a) Probability that all four have type B+ blood = 0.00031213
b) Probability that none of the four have type B+ blood = 0.57289761
c) Probability that at least one of the four has type B+ blood = 0.42710239
d) B. The event in part (a) is unusual because its probability is less than or equal to 0.05.
To solve these probability problems, we'll use the binomial probability formula:
[tex]P(X=k) = (n, k) \times p^k \times (1-p)^{(n-k)[/tex]
Where:
P(X=k) is the probability of having exactly k successes in n trials.
(n choose k) is the number of ways to choose k successes from n trials (n! / (k! (n-k)!), where n! is the factorial of n).
p is the probability of success (having type B+ blood in this case).
q = 1 - p is the probability of failure (not having type B+ blood).
n is the number of trials.
Given:
p = 0.13 (probability of having type B+ blood)
q = 1 - p = 0.87 (probability of not having type B+ blood)
n = 4 (number of trials)
Let's solve each part step by step:
(a) Probability that all four have type B+ blood:
[tex]P(X=4) = (4, 4) \times 0.13^4 \times 0.87^{(4-4)[/tex]
[tex]P(X=4) = 1 \times 0.00031213 \times 1 \\\\= 0.00031213[/tex]
(b) Probability that none of the four have type B+ blood:
[tex]P(X=0) = (4, 0) \times 0.13^0 \times 0.87^4 \\\\P(X=0) = 1 \times 1 \times 0.57289761 \\\\= 0.57289761[/tex]
(c) Probability that at least one of the four has type B+ blood:
P(at least one) = 1 - P(none)
P(at least one) = 1 - 0.57289761
= 0.42710239
Now, let's determine which events are considered unusual. Generally, an event with a probability less than or equal to 0.05 is considered unusual.
Let's compare the probabilities:
Probability in part (a): 0.00031213 (less than 0.05)
Probability in part (b): 0.57289761 (greater than 0.05)
Probability in part (c): 0.42710239 (greater than 0.05)
Based on the comparison, the only event that can be considered unusual is the event in part (a) because its probability is less than 0.05. Therefore, the correct answers are:
B. The event in part (a) is unusual because its probability is less than or equal to 0.05.
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The calculations show that the probability of all four people having B+ blood is 0.000028561, and the likelihood of none of them having B+ blood is 0.569532. The chance of at least one of them having B+ blood is 0.430468. Thus, only the event in part (a) is considered unusual due to its low probability.
Explanation:This question is about using probability principles to figure out the likelihood of having certain blood types in a population.
(a) To find the probability that all four individuals have type B+ blood, we need to multiply the individual probabilities together. The probability that one person has B+ blood is given as 13% or 0.13. So, the probability that all four have B+ blood is 0.13*0.13*0.13*0.13 = 0.000028561.
(b) The probability that none of the four have type B+ blood is the complement of the probability that one person has B+ blood. This is 1 - 0.13 = 0.87. We now raise this to the power of four to find the probability that all four selected people do not have B+ blood: 0.87*0.87*0.87*0.87 = 0.569532.
(c) The probability that at least one has type B+ blood is the complement of the result in part b. We subtract our answer from part b from 1: 1 - 0.569532 = 0.430468.
(d) An event is considered unusual if its probability is less than or equal to 0.05. Here, the event in part (a) is unusual because its probability (0.000028561) is less than or equal to 0.05.
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A major television manufacturer has determined that its 44 inch screens have a mean service life that can be modeled by a normal distribution with a mean of 6 years and a standard deviation of one-half year (6 months). What is the probability that the service life of that product is between 5 and 7 years
Answer:
[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]
And we can find this probability with thie difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(6,0.5)[/tex]
Where [tex]\mu=6[/tex] and [tex]\sigma=0.5[/tex]
We are interested on this probability
[tex]P(5<X<7)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]
And we can find this probability with thie difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]
Final answer:
The probability that a 44 inch television screen will have a service life between 5 and 7 years, given a normal distribution with a mean of 6 years and a standard deviation of 0.5 years, is approximately 95.45%.
Explanation:
To calculate the probability that the service life of a 44 inch television screen is between 5 and 7 years, we use the properties of the normal distribution where the mean (μ) is 6 years and the standard deviation (σ) is 0.5 years. We need to find the z-scores for 5 and 7 years and then use the standard normal distribution table or a calculator to find the probability that the service life falls between these two z-scores.
First, calculate the z-score for 5 years:
z = (X - μ) / σ
z = (5 - 6) / 0.5
z = -2.0
Next, calculate the z-score for 7 years:
z = (7 - 6) / 0.5
z = 2.0
Once we have the z-scores, we look up the corresponding probabilities in the normal distribution table or use a calculator with normal distribution functions. The probability of z being between -2.0 and 2.0 in a standard normal distribution is approximately 0.9545.
Therefore, the probability that a television will last between 5 and 7 years is approximately 95.45%.
) Let y(1) = y0, y 0 (1) = v0. Solve the initial value problem. What is the longest interval on which the initial value problem is certain to have a unique twice differentiable solution?
