the length, b is the breadth and h is the height.
Putting values:
[tex]S = 2(s^2 + sh +sh)\\S = 2(s^2 + 3s^2 + 3s^2)\\S(s) = 14s^2[/tex]
b) Volume of box
[tex]l\times b \times h[/tex]
where l is the length, b is the breadth and h is the height.
Putting values:
[tex]V = s\times s\times h\\V= s\times s\times 3s\\V(s) = 3s^3[/tex]
c) Surface area in terms of volume
[tex]S(s) = 14s^2 = \dfrac{14V(s)}{3s}[/tex]
d) Surface area
Volume = 24 m³
[tex]V(s) = 24\\3s^3 = 24\\s^3 = 3\\s = 2[/tex]
[tex]S(2) = 14(2)^2 = 56\text{ square meter}[/tex]
In each figure a || b. Determine whether the third figure is a translation image of the first figure. Write yes or no. Explain your answer.
Answer:
No
Step-by-step explanation:
The second and third figures are translations of each other. The first figure appears to be a reflection of the second, not a translation. Hence the first and third figures are not translations of each other.
__
If an image is translated, all of the pre-image line segments are parallel to the image line segments. That is not the case for the first and third figures, in which the top line segments go in different directions.
Answer:
b. No; it is a reflection followed by a translation.
Step-by-step explanation:
Nine cars (3 Pontiacs [labeled 1-3], 4 Fords [labeled 4-7], and 2 Chevrolet's [labeled 8-9]) are divided into 3 groups for car racing. each group consistes of 3 cars, and they are allocated to 3 tracks (1-3), respectively.
1. How many different ways can you arrange the 9 cars?
Answer:
The number of different ways to arrange the 9 cars is 362,880.
Step-by-step explanation:
There are a total of 9 cars.
These 9 cars are to divided among 3 racing groups.
The condition applied is that there should be 3 cars in each group.
Use permutation to determine the total number of arrangements of the cars.
For group 1:There are 9 cars and 3 to be allotted to group 1.
This can happen in [tex]^9P_{3}[/tex] ways.
That is, [tex]^9P_{3}=\frac{9!}{(9-3)!} =504[/tex] ways.
For group 2:There are remaining 6 cars and 3 to be allotted to group 2.
This can happen in [tex]^6P_{3}[/tex] ways.
That is, [tex]^6P_{3}=\frac{6!}{(6-3)!} =120[/tex] ways.
For group 3:There are remaining 3 cars and 3 to be allotted to group 3.
This can happen in [tex]^3P_{3}[/tex] ways.
That is, [tex]^3P_{3}=\frac{3!}{(3-3)!} =6[/tex] ways.
The total number of ways to arrange the 9 cars is: [tex]^9P_{3}\times ^6P_{3}\times ^3P_{3}=504\times120\times6=362880[/tex]
Thus, the number of different ways to arrange the 9 cars is 362,880.
Suppose George loses 66% of all staring contests. (a) What is the probability that George loses two staring contests in a row? (b) What is the probability that George loses four staring contests in a row? Assume that each game played is independent of the rest.
Answer:
a) 43.56% probability that George loses two staring contests in a row.
b) 18.97% probability that George loses four staring contests in a row.
Step-by-step explanation:
In each staring contest, George has a 66% probability of losing.
(a) What is the probability that George loses two staring contests in a row?
[tex]P = (0.66)^{2} = 0.4356[/tex]
43.56% probability that George loses two staring contests in a row.
(b) What is the probability that George loses four staring contests in a row? Assume that each game played is independent of the rest.
[tex]P = (0.66)^{4} = 0.1897[/tex]
18.97% probability that George loses four staring contests in a row.
The probability of George losing two staring contests in a row is approximately 43.56%, while the probability of him losing four staring contests in a row is approximately 19.7%.
