Answer: d) 6,534,385
Therefore, the number of ways of selecting 9 hearts is 6,534,385
Step-by-step explanation:
Given;
Number of cards to be chosen = 12
Number of hearts to be chosen = 9
Number of non-hearts to be selected = 3
Total number of cards = 52
Number of hearts total = 13
Number of non-hearts total= 39
The number of ways of selecting hearts can be given by the illustration below.
Number of ways of selecting hearts × number of ways of selecting non-hearts
N = 13C9 × 39C3 (it's combination since order is not important)
N = 13!/(9! ×4!) × 39!/(36! × 3!)
N = 6,534,385
Therefore, the number of ways of selecting 9 hearts is 6,534,385
The number of ways to choose 9 hearts from 12 cards is 220, option (b) is correct.
To solve this problem, we'll use the combination formula, which is used when selecting objects from a larger set without regard to the order. The formula for combinations is:
[tex]\[ \text{C}(n, k) = \frac{n!}{k!(n - k)!} \][/tex]
where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and (!) denotes factorial.
Given that we need to choose 12 cards from a deck of 52, and we want to know the number of ways to choose 9 hearts, which is [tex]\( k = 9 \)[/tex] hearts out of a total of [tex]\( n = 12 \)[/tex] cards.
Step 1:
Calculate the number of ways to choose 9 hearts from the 12 chosen cards:
[tex]\[ \text{C}(12, 9) = \frac{12!}{9!(12 - 9)!} \][/tex]
Step 2:
Simplify the expression:
[tex]\[ \text{C}(12, 9) = \frac{12!}{9!3!} \][/tex]
[tex]\[ \text{C}(12, 9) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \][/tex]
[tex]\[ \text{C}(12, 9) = \frac{1320}{6} \][/tex]
[tex]\[ \text{C}(12, 9) = 220 \][/tex]
Therefore, there are 220 ways to choose 9 hearts from 12 cards. So, the correct answer is option (b).
Researchers continue to find evidence that brains of adolescents behave quite differently than either brains of adults or brains of children. In particular, adolescents seem to hold on more strongly to fear associations than either children or adults, suggesting that frightening connections made during the teen years are particularly hard to unlearn. In one study,1 participants first learned to associate fear with a particular sound. In the second part of the study, participants heard the sound without the fear-causing mechanism, and their ability to "unlearn" the connection was measured. A physiological measure of fear was used, and larger numbers indicate less fear. We are estimating the difference in mean response between adults and teenagers. The mean response for adults in the study was 0.225 and the mean response for teenagers in the study was 0.059. We are told that the standard error of the estimate is 0.091. Let group 1 be adults and group 2 be teenagers.
(a) Give notation for the quantity that is being estimated.
Answer:
a) [tex]\mu_1 -\mu_2[/tex] parameter of interest.
Where [tex]\mu_1[/tex] represent the mean response for adults
[tex]\mu_2[/tex] represent the mean response for teenegers
b) The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]
Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]
c) The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]
d) The 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]
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".
Let group 1 be adults and group 2 be teenagers.
[tex]\bar X_1 =0.225[/tex] represent the sample mean 1
[tex]\bar X_2 =0.059[/tex] represent the sample mean 2
n1 represent the sample 1 size
n2 represent the sample 2 size
[tex]s_1 [/tex] sample standard deviation for sample 1
[tex]s_2 [/tex] sample standard deviation for sample 2
SE =0.091 represent the standard error for the estimate
(a) Give notation for the quantity that is being estimated.
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
(b) Give notation for the quantity that gives the best estimate.
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
The best estimate is given by [tex]\bar X_1 -\bar X_2[/tex]
Since the best estimator for the true mean is the sample mean [tex]\hat \mu = \bar X[/tex]
(c) Give the value for the quantity that gives the best estimate.
The best estimate is given by [tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]
(d) Give a confidence interval for the quantity being estimated. Assuming 95% of confidence
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =0.225-0.059=0.166[/tex]
We can assume that since we know the standard error the deviations are known and we can use the z distribution instead of the t distribution for the confidence interval.
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]
The standard error is given by the following formula:
[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=0.091[/tex]
Given by the problem
Now we have everything in order to replace into formula (1):
[tex]0.166-1.96(0.091)=-0.012[/tex]
[tex]0.166+1.96(0.091)=0.344[/tex]
So on this case the 95% confidence interval would be given by [tex]-0.012 \leq \mu_1 -\mu_2 \leq 0.344[/tex]
Final answer:
The quantity being estimated in the study is the difference in mean response to unlearn fear associations between adults and teenagers, denoted by Δμ = μ1 - μ2, where μ1 and μ2 represent the mean responses for adults and teenagers, respectively. This study contributes to understanding how fear associations are formed and unlearned, with implications on evolutionary predisposition towards certain fears.
Explanation:
The quantity being estimated in the study between adolescents and adults regarding their ability to unlearn fear associations tied to a specific sound is captured by the notation Δμ = μ1 - μ2. Here, μ1 represents the mean response for adults, and μ2 represents the mean response for teenagers. In this context, a higher physiological measure indicates less fear, with adults showing a mean response of 0.225 and teenagers showing a mean response of 0.059. The standard error of the estimate provided is 0.091, which helps in understanding the variability or precision of our estimated difference between the two groups' mean responses.
This study hints at the broader theory of preparedness, suggesting that humans are evolutionarily predisposed to easily associate certain stimuli with fear. Notably, the differentiation in fear response unlearning between age groups aligns with observations in social and developmental psychology about the specificity of fear acquisition and the challenges in modify these responses once established, especially during the teenage years.