QUESTION IS INCOMPLETE.
Nevertheless, I will explain how to find, without solving, the longest interval in which an initial value problem is certain to have a unique twice differentiable solution.
Step-by-step explanation:
Consider the Existence and uniqueness theorem:
Let p(t) , q(t) and r(t) be continuous on an interval a ≤ t ≤ b, then the differential equation given by:
y''+ p(t) y' +q(t) y = r(t) ;
y(t_0) = y_0, y'(t_0) = y'_0
has a unique solution defined for all t in the stated interval.
Example:
Consider the differential equation
ty'' + 9y = t
y(1) = y_0,
y'(1) = v_0
ty'' + 9y = t .................................(1)
First, write the differential equation (1) in the form:
y'' + p(t)y' + q(t)y = r(t) ..................(2)
by dividing (1) by t
So
y''+ (9/t)y = 1 ....................................(3)
Comparing (3) with (2)
p(t) = 0
q(t) = 9/t
r(t) = 1
For t = 0, p(t) and r(t) are continuous, but q(t) is undefined.
q(t) is continuous everywhere apart from the point t = 0.
We say (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous.
But t = 1, which is contained in the initial conditions y(1) = y_0 and y'(1) = v_0 is found in (0,∞).
So, we conclude that this interval is the longest interval in which the initial value problem has a unique twice differentiable solution.
The Whitt Window Company, a company with only three employees, makes two different kinds of hand-crafted windows: a wood-framed and an aluminum-framed window. The company earns $300 profit for each wood-framed window and $150 profit for each aluminum-framed window. Doug makes the wood frames and can make 6 per day. Linda makes the aluminum frames and can make 4 per day. Bob forms and cuts the glass and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass.
The company wishes to determine how many windows of each type to produce per day to maximize total profit.
(a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources.
(b) Formulate a linear programming model for this problem.(c) Use the graphical method to solve this model.
Answer:
Maximize Z = 6x1 + 3x2
other answers are as follows in the explanation
Step-by-step explanation:
Employee Glass Needed per product(sq feet) Glass available per production
Product
Wood framed glass Aluminium framed glass
doug 6 0 36
linda 0 8 32
Bob 6 8 48
profit $300 $150
per batch
Z = 6x1 + 3x2,
with the constraint
6x1 ≤ 36 8x2 ≤ 32 6x1 + 8x2 ≤ 48
and x1 ≥ 0, x2 ≥ 0
Maximize Z = 6x1 + 3x2
to get the points of the boundary on the graph we say
when 6x1= 36
x1=6
when
8x2= 32
x2=4
to get the line of intersect , we go to
6x1 + 8x2 ≤ 48
so, 6x1 + 8x2 = 48
When X1=0
8x2=48
x2=6
when x2=0
x1=8
the optimal point can be seen on the graph as attached
In this exercise we have to write the maximum function of a company, in this way we find that:
A)[tex]M(Z) = 6x_1+ 3x_2[/tex]
B)[tex]X_1= 8 \ and \ x_2 = 6 \ or \ 0[/tex]
A)So to calculate the maximum equation we have:
[tex]Z = 6x_1 + 3x_2\\6x_1 \leq 36 \\ 8x_2 \leq 32 \\ 6x_1 + 8x_2 \leq 48[/tex]
B) To calculate the limits of the graph we have to do:
[tex]6x_1= 36\\x_1=6\\8x_2= 32\\x_2=4\\6x_1 + 8x_2 = 48\\X_1=0\\8x_2=48\\x_2=6\\x_2=0\\x_1=8[/tex]
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A product has a 4 week lead time. The standard deviation of demand for each of the week is given below. What is the standard deviation of demand over the lead time? (Answer to 2 decimal places) Week Standard deviation of demand 1 16 2 15 3 17 4 13
÷Answer:
Standard Deviation = 176.5
Step-by-step explanation:
To calculate the standard deviation, calculate the mean score for the 4 standard deviation scores:
mean, m = Σx ÷ n
where Σx represents summation of each value = 162 + 153 + 317 + 413
= 1045
n = number of samples to be considered = 4
mean, m = 1045 ÷4
= 261.25
To calculate the standard deviation, use the formula below
SD = [tex]\sqrt{\frac{Σ(x-m)}{n} ^{2} }[/tex]
where x = each value from the week lead time
m = mean = 261.25
n = the size = 4
The Standard deviation formula can be simplified further
when x = 162
[tex]\sqrt{\frac{(x1-m)}{n} ^{2} }[/tex] = 49.625
when x = 153
[tex]\sqrt{\frac{(x2-m)}{n} ^{2} }[/tex] = 23.125
when x = 317
[tex]\sqrt{\frac{(x3-m)}{n} ^{2} }[/tex]= 27.875
when x = 413
[tex]\sqrt{\frac{(x4-m)}{n} ^{2} }[/tex]= 75.875
Note that the above 4 equations can be lumped up into one giant equation by applying a big square root function instead of breaking it down
SD = 49.625 + 23.125 + 27.875 + 75.875
SD = 176.5
The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays. Assuming the number of accidents on a day follows a Poisson distribution.
What is the probability there are no car accidents on that stretch on Monday?
Answer:
1.83% probability there are no car accidents on that stretch on Monday
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays.
This means that [tex]\mu = 4[/tex]
What is the probability there are no car accidents on that stretch on Monday?
This is P(X = 0).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183[/tex]
1.83% probability there are no car accidents on that stretch on Monday
The probability of there being no car accidents on Monday on a certain section of I-40 can be calculated using the Poisson distribution. In this case, with an average of 4 accidents per weekday, the probability is approximately 1.83%.
Explanation:To calculate the probability of there being no car accidents on Monday, we can use the Poisson distribution. In this case, the average number of accidents per weekday is given as 4. The Poisson distribution can be used to calculate the probability of a specific number of events occurring in a given time period.
The formula for calculating the probability of x events occurring in a Poisson distribution is:
P(x) = (e^-λ * λ^x) / x!
Where λ is the average number of events, and x is the number of events we want to calculate the probability for.
In this case, the average number of accidents per weekday is 4, so λ = 4. And we want to calculate the probability of there being no accidents, so x = 0.
Using the formula, we can calculate:
P(0) = (e^-4 * 4^0) / 0! = (e^-4 * 1) / 1 = e^-4, approximately 0.0183Therefore, the probability of there being no car accidents on Monday is approximately 0.0183, or 1.83%.
Consider an experiment with sample space S 5 50, 1, 2, 3, 4, 5, 6, 7, 8, 96 and the events A 5 {0, 2, 4, 6, 8} B 5 {1, 3, 5, 7, 9} C 5 {0, 1, 2, 3, 4} D 5 {5, 6, 7, 8, 9} Find the outcomes
Answer with Step-by-step explanation:
S={0,1,2,3,4,5,6,7,8,9}
A={0,2,4,6,8}
B={1,3,5,7,9}
C={0,1,2,3,4}
D={5,6,7,8,9}
a.A'=S-A
A'={0,1,2,3,4,5,6,7,8,9}-{0,2,4,6,8}
A'={1,3,5,7,9}
b.C'=S-C
C'={0,1,2,3,4,5,6,7,8,9}-{0,1,2,3,4}
C'={5,6,7,8,9}
c.D'=S-D
D'={0,1,2,3,4,5,6,7,8,9}-{5,6,7,8,9}
D'={0,1,2,3,4}
d.[tex]A\cup B=[/tex]{0,2,4,6,8}[tex]\cup[/tex]{1,3,5,7,9}
[tex]A\cup B=[/tex]{0,1,2,3,4,5,6,7,8,9}=S
e.[tex]A\cup C[/tex]={0,2,4,6,8}[tex]\cup[/tex]{0,1,2,3,4}
[tex]A\cup C[/tex]={0,1,2,3,4,6,8}
f.[tex]A\cup D[/tex]={0,2,4,6,8}[tex]\cup[/tex]{5,6,7,8,9}
[tex]A\cup D[/tex]={0,2,4,5,6,7,8,9}
In one study on preferences, researchers formed different displays of grills by rearranging the position (left, center and right) of 3 different grills (A, B, and C), and asked participants to rank the displays from favorite to least favorite.
If, prior to the experiment, all displays were expected to have equivalent levels of preference, what is the probability that a given participant would rank as the favorite a display that had grill A on the left and grill B on the right?
Answer:
⅙
Step-by-step explanation:
Total possibilities: 3×2×1 = 6
Listed below are the numbers of manatee deaths caused each year by collisions with watercraft. The data are listed in order for each year of the past decade.
(a) Find the range, variance, and standard deviation of the data set.
(b) What important feature of the data is not revealed through the different measures of variation?
80 68 71 72 95 89 97 72 75 81
Answer:
Range = 29
Variance= (X₁- U) ² / N= 973/10 = 97.3
Standard Deviation= √variance= √97.3= 9.864
Step-by-step explanation:
Range = Difference between the highest and lowest value = 97-68= 29
Variance
X₁ X₁-U (X₁- U) ²
80 0 zero
68 -12 144
71 -9 81
72 -8 64
95 15 225
89 9 81
97 17 289
72 -8 64
75 -5 25
81 1 1
∑ 800 ZERO 973
u= ∑X₁ /10=800/10=80
Variance= (X₁- U) ² / N= 973/10 = 97.3
Standard Deviation= √variance= √97.3= 9.864
(b) The important feature of the data is not revealed through the different measures of variation is that the variability of two or more than two sets of data cannot be compared unless a relative measure of dispersion is used .
In 2001, a total of 15,555 homicide deaths occurred among males and 4,753 homicide deaths occurred among females. The estimated 2001 midyear populations for males and females were 139,813,000 and 144,984,000 respectively
a) Calculate the homicide-related death rates for males per 100,000.
b) Calculate the homicide-related death rates for females per 100,000.
c) What type(s) of mortality rates did you calculate in Questions 17and 18?
d) Calculate the ratio of homicide-mortality rates for males compared to females.
e) Interpret the rates you calculated in Question 20 as if you were presenting information to a policymaker.
