If the atomic radius of a metal that has the face-centered cubic crystal structure is 0.137 nm, calculate the volume of its unit cell.

Answer:

1. (-∞,1) 2. (-∞,4]

Step-by-step explanation:

-2x-4 > -6

-2x > -2

x < 1

3(x+2) ≤ 18

3x+6 ≤ 18

3x ≤ 12

x ≤ 4

Answer:

48

Step-by-step explanation:

to find erea you just multiply one number by the other

Answer:48

Step-by-step explanation:

Answer:

$37,500

Step-by-step explanation:

We have been given that a house worth $180,000 has a coinsurance clause of 75 percent. The owners insure the property for $101,250. They then have a loss that results in a $50,000 claim.

We will use loss settlement formula to solve our given problem.

[tex]\text{Loss settlement}=\frac{\text{Loss}\times\text{Limit of insurance}}{\text{Actual cash value}\times \text{Coinsurance}\%}[/tex]

Upon substituting our given values, we will get:

[tex]\text{Loss settlement}=\frac{\$50,000\times\$101,250}{\$180,000\times 75\%}[/tex]

[tex]\text{Loss settlement}=\frac{\$50,000\times\$101,250}{\$180,000\times 0.75}[/tex]

[tex]\text{Loss settlement}=\frac{\$5,062,500,000}{\$135,000}[/tex]

[tex]\text{Loss settlement}=\$37,500[/tex]

Therefore, they will receive $37,500 from insurance.

The correct answer is $37,500. A house worth $180,000 has a coinsurance clause of 75 percent. The owners insure the property for $101,250. They then have a loss that results in a $50,000 claim. They will receive $37,500.00 from insurance.

A house worth $180,000 has a coinsurance clause of 75 percent. This means the owners must insure the house for at least 75% of its value to receive full coverage on claims. The required coverage amount is calculated as follows:

Required Insurance Coverage = 75% of $180,000 = 0.75 * $180,000 = $135,000

The owners insured the property for only $101,250. When a loss occurs, the amount received will be proportionate to the actual coverage relative to the required coverage:

[tex]Payout Ratio = \frac{Actual\ Insurance}{Required\ Insurance}[/tex]
[tex]Payout\ Ratio = \frac{\$ 101,250}{\$135,000} \approx 0.75[/tex]

Since the claim amount is $50,000, the actual payout from the insurance will be:

Insurance Payout = Payout Ratio * Claim Amount
Insurance Payout [tex]\approx[/tex] 0.75 * $50,000 = $37,500

Therefore, the owners will receive $37,500.00 from the insurance.

Answer: 8648640 ways

Step-by-step explanation:

Number of positions = 7

Number of eligible candidates = 13

This can be done by solving the question using the combination Formula for selection in which we use the combination formula to choose 7 candidates amomg the possible 13.

The combination Formula is denoted as:

nCr = n! / (n-r)! * r!

Where n = total number of possible options.

r = number of options to be selected.

Hence, selecting 7 candidates from 13 becomes:

13C7 = 13! / (13-7)! * 7!

13C7 = 1716.

Considering the order they can come in, they can come in 7! Orders. We multiply this order by the earlier answer we calculated. This give: 1716 * 7! = 8648640

Answer:

False

Step-by-step explanation:

We are given the following information:

We treat adult who prefer one child to be a boy as a success.

P(prefer one child to be a boy) = 40% = 0.4

Then the number of adults follows a binomial distribution, where

[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15 and x = 8

We have to evaluate:

[tex]P(x \geq 8)\\= P(x = 8) + P(x = 9)+...+ P(x = 14) + P(x =15)\\\\= \binom{15}{8}(0.4)^{8}(1-0.4)^{7} +\binom{15}{9}(0.4)^{9}(1-0.4)^{6}+...\\\\...+\binom{15}{14}(0.4)^{14}(1-0.4)^{1} +\binom{15}{8}(0.4)^{15}(1-0.4)^{0}\\\\= 0.2131[/tex]

Since the probability of 8 or more is 0.2131 is not very small, thus, it is not a rare event.

Thus, the given statement is false.

Additional information to complete the question:

How long does it take for a short pulse of light to travel from one end of the glass to the other?

Express your answer in terms of some or all of the variables f and L. Use the numeric value given for n in the introduction.