True or False? Tell whether the pair of ratios form a proportion. 4/5 and 5/6 Please explain why you chose what you chose
Answer:
False. The products from cross multiplication are different.
Step-by-step explanation:
To know if a pair of ratios form a proportion, cross multiply. If the products are equal, they are a proportion.
Write like this to see top (numerator) and bottom (denominator) clearly.
[tex]\frac{4}{5} =\frac{5}{6}[/tex]
Multiply each numerator with the other side's denominator:
4 X 6 = 24
5 X 5 = 25
Are they equal? No. 24 ≠ 25
Therefore it's not a proportion.
A scientist measured the speed of light. His values are in km/sec and have 299,000 subtracted from them. He reported the results of 25 trials with a mean of 756.22 and a standard deviation of 100.89.
(a) Find a 90% confidence interval for the true speed of light from these statistics.
(b) State in words what this interval means. Keep in mind that the speed of light is a physical constant that, as far as we know, has a value that is true throughout the universe.
(c) What assumptions must you make in order to use your method?
Answer:
a) The 90% confidence interval would be given by (721.716;790.724)
b) We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)
c) We assume the following conditions:
RandomizationIndependenceDeviation unknown [tex]\sigma[/tex]Step-by-step explanation:
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".
Part a
[tex]\bar X=756.22[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=100.89[/tex] represent the sample standard deviation
n=25 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=25-1=24[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,24)".And we see that [tex]t_{\alpha/2}=1.71[/tex]
Now we have everything in order to replace into formula (1):
[tex]756.22-1.71\frac{100.89}{\sqrt{25}}=721.716[/tex]
[tex]756.22+1.71\frac{100.89}{\sqrt{25}}=790.724[/tex]
So on this case the 90% confidence interval would be given by (721.716;790.724)
Part b
We are 90% confident that the true mean for the true speed of light is between (721.716;790.724)
Part c
We assume the following conditions:
RandomizationIndependenceDeviation unknown [tex]\sigma[/tex]Final answer:
The 90% confidence interval for the true speed of light is between 299,723.02683 km/sec and 299,789.41317 km/sec. This interval suggests we can be 90% confident that the constant speed of light falls within this range, with the understanding that the true speed of light is approximately 299,792,458 m/s.
Explanation:
The sample mean is 756.22, and the standard deviation is 100.89 with 25 trials.
First, add 299,000 km/sec to the sample mean to revert to the actual speed of light. Adjusted mean = 756.22 + 299,000 = 299,756.22 km/sec.
Since the sample size (25) is greater than 30, we use the z-score for a 90% confidence interval, which is 1.645.
To find the margin of error (ME), use the formula ME = z * (σ/√n), where σ is the standard deviation and n is the sample size. ME = 1.645 * (100.89/√25)= 1.645 * 20.178 = 33.19317 km/sec.
The confidence interval is then mean ± ME. That gives us the interval: [299,756.22 - 33.19317, 299,756.22 + 33.19317] or [299,723.02683, 299,789.41317] km/sec.
Interpretation: We are 90% confident that the true speed of light lies within the interval of 299,723.02683 to 299,789.41317 km/sec.
The samples are independent and randomly selected.
The data reported is accurate and measured without systematic errors.
The data is normally distributed or the sample size is large enough for the Central Limit Theorem to apply.
These results align with the known fact that the speed of light is a constant at approximately 299,792,458 meters/second, and any deviations observed in the experiment are likely due to measurement error or experimental uncertainties.
Weatherwise is a magazine published by the American Meteorological Society. One issue gives a rating system used to classify Nor'easter storms that frequently hit New England and can cause much damage near the ocean. A severe storm has an average peak wave height of μ = 16.4 feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. Peak wave heights are usually measured from land (using binoculars) off fixed cement piers. Suppose that a reading of 38 waves showed an average wave height of x= 17.3 feet. Previous studies of severe storms indicate that σ = 3.3 feet. Does this information suggest that the storm is (perhaps temporarily) increasing above the severe rating? Use α = 0.01. Solve the problem using the critical region method of testing (i.e., traditional method). (Round your answers to two decimal places.)test statistic = critical value = State your conclusion in the context of the application.Reject the null hypothesis, there is sufficient evidence that the average storm level is increasing.Reject the null hypothesis, there is insufficient evidence that the average storm level is increasing. Fail to reject the null hypothesis, there is sufficient evidence that the average storm level is increasing.Fail to reject the null hypothesis, there is insufficient evidence that the average storm level is increasing.Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?
Answer:
Step-by-step explanation:
The formula Upper A equals 23.1 e Superscript 0.0152 tA=23.1e0.0152t models the population of a US state, A, in millions, t years after 2000.
a. What was the population of the state in 2000?
b. When will the population of the state reach 28.328.3 million?
Answer:
a) [tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]
b) [tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]
So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014
Step-by-step explanation:
For this case we assume the following model:
[tex]A(t)= 23.1 e^{0.0152 t}[/tex]
Where t is the number of years after 2000/
Part a
For this case we want the population for 2000 and on this case the value of t=0 since we have 0 years after 2000. If we rpelace into the model we got:
[tex]A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1[/tex]
So then the initial population at year 2000 is 23.1 million of people.
Part b
For this case we want to find the time t whn the population is 28.3 million.