Answer:
a. 11
b. 3
c. homicide mortality rate
d. 11:3
Step-by-step explanation:
a.) If 15,555 homicide cases were recorded among males of 139,813,000 population, then in every 100,000 males, the number of homicides cases will become:
= [15,555/139,813,000] * 100,000
= 11.13 aproximately 11 homicide cases in every 100,000 males.
b.) If 4,753 homicide cases were recorded among Females of 144,984,000 population, then in every 100,000 Females, the number of homicides cases will become:
= [4753/144,984,000] * 100,000
= 3.28 approximately 3 homicide cases in every 100,000 females
C.) Homicide-mortality rate
d.) ratio of male to female homicide rate = 11 : 3
e.) What those rates means is that in every 100,000
Males in 2001, 11 of them were Victims of homicide and in every 100,000 females, 3 of them are victims of homicide.
One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 93% detection rate for carriers and a 2% false positive rate. Suppose that an individual is tested. What is the probability that an individual who tests negative does not carry the disease? What is the specificity of the test?
Answer:
(1) The probability that an individual who tests negative does not carry the disease is 0.9709.
(2) The specificity of the test is 98%.
Step-by-step explanation:
Denote the events as follows:
X = a person carries the disease
Y = the test detected the disease.
Given:
[tex]P(X) = 0.01\\P(Y|X)=0.93\\P(Y|X^{c})=0.02[/tex]
The probability of a person not carrying the disease is:
[tex]P(X^{c})=1-P(X)=1-0.01=0.99[/tex]
The probability that the test does not detects the disease when the person is carrying it is:
[tex]P(Y^{c}|X)=1-P(Y|X)=1-0.93=0.07[/tex]
The probability that the test does detects the disease when the person is not carrying it is:
[tex]P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98[/tex]
(1)
Compute the probability that an individual who tests negative does not carry the disease as follows:
[tex]P(X^{c}|Y^{c})=\frac{P(Y^{c}|X^{c})P(X^{c})}{P(Y^{c}|X^{c})P(X^{c})+P(Y^{c}|X)P(X)} \\=\frac{(0.98\times 0.99)}{(0.98\times 0.99)+(0.07\times 0.01)} \\=0.9709[/tex]
Thus, the probability that an individual who tests negative does not carry the disease is 0.9709.
(2)
By specificity it implies that how accurate the test is.
Compute the probability of negative result when the person is not a carrier as follows:
[tex]P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98[/tex]
Thus, the specificity of the test is 98%.
plz help me answer this question
Answer:
The answer to your question is distance = 3500000 km
Step-by-step explanation:
Data
Scale 1 : 500000
7 cm
Process
1.- To solve this problem use direct proportions or rule of three.
1 : 500 000 :: 7 : x
2.- Multiply the middle numbers and the result divide it by the edge.
x = (7 x 500000) / 1
Simplification
x = 3500000 km
A card is drawn from a standard deck of 5252 playing cards. What is the probability that the card will be a heart or a face card? Express your answer as a fraction or a decimal number rounded to four decimal places.
Answer:
The probability that the card will be a heart or a face card is P=0.4231.
Step-by-step explanation:
We have a standard deck of 52 cards.
In this deck, we have 13 cards that are a heart.
We also have a total of 12 face cards (4 per each suit). So there are 3 face cards that are also a heart.
To calculate the probability that a card be a heart or a face card, we sum the probability of a card being a heart and the probability of it being a face card, and substract the probability of being a heart AND a face card.
We can express that as:
[tex]P(H\,or\,F)=P(H)+P(F)-P(H\&F)\\\\P(H\,or\,F)=13/52+12/52-3/52=22/52=0.4231[/tex]
Given that x is a normal variable with mean μ = 49 and standard deviation σ = 6.7, find the following probabilities. (Round your answers to four decimal places.) (a) P(x ≤ 60) (b) P(x ≥ 50) (c) P(50 ≤ x ≤ 60)
Answer:
0.9500, 0.4407, 0.3904
Step-by-step explanation:
(a) P(x ≤ 60). We need to find the area under the standard normal curve to the left of x = 60. The appropriate command when using a TI-83 Plus calculator with statistical functions is normcdf(-1000, 60, 49, 6.7). This comes out to 0.9500. P(x ≤ 60) = 0.9500
(b) P(x ≥ 50) would be normcdf(50, 1000,49, 6.7), or 0.4407
(c) P(50 ≤ x ≤ 60) would be normcdf(50,60,49,6.70, or 0.3904
The question involves finding probabilities for a normal distribution given mean μ and standard deviation σ. By converting x values to z-scores and using the standard normal distribution, we can calculate the desired probabilities.
Explanation:The student is asking about probabilities related to a normally distributed random variable with a given mean (μ) and standard deviation (σ). To find these probabilities, we convert the x values to z-scores and use the standard normal distribution.
P(x ≤ 60): Subtract the mean from 60 and divide by the standard deviation to get the z-score. Then use the standard normal distribution table or a calculator's normalcdf function to find the probability.P(x ≥ 50): Find the z-score for x = 50, then calculate 1 minus the cumulative probability up to that z-score to obtain the probability that x is greater than or equal to 50.P(50 ≤ x ≤ 60): Calculate the z-scores for x = 50 and x = 60, then find the cumulative probability for each. The desired probability is the difference between these two cumulative probabilities.Calculations here are based on the normal distribution parameters provided and the standard normal distribution. The z-score is the key to converting any normal distribution to the standard normal distribution, enabling the use of standard tables or software functions for probability calculations.
An aerospace company has submitted bids on two separate federal government defense contracts. The company president believes that there is a 40% probability of winning the first contract. If they win the first contract, the probability of winning the second is 65%. However, if they lose the first contract, the president thinks that the probability of winning the second contract decreases to 49%.
What is the probability that they win both contracts?
Answer:
26% probability that they win both contracts.
Step-by-step explanation:
These following probabilities are important to solve this question:
0.4 = 40% probability of winning the first contract.
0.65 = 65% probability of winning the second contract if the first contract is won.
What is the probability that they win both contracts?
[tex]P = 0.4*0.65 = 0.26[/tex]
26% probability that they win both contracts.
Consider the problem of shrinkage in a supply chain. Use this data: Expected Consumer Demand = 5,000 Retail: Theft and Damage - 5% Distribution Center: Theft and Damage - 4% Packaging Center: Damage - 3% Manufacturing: Defect rate - 4% Materials: Supplier defects - 5% How many units should the materials plan account for in order to meet the expected consumer demand? (Choose the closest answer.)
Answer: units = 6198
Step-by-step explanation:
expected consumer demand is 5000, the units that must be planned given shrinkage percentage levels in different stages of the supply chain we have to use the trial and error method.
lets try 6200 units
6200 x 95% = 5890, 5890 x 96% = 5654.4, 5654.40 x 97% = 5484.768, 5484.768 x 96% = 5265.37728, 56265 x 95% = 5002.108416 ≈ 5002
6198 units
(6198 x 95% = 5888.10) (5888.10 x 96% = 5652.576) (5652.576 x 97% = 5482.99872) (5482.99872 x 96% = 5263.6787712) (5263.6787712 x 95% = 5000.4948326) ≈ 5000
using the same procedure for 6197 units the answer will be 4999.688041
units that should be produced to cover the demand of 5000 = 6198
To meet the 5,000 unit consumer demand, the materials plan should account for approximately 6,195 units, considering the cumulative loss percentages at each stage of the supply chain.
Explanation:To meet the expected consumer demand of 5,000 units while accounting for shrinkage at various stages of the supply chain, we need to calculate the cumulative effect of theft, damage, and defects on the number of units. We need to work backwards from the consumer to the materials supplier to determine the initial quantity needed.
Start with the expected consumer demand: 5,000 units.Account for retail theft and damage: 5% loss means we need 5,000 / (1 - 0.05) = 5,263 units from the distribution centers.Account for distribution center theft and damage: 4% loss means we need 5,263 / (1 - 0.04) ≈ 5,482 units from the packaging center.Account for packaging center damage: 3% loss means we need 5,482 / (1 - 0.03) ≈ 5,650 units from manufacturing.Account for manufacturing defects: 4% loss means we need 5,650 / (1 - 0.04) ≈ 5,885 units from the materials.Finally, account for supplier defects: 5% loss means we need 5,885 / (1 - 0.05) ≈ 6,195 units.The materials plan should account for approximately 6,195 units to meet the expected consumer demand, accounting for expected losses due to theft, damage, and defects throughout the supply chain stages.
In ΔABC, b = 68 inches, ∠B=65° and ∠C=93°. Find the length of a, to the nearest inch.
Answer:
28 inches
Step-by-step explanation:
∠A + ∠B + ∠C = 180°
∠A + 65° +93° = 180°
∠A + 158° = 180°
∠A= 180°-158° = 22°
Using Law of Sines
a/sinA= b/sinB
a/sin22= 68 inches/sin65
a/sin22 = 68/0.9063 = 75.03
a = 75.03 x sin 22
a = 75.03 x 0.3746 = 28.106238≈28 inches
Find each difference in the photo below
Answer:
[tex]= - 2 x^{3} + 4x -8[/tex]
Step-by-step explanation:
The first step is to open the parenthesis,
Since there is a negative sign before the second parenthesis, so the sign of all the values in second parenthesis will be changed and the equation will look something like this
[tex]= 2x^{3} + 4x -2 - 4 x^{3} + 6[/tex]
The second step is to re arrange the equation
[tex]= 2x^{3} - 4 x^{3} + 4x - 2 + 6[/tex]
The last and final step is to solve the equation
[tex]= - 2 x^{3} + 4x -8[/tex]
This is our answer
Answer:
The difference is -2x³ + 4x + 4.