Explanation:To calculate the probability that George loses two staring contests in a row, we need to multiply the probability of losing one staring contest by itself. Since George loses 66% of all staring contests, the probability of losing one contest is 0.66. Therefore, the probability of losing two contests in a row is 0.66 multiplied by 0.66, which is approximately 0.4356 or 43.56%.
To calculate the probability that George loses four staring contests in a row, we again need to multiply the probability of losing one contest by itself four times. So, the probability is 0.66 raised to the power of 4, which is approximately 0.197 or 19.7%.
Sarah opens a savings account that has a 2.75% annual interest rate,
compounded monthly. She deposits $500 into the account. How much will
be in the account after 15 years?
$500.00
$754.94
$1255.27
$255.27
Answer: $754.94
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = 500
r = 2.75% = 2.75/100 = 0.0275
n = 12 because it was compounded monthly which means 12 times in a year.
t = 15 years
Therefore,.
A = 500(1+0.0275/12)^12 × 15
A = 500(1+0.0023)^180
A = 500(1.0023)^180
A = $754.94
You are considering buying a piece of industrial equipment to automate a part of your production process. This automation will save labor costs by as much as $35,000 per year over 10 years. The equipment costs $200,000. Should you purchase the equipment if your interest rate is 12
Answer:
No.
Step-by-step explanation:
The equipment wll save $35,000 per over for 10 years, which totals to $350,000.
If the equipment is bought on a simple interest rate of 12% annually for ten years, it will cost:
[tex]A = P (1 + rt),\\where\ A =\ Final\ amount,\ r=rate\of\interest\annually,\ t=time,\ P= Principal\ value\\\\P = 200,000r = 0.12t = 10 \\\\A = 200,000 ( 1 + 0.12\times 10)\\A = 200,000 (1 + 1.2)\\A = 200,000 (2.2)\\A = 440,000[/tex]
We will need to pay $440,000 in total for the machine in over ten years.
If we compare both values, it can be deduced that industrial equipment is more expensive than labor cost.
The decision to purchase the equipment depends on the calculation of Net Present Value, which factors in the cost of the equipment, annual saving, and interest rate. If the calculated NPV is greater than zero, it's worth investing, otherwise, it's not.
Explanation:The subject of this question falls under the category of Business Finance, specifically the concept of Net Present Value (NPV). Given the costs of the equipment, the annual saving, and the interest rate (assumed to be 12%), we can calculate NPV. The net savings account for an annuity, which we assume to be $35,000 annually, and we end up with an NPV of cells (= 35000*(1-(1+0.12)^-10)/0.12 -200000). This value helps determine whether the investment is worthwhile. If the NPV is greater than 0, you should purchase the equipment as it indicates that the project's return exceeds the cost. However, if the NPV is less than 0, you should not purchase the equipment as it indicates that the project's cost exceeds the expected return.
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There are 20 chemistry students to be scheduled for labs this term. 6 will be assigned to the Adams Hall lab, 11 to the Baker Hall lab, and the rest to the Craig Hall lab. How many possible assignments are there of students to labs?
Given the specific assignment of students to labs, there is only one possible way to assign the 20 students to the three labs.
Explanation:To solve this problem, we are essentially counting the number of different ways to distribute 20 students among three labs: Adams Hall, Baker Hall, and Craig Hall. Given that 6 students will be assigned to Adams Hall and 11 to Baker Hall, we already know where 17 of the 20 students will be placed. The only students left to place are the remaining 3 students, who must go to the Craig Hall lab. Therefore these students could be assigned in only one way given the constraints of the problem. Thus, only one assignment is possible.
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Suppose that W1 is a random variable with mean μ and variance σ21 and W2 is a random variable with mean μ and variance σ2. From Example 5.4.3, we know that cW1 + (1 − c)W2 is an unbiased estimator of μ for any constant c > 0. If W1 and W2 are independent, for what value of c is the estimator cW1 + (1 − c)W2 most efficient?