So we need to solve this equation:
[tex]28.3= 23.1 e^{0.0152(t)}[/tex]
We can divide both sides by 23.1 and we got:
[tex]\frac{28.3}{23.1}= e^{0.0152t}[/tex]
Now we can apply natural log on both sides and we got:
[tex]ln(\frac{28.3}{23.1})= 0.0152 t[/tex]
And then for t we got:
[tex]t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years[/tex]
So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014
Solve the inequality. Graph the solution. 4(n-3) -6>18
Answer:
n>9
Step-by-step explanation:
4(n-3)-6>18
4n-12-6>18
4n-18>18
4n>18+18
4n>36
n>36/4
n>9
For each of the given situations, write out the alternative hypothesis, being sure to state whether it is one-sided or two-sided.a) A consumer magazine discovered that 30% of a certain computer model had warranty problems over the first three months. From a random sample, the manufacturer wants to know if a new model has improved that rate.Complete the alternative hypothesis and determine whether the alternative hypothesis is one-sided or two-sided.
Answer:
Null hypothesis: [tex]p\leq 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
Step-by-step explanation:
1) Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
2) Solution to the problem
On this case we want to test is [tex]p>0.3[/tex] since we want to check if the new model has improved the warranty rate, we can express it like this:
[tex]p-0.3<0[/tex] since are equivalent expressions.
And the alternative hypothesis should be the complement:
Null hypothesis: [tex]p\leq 0.3[/tex] or [tex]p=0.3[/tex]
So the correct system of hypothesis for this case would be:
Null hypothesis: [tex]p\leq 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
The alternative way should be:
Null hypothesis: [tex]p = 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
Location is known to affect the number, of a particular item, sold by HEB Pantry. Two different locations, A and B, are selected on an experimental basis. Location A was observed for 18 days and location B was observed for 13 days. The number of the particular items sold per day was recorded for each location. On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4. Select a 99% confidence interval for the difference in the true means of items sold at location A and B. a) O [-1242,-7582]
b) O132.76, 45.24]
c)。8 1.76, 94.24]
d) 0-1 6.03,-3.97]
e)。[42.76, 55.24]
F. None of the above
Answer:
d) [-16.03,-3.97]
[tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].
Step-by-step explanation:
Notation and 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]n_A=18[/tex] represent the sample of A
[tex]n_B =13[/tex] represent the sample of B
[tex]\bar x_A =39[/tex] represent the mean sample for A
[tex]\bar x_B =49[/tex] represent the mean sample for B
[tex]s_A =8[/tex] represent the sample deviation for A
[tex]s_B =4[/tex] represent the sample deviation for B
[tex]\alpha=0.01[/tex] represent the significance level
Confidence =99% or 0.99
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex] (1)
The point of estimate for [tex]\mu_A -\mu_B[/tex] is just given by:
[tex]\bar X_A -\bar X_B =39-49=-10[/tex]
The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2=18+13-2=29[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.756[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex]
And replacing we have:
[tex]SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03[/tex]
[tex]-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97[/tex]
So on this case the 99% confidence interval for the differences of means would be given by [tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].
d) [-16.03,-3.97]
The 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], therefore the correct answer is F. None of the above.
Explanation:This question is about computing a confidence interval for the difference of two sample means. The formula for the 99% confidence interval for the difference between two means is:
(X1 - X2) ± Z * sqrt [s1^2/n1 + s2^2/n2]
Where X1 and X2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. We plug the given values into the equation to calculate:
(39 - 49) ± 2.576 * sqrt [(8^2 / 18) + (4^2 / 13)] => -10 ± 2.576 * sqrt [3.56 + 1.23] => -10 ± 2.576 * sqrt [4.79] => -10 ± 2.576 * 2.19 => -10 ± 5.64
This means the 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], which is not among the given options, so the correct answer is F. None of the above.
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In order to comply with the Environmental protection Agency (EPA) regulations of the Clean Water Act, a large agricultural company wants to know the average nitrogen concentration in the soil of an agricultural region it plans to purchase. The seller claims that the average nitrogen level does not exceed 0.49 units. To test this claim at 0.05 level of significance, nitrogen concentration of soil samples were recorded at 51 sites in that agricultural region. The sample mean was found to be 0.505 and the sample standard deviation 0.12.
Answer:
[tex]t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893[/tex]
[tex]p_v =P(t_{50}>0.893)=0.1881[/tex]
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.
Step-by-step explanation:
Data given and notation
[tex]\bar X=0.505[/tex] represent the sample mean
[tex]s=0.12[/tex] represent the standard deviation for the sample
[tex]n=51[/tex] sample size
[tex]\mu_o =0.49[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses to be tested
We need to conduct a hypothesis in order to determine if the average nitrogen level dos not exced 0.49 units, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 0.49[/tex]
Alternative hypothesis:[tex]\mu > 0.49[/tex]
Compute the test statistic
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
We can replace in formula (1) the info given like this:
[tex]t=\frac{0.505-0.49}{\frac{0.12}{\sqrt{51}}}=0.893[/tex]
Now we need to find the degrees of freedom for the t distirbution given by:
[tex]df=n-1=51-1=50[/tex]
What do you conclude?
Compute the p-value
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{50}>0.893)=0.1881[/tex]
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean is not significantly higher than 0.49 units.
According to a recent study, 1 in every 9 women has been a victim of domestic abuse at some point 19) in her life. Suppose we have randomly and independently sampled twenty-five women and asked each whether she has been a victim of domestic abuse at some point in her life.
1. Find the probability that at least 2 of the women sampled have been the victim of domestic abuse. Round to six decimal places.
