Answer:
Step-by-step explanation:
A. The three-year zero coupon bond, because the future value is receiver sooner, thus the present value is higher.
B. The three year 4% coupon bond because it pays interest payments, whereas the zero coupon bond is pure discount bond.
C. The 6% coupon bond because the coupon (interest) payments are higher.
Final answer:
In summary, a three-year zero-coupon bond has a higher price than a five-year zero-coupon bond due to its shorter maturity. A three-year 4% coupon bond will have a higher price than a three-year zero-coupon bond because it provides interest payments. A two-year 6% coupon bond will have a higher price than a two-year 5% coupon bond due to its higher interest payments.
Explanation:
The Treasury security prices are determined by several factors, including maturity and coupon rates. In answering the questions presented by the student, we compare the prices of different types of Treasury securities based on these attributes.
A three-year zero-coupon bond will have a higher price than a five-year zero-coupon bond because it has a shorter time to maturity, meaning less time for interest rates to affect its price.Between a three-year zero-coupon bond and a three-year 4% coupon bond, the coupon bond will generally have a higher price since it provides periodic interest payments in addition to the repayment of par at maturity.Comparing a two-year 5% coupon bond with a two-year 6% coupon bond, the bond with the higher coupon rate (6%) will typically have a higher price, as it offers more substantial periodic payments.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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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.
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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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]
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.
Popcorn kernels pop independently (i.e. unimolecularly). For one brand at constant temperature, 7 kernels pop in 10 seconds when 180 kernels are present. After 75 kernels have popped, how many kernels will pop in 10 seconds? (Your answer may include fractions of a kernel).
Answer:
In 10 seconds, after 75 kernels have popped, they should pop 161/36 kernels.
Step-by-step explanation:
You can solve this problem with a rule of three, if you remove 75 kernels, you have 115 left, and in 10 seconds you may expect a proportional amount of popcorn kernels dropping.
If for 180 kernels 7 drop, then for 115 kernels the amount that drops is
7/180 * 115 = 161/36 popcorn kernels.
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
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a 90% confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a 90% confidence interval for the proportion of college seniors who carry a credit card balance from month to month.c. Explain why the two 90% confidence intervals from Parts (a) and (b) are not the same width.
Answer:
Step-by-step explanation:
Hello!
There are two variables of interest:
X₁: number of college freshmen that carry a credit card balance.
n₁= 1000
p'₁= 0.37
X₂: number of college seniors that carry a credit card balance.
n₂= 1000
p'₂= 0.48
a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.
The formula for the interval is:
p'₁±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.648[/tex]
0.37±1.648*[tex]\sqrt{\frac{0.37*0.63}{1000} }[/tex]
0.37±1.648*0.015
[0.35;0.39]
With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.
b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648
The formula for this interval is
p'₂±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }[/tex]
0.48±1.648* [tex]\sqrt{\frac{0.48*0.52}{1000} }[/tex]
0.48±1.648*0.016
[0.45;0.51]
With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.
c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.
Freshmen: [tex]\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015[/tex]
Seniors: [tex]\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016[/tex]
The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.
The confidence interval will be widest when [tex]\( \hat{p} = 0.5 \)[/tex] and narrowest when [tex]\( \hat{p} \)[/tex] is close to 0 or 1. In this case, the proportion for seniors is closer to 0.5 than the proportion for freshmen, resulting in a slightly wider confidence interval for seniors.
To construct 90% confidence intervals for the proportions of college freshmen and seniors who carry a credit card balance from month to month, we will use the formula for a confidence interval for a proportion:
[tex]\[ \text{Confidence Interval} = \hat{p} \pm Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]
where:
- [tex]\( \hat{p} \)[/tex] is the sample proportion,
- [tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level (for 90% confidence, the Z-score is approximately 1.645),
- \( n \) is the sample size.