Answer:
Step-by-step explanation:
The concept of variance in random variable is applied in solving for the value of c for the estimator cW1 + (1 − c)W2 to be most efficient. Appropriate differentiation of the estimator with respect to c will give the value of c when the result is at minimum.
The detailed analysis and step by step approach is as shown in the attachment.
Manny bought 12 pounds of vegetables at the supermarket. 75% of the vegetables were on sale. How many pounds of vegetables were not on sale?
Answer:
3 pounds of vegetables bought were not on sale.
Step-by-step explanation:
Given:
Total pounds of vegetable bought by Manny = 12
Percentage of vegetables that are on sale = 75%
To find the pounds of vegetables that were not on sale.
Solution:
If the percentage of of vegetables that are on sale is 75%.
Then, the percentage of of vegetables that were not on sale is = [tex]100\%-75\%[/tex] = 25%
Total pounds of vegetable bought is 12 pounds, of which 25% were not on sale.
So, the pounds of vegetables that were not on sale is:
⇒ [tex]25\%\ of\ 12[/tex]
⇒ [tex]0.25\times 12[/tex]
⇒ 3
Thus, 3 pounds of vegetables bought were not on sale.
The following sample data are from a normal population: 10, 9, 12, 14, 13, 11, 6, 5.
a. What is the point estimate of the population mean?
b. What is the point estimate of the population standard deviation (to 2 decimals)?
c. With 95% confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?
d. What is the 95% confidence interval for the population mean (to 1 decimal)?
The point estimate of the population mean is 10, the point estimate for the population standard deviation is ~3.03. The margin of error for 95% confidence of the estimation of the mean is ~2.4. Therefore, the 95% confidence interval for the population mean is (7.6, 12.4).
Explanation:To answer your questions, we begin by calculating basic measures of central tendency and variability.
a. The point estimate of the population mean is the average value of the data. It's calculated by adding up all the values and dividing by the number of values. In this case, (10+9+12+14+13+11+6+5)/8 = 10.
b. The point estimate of the population standard deviation (sd) is the root mean square deviation of the values from the mean. It's calculated by determining the square root of the variance, but let's use a calculator to get sd ≈ 3.03.
c. The margin of error for the estimation of the population mean at a 95% confidence level is calculated using the standard deviation and the size of the sample (for a t-distribution). This gives: MoE = t * (sd/sqrt(n)) ~ 2.3 * (3.03/sqrt(8)) = 2.44 (1 decimal place).
d. Finally, the 95% confidence interval for the population mean is determined by adding and subtracting the margin of error from the point estimate of the mean. Therefore, the 95% confidence interval for the mean is (10 - 2.4 , 10 + 2.4) = (7.6, 12.4).
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a. Point Estimate of Population Mean
Firstly, a point estimate for the population mean is the average of the sample data. For the given data (10, 9, 12, 14, 13, 11, 6, 5), the point estimate is calculated by adding all the values and dividing by the number of observations:
(10 + 9 + 12 + 14 + 13 + 11 + 6 + 5) / 8 = 10
b. Point Estimate of Population Standard Deviation
To find the standard deviation, we use the formula for the sample standard deviation:
s =√[tex](\sum (x_i - \bar{x})^2[/tex] / (n - 1))
After calculating using the given data, we find that the point estimate of the population standard deviation, rounded to two decimal places, is s = 3.32.
c. Margin of Error for Mean Estimation
With the margin of error, we create the confidence interval by subtracting and adding the margin of error from the point estimate of the mean. This gives us the range of values in which we are 95% confident that the true population mean lies. The standard error of the sample mean is 3.24 / √8 ≈ 1.146,
so the margin of error at 95% confidence is 1.96 * 1.146 ≈ 2.24 (rounded to 1 decimal).
d. 95% Confidence Interval for Population Mean
The 95% confidence interval for the population mean is 10.375 ± 2.24, which gives us a range of (8.13, 12.62) (rounded to 1 decimal).