T = ___________ s

Answer:

 [tex]T = \frac{15}{f}[/tex]

Step-by-step explanation:

Given:

Thickness og glass = L

Index of refraction n=1.5

Frequency = f

[tex]Wavelength = \frac{L}{10}[/tex]

λ(air) [tex]= \frac{L}{10}[/tex]

λ(glass) = λ(air) / n

             = [tex]\frac{\frac{L}{10}}{1.5}[/tex]

             = [tex]\frac{L}{10} * \frac{1}{1.5}[/tex]

             = [tex]\frac{L}{15}[/tex]

V(glass) = fλ(glass)

              [tex]= f * \frac{L}{15}[/tex]

          [tex]T = \frac{L}{V_{glass}} = \frac{15}{f}[/tex]

Answer:

a) There is a 48% probability that both Alice and Betty watch TV tomorrow.

b) There is a 48% probability that Betty watches TV tomorrow.

c) There is a 12% probability that only Alice watches TV tomorrow.

Step-by-step explanation:

We have these following probabilities:

A 60% probability that Alice watches TV.

If Alice watches TV, an 80% probability Betty watches TV.

If Alice does not watch TV, a 0% probability that Betty watches TV, since she cannot turn the TV on by herself.

a) What is the probability that both Alice and Betty watch TV tomorrow?

Alice watches 60% of the time. Betty watches in 80% of the time Alice watches. So:

[tex]P = 0.6*0.8 = 0.48[/tex]

There is a 48% probability that both Alice and Betty watch TV tomorrow.

b) What is the probability that Betty watches TV tomorrow?

Since Betty only watches when Alice watches(80% of the time), this probability is the same as the probability of both of them watching. So

[tex]P = 0.6*0.8 = 0.48[/tex]

There is a 48% probability that Betty watches TV tomorrow.

c) What is the probability that only Alice watches TV tomorrow?

There is a 60% probability that Alice watches TV tomorrow. If she watches, there is an 80% probability that Betty watches and a 20% probability she does not watch.

So

[tex]P = 0.6*0.2 = 0.12[/tex]

There is a 12% probability that only Alice watches TV tomorrow.

Answer:

No. It is not true.

Step-by-step explanation:

p = There is no pollution in New Jersey

¬p = There is pollution in New Jersey.

Removing the 'no' in the statement yield the negation.

The given statement "The whole world is polluted" is not correct because it has gone beyond it context/domain. Statement p is about New Jersey, so the negation should be about New Jersey.

The negation can also be written as: "New Jersey is polluted".

In logic, the negation of a statement is its direct contradiction. In this case, 'The whole world is polluted' is not the negation of 'There is no pollution in New Jersey'.

In the field of logic and reasoning, the negation of a statement is a statement which contradicts or denies the original one. If the statement 'p' is 'There is no pollution in New Jersey', its negation would be 'There is pollution in New Jersey' because it is the exact opposite of the original statement. However, the statement 'The whole world is polluted' is not the direct negation of 'There is no pollution in New Jersey'. While it implies that there is pollution in New Jersey, it goes beyond that by including every other part of the world.

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Answer:

a) 0.80

b) 0.45

c) 0.55

Step-by-step explanation:

Given P(A) = 0.20 and P(B) = 0.35

Applying probability of success and failure; P(success) + P( failure) = 1

a) probability that the component does not fail the test = The component does not fail a particular test [P(success)] = 1 - P(A)

= 1 - 0.20 = 0.80

b) probability that the component works perfectly well

= P( the component works perfectly well) - P(component shows strain but does not fail test)

= 0.80 - 0.35 = 0.45

c) probability that the component either fails or shows strain in the test = 1 - P(the component works perfectly well)

= 1 - 0.45 = 0.55

This question is based on the concept of probability. Therefore, the answers are, (a) 0.80, (b) 0.45 and (c) 0.55.

Given:

Event A occurs with probability P(A) =  0.20, and event B occurs with probability  P(B) = 0.35.

According to the question,

Given P(A) = 0.20 and P(B) = 0.35,

As we know that,  probability of success and failure,

⇒ P(success) + P( failure) = 1

a) Probability that the component does not fail the test = The component does not fail a particular test

= P(success) = 1 - P(A)

= 1 - 0.20 = 0.80

b) Probability that the component works perfectly well

= P( the component works perfectly well) - P(component shows strain but does not fail test)

= 0.80 - 0.35 = 0.45

c) Probability that the component either fails or shows strain in the test = 1 - P(the component works perfectly well)

= 1 - 0.45 = 0.55

Therefore, the answers are, (a) 0.80, (b) 0.45 and (c) 0.55.

