There are 100 hours of labor, 500 lbs of material and 1000 gallons of water available. If the goal is to maximize the total profit then the objective function is: (the variables A,B,C & D are the number of widgets of each type produced)
a.min 10A +1268C9D
b.min 10A-15B 7CBD
c.max 10A +15B +7C+8D
d.max A B C D

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]

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]

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 .

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%.

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.

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                            

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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}

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.

Answer:(22,0)

Step-by-step explanation: you have to find when y equals 0

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.

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

https://brainly.com/question/24466382

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Answer:

⅙

Step-by-step explanation:

Total possibilities: 3×2×1 = 6

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

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.

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

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%

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.

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.

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]

See more about graphs at brainly.com/question/14375099

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.

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.

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

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.