Step-by-step explanation:
Subtract the two expression as follows:
[tex](2x^{3}+4x-2)-(4x^{3}-6)=2x^{3}+4x-2-4x^{3}+6\\[/tex]
Combine the like terms together:
[tex]=2x^{3}-4x^{3}+4x-2+6[/tex]
Simplify as follows:
[tex]=-2x^{3}+4x+4[/tex]
Thus, the difference is -2x³ + 4x + 4.
Agee Storage issued 35 million shares of its $1 per common stock at $16 per share several years ago. Last year, for the first time, Agee reacquired 1 million shares at $14 per share. If Agee now retires 1 million shares at $19 per share. By what amount will Agee's total paid-in capital decline?
Answer:
$18 Millions
Step-by-step explanation:
Decline in total paid in capital=Total value of reacquisition - Decline in retained earnings
$19-$1=$18 millions
Answer:
$18 000 000
Step-by-step explanation:
The first step is to calculate the par value of shares
$1*35000000=35000000
=$35000000
Issued shares of the the par
amount over par
=$16-$1=$15
Then paid capital in excess of par
$15*1000000
=$525000000
The first reacquire remember there were issued at $16
so for first reacquire
$16-$14=$2
$2*35000000=$70 000000
so Agee total paid-in capital will decline by
$15+$2+$1=$18 per share
This table gives a few (x,y) pairs of a line in the coordinate plane
x (48) (61) (74)
y (-30) (-45) (-60)
what is the x-intercept of the line?
Answer:(22,0)
Step-by-step explanation: you have to find when y equals 0
If the Durbin-Watson statistic has a value close to 0, which assumption is violated? In other words, which assumption is the Durbin-Watson statistic checking to see is violated?
a - Independence of errors
b- Normality of errors
c- Homoscedasticity
d- none of the above
Answer:
Null Hypothesis: No first order autocorrelation
Alternative hypothesis: first order correlation exists
The assumptions to run this test are:
1) Errors are normally distributed with mean 0
2) Errors follows an stationary process
3) Independence condition between the erros
The statistic is defined as:
[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]
And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.
Step-by-step explanation:
The Durbin Watson test is a way to check autocorrelation in residuals for a time seeries or a regression.
We need to remember that the autocorrelation is the similarity of the time series in successive intervals. When we conduct this type of test we are checking if the time series can be modeled with and AR(1) process autoregressive.
The system of hypothesis on this case are:
Null Hypothesis: No first order autocorrelation
Alternative hypothesis: First order correlation exists
The assumptions to run this test are:
1) Errors are normally distributed with mean 0
2) Errors follows an stationary process
3) Independence condition between the errors
The statistic is defined as:
[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]
And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.
Consider two people being randomly selected. (For simplicity, ignore leap years.)
(a) What is the probability that two people have a birthday on the 9th of any month?
(b) What is the probability that two people have a birthday on the same day of the same month?
Answer:
[tex](a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}[/tex]
Step-by-step explanation:
Part (a) the probability that two people have a birthday on the 9th of any month.
Neglecting leap year, there are 365 days in a year.
There are 12 possible 9th in months that make a year calendar.
If two people have birthday on 9th; P(1st person) and P(2nd person).
[tex]=\frac{12}{365} X\frac{12}{365} = \frac{144}{133225}[/tex]
Part (b) the probability that two people have a birthday on the same day of the same month
P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on same day of same month) = 1
P(2 people selected not having birthday on same day of same month):
[tex]= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}[/tex]
P(2 people selected have birthday on the same day of same month) [tex]= 1-\frac{364}{365} \\\\= \frac{1}{365}[/tex]
Final answer:
The probability that two people have a birthday on the 9th of any month is 1/133,225. The probability that two people have a birthday on the same day of the same month is also 1/133,225.
Explanation:
To calculate the probability that two people have a birthday on the 9th of any month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year, so the number of possible outcomes is 12. The probability of each person having a birthday on the 9th is 1/365. Therefore, the probability that two people have a birthday on the 9th of any month is (1/365) x (1/365) = 1/133,225.
To calculate the probability that two people have a birthday on the same day of the same month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year and each month has 30 or 31 days. So the number of possible outcomes is 12 x 31 = 372. The probability of each person having a birthday on a specific day is 1/365. Therefore, the probability that two people have a birthday on the same day of the same month is (1/365) x (1/365) = 1/133,225.
what are some questions to ask about a function when sketching its graph?
Match the name of the sampling method descriptions given.Situations 1. divide the population by age and select 5 people from each age2. writing everyones name on a playing card, shuffling the deck, then choosing the top 20 cards3. choosing every 5th person on a list 4. randomly select two tables in the cafeteria and survey all the people at those two tables 5. asking people on the streetSampling Method a. Simple Random b. Stratified c. Systematicd. Convenience e. Cluster
Answer:
1. ⇔ b.
2. ⇔ a.
3. ⇔ c.
4. ⇔ e.