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Answer:

0.018 is  the probability that a randomly selected adult has an IQ greater than 131.5                                

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 15

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(IQ greater than 131.5)

P(x > 131.5)

[tex]P( x > 131.5) = P( z > \displaystyle\frac{131.5 - 100}{15}) = P(z > 2.1)[/tex]

[tex]= 1 - P(z \leq 2.1)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x > 131.5) = 1 - 0.982 = 0.018 = 1.8\%[/tex]

0.018 is  the probability that a randomly selected adult has an IQ greater than 131.5

Answer:

in a multicriteria decision problem the decision maker must evaluate each alternative with respect to each criterion.

Step-by-step explanation:

Multiple-criteria decision-making (MCDM)  is a sub-discipline of operations research that explicitly evaluates multiple conflicting criteria in decision making (both in daily life and in settings such as business, government and medicine e.t.c)

In multicriteria decision problems, each alternative must be evaluated separately based on each criterion. It's not impossible to choose a single decision path, but it can be complex and might require multiple stages of decision-making over time. All statements in the question are relatively true.

In multicriteria decision problems, various factors or criteria come into play. It's important to note that all the provided statements have some truth in them. From a decision-maker's perspective, one must evaluate each alternative against each criterion. This ensures that the pros and cons of each option are meticulously considered. Contrary to option a, it's not impossible to select a single decision alternative. However, the decision-making process could become complicated due to differing priorities and preferences, ultimately delaying the selection of a single alternative. On the other hand, statement c is partially true; depending on the complexity and scale of the decision, it might require several rounds of decision-making over time. Hence, all of these statements are indeed relatively accurate.

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Answer:

1) 0.375

2) 0.375

3) 0.5

4) 0.5

5) 0.875

6) 0.5                          

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

[tex]\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375[/tex]

2. The probability of getting exactly two tails

P(Exactly two tail)

Favorable outcomes ={ HTT,THT, TTH}

[tex]\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375[/tex]

3. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

[tex]\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

4. The probability of getting a tail on the last toss

P(tail on the last toss)

Favorable outcomes ={HHT,HTT,THT,TTT}

[tex]\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

5. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

[tex]\text{P(at least one head)} = \dfrac{7}{8} = 0.875[/tex]

6. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

[tex]\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

Answer:

There is a 51% probability they will make it to the super bowl.

Step-by-step explanation:

We have these following probabilities:

A 70% probability that the Patriots get homefield advantage.

A 30% probability that the Patriots does not get homefield advantage.

If they get homefield advantage, a 60% probability of making the Super Bowl.

If they do not get homefield advantage, a 30% probability of making the Super Bowl.

What is the probability they will make it to the Super Bowl?

This is 60% of 70%(when they get homefield advantage and make the super bowl), and 30% of 30%(no homefield, no super bowl). So

[tex]P = 0.6*0.7 + 0.3*0.3[/tex]

[tex]P = 0.51[/tex]

There is a 51% probability they will make it to the super bowl.

Answer:

a) [tex]3250-1.96\frac{31.623}{\sqrt{18}}=3232.108[/tex]    

[tex]3250+1.96\frac{31.623}{\sqrt{18}}=3267.892[/tex]    

So on this case the 95% confidence interval would be given by (3232.108;3267.892)    

b) For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

[tex]\bar X= 3250[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma^2 =1000[/tex]represent the sample standard variance

[tex] s = \sqrt{1000}[/tex] represent the sample deviation

n=12 represent the sample size  

Part a

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

Now we have everything in order to replace into formula (1):

[tex]3250-1.96\frac{31.623}{\sqrt{18}}=3232.108[/tex]    

[tex]3250+1.96\frac{31.623}{\sqrt{18}}=3267.892[/tex]    

So on this case the 95% confidence interval would be given by (3232.108;3267.892)    

Part b

For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.

Answer:

8 rooms

Step-by-step explanation:

39 - 7 = 32. 32/4 per room is 8 rooms.

Final answer:

The equation to find the number of backstage rooms needed is (Total volunteers - Stage volunteers) / Volunteers per room = Number of rooms. Using the values provided (39 - 7) / 4, we find that there were eight backstage rooms at the dance recital.