Answer:
[tex]P(X\geq 2)=1-P(X\leq 1)=1-[0.054294+0.167762]=0.777944[/tex]
Step-by-step explanation:
Previous concepts
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
Find the probability that at least 2 of the women sampled have been the victim of domestic abuse.
On this case we want to find this probability
[tex]P(X\geq 2) =1-P(X<2)=1-P(X\leq 1)= 1-[P(X=0)+P(X=1)][/tex]
And we can find the individual probabilities like this:
[tex]P(X=0)=(25C0)(0.11)^0 (1-0.11)^{25-0}=0.054294[/tex]
[tex]P(X=1)=(25C1)(0.11)^1 (1-0.11)^{25-1}=0.167762[/tex]
[tex]P(X\geq 2)=1-P(X\leq 1)=1-[0.054294+0.167762]=0.777944[/tex]
Using the binomial distribution, it is found that there is a 0.287825 = 28.7825% probability that at least 2 of the women sampled have been the victim of domestic abuse.
What is the binomial distribution formula?The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
x is the number of successes.n is the number of trials.p is the probability of a success on a single trial.In this problem:
1 in every 9 women has been a victim of domestic abuse at some point in her life, hence p = 1/9 = 0.1111.25 women are sampled, hence n = 25.The probability that at least 2 of the women sampled have been the victim of domestic abuse is given by:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which:
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
Hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.1111)^{0}.(0.8889)^{25} = 0.052641[/tex]
[tex]P(X = 1) = C_{25,1}.(0.1111)^{1}.(0.8889)^{24} = 0.164484[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.052641 + 0.164484 = 0.217125[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.217125 = 0.782875[/tex]
0.287825 = 28.7825% probability that at least 2 of the women sampled have been the victim of domestic abuse.
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Write an equation in slope-intercept form of the line having the given slope and y-intercept. m:-4/6, (0,-4)
Answer:
y = -4/6x - 4
Step-by-step explanation:
y = m(x - x₁) + y₁
You're given m=-4/6 and (0,-4) ←x₁=0, y₁=-4
so just plug it into the point-slope equation.
y = (-4/6)(x - (0)) + (-4)
y = (-4/6)(x) + (-4)
y = -4/6x - 4
Answer:y = -4x/6 - 4
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The slope,m of the given line is -4/6
To determine the intercept, we would substitute m = -4/6, x = 0 and y = -4 into y = mx + c. It becomes
- 4 = -4/6 × 0 + c = 0 + c
c = - 4
The equation becomes
y = -4x/6 - 4
There are 4 suits (heart, diamond, clover, and spade) in a 52-card deck, and each suit has 13 cards. Suppose your experiment is to draw one card from a deck and observe what suit it is. Express the probability in fraction format. (Show all work. Just the answer, without supporting work, will receive no credit.)
Answer:
The probability of drawing a heart or diamond is 1/2 or 0.5
The probability that the card is not a spade is 3/4 or 0.75
Step-by-step explanation:
Consider the provided information.
Part (a) Find the probability of drawing a heart or diamond.
There are 13 cards of heart and 13 cards of diamond.
We need to find the probability of drawing a heart or diamond.
[tex]P(\text{Heart or Diamond})=P(\text{Heart card Drawn})+P(\text{Diamond card Drawn})[/tex]
[tex]P(\text{Heart or Diamond})=\frac{13}{52}+\frac{13}{52}[/tex]
[tex]P(\text{Heart or Diamond})=\frac{26}{52}=\frac{1}{2}=0.5[/tex]
Hence, the probability of drawing a heart or diamond is 1/2 or 0.5
(b) Find the probability that the card is not a spade.
Out of 52 cards 13 are spade,
That means 52 - 13 = 39 cards are not a spade.
[tex]P(\text{Not spade})=\frac{39}{52}=\frac{3}{4}=0.75[/tex]
Hence, the probability that the card is not a spade is 3/4 or 0.75
A 20 year par value bond with semi-annual coupons at a nominal annual rate of 8% convertible semi-annually is purchased at a price of 1783.27. The bond can be called at par value X on any coupon date starting at the end of year 12 after the coupon is paid. The price guarantees a nominal annual rate of interest convertible semi-annually of at least 6%. Calculate X.
Answer:
3.216%
Step-by-step explanation:
This bond sells at a higher price or value, which means that its coupon is bogus of market interest rate. Therefore, the minimum yield rate that accounts for the possibility of the bond being called is calculated at the earliest possible call date. Let say exactly 15 years from the date of purchase, because that would be the most disadvantageous date for the bondholder for the call to occur.
The minimum semiannual yield:
j= i²/2
i² = 2j
which therefore satisfies the expression below for the worst possible case scenario yield:
1722.25 = 0.04*1100*[tex]a]_30[/tex]+[tex]\frac{1100}{(1+j)^30}[/tex]
Also, with the use of a financial calculator (making sure that the calculator is not in BGN mode)
1722.25 PV, -44 PMT, -1100 FV, 30 N, CPT 1/Y.
j can be found to be 1.608245%. The corresponding nominal annual rate compounded semiannually is (X) = i² = 2j =3.216%
Antonette gets $70\%$ on a 10-problem test, $80\%$ on a 20-problem test and $90\%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is her overall score, rounded to the nearest percent?