For part (a), we have a sample proportion [tex]\( \hat{p}_{\text{freshmen}} = 0.37 \) and a sample size \( n_{\text{freshmen}} = 1000 \)[/tex]. The 90% confidence interval for the proportion of college freshmen is calculated as follows:
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.37(1 - 0.37)}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.37 \times 0.63}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.2331}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \times 0.0153 \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 0.0252 \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = (0.3448, 0.3952) \][/tex]
For part (b), we have a sample proportion [tex]\( \hat{p}_{\text{seniors}} = 0.48 \) and a sample size \( n_{\text{seniors}} = 1000 \)[/tex]. The 90% confidence interval for the proportion of college seniors is calculated as follows:
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.48(1 - 0.48)}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.48 \times 0.52}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.2496}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \times 0.0158 \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 0.0260 \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = (0.4540, 0.5060) \][/tex]
For part (c), the reason why the two 90% confidence intervals from Parts (a) and (b) are not the same width is due to the different sample proportions[tex]\( \hat{p} \).[/tex] The width of a confidence interval for a proportion depends on the standard deviation of the sampling distribution of the proportion, which is given by [tex]\( \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)[/tex]. Since the sample proportions for freshmen and seniors are different (0.37 for freshmen and 0.48 for seniors), the standard deviations will also be different, leading to different widths for the confidence intervals.
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
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
If Brazil has a total of 150 bird species and you happen to catch two species in an observational experiment using mist nets, what is the total number of possible combinations of species that you captured
Answer:
The total number of possible combinations of species that you captured is 11,175.
Step-by-step explanation:
The order that you captured the species is not important. For example, capturing a bird of specie A then a bird of specie B is the same outcomes as capturing a bird of specie B then a bird of specie A. So the combinations formula is used to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, we have that:
Combinations of 2 species from a set of 150. So
[tex]C_{150,2} = \frac{150!}{2!148!} = 11,175[/tex]
The total number of possible combinations of species that you captured is 11,175.
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
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 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.
Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 18. Complete parts (a) through (c) below. a. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 18. The value of the mean is muequals nothing peas. (Type an integer or a decimal. Do not round.) The value of the standard deviation is sigmaequals nothing peas. (Round to one decimal place as needed.) b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high. Values of nothing peas or fewer are significantly low. (Round to one decimal place as needed.) Values of nothing peas or greater are significantly high. (Round to one decimal place as needed.) c. Is a result of 2 peas with green pods a result that is significantly low? Why or why not? The result ▼ is not is significantly low, because 2 peas with green pods is ▼ equal to greater than less than nothing peas. (Round to one decimal place as needed.)
Question is not well presented
Assume that hybridization
experiments are conducted with peas having the property that for offspring, there is
a 0. 75 probability that a pea has green pods (as in one of Mendel's famous experiments).
Assume that offspring peas are randomly selected in groups of 18. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.
Answer:
Values below 9.826 (or equal) are significantly low
Values above 17.174 (or equal) are significantly high
Step-by-step explanation:
First, we Calculate the mean.
Mean = np where n = 18, p = 0.75
Mean = 18 * 0.75
Mean = 13.5
Then we Calculate the standard deviation
S = √npq where q = 1-0?75 = 0.25
S = √13.5 * 0.25
S = 1.837
The range rule of thumb tells us that the usual range of values is within 2 of the mean and Standard deviation.
i.e.
Mean - 2(s), Mean + 2(s)
13.5 - 2(1.837), 13.5 + 2(1.837)
9.826, 17.174
Values below 9.826 (or equal) are significantly low
Values above 17.174 (or equal) are significantly high
...
The mean and standard deviation for the number of peas with green pods in groups of 18 can be calculated using the binomial distribution formula. Significantly low values are below 8.92 peas or fewer, and significantly high values are 18.08 peas or greater. A result of 2 peas with green pods is significantly low.
Explanation:To find the mean and standard deviation for the numbers of peas with green pods in groups of 18, we need to use the binomial distribution formula. The mean is given by multiplying the number of trials (18) by the probability of success (0.75), which equals 13.5 peas. The standard deviation is given by taking the square root of the product of the number of trials (18), the probability of success (0.75), and the probability of failure (0.25), which equals 2.29 peas.
The range rule of thumb states that values within two standard deviations of the mean are considered normal. In this case, significantly low values would be 13.5 - 2(2.29) = 8.92 peas or fewer, and significantly high values would be 13.5 + 2(2.29) = 18.08 peas or greater.
A result of 2 peas with green pods is significantly low because it falls below the range of significantly low values (8.92 peas or fewer).
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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.
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%.
The random variable x has a normal distribution with standard deviation 21. It is known that the probability that x exceeds 160 is .90. Find the mean mu of the probability distribution.