Explanation:

Ms. Deutsch needs seven parent volunteers to help students get on and off stage and four parent volunteers per backstage room. The total number of parent volunteers used on the day of the recital is 39. To find the number of backstage rooms, we subtract the seven parent volunteers required for stage assistance from the total, leaving us with 32 volunteers. We then divide this number by the four parent volunteers per room, resulting in eight backstage rooms.

Initial number of volunteers required for stage assistance: 7

Each room requires: 4 volunteers

Total volunteers: 39

Volunteers for rooms: 39-7 = 32

Number of rooms: 32 / 4 = 8

Answer:

Chi-square value = 2

Step-by-step explanation:

Given data:

Frequency:      Tall     Short

                         30       20

Null hypothesis ( as it is already given so there is no difference between observed and expected values)

Ratio:    1:1 (50% expected frequency of each tall and short pea-plant)

Solution:

Phenotype  Observed  Expected O-E  (O-E)²  (O-E)[tex]^{2/E}[/tex]

                          O                  E

Short                20               25        -5       25        1                          

TALL                  30               25         5        25       1

TOTAL                                                                       2

So, from all these calculations using expected and observed values we get chi-square value equal to 2.

Final answer:

To test the null hypothesis that the deviation from a 1:1 phenotypic ratio is due to chance in a cross between a heterozygous tall and a homozygous short pea plant, the Chi-square value is calculated to be 2.

Explanation:

The question relates to a cross between a heterozygous tall pea plant and a homozygous short pea plant, with the intention to calculate the Chi-square value to test the null hypothesis that the observed deviation from a 1:1 ratio is due to chance. In this scenario, the expectation is a 1:1 phenotypic ratio, meaning 25 tall and 25 short plants out of 50 offspring.

To calculate the Chi-square (χ²) value, the formula is χ² = Σ  ( (observed - expected)² / expected ), where Σ symbolizes the sum of calculations for each category. For tall plants, the calculation is ((30-25)² / 25) = (5² / 25) = 1 and for short plants, the calculation is ((20-25)² / 25) = (5² / 25) = 1. Therefore, the total χ² value is 1 + 1 = 2.

Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 29858, \sigma = 5600[/tex]

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{20000 - 29858}{5600}[/tex]

[tex]Z = -1.76[/tex]

[tex]Z = -1.76[/tex] has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{30000 - 29858}{5600}[/tex]

[tex]Z = 0.02[/tex]

[tex]Z = 0.02[/tex] has a pvalue of 0.5080.

X = 20000

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{20000 - 29858}{5600}[/tex]

[tex]Z = -1.76[/tex]

[tex]Z = -1.76[/tex] has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.645 = \frac{X - 29858}{5600}[/tex]

[tex]X - 29858 = 5600*1.645[/tex]

[tex]X = 39070[/tex]

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

Answer:

8[tex]\sqrt{2}[/tex]

Step-by-step explanation:

64/4=16

so that is 16 on each side.

The diagonal of the rhombus create a right triangle.

We then use the Pythagorean theorem.

[tex]a^{2} +b^{2} =c^{2}[/tex]

[tex]16^{2} +16^{2} =c^{2}[/tex]

[tex]256+256=c^{2}[/tex]

[tex]512 =c^{2}[/tex]

[tex]\sqrt{512} =c[/tex]

8[tex]\sqrt{2}[/tex]

The length of the diagonals of a rhombus with perimeter of 64 and one of its angles as 120 degrees are 16 units and 27.71 units

Therefore,

perimeter = 4l

where

l = length

64 = 4l

l = 64 / 4

length = 16

One of its angle is 120°. Therefore, let's use the angle to find the length of the diagonal.

Using trigonometric ratio,

cos 60° = adjacent  / hypotenuse

cos 60° = x / 16

x = 16 × cos 60

x = 8

2(x) = diagonal

diagonal = 16 units

The second diagonal length

sin 60° = opposite / hypotenuse

sin 60 = y / 16

y = 16 × sin 60

y = 13.8564064606

y = 13.85

Therefore,

diagonal = 2(13.85) = 27.71 units

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Answer:

(a) Type I error in our context is that our test indicates that the proportion of defective products has increased after inspecting but in actual the proportion of defective products was small.

(b) Type II error in our context is that our test indicates that the proportion of defective products has remained small after inspecting but in actual the proportion of defective products was increased.

(c) Factory owner would consider Type 1 error more serious.

(d) Customers will consider Type II error more serious.

Step-by-step explanation:

     Let  [tex]H_0[/tex] = Proportion of defective products remains small

            [tex]H_1[/tex] = Proportion of defective products increases

(a) Type I error represents that we have rejected our null hypothesis given the fact that null hypothesis is True.

Interpretation of this Type I error in our context is that our test indicates that the proportion of defective products has increased after inspecting but in actual the proportion of defective products was small.

(b) Type II error represents that we have accepted our null hypothesis given the fact that null hypothesis is False.

Interpretation of this Type II error in our context is that our test indicates that the proportion of defective products has remained small after inspecting but in actual the proportion of defective products was increased.

(c) Factory owner would consider Type 1 error more serious because after inspecting and testing he assumed that the proportion of defective products  has increased due to which he will halt the assembly process till the time the problem is identified and is repaired but in actual he should continue his  assembly process as in actual the proportion of defective products was small.

(d) Customers will consider Type II error more serious because after inspecting and testing factory owner assumed that the proportion of defective products is small and he will keeps on producing products and assembly process will keeps on going but in actual the proportion of defective products was increased and due to which customers will not get good quality products and they will not be able to purchase the products further.

a. Type I Error: Incorrectly concluding there's a significant increase in defective items when there isn't, leading to unnecessary halting of the assembly line.

b. Type II Error: Failing to detect a real increase in defective items, allowing the assembly to continue with actual defects.

c. The factory owner would consider a Type II error more serious.

d. Customers might find a Type I error more serious due to potential delays and disruptions in product availability.

In the context of the assembly line production process:

a. Type I error: Rejecting a null hypothesis (assuming there is no significant increase in defective products) when it is actually true.

This means that the production line is falsely halted due to the mistaken belief that there is a problem when there actually isn't. This can lead to unnecessary downtime, lost productivity, and increased costs.

b. Type II error: Failing to reject a null hypothesis (assuming there is no significant increase in defective products) when it is actually false.

This means that the production line continues to operate despite the presence of a problem that is causing an increase in defective products. This can lead to subpar products being shipped to customers, damaging the company's reputation and potentially leading to recalls or lawsuits.

c. The factory owner would consider a Type II error to be more serious.

A Type II error allows defective products to reach customers, which can damage the company's reputation, lead to recalls or lawsuits, and erode customer trust. While a Type I error can cause some inconvenience and expense, it is ultimately better to err on the side of caution and halt production if there is any suspicion of a problem.

d. Customers might consider a Type I error to be more serious.

Customers would prefer to receive products that are free of defects, even if it means that production is occasionally halted unnecessarily. A Type I error ensures that defective products are not shipped to customers, while a Type II error allows defective products to reach customers, which can cause inconvenience, frustration, and even safety hazards.

Answer:

The balance will be $7,164.31.

Step-by-step explanation:

The compound interest formula is given by:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that:

[tex]P = 6000, r = 0.06[/tex]

Semi-annually is twice a year, so [tex]n = 2[/tex]

We want to find A when [tex]t = 3[/tex]

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

[tex]A = 6000(1 + \frac{0.06}{2})^{2*3}[/tex]

[tex]A = 7164.31[/tex]

The balance will be $7,164.31.

Answer:

1.        t = 0.995 s

2.       h = 15.92  ft

Step-by-step explanation:

First we have to look at the following formula

Vf = Vo + gt

then we work it to clear what we want

Vo + gt = Vf

gt = Vf - Vo

t = (Vf-Vo)/g

Now we have to complete the formula with the real data

Vo = 32 ft/s      as the statement says

Vf = 0     because when it reaches its maximum point it will stop before starting to lower

g = -32,16 ft/s²        it is a known constant, that we use it with the negative sign because it is in the opposite direction to ours

t = (0 ft/s - 32 ft/s) / -32,16 ft/s²

we solve and ...

t = 0.995 s

Now we will implement this result in the following formula to get the height at that time

h = (Vo - Vf) *t /2

h = (32 ft/s - 0 ft/s) * 0.995 s / 2

h = 32 ft/s * 0.995 s/2

h = 31.84 ft / 2

h = 15.92  ft

The ball takes 2 seconds to reach its highest point and the height at this point is 32 feet.

To find the time it takes for the ball to reach its highest point, we can use the equation h = -16t^2 + 32t, where h is the height and t is the time. The maximum height is reached when the ball is at its highest point, which occurs when the ball is not moving vertically.

