Answer:
0% probability that at least 1,500 will agree to respond
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
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.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 15000, p = 0.09[/tex]
So
[tex]\mu = E(X) = np = 15000*0.09 = 1350[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05[/tex]
What is the probability that at least 1,500 will agree to respond
This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1499 - 1350}{35.05}[/tex]
[tex]Z = 4.25[/tex]
[tex]Z = 4.25[/tex] has a pvalue of 1.
1 - 1 = 0
0% probability that at least 1,500 will agree to respond
The probability that at least 1,500 will agree to respond is 0.000009.
How to calculate the probabilityFrom the information given,
n = 15000
p = 0.09
Therefore, np = 15000 × 0.09
= 1350
nq = 15000 × (0.91)
= 13650
The probability that at least 1,500 will agree to respond will be:
= 1 - P (Z < (1500 - 1350)/35.05]
= 1 - P(Z < 4.2796)
= 0.000009
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The temperature at a point (x,y,z) is given by T(x,y,z)=200e^-x^2-3y^-9z^2, where T measured in degrees Celsius and x,y,z in meters. Find the rate of change of temperature at the point P(2, -1, 2) in the direction toward the point (3, -3, 3)
Answer:
Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m.
Step-by-step explanation:
Given that, the temperature at a point (x,y,z) is
[tex]T(x,y,z)=200e^{-x^2-3y^2-9z^2}[/tex].
Rate change of temperature at the point P(2,-1,2) in the direction toward the point Q (3,-3,3) is [tex]D_uT(2,-1,2)[/tex]
[tex]T_x= 200(-2x)e^{-x^2-3y^2-9z^2}[/tex][tex]=-400x200.2xe^{-x^2-3y^2-9z^2}[/tex]
[tex]T_y = 200.(-6y)e^{-x^2-3y^2-9z^2}[/tex][tex]=-1200ye^{-x^2-3y^2-9z^2}[/tex]
[tex]T_z=200(-18z)e^{-x^2-3y^2-9z^2}[/tex][tex]=-3600ze^{-x^2-3y^2-9z^2}[/tex]
The gradient of the temperature
[tex]\bigtriangledown T(x,y,z)= (-400x200.2xe^{-x^2-3y^2-9z^2},-1200ye^{-x^2-3y^2-9z^2},-3600ze^{-x^2-3y^2-9z^2})[/tex] [tex]=-400e^{-x^2-3y^2-9z^2}(x,3y,9z)[/tex]
[tex]\bigtriangledown T(2,-1,2) = -400e^{-2^2-3(-1)^2-9.2^2}(2,3.(-1),9.2)[/tex]
[tex]=-400 e^{-43}(2,-3,18)[/tex]
V=[tex]\overrightarrow {PQ}= \vec{Q}-\vec{P}[/tex]=(3,-3,3)-(2,-1,2)=(1,-2,1)
The unit vector of V is [tex]\frac{(1,-2,1)}{\sqrt{1^2+(-2)^2+1^2}}[/tex]
[tex]=\frac{1}{\sqrt6}(1,-2,1)[/tex]
Therefore,
[tex]D_uT(2,-1,2) = \bigtriangledown T(2,-1,2).\frac{1}{\sqrt6}(1,-2,1)[/tex]
[tex]=-400 e^{-43}(2,-3,18).\frac{1}{\sqrt6}(1,-2,1)[/tex]
[tex]=-\frac{400}{\sqrt6}e^{-43}(2.1+(-3).(-2)+18.1)[/tex]
[tex]=-\frac{10400}{\sqrt6}e^{-43}[/tex]
[tex]=-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m
Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m.
Which of the following represents a null hypothesis? Group of answer choices Class A high school basketball teams who employ a sports psychologist will have a higher proportion ofwins over the course of the season than comparable teams who do not employ a sports psychologist. There will be no difference in rate of skill improvement between college gymnasts who practice meditation and those who do not. Does incorporating relaxation exercises into the daily practice routine of college vocal majors enhance their performance confidence? None of the above
In factual tests, an invalid speculation recommends that there is no importance between factors in a bunch of noticed information. In the given decisions, the assertion shows no distinction in the pace of ability improvement between school gymnasts who practice contemplation and the people who don't address the invalid speculation.
Explanation:In factual theory testing, an invalid speculation is an assertion recommending that no measurable relationship and importance exists in a bunch of noticed information between factors. It expects that any noticed contrasts are because of possibility. So in the given choices, the invalid speculation is: 'There will be no distinction in the pace of expertise improvement between school gymnasts who practice contemplation and the people who don't.' This assertion proposes no effect of the variable (reflection) on the result (ability improvement), which is what an invalid theory addresses in a trial of importance.
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please someone help me!!!! SHOW ALL WORK! ive been stuck on this question for 2 days!! :((((((
Step-by-step explanation:
x² + y² − 8x + 10y − 8 = 0
Rearrange:
x² − 8x + y² + 10y = 8
To complete the square, take half of the x and y coefficients, square it, then add the result to both sides.
(-8/2)² = 16
(10/2)² = 25
x² − 8x + 16 + y² + 10y + 25 = 8 + 16 + 25
x² − 8x + 16 + y² + 10y + 25 = 49
Factor the squares:
(x − 4)² + (y + 5)² = 49
The center is (4, -5) and the radius is 7.
The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability that there are 3 or less errors in 100 pages? Round your answer to four decimal places (e.g. 98.7654).
Answer:
Therefore the required probability is 0.6472.
Step-by-step explanation:
Poisson distribution: A Poisson distribution is discrete distribution.
Let X be a discrete variable the number of event. Let λ be the mean of X.