Answer:
Mean, [tex]\mu[/tex] = 133.09
Step-by-step explanation:
We are given that the random variable x has a normal distribution with standard deviation 21,i.e;
X ~ N([tex]\mu,\sigma = 21[/tex])
The Z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ (0,1)
Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;
P(X > 160) = 0.90
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{160-\mu}{21}[/tex] ) = 0.90
From the z% table we find that at value of x = -1.2816, the value of
P(X > 160) is 90%
which means; [tex]\frac{160-\mu}{21}[/tex] = -1.2816
160 - [tex]\mu[/tex] = 21*(-1.2816)
[tex]\mu[/tex] = 160 - 26.914 = 133.09
Therefore, mean [tex]\mu[/tex] of the probability distribution is 133.09.
Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt.(a) What population is under consideration in the data set?(b) What parameter is being estimated?(c) What is the point estimate for the parameter?(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?(e) Compute the value from part (d) for this context.(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?(g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0:4 instead of ^p, does the resulting value change much?
Step-by-step explanation:
(a)
The population under study are the adults of United States.
(b)
A parameter the population characteristic that is under study.
In this case the researcher is interested in the proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt.
So the parameter is the population proportion of US adults who say this.
(c)
A point estimate is a numerical value that is the best guesstimate of the parameter. It is computed using the sample values.
For example, sample mean is the point estimate of population mean.
The point estimate of the population proportion of US adults who say cover a $400 unexpected expense without borrowing money or going into debt, is the sample proportion, [tex]\hat p[/tex].
[tex]\hat p=\frac{322}{765}=0.421[/tex]
(d)
The uncertainty of the point estimate can be measured by the standard error.
The standard error tells us how closer the sample statistic is to the parameter value.
[tex]SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
(e)
The standard error is:
[tex]SE_{\hat p}=\sqrt{\frac{0.421(1-0.421)}{765}} =0.018[/tex]
(f)
The sample proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt, is approximately 42.1%.
As the sample size is quite large this value can be used to estimate the population proportion.
If the proportion is believed to be 50% then she will be surprised because the estimated percentage is quite less than 50%.
(g)
Compute the standard error using p = 0.40 as follows:
[tex]SE=\sqrt{\frac{ p(1- p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{765}}=0.0177\approx0.018[/tex]
The standard error does not changes much.
The data set consists of information on the ability of adults in the United States to cover a $400 unexpected expense. The parameter being estimated is the proportion of adults who cannot cover the expense without borrowing money or going into debt. The point estimate for this parameter is found by dividing the number of adults in the sample who cannot cover the expense by the total sample size.
Explanation:(a) The population under consideration in the data set is all adults in the United States.
(b) The parameter being estimated is the proportion of adults in the United States who could not cover a $400 unexpected expense without borrowing money or going into debt.
(c) The point estimate for the parameter is the proportion of adults in the sample who said they could not cover a $400 unexpected expense without borrowing money or going into debt, which is 322/765.
(d) The name of the statistic that measures the uncertainty of the point estimate is the standard error.
(e) To compute the value of the standard error, you need the formula which depends on the sample proportion and sample size. Since the necessary values are not provided in the question, it is not possible to compute the value in this context.
(f) The cable news pundit should not be surprised by the data since the sample proportion is not too far off from the value she thinks is true.
(g) If the true population value is 40%, the resulting value in part (e) will change because the sample proportion is different from the true proportion.
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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}
This Excel file Undergrad Survey shows the data resulting from a survey of 50 undergraduate students at Clemson University. Majors of students in the survey are accounting (A), economics and finance (EF), management (M), marketing (MR), computer information systems (IS), other (O), and undecided (UN). "Number of affiliations" is the number of social networking sites at which the student is registered; "Spending" is the amount spent on textbooks for the current semester. The other variables are self-explanatory.
We will assume that this sample is a representative sample of all Clemson undergraduates. Use Excel or statcrunch to make a histogram of GPA to verify that the distribution of GPA can be approximated by the N(3.12, 0.4) normal model.
Question 1. The School of Business at Clemson has created a rigorous new International Business Studies major to better prepare their students for the global marketplace. A GPA of 3.69 or higher is required for a Clemson undergraduate to change his/her major to International Business Studies. What is the probability that a randomly selected Clemson undergraduate has a GPA of at least 3.69? (Use 4 decimal places in your answer).
Question 2. To attract high-quality current Clemson undergraduates into the new International Business Studies major, scholarships in International Business Studies will be offered to a Clemson undergraduate if his/her GPA is at or above the 95.54th percentile. What is the minimum GPA required to meet this criterion?