At this point, the velocity of the ball is 0, so the equation v = -16t + 32 can be used to find the time. Setting v = 0 and solving for t, we get t = 2 seconds.

Substituting this time value into the equation for height, we can find the height at this point. h = -16(2)^2 + 32(2) = 32 feet.

Therefore, the ball takes 2 seconds to reach its highest point and the height at this point is 32 feet.

Answer:

99.065% probability that at least two of the three will be available at any given time.

Step-by-step explanation:

We have these following probabilities:

99% probability of the first workstation being available

95% probability of the second workstation being available

85% probability of the third workstation being avaiable

Two being available:

We can have three outcomes

First and second available, third not. So

0.99*0.95*0.15 = 0.141075

First and third available, second not. So

0.99*0.05*0.85 = 0.042075

Second and third available, first not. So

0.01*0.95*0.85 = 0.008075

Adding them all

P(2) = 0.141075 + 0.042075 + 0.008075 = 0.191225

Three being available:

P(3) = 0.99*0.95*0.85 = 0.799425

What is the probability that at least two of the three will be available at any given time?

P = P(2) + P(3) = 0.191225 + 0.799425 = 0.99065

99.065% probability that at least two of the three will be available at any given time.

The probability that at least two out of the three workstations are available is 0.967.

We can find the probability that at least two out of the three workstations are available using the concept of complementary events. The probability of at least two workstations being available is equal to 1 minus the probability of none or only one workstation being available.

Let's calculate the probability of none or only one workstation being available:

Probability of none being available: 0.01 * 0.05 * 0.15 = 0.00075
Probability of only one being available: (0.99 * 0.05 * 0.15) + (0.01 * 0.95 * 0.15) + (0.01 * 0.05 * 0.85) = 0.03225

Now, subtracting this from 1:

1 - (0.00075 + 0.03225) = 0.967

Therefore, the probability that at least two out of the three workstations are available at any given time is 0.967.

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Answer:

The probability that the hand drawn is a full house is 0.00144.

Step-by-step explanation:

In a full house we have a hand that consists of two of one kind and three of another kind, i.e 5 cards are selected.

The number of ways of selecting 5 cards from 52 cards is:

                          [tex]{52\choose 5} = \frac{52!}{5!(52-5)!} \\=\frac{52!}{5!\times47!} \\=2598960[/tex]

In a deck of 52 cards there are 13 kind of cards, namely{K, Q, J, A, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Two kinds can be selected in, [tex]{13\choose 2}=\frac{13!}{2!\times(13-2)!} =\frac{13!}{2!\times11!} =78[/tex] ways

One of the two kinds can be selected for 3 cards combination in [tex]{2\choose 1} = 2[/tex] ways.

There are 4 cards of each kind.

So 3 cards combination can be selected from any of the two kinds in [tex]{4\choose 3} =\frac{4!}{3!(4-3)!} =4[/tex] ways.

And 2 cards combination can be selected from any of the two kinds in [tex]{4\choose 2} =\frac{4!}{2!(4-2)!} =6[/tex] ways.

Thus, total number of ways to select a full house is:

[tex]{13\choose 2}\times{2\choose 1}\times{4\choose 3}\times{4\choose 2}\\=78\times2\times4\times6\\=3744[/tex]

The probability that the hand drawn is a full house is:

[tex]\frac{Number\ of\ ways\ of\ Drawing\ a\ Full\ house)}{Number\ of\ ways\ of\ Selecting\ 5\ cards } =\frac{3744}{2598960} =0.00144[/tex]

Thus, the probability of playing a full house is 0.00144.

Answer: [tex]9.616 ft[/tex]

Step-by-step explanation:

The last part of the question is: Find the value of the width

If the Texas Pee Wee football field has a rectangular shape (as shown in the figure), where the width is [tex]w[/tex] and the length is [tex]218 ft w[/tex]; its area [tex]A[/tex] is:

[tex]A=20,160 ft^{2}=(length)(width)[/tex]

[tex]20,160 ft^{2}=(218 w)(w)[/tex]

Isolating [tex]w[/tex]:

[tex]w=\sqrt{\frac{20,160 ft^{2}}{218}}[/tex]

Finally:

[tex]w=9.616 ft[/tex] This is the widht of the Texas Pee Wee football field

The Texas Pee Wee football field has a width of approximately 70 feet and a length of 288 feet, calculated by solving a quadratic equation derived from the given area and the relationship between length and width.

Step-by-Step Explanation:

Let the width of the field be denoted as w.

Therefore, the length of the field is w + 218 feet.

The area of the rectangle (football field) is given by the formula:

Area = length x width.

Substituting the given values into the formula, we have:

20,160 = w × (w + 218).

This results in a quadratic equation:

w² + 218w - 20,160 = 0.

We will solve this quadratic equation using the quadratic formula:

w = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = 218, and c = -20,160.

Calculate the discriminant:

b² - 4ac = 218² - 4 × 1 × (-20,160) = 47524 + 80640 = 128164.

Find the square root of the discriminant:

√128164 ≈ 357.94.

Substitute back into the quadratic formula:

w = (-218 ± 357.94) / 2.

This results in two potential solutions:

w = (357.94 - 218) / 2 ≈ 69.97 (approximately 70 feet) and w = (-218 - 357.94) / 2 (a negative value, which is not possible for width).

Thus, the width w is approximately 70 feet.

Substitute the width back into the length formula:

length = 70 + 218 = 288 feet.

Therefore, the dimensions of the Texas Pee Wee football field are approximately 70 feet in width and 288 feet in length.

Answer:

A= {M1,M2},{M2,M3}, {M2,M3}

A U B = S

A n B = 0

A n B'= A

Step-by-step explanation:

A= ( Two males) = { (M1,M2), (M2,M3), (M2,M3)

B= (Atleast one female) = {M1,W1}, {M,W1}, {M3,W1}, {M1,W2} , {M2,W2}, {M3,W2}

Following are the solution to the required function:

Given that there are five applicants with three men and two women.

Let S be the subset of the set of all possible outcomes,

[tex]\{M_1, M_2\}, \{M_2, M_3\},\{M_3,M_1\},\{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}[/tex]

Let A denote the subset of outcomes corresponding to the selection of two men.

The possible outcomes of A are,

[tex]\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}[/tex]

Let B be the subset corresponding to the selection of at least one woman.

[tex]\{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}[/tex]

Then [tex]\bar{B} =[/tex]

[tex]\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}[/tex]

Find [tex]A\cup B\\\\[/tex]

[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\} \cup \{ \{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}, \{W_1,M_1\},\{W_2,M_1\},\{W_1, M_2\},\\\{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\[/tex]

Find [tex]A\cap B\\\\[/tex]

[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\} \cap \{ \{W_1,M_1\},\{W_2,M_1\}, \{W_1, M_2\}, \\ \{W_2,M_2\},\{W_1, M_3\},\{W_2, M_2\}, \{W_1,W_2\}\}\\\\ =\{\phi\}[/tex]

Find [tex]A\cap \bar{B}\\\\[/tex]

[tex]=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\cap\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\\\\=\{\{M_1,M_2\}, \{M_2,M_3\},\{M_3,M_1\}\\\\[/tex]

Learn more about the set function here:

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Answer:

Step-by-step explanation:

There are 200 dice out of which 199 are fair

Prob for 6 in one special die = 1 and

Prob for 6 in other die = 1/6

A1- drawing special die and A2 = drawing any other die

A1 and A2 are mutually exclusive and exhaustive

P(A1) = 1/200 and P(A2) = 199/200

B = getting 6

i) Required probability

= P(A1/B) = [tex]\frac{P(A1B)}{P(A1B)+P(A2B)} \\[/tex]

P(A1B) = [tex]\frac{1}{200} *1 = \frac{1}{200}[/tex]

P(A2B) = [tex]\frac{199}{200}*\frac{1}{6}=\frac{199}{1200}[/tex]

P(B) = [tex]\frac{205}{1200} =\frac{41}{240}[/tex]

P(A1/B) = [tex]\frac{1/200}{41/240} =\frac{7}{205}[/tex]

P(A2/B) = [tex]\frac{199/1200}{41/240} =\frac{199}{205}[/tex]

Answer:

2,136 yards

Step-by-step explanation:

Since the roads met at a right angle, the distance between Dale and Betty can be interpreted as the hypotenuse of a right triangle with sides measuring 2000 yd and 750 yd. The distance between them is:

[tex]d^2=2000^2+750^2\\d=\sqrt{2000^2+750^2}\\d=2,136\ yards[/tex]

Dale and Betty are 2,136 yards away from each other after two minutes.