Answer:
Percentage score will be 83.33 %
Step-by-step explanation:
We have given Antonette gets 70 % on 10 problem test
Let consider here here total problem = total marks
So marks get 10 10 problem test = 10×0.7 = 7
Marks get in 20 problem test = 20×0.8 = 16
And marks get in 30 problem test = 30×0.9 = 27
Now total marks get get by Antonette = 7 +16 + 27 = 50
And total marks = 60
So percentage score of Antonette [tex]=\frac{50}{60}\times 100=83.33[/tex] %
As risk management officer at your firm, you are in charge of analyzing the data on personal injury claims filed against your firm. Some summary statistics for a random sample ofthe costs of 100 claims filed in the recent past are below.
Mean = $1,040.47 1st Quartile = $989.72
Median $1,039.71 3rd Quartile $1,088.18 .
Standard deviation = $89.50 !)o.:,
1. Which statement is correct?
The middle 50% ofthe costs are between $989.72 and $1,088.18."
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The correct statement is that the middle 50% of personal injury claim costs fall between $989.72 and $1,088.18, which represents the interquartile range. This range and the standard deviation are key in evaluating the distribution of claim costs.
The statement that the middle 50% of the costs are between $989.72 and $1,088.18 is correct in reference to the provided summary statistics of personal injury claims. This range is defined by the first and third quartiles, also known as the interquartile range (IQR). The IQR is a measure of variability and represents the span between the 25th percentile (first quartile) and the 75th percentile (third quartile), which indeed encompasses the middle 50% of data in a given sample.
In the context of personal injury claims costs at your firm, this means that half of the claim costs fall within that range, with fewer costs being less than $989.72 (the lower 25%) and fewer costs being more than $1,088.18 (the upper 25%). This can be useful information for assessing claims costs and preparing for future claims expenses. The provided standard deviation of $89.50 indicates the average amount that claim costs vary from the mean ($1,040.47).
Find the coordinates of the orthocenter of ABC. A(-1,0) B(0,4) C(3,0)
Answer:
[tex](0,0.75) \:or\:(0,\frac{3}{4})[/tex]
Step-by-step explanation:
Hi there!
1) Firstly, connect the points to draw a triangle.
2) From each vertex either with a pair of square or with a software trace a perpendicular line to the opposite side.
3) The concurrent point, i.e. the intersection point of these three altitudes is the orthocenter. Orthocenter means the the right center.
In equilateral triangles the Orthocenter coincides with the Centroid.
4) Finally, the coordinates of the Orthocenter found is (0,0.75)
drug that is used for treating cancer has potentially dangerous side effects if it is taken in doses that are larger than the required dosage for the treatment. The pharmaceutical company that manufactures the drug must be certain that the standard deviation of the drug content in the tablet is not more than 0.1 mg. Twenty-five tablets are randomly selected and the amount of drug in each tablet is measured. The sample has a mean of 20 mg and a variance of 0.02 mg. The hypotheses for the test are H0: ?2 ? 0.01 vs Ha: ?2 > 0.01.
Step 1 of 2:
Calculate the test statistic. Round your answer to two decimal places.
Answer:
[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]
Step-by-step explanation:
Data given
[tex]\bar X=20[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]s^2=0.02[/tex] represent the sample variance
[tex]s=0.141[/tex] represent the sample deviation
n=25 represent the sample size
State the null and alternative hypothesis
On this case we want to check if the population standard deviation is more than 0.01, so the system of hypothesis are:
H0: [tex]\sigma \leq 0.1[/tex]
H1: [tex]\sigma >0.1[/tex]
In order to check the hypothesis we need to calculate the statistic given by the following formula:
[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]
This statistic have a Chi Square distribution distribution with n-1=25-1=24 degrees of freedom.
What is the value of your test statistic?
Now we have everything to replace into the formula for the statistic and we got:
[tex] t=(25-1) [\frac{0.141}{0.1}]^2 =47.71[/tex]
What is the critical value for the test statistic at an α = 0.05 significance level?
Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 24 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.
We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,24)". And our critical value would be [tex]\chi^2 =36.415[/tex]
Since our calculated value is higher than the critical value we reject the null hypothesis at 5% of significance.
The variance hypothesis test for a cancer treatment drug with a sample mean of 20 mg and sample variance of 0.02 mg results in a chi-square test statistic of 48. This test statistic will be used to determine if the drug's variance exceeds the acceptable limit.
Explanation:The question at hand is concerning a hypothesis test of the variance in dosage of a cancer treatment drug. The null hypothesis (H0) claims that the standard deviation of the drug content is not more than 0.1 mg, which corresponds to a variance of 0.01 mg² since variance = standard deviation². The alternative hypothesis (Ha) is that the variance is greater than 0.01 mg². Given the sample variance as 0.02 mg² and a sample size of 25, the test statistic for the chi-square test can be calculated using the formula:
Test statistic (chi-square) = (n - 1)*sample variance / hypothesized variance
Test statistic = (25 - 1) * 0.02 / 0.01 = 24 * 2 = 48
The calculated test statistic is 48. Since the sample variance is greater than the hypothesized variance, we have a test statistic that would fall in the rejection region based on the selected significance level in a Chi-square distribution, suggesting that the drug dosage may indeed have greater variability than the company's standard.
Teachers in a medium-sized suburban school district have an average salary of $47,500 per year, with a standard deviation of $4,600. After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. What are the new mean and standard deviation of the teacher's salaries during the year in which they recieve bonus?A. $50,125; $4,960B. $49,875; $4,830C. $49,875; $5,330D. $50,375; $4,830E. $50,375; $5,330
Answer: $49,875; $4,830C.
Step-by-step explanation:
The average salary of Teachers in a medium-sized suburban school district is $47,500 per year.
The standard deviation is $4,600
After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. The bonus of $500 will not alter the mean and standard deviation because equal amount is added for each teacher.
5% increase in each teacher's salary would differ. Therefore, it will affect the mean and standard deviation by 5%. Therefore, the new mean would be
47500 + (5/100 × 47500) = $49875
The new standard deviation would be
4600 + (5/100 × 4600) = $4830
Answer:
$50,375; $4,830 (I just answered it on one of my quizzes)
Step-by-step explanation:
Keitaro walks at a pace of 3 miles per hour and runs at a pace of 6 miles per hour. Each month, he wants to complete at least 36 miles but not more than 90 miles. The system of inequalities represents the number of hours he can walk, w, and the number of hours he can run, r, to reach his goal.3w + 6r ≥ 363w + 6r ≤ 90Which combination of hours can Keitaro walk and run in a month to reach his goal?A. 2 hours walking; 12 hours runningB. 4 hours walking; 3 hours runningC. 9 hours walking; 12 hours runningD. 12 hours walking; 10 hours running
Answer:
A. 2 hours walking; 12 hours running
Step-by-step explanation:
The combination of hours walking and running has to respect both these inequalities:
[tex]3w + 6r \geq 36[/tex]
[tex]3w + 6r \leq 90[/tex]
A. 2 hours walking; 12 hours running
3w + 6r = 3*2 + 6*12 = 6+72 = 78.
Ok, it is larger than 35 and smaller than 91.
B. 4 hours walking; 3 hours running
3w + 6r = 3*4 + 6*3 = 12 + 18 = 30.
Invalid. Lesser than 36.
C. 9 hours walking; 12 hours running
3w + 6r = 3*9 + 6*12 = 27 + 72 = 99
Larger than 90. Invalid
D. 12 hours walking; 10 hours running
3w + 6r = 3*12 + 6*10 = 96
Larger than 90. Invalid
The combination of hours Keitaro can walk and run in a month to reach his goal is 2 hours walking; 12 hours running
3w + 6r ≥ 36. (1)
3w + 6r ≤ 90. (2)
substitute each option into the equation
A. 2 hours walking; 12 hours running
3w + 6r ≥ 36
3(2) + 6(12) ≥ 36
6 + 72 ≥ 36
78 ≥ 36
True
3w + 6r ≤ 90
3(2) + 6(12) ≤ 90
6 + 72 ≤ 90
78 ≤ 90
True
B. 4 hours walking; 3 hours running
3w + 6r ≤ 90
3(4) + 6(3) ≤ 90
12 + 18 ≤ 90
30 ≤ 90
True
B. 4 hours walking; 3 hours running
3w + 6r ≥ 36
3(4) + 6(3) ≥ 36
12 + 18 ≥ 36
30 ≥ 36
False
C. 9 hours walking; 12 hours running
3w + 6r ≥ 36
3(9) + 6(12) ≥ 36
27 + 72 ≥ 36
99 ≥ 36
True
3w + 6r ≤ 90
3(9) + 6(12) ≤ 90
27 + 72 ≤ 90
99 ≤ 90
False
D. 12 hours walking; 10 hours running
3w + 6r ≥ 36
3(12) + 6(10) ≥ 36
36 + 60 ≥ 36
96 ≥ 36
True
3w + 6r ≤ 90
3(12) + 6(10) ≤ 90
36 + 60 ≤ 90
96 ≤ 90
False.
Therefore, the combination of hours Keitaro can walk and run in a month to reach his goal is 2 hours walking; 12 hours running
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A consumer activist group wants to determine the mean lifetime of the Amazon Kindle DX. The consumer activist groups randomly selects 25 Kindles and finds that the average lifespan was 38 months with standard deviation 12 months. Find a 95% confidence interval for the population mean lifetime of the Amazon Kindle DX.
Answer:
Confidence interval for the population mean lifetime of the Amazon Kindle DX is (33.30 months to 42.70 months)
Step-by-step explanation:
Given;
Mean lifespan x = 38 months
Standard deviation r = 12 months
Number of kindle selected n = 25
Confidence range = 95%
Z*(95%) = 1.96
Confidence interval = x+/-Z*(r/√n)
= 38 +/- 1.96(12/√25)
= 38 +/- 4.70
Confidence interval = (33.30 months to 42.70 months)
Devise the exponential growth function that fits the given data, then answer the accompanying question. Be sure to identify the refernce point (t=0) and units of time.Between 2003 and 2008, the average rate of inflation in a certain country was about 4% per year. If a cart of groceries cost $120 in 2003, what will it cost in 2013 assuming the rate of inflation remains constant?
Answer:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
Step-by-step explanation:
For this case we need to define some notation first.
A= population , t= represent the years after 2003, C= constant for the exponential model.
The starting point t=0 correspond to the year of 2003.
On this case we are assuming the following exponential model:
[tex]A(t) = A_o e^{Ct}[/tex]
The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:
[tex]120=A_o e^{C(0)}=A_o e^0 = A_o[/tex]
And then the model is:
[tex]A(t) =120 e^{Ct}[/tex]
Now we need to determine the value for C. Since we know that inflation increase 4% per year we have that after one year we have 1.04 times the value of the original value, and we have this equation:
[tex]1.04 A_o= A_o e^{C(1)}= A_o e^C[/tex]
And we got this:
[tex]1.04= e^C [/tex]
Applying ln on both sides we got:
[tex]ln(1.04)= C=0.0392207[/tex]
So then our model is given by:
[tex]A(t) = 120 e^{ln(1.04)t}[/tex]
For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
The cost of groceries in 2013, after applying an annual inflation rate of 4% for 10 years, will be approximately $177.63.
Explanation:To calculate the annual rate of inflation and the cost in 2013, you can use the exponential growth function, where cost =[tex]initial_{cost} * (1 + rate)^{time[/tex]. In this case, the initial cost in 2003 (t=0) is $120 and the annual rate of inflation is 4%, or 0.04. To calculate the cost in 2013, which is 10 years after 2003, apply the exponential growth function:
Cost in 2013 = $120 * (1 + 0.04)¹⁰
This calculation yields:
Cost in 2013 = $120 * (1.04)¹⁰
Cost in 2013 = $120 * 1.48024
Cost in 2013 = $177.63
Therefore, in 2013, assuming the rate of inflation remains constant, the cart of groceries will cost approximately $177.63.
Find the area of the surface generated when the given curve is revolved about the x-axis y=4x+5 [0,2 ]a. 36√17.xb. 36πc. 36π/√17d. 32√17.π
Answer:
The area of this revolted surface is 36π
Step-by-step explanation:
To obtain the area of a revolted surface, you have to define:
1) which is the axis on which the surface is revolted: this defines the limits on that axis or hight of the surface. In this case x∈[0;2]
2) which is the expression of the radius of the revolted surface and its dependence with the hight. In this case, the radius expression could be Y=4x+5
3) Define the angular variable: If this is a fully revolted surface, the angular variable will go from 0 to 2π
Now we can obtain the area with a double integral:
[tex]A=\int\limits^{2}_0 { \int\limits^{2\pi}_0 {r} \, d \varphi } \, dx =\int\limits^{2}_0 { \int\limits^{2\pi}_0 {4x+5} \, d \varphi } \, dx =\int\limits^{2}_0 { (2\pi)(4x+5)} \, dx=36\pi[/tex]
The incorrect work of a student to solve an equation 2(y + 8) = 4y is shown below:
Step 1: 2(y + 8) = 4y
Step 2: 2y + 10 = 4y
Step 3: 2y = 10
Step 4: y = 5
Which of the following explains how to correct Step 2 and shows the correct value of y? (5 points)
Group of answer choices
Answer:
Step 2 involved distributive property and the value of y is equal to 8.
Step-by-step explanation:
In step 2, there has been an error in applying the Distributive Property correctly.
Distributive Property- a(b+c)
= a x b + a x c
Step 2: [tex]2y+16=4y[/tex]
Step 3:[tex]2y=16[/tex]
Step 4: [tex]y=8[/tex]
y=8
Answer:
2 should be distributed as 2y + 16; y + 8
Consider the following data collected in two recent surveys of whether voters in cities A and B favor a ballot proposition in the next election. City Sample Size In Favor A 615 463 B 585 403 Suppose you're going to find a confidence interval for the difference between the population proportions in the two cities. What's the standard error of the estimate of the difference between the two proportions?
Answer:
Standard error of the estimate of the difference between the two proportions=0.0259
Step-by-step explanation:
Given that the following data collected in two recent surveys of whether voters in cities A and B favor a ballot proposition in the next election.
City A B Total
Sample size 615 585 1200
Favour X 463 403 866
Proportion p 0.7528 0.6889 0.7217
Std error for difference
= [tex]\sqrt{p(1-p)(\frac{1}{n_1} }+ \frac{1}{n_2} \\[/tex]
p =0.7217
1-p = 0.2783
by substituting p and n1 = 615 and n2 = 585 we get
Std error = 0.0259
Standard error of the estimate of the difference between the two proportions=0.0259
In a fund-raising game for your school, you bet $1 to roll two dice. If your total is 8,9,10 or 11 you win $2. If your total is 12, you win $6. If your total is 7 or less, you lose the dollar you bet. How much, on average do you expect to win or lose with each dollar bet?
A. You will lose 56 cents
B. You will 5.6 cents.
C. On average, you will break even.
D. You will win 2/36
E. You will lose 5.6 cents
Answer: E , You will lose 5.6 cents
Step-by-step explanation:
Because with two dice, there are 36 possible outcomes, 21 are 7 or less, 14 are 8 through 11, and 1 is twelve.
Also when you have a total of 8,9,10 or 11, you gain $1 deducing the $1 you bet. The same with when you have 12 you gain $5.
Average $ to gain when total is 8,9,10 or 11 = P(8,9,10,11)
P(8,9,10,11) = ($2-$1)(14/36) = $14/36 gain
P(12) = ($6-$1)(1/36) = $5/36 gain
P(7 or less) = (0-$1)(21/36) = -$21/36 loss
P(loss or gain)= P(8,9,10,11) + P(12) + P(7 or less)
P(loss or gain) = $( 14/36 + 5/36 - 21/36) = -$2/36
P(loss or gain) = -$0.056 = -5.6 cents loss
Therefore, For every $1 bet you will lose 5.6 cents.
Tara wants to weigh her three stuffed animals. They will only fit on the scale two at a time. Together Addie and Missy weight 18 ounces. Missy and Corky weigh 22 ounces, and Addie and Corky weigh 12 ounces. How much does each animal weigh?
Answer: Addie weighs 4 ounces
Missy weighs 14 ounces
Corky weighs 8 ounces
Step-by-step explanation:
Let a represent the weight of Addie.
Let m represent the weight of Missy.
Let c represent the weight of Corky.
Together Addie and Missy weigh 18 ounces. This means that
a + m = 18 - - - - - - - - - 1
Missy and Corky weigh 22 ounces. This means that
m + c = 22
m = 22 - c - - - - - - - - - - 2
Addie and Corky weigh 12 ounces. This means that
a + c = 12
a = 12 - c - - - - - - - - - - - 3
Substituting equation 2 and equation 3 into equation 1, it becomes
22 - c + 12 - c = 18
34 - 2c = 18
- 2c = 18 - 34 = - 16
c = - 16/ - 2 = 8
Substituting c = 8 into equation 2, it becomes
m = 22 - 8
m = 14
Substituting c = 8 into equation 3, it becomes
a = 12 - 8
a = 4
Consider the following function. Without finding the inverse, evaluate the derivative of the inverse at the given point. f(x)=ln(8x+e); (1,0)
We can use the inverse function derivative theorem:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.[/tex]
In this case, we want to evaluate [tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}[/tex], so:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.[/tex]
The derivative is:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.[/tex]
The ordinate of the point is [tex]f^{-1}(1) = 0[/tex], so we evaluate:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.[/tex]
Finally:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.[/tex]
We can check the answer by finding the inverse:
[tex]y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},[/tex]
so that
[tex]f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.[/tex]
Therefore:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.[/tex]
Which finally gives the same answer as before:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.[/tex]
Answer: [tex]\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.[/tex]
7. Solving for dominant strategies and the Nash equilibrium Suppose Nick and Rosa are playing a game in which both must simultaneously choose the action Left or Right. The payoff matrix that follows shows the payoff each person will earn as a function of both of their choices. For example, the lower-right cell shows that if Nick chooses Right and Rosa chooses Right, Nick will receive a payoff of 6 and Rosa will receive a payoff of 5. Rosa Left Right Nick Left 8, 4 4, 5 Right 5, 4 6, 5
In a game of choice and payoff, Nick's dominant strategy is to choose 'Right', while Rosa lacks a dominant strategy. The Nash equilibrium is when Nick chooses 'Right' and Rosa chooses either 'Left' or 'Right' because changing their decisions would not lead to higher payoff.
Explanation:The subject of this question is about a concept from game theory known as dominant strategies and the Nash equilibrium. Nick and Rosa are playing a game where they each simultaneously choose an action (Left or Right) and receive a payoff that depends on both their choices. To find the dominant strategy for each player, we need to identify what action that player would take, regardless of the other player's choice.
For Nick, the dominant strategy is to choose Right because his payoff (5 when Rosa picks Left, 6 when Rosa picks Right) is higher than when he picks Left (8 when Rosa picks Left, 4 when Rosa picks Right). For Rosa, she has no dominant strategy because her payoff is the same (4) whether she chooses Left or Right if Nick is choosing Left, and the same holds if Nick is choosing Right.
The Nash equilibrium is a situation where neither player can benefit by changing their strategy, assuming the other player stays the same. Here, the Nash equilibrium occurs when Nick chooses Right and Rosa chooses Left or Right, because neither player can gain a higher payoff by unilaterally changing their strategy.
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there are 24 panes of glass in 8 windows. use ratio to complete the table below
Answer: It's attached.
Step-by-step explanation:
The table is attached.
The ratio is:
[tex]ratio=\frac{24}{8}\\\\ratio=3[/tex]
Knowing tha ratio, you can complete the table.
The steps are:
1. Multiply the number of panes given in the table by the ratio find above, in order to find the number of windows.
3. Divide the number of windows given in the table by the ratio find above, in order to find the number of panes.
Given [tex]Panes=3[/tex]:
[tex]Windows=3*3=9[/tex]
Given [tex]Windows=3[/tex]:
[tex]Panes=\frac{3}{3}=1[/tex]
Given [tex]Windows=5[/tex]:
[tex]Panes=\frac{5}{3}[/tex]
Given [tex]Panes=18[/tex]:
[tex]Windows=18*3=54[/tex]
Answer:windows x6
Step-by-step explanation:
Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 ft high? (Round your answer to two decimal places.)
Answer:
The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.
Step-by-step explanation:
The volume of a cone is given by the following formula:
[tex]V = \frac{\pi r^{2}h}{3}[/tex]
We have that the diameter and the height are equal, so [tex]r = \frac{h}{2}[/tex]
So
[tex]V = \frac{\pi h^{3}}{12}[/tex]
Let's derivate this equation, using implicit derivatives.
[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]
In this problem, we have to:
Find [tex]\frac{dh}{dt}[/tex], when [tex]\frac{dV}{dt} = 20, h = 14[/tex]. So
[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]
[tex]20 = \frac{196\pi}{4}\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{20}{49\pi}[/tex]
The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.
This involves relationship between rates using Calculus.
dh/dt = 0.13 ft/min
We are given;Volumetric rate; dv/dt = 20 ft³/min
height of pile; h = 14 ft
We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.
Thus; diameter; d = 14 ft
radius; r = h/2 = d/2 = 14/2
radius; r= 7 ft
Formula for volume of a cone is; V = ¹/₃πr²h We want to find how fast the height is increasing and this is dh/dt. Thus, we will need to express r in the volume formula in terms of h; V = ¹/₃π(h/2)²h V = ¹/₃π(h²/4)h V = ¹/₁₂πh³ differentiating both sides with respect to time t gives; dV/dt = 3(¹/₁₂πh²)dh/dt dV/dt = ¹/₄πh²(dh/dt)Plugging in the relevant values, we have;
20 = ¹/₄π × 14² × (dh/dt)
dh/dt = (20 × 4)/(π × 14²)
dh/dt = 0.13 ft/min
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