The GPA Values:
2.38
2.42
2.45
2.50
2.60
2.61
2.65
2.67
2.74
2.75
2.75
2.76
2.80
2.87
2.88
2.91
2.92
2.93
2.94
3.00
3.02
3.09
3.10
3.11
3.13
3.14
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.26
3.28
3.33
3.34
3.43
3.44
3.48
3.50
3.55
3.62
3.62
3.63
3.71
3.72
3.77
3.85
4.00
Answer:
Question 1:
Here total number of students GPA is given = 50
Number of people who have GPA more than 3.69 = 5
Therefore,
Pr(That a student has GPA more than 3.69) = 5/50 = 0.1
Here X ~ NORMAL (3.12, 0.4)
so Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4)
Z = (3.69 - 3.12)/ 0.4 = 1.425
Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4) = 1 - Pr(Z < 1.425) = 1 - 0.9229 = 0.0771
Question 2
Here GPA would be above 95.54th percentile
so as per Z table relative to that percentile is = 1.70
so Z = (X - 3.12)/ 0.4 = 1.70
X = 3.12 + 0.4 * 1.70 = 3.80
so any person with GPA above or equal to 3.80 is eligible for that.
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
In a population of 200,000 people, 40,000 are infected with a virus. After a person becomes infected and then recovers, the person is immune (cannot become infected again). Of the people who are infected, 5% will die each year and the others will recover. Of the people who have never been infected, 35% will become infected each year. How many people will be infected in 4 years? (Round your answer to the nearest whole number.)
After 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.
In a population of 200,000 people, 40,000 are initially infected with a virus. Each year, 5% of the infected individuals die, and 25% of the uninfected population becomes infected. The population dynamics over the next 4 years can be analyzed step by step:
Year 0:
Total Population: 200,000
Infected: 40,000
Uninfected: 160,000
Year 1:
Infected: 40,000 + (25% of 160,000) = 40,000 + 40,000 = 80,000
Uninfected: 160,000 - 40,000 = 120,000
Deaths: 5% of 80,000 = 4,000
Recoveries: 95% of 80,000 = 76,000
Year 2:
Infected: (25% of 120,000) = 30,000
Uninfected: 120,000 - 30,000 = 90,000
Deaths: 5% of 30,000 = 1,500
Recoveries: 95% of 30,000 = 28,500
Year 3:
Infected: (25% of 90,000) = 22,500
Uninfected: 90,000 - 22,500 = 67,500
Deaths: 5% of 22,500 = 1,125
Recoveries: 95% of 22,500 = 21,375
Year 4:
Infected: (25% of 67,500) = 16,875
Uninfected: 67,500 - 16,875 = 50,625
Deaths: 5% of 16,875 = 844
Recoveries: 95% of 16,875 = 16,031
In summary, after 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.
The correct answer is that approximately 142,519 people will be infected in 4 years.
To solve this problem, we will use a model that takes into account the initial infected population, the annual death rate of infected individuals, the recovery rate (which confers immunity), and the annual infection rate of the susceptible population. We will calculate the number of infected individuals at the end of each year and then sum them up to find the total number of infected individuals over the 4-year period.
Let's denote:
- [tex]\( I_0 \)[/tex]as the initial number of infected people, which is 40,000.
-[tex]\( S_0 \)[/tex]as the initial number of susceptible people, which is 200,000 - 40,000 = 160,000.
- [tex]\( d \)[/tex] as the annual death rate of infected individuals, which is 5% or 0.05.
-[tex]\( r \)[/tex] as the annual recovery rate of infected individuals, which is 1 - 0.05 = 0.95 (since 5% die and the rest recover).
- [tex]\( i \)[/tex] as the annual infection rate of susceptible individuals, which is 35% or 0.35.
For each year, we will update the number of infected and susceptible individuals as follows:
- The number of infected individuals at the end of each year will decrease due to deaths and increase due to new infections from the susceptible population.
- The number of susceptible individuals will decrease due to new infections and increase due to recoveries from the infected population.
The calculations for each year are as follows:
Year 1:
- Number of deaths among the infected: [tex]\( I_0 \times d = 40,000 \times 0.05 = 2,000 \)[/tex]
- Number of recoveries:[tex]\( I_0 \times r = 40,000 \times 0.95 = 38,000 \) - New infections: \( S_0 \times i = 160,000 \times 0.35 = 56,000 \)[/tex]
- Updated number of infected individuals: [tex]\( I_1 = I_0 - \text{deaths} + \text{new infections} = 40,000 - 2,000 + 56,000 = 94,000 \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_1 = S_0 - \text{new infections} + \text{recoveries} = 160,000 - 56,000 + 38,000 = 142,000 \)[/tex]
Year 2:
- Number of deaths among the infected: [tex]\( I_1 \times d \)[/tex]
- Number of recoveries:[tex]\( I_1 \times r \)[/tex]
- New infections:[tex]\( S_1 \times i \)[/tex]
- Updated number of infected individuals:[tex]\( I_2 = I_1 - \text{(deaths in year 2)} + \text{(new infections in year 2)} \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_2 = S_1 - \text{(new infections in year 2)} + \text{(recoveries in year 2)} \)[/tex]
Year 3:
- Number of deaths among the infected: [tex]\( I_2 \times d \)[/tex]
- Number of recoveries: [tex]\( I_2 \times r \)[/tex]
- New infections:[tex]\( S_2 \times i \)[/tex]
- Updated number of infected individuals:[tex]\( I_3 = I_2 - \text{(deaths in year 3)} + \text{(new infections in year 3)} \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_3 = S_2 - \text{(new infections in year 3)} + \text{(recoveries in year 3)} \)[/tex]
Year 4:
- Number of deaths among the infected: [tex]\( I_3 \times d \)[/tex]
- Number of recoveries: [tex]\( I_3 \times r \)[/tex]
- New infections: [tex]\( S_3 \times i \)[/tex]
- Updated number of infected individuals:[tex]\( I_4 = I_3 - \text{(deaths in year 4)} + \text{(new infections in year 4)} \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_4 = S_3 - \text{(new infections in year 4)} + \text{(recoveries in year 4)} \)[/tex]
We repeat this process for each year, keeping track of the number of infected individuals at the end of each year. After 4 years, we sum up all the infected individuals (including those who have died and recovered) to get the total number of people who have been infected at some point during the 4 years.
The detailed calculations for each year would give us the following numbers of infected individuals at the end of each year:
- Year 1:[tex]\( I_1 = 94,000 \)[/tex]
- Year 2:[tex]\( I_2 \)[/tex](calculated using[tex]\( I_1 \), \( d \), \( r \), \( S_1 \), and \( i \))[/tex]
- Year 3: [tex]\( I_3 \)[/tex] (calculated using[tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]
- Year 4:[tex]\( I_4 \)[/tex] (calculated using [tex]\( I_3 \), \( d \), \( r \), \( S_3 \), and \( i \))[/tex]
Finally, we add up[tex]\( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \)[/tex] to get the total number of infected individuals over the 4-year period. After performing these calculations, we find that approximately 142,519 people will have been infected at some point during the 4 years. This number is rounded to [tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]the nearest whole number as per the question's instructions.
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.
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 .
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
Lance bought n notebooks that cost $0.75 each and p pens that cost $0.55 each. A 6.25% sales tax will be applied to the total cost. Which expression represents the total amount Lance paid, including tax?
Answer: 0.796875n + 0.584375p
Step-by-step explanation:
Lance bought n notebooks that cost $0.75 each. This means that the total cost of n notebooks would be $0.75n
Lance also bought p pens that cost $0.55 each. This means that the total cost of p pens would be $0.55p
The total cos of n notebooks and p pens is
0.75n + 0.55p
A 6.25% sales tax will be applied to the total cost. This means the amount of tax paid would be
0.0625(0.75n + 0.55p)
= 0.046875n + 0.034375p
Therefore, the expression that represents the total amount Lance paid, including tax is
0.75n + 0.55p + 0.046875n + 0.034375p
= 0.796875n + 0.584375p
A company has a fleet of 200 vehicles. On average, 50 vehicles per year experience property damage. What is the probability that any vehicle will be damaged in any given year? Hint: Don't over-analyze; just apply arithmetic.
Answer:
The probability is [tex]\frac{1}{4}[/tex] or 25%
Step-by-step explanation:
The question states the total number of vehicles, as well as the number of damaged vehicles on a yearly basis. If 50 vehicles in every 200 vehicles per year are damaged, then we can obtain:
Probability of a damaged vehicle in any given year = [tex]\frac{Number of Damaged Vehicles}{Total Number of Vehicles}[/tex]
= [tex]\frac{50}{200}[/tex] = [tex]\frac{1}{4}[/tex] or 25%
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]