Answer:

the amount of water that is left in the tank after 10 min is 98 L

Step-by-step explanation:

since the water drains off at rate that is proportional to the water present

(-dV/dt) = k*V , where k= constant

(-dV/V) = k*dt

-∫dV/V) = k*∫dt

-ln V/V₀=k*t

or

V= V₀*e^(-k*t) , where V₀= initial volume

then since V₁=0.7*V₀ at t₁= 3 min

-ln V₁/V₀=k*t₁

then for t₂= 10 min we have

-ln V₂/V₀=k*t₂

dividing both equations

ln (V₂/V₀) / ln (V₁/V₀) =(t₂/t₁)

V₂/V₀ = (V₁/V₀)^(t₂/t₁)

V₂=V₀ *  (V₁/V₀)^(t₂/t₁)

replacing values

V₂=V₀ *  (V₁/V₀)^(t₂/t₁)  = 200 L * (0.7)^(10min/5min) = 98 L

then the amount of water that is left in the tank after 10 min is 98 L

The problem represents a case of exponential decay. Initially, 30% of water, or 60 liters, leaks out in 5 minutes, leaving 140 liters in the tank. Assuming the same rate of leakage, another 30% of water or 42 liters will leak out in the next 5 minutes, leaving 98 liters in the tank after 10 minutes.

In this problem, we are dealing with a situation involving exponential decay due to the water leakage which happens at a rate proportional to the amount of water present in the tank.

First, let's consider the 30% of water that leaks out in the first 5 minutes from a 200-liter tank. This amounts to 60 liters (200 * 0.30), which leaves 140 liters (200 - 60) of water in the tank after 5 minutes.

Now, since the decrease of water is proportional to the amount of water present, this implies an exponential decay over time. Given that the amount of water decreased by 30% in the first 5 minutes, it's reasonable to assume that it will decrease by the same percentage in the next 5 minutes as well.

So, the amount of water left in the tank after 10 minutes would be 98 liters (140 * 0.70).

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Final answer:

The product of x^2 and x^3 equals x^6 for any real number x is true. The statement that x^2 > 0 for any real number x is false, as it should state x^2 >= 0. 315 - 8 being even is false since it results in an odd number. The sum of two odd integers being even is true.

Explanation:

Determine the truth values of these statements:

The product of x2 and x3 is x6 for any real number x. This statement is true because according to the laws of exponents, when multiplying powers with the same base, you add the exponents. Therefore, x2 * x3 = x2+3 = x6.

x2 > 0 for any real number x. This statement is false because when x = 0, x2 = 0, not greater than 0. The correct statement should be x2 >= 0 for any real number x.

The number 315 − 8 is even. This statement is true because 315 − 8 = 307, and any number ending in 7 is odd. Therefore, the statement is false.

The sum of two odd integers is even. This statement is true because when you add two odd numbers, the sum is always even. For example, 3 + 5 = 8.

Answer:

30.56%

Step-by-step explanation:

Let the sides on each dice be labeled as 1, 2a, 2b, 2c, 3, 4.

The sample space for the sum of the values being 4 is:

S={1,3; 3,1; 2a,2a; 2a,2b; 2a,2c; 2b,2a; 2b,2b; 2b,2c; 2c,2a; 2c,2b; 2c,2c}

There are 11 possible sums out of the 36 possible outcomes that result in a sum of 4. Therefore, the probability their total is 4 is:

[tex]P(S) =\frac{11}{36}=0.3056 =30.56\%[/tex]

There is a 30.56% probability that their sum is 4.

Final answer:

The probability of getting a sum of 4 with two modified dice numbered 1,2,2,2,3,4 is 5/36. This is found by adding all possible combinations that total 4, considering the multiplicity of the number 2 on the dice.

Explanation:

To calculate the probability that the sum of two modified dice is 4, we must consider all possible combinations of rolls that could result in a total of 4. Each die is numbered with 1,2,2,2,3,4, so the outcomes that give us a sum of 4 are (1,3), (2,2), (3,1), and there are three different 2s on each die that can contribute to the sum.

Therefore, the probability of getting a sum of 4 with one die already showing 2 is the probability of rolling either a 1 or another 2 on the second die.

The total number of outcomes for one die is 6. To find the sum of 4:

This results in 1 + 3 + 1 = 5 favorable outcomes. Since there are a total of 36 possible outcomes when rolling two dice, the probability is 5/36.