5. ⇔ d.
Step-by-step explanation:
We conclude that:
5. asking people on the street Sampling is (d.) Convenience.
4. randomly select two tables in the cafeteria and survey all the people at those two tables is (e.) Cluster.
3. choosing every 5th person on a list is (c.) Systematic.
1. divide the population by age and select 5 people from each age is (b.) Stratified.
2. writing everyones name on a playing card, shuffling the deck, then choosing the top 20 cards is (a.) Simple Random.
Employing appropriate sampling methods is crucial for obtaining accurate and representative data. Stratified, simple random, cluster, systematic, and convenience sampling techniques serve specific purposes and should be selected based on the research objectives and population characteristics.
Here are the matching sampling methods for the given situations:
Divide the population by age and select 5 people from each age group.
Sampling Method: Stratified (b)
In this method, the population is divided into strata (age groups in this case) and a random sample is taken from each stratum.
Write everyone's name on a playing card, shuffle the deck, then choose the top 20 cards.
Sampling Method: Simple Random (a)
This method involves selecting individuals randomly, ensuring every person has an equal chance of being chosen.
Sampling Method: Systematic (c)
Systematic sampling involves selecting every kth individual from a list after selecting a random starting point.
Randomly select two tables in the cafeteria and survey all the people at those two tables.
Sampling Method: Cluster (e)
Cluster sampling involves dividing the population into clusters (cafeteria tables in this case), randomly selecting some clusters, and then surveying all individuals within those clusters.
Sampling Method: Convenience (d)
Convenience sampling involves selecting individuals who are convenient to reach, which may not represent the entire population accurately due to potential bias.
To learn more about sampling method
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Pre Calculus, Trigonometry Help
Answer:
[tex]\displaystyle cos\theta=\frac{36}{164}=\frac{9}{41}[/tex]
[tex]\displaystyle tan\theta=\frac{160}{36}=\frac{40}{9}[/tex]
[tex]\displaystyle csc\theta=\frac{164}{160}=\frac{41}{40}[/tex]
[tex]\displaystyle sec\theta=\frac{164}{36}=\frac{41}{9}[/tex]
[tex]\displaystyle cot\theta=\frac{36}{160}=\frac{9}{40}[/tex]
Step-by-step explanation:
Trigonometric ratios in a Right Triangle
Let ABC a right triangle with the right angle (90°) in A. The longest length is called the hypotenuse and is the side opposite to A. The other sides are called legs and are shorter than the hypotenuse.
Some trigonometric relations are defined in a right triangle. Being [tex]\theta[/tex] one of the angles other than the right angle, h the hypotenuse, x the side opposite to [tex]\theta[/tex] and y the side adjacent to [tex]\theta[/tex], then
[tex]\displaystyle sin\theta=\frac{x}{h}[/tex]
[tex]\displaystyle cos\theta=\frac{y}{h}[/tex]
[tex]\displaystyle tan\theta=\frac{x}{y}[/tex]
[tex]\displaystyle csc\theta=\frac{h}{x}[/tex]
[tex]\displaystyle sec\theta=\frac{h}{y}[/tex]
[tex]\displaystyle cot\theta=\frac{y}{x}[/tex]
We are given the values of h=164 and x=160, let's find y
[tex]y=\sqrt{164^2-160^2}=36[/tex]
Now we compute the rest of the ratios
[tex]\displaystyle cos\theta=\frac{36}{164}=\frac{9}{41}[/tex]
[tex]\displaystyle tan\theta=\frac{160}{36}=\frac{40}{9}[/tex]
[tex]\displaystyle csc\theta=\frac{164}{160}=\frac{41}{40}[/tex]
[tex]\displaystyle sec\theta=\frac{164}{36}=\frac{41}{9}[/tex]
[tex]\displaystyle cot\theta=\frac{36}{160}=\frac{9}{40}[/tex]
An article in Medicine and Science in Sports and Exercise "Maximal Leg-Strength Training Improves Cycling Economy in Previously Untrained Men," (2005, Vol. 37 pp. 131–1236) studied cycling performance before and after eight weeks of leg-strength training. Seven previously untrained males performed leg-strength training three days per week for eight weeks (with four sets of five replications at 85% of one repetition maximum). Peak power during incremental cycling increased to a mean of 315 watts with a standard deviation of 16 watts. Construct a 99% two-sided confidence interval for the mean peak power after training. Assume population is approximately normally distributed.
The 99% confidence interval for the mean peak power after training is approximately [tex](292.61, 337.39)[/tex] watts.
Identify the given data:
Mean peak power after training: [tex]\bar{x} = 315[/tex] watts
Standard deviation: [tex]s = 16[/tex] watts
Sample size: [tex]n = 7[/tex]
Confidence level: 99%
Find the critical value:
Since the sample size is small (n < 30) and the population standard deviation is not known, we use the t-distribution. For a 99% confidence interval with [tex](n-1) = 6[/tex] degrees of freedom, the critical value (t-value) can be found from the t-table. Using the t-table, [tex]t_{\frac{\alpha}{2},6} = 3.707[/tex].
Calculate the standard error (SE):
[tex]SE = \frac{s}{\sqrt{n}} = \frac{16}{\sqrt{7}} = \frac{16}{2.6458} \approx 6.04[/tex] watts
Compute the margin of error (ME):
[tex]ME = t_{\frac{\alpha}{2}} \times SE = 3.707 \times 6.04 \approx 22.39[/tex] watts
Construct the confidence interval:
Lower bound: [tex]\bar{x} - ME = 315 - 22.39 \approx 292.61[/tex] watts
Upper bound: [tex]\bar{x} + ME = 315 + 22.39 \approx 337.39[/tex] watts
On a certain airline, the chance the early flight from Atlanta to Chicago is full is 0.8. The chance the late flight is full is 0.7. The chance both flights are full is 0.6. Are the two flights being full independent events?
Answer:
No
Step-by-step explanation:
A- full early flight atlanta to chicago
P(a)=0,8
B- full late night flightt
P(b)=0,7
A&b- both are full
P(a&b) probability that both flights are full
Suppose that they are independet, then we have:
P(a&b)=p(a)*p(b)=0,8*0,7=0,56.
So if they are independet then p(a&b)=0,56, and that is not true.
Using the concepts of marginal social benefit and marginal social cost, explain how the optimal combination of goods can be determined in an economy that produces only two goods.
Answer:
Optimal combination of goods can be determined in an economy that produces only two goods, with production of extra units of the two goods at a minimal marginal social cost. The consumption of the additional units of the two goods being produced will be benefitted by the consumers. This is known as marginal social benefit.
Step-by-step explanation:
Marginal social cost is the change in society's total cost brought about by the production of an additional unit of a good or service. It includes both marginal private cost and marginal external cost.
Marginal social benefit is the change in benefits associated with the consumption of an additional unit of a good or service. It is measured by the amount people are willing to pay for the additional unit of a good or service.
The ages (in years) of the 5 doctors at a local clinic are the following. 40, 44, 49, 40, 52 Assuming that these ages constitute an entire population, find the standard deviation of the population. Round your answer to two decimal places.
Answer:
The standard deviation of the population is 4.82 years.
Step-by-step explanation:
Mean = summation of all ages ÷ number of doctors = (40+44+49+40+52) ÷ 5 = 225 ÷ 5 = 45 years
Population standard deviation = sqrt[sum of squares of the difference between each age and mean ÷ number of doctors] = sqrt[((40 - 45)^2 + (44 - 45)^2 + (49 - 45)^2 + (40 - 45)^2 + (52 - 45)^2) ÷ 5] = sqrt[(25+1+16+25+49) ÷ 5] = sqrt[116 ÷ 5] = sqrt(23.2) = 4.82 years
Final answer:
The standard deviation of the population of doctors' ages is determined by calculating the mean, finding the squared differences from the mean, averaging these, and taking the square root, resulting in approximately 4.82 years.
Explanation:
To find the standard deviation of the population of doctors' ages, you first need to calculate the mean (average) age. Then, you compute the variance by finding the squared differences from the mean for each age, and average those values. Finally, the standard deviation is the square root of the variance.
Calculate the mean age: (40 + 44 + 49 + 40 + 52) / 5 = 225 / 5 = 45 years.
Find the squared differences from the mean: (40-45)², (44-45)², (49-45)², (40-45)², (52-45)².
Sum the squared differences: 25 + 1 + 16 + 25 + 49 = 116.
Calculate the variance: 116 / 5 = 23.2 years² (because we're dealing with a population, not a sample).
Find the standard deviation: sqrt(23.2) ≈ 4.82 years.
The standard deviation of the population of doctors' ages is approximately 4.82 years or rounding off to nearest number will be 5 years.
According to the Bureau of Labor Statistics it takes an average of 16 weeks for young workers to find a new job. Assume that the probability distribution is normal and that the standard deviation is two weeks. What is the probability that 20 young workers average less than 15 weeks to find a job?
Answer:
1.25% probability that 20 young workers average less than 15 weeks to find a job
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
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.
Central Limit Theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 16, \sigma = 2, n = 20, s = \frac{2}{\sqrt{20}} = 0.4472[/tex]
What is the probability that 20 young workers average less than 15 weeks to find a job?
This is the pvalue of Z when X = 15. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15 - 16}{0.4472}[/tex]
[tex]Z = -2.24[/tex]
[tex]Z = -2.24[/tex] has a pvalue of 0.0125.
1.25% probability that 20 young workers average less than 15 weeks to find a job
A market analyst is developing a regression model to predict monthly household expenditures on groceries as a function of family size, household income, and household neighborhood (urban, suburban, and rural). The response variable in this model is _____.
Answer:
Monthly household expenditures on groceries
Step-by-step explanation:
The response variable is the one for which measurements are desired and that depends on other variables.
In this case, family size, household income, and household neighborhood are independent variables, while the response variable is the monthly household expenditures on groceries.