Then the probability of k event is
[tex]P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}[/tex]
Here mean of each page is =0.03
Mean of 100 pages = (0.03×100)= 3
λ = 3
The required probability
P(X≤3)
= P(X=0)+P(X=1)+P(X=2)+P(X=3)
[tex]=\frac{3^0e^{-3}}{0!}[/tex] [tex]+\frac{3^1e^{-3}}{1!}[/tex][tex]+\frac{3^2e^{-3}}{2!}[/tex][tex]+\frac{3^3e^{-3}}{3!}[/tex]
[tex]=e^{-3}(1+3+\frac{9}{2}+\frac{27}{6})[/tex]
[tex]=e^{-3}(13)[/tex]
≈ 0.6472
Therefore the required probability is 0.6472
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25∘ north of east. The road then turns due north for another 3.0 mi to the finish line. In miles, what is the straight-line distance from the starting point to the end of the race? Express your answer in miles.
Answer:
6.79 miles
Step-by-step explanation:
Consider the triangle ABC attached where A is the starting point and C is the finish point.
|AB|=5.2 miles
|BC|=3.0 Miles
≺ABC = 90+75 =165°
We are given the length of two sides and an angle not opposite any of the given sides.
The rule for solving this scenario is referred to as the Cosine Rule.
Note that the side opposite each angle is labelled using the corresponding small letter.
The Cosine Rule States that:
b²=a²+c²-2acCosB
|AC|²=3²+5.2²-(2X5.2XCos165°)
|AC|²=9+27.04-(-10.05)
|AC|²=9+27.04+10.05=46.09
|AC|=√46.09=6.79 miles
Therefore the distance on a straight line from A to C is 6.79 miles
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack _____________.
Answer:
Reliability
Step-by-step explanation:
Dylan took the Whaddayanno IQ Test today, and his IQ score was 130. Last week, his IQ score on the same test was 70. The Whaddayanno IQ Test appears to lack reliability.
Reliability can be defined as the extent to which an experiment or test provides the same result on repeated trials.
The huge difference in Dylan's IQ test scores in a short period of time suggests that the Whaddayanno IQ test is not reliable.
Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. Based on your knowledge of the empirical rule, within what limits would you expect to find the number of infested fields, with probability approximately 95%? (Round your answers to three decimal places.) 4.29 X fields to 25.71 x fields What might you conclude if you found that x = 45 fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.
A. Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent. O
B. Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that the trials have more than two possible outcomes.
C. Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that there are an indefinite number of trials.
D. Based on the limits above, it is likely that we would see x = 45, so all of the characteristics of a binomial experiment are satisfied.
Answer:
a.[tex]\mu=15[/tex]
b.[tex]\mu=7.8586 \ and \ \mu=22.1414[/tex]
c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.
Step-by-step explanation:
a.Binomial distribution is defined by the expression
[tex]P(X=k)=C_k^n.p^k.(1-p)^{n-k}[/tex]
Let n be the number of trials,[tex]n=100[/tex]
and p be the probability of success,[tex]p=15\%[/tex]
The mean of a binomial distribution is the probability x sample size.
[tex]\mu=np=100\times0.15=15[/tex]
b.Limits within which p is approximately 95%
sd of a binomial distribution is given as:[tex]\sigma=\sqrt npq\\q=1-p[/tex]
Therefore, [tex]\sigma=\sqrt(100\times0.015\times0.85)=3.5707[/tex]
Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.
[tex]sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414[/tex]
Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).
c. [tex]x=45[/tex] is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.
Final answer:
Using the empirical rule, we can determine the limits within which we would expect to find the number of infested fields with a 95% probability. If we found a number of infested fields that is higher than expected, it suggests that one of the characteristics of a binomial experiment may not be satisfied. Correct answer is B.
Explanation:
In this problem, we are given that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. We are asked to determine the limits within which we would expect to find the number of infested fields with a probability of approximately 95%. This situation can be modeled using the binomial distribution, as each field can be considered as a separate trial with two possible outcomes, infested or not infested.
According to the empirical rule, for a binomial distribution, we can expect about 95% of the outcomes to fall within two standard deviations of the mean. The mean number of infested fields can be calculated as 15% of the total number of fields, which is 15. The standard deviation of the number of infested fields can be determined using the formula σ = [tex]\sqrt{(npq)}[/tex], where n is the number of trials, p is the probability of success, and q is the probability of failure. In this case, n = 100, p = 0.15, and q = 0.85.
Using these values, we can calculate the standard deviation as σ = [tex]\sqrt{(100 * 0.15 * 0.85)}[/tex] ≈ 3.150. Therefore, we would expect to find about 95% of the number of infested fields within two standard deviations of the mean, which is between 15 - (2 * 3.150) = 8.700 and 15 + (2 * 3.150) = 21.300.
If we found that x = 45 fields were infested, we can conclude that this number is higher than what we would expect based on the binomial distribution. It is unlikely to observe such a high number of infested fields if the trials were independent and had only two possible outcomes. Therefore, we might suspect that one of the characteristics of a binomial experiment is not satisfied in this experiment.
Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that the trials have more than two possible outcomes. Therefore, the correct answer is B.
Functionally important traits in animals tend to vary little from one individual to the next within populations, possible because individuals that deviate too much from the mean die sooner or leave fewer offspring in the long run. If so, does variance in a trait rise after it becomes less functionally important? Billet et al. (2012) investigated this question with the semicircular canals (SC) of the inner ear of the three-toed sloth (Bradypus variegatus). Sloths move very slowly and infrequently, and the authors suggested that this behavior reduces the functional demands on the SC, which usually provide information on angular head movement to the brain. Indeed, the motion signal from the SC to the brain may be very weak in sloths as compared to faster-moving animals. The following numbers are measurements of the length to the width of the anterior semicircular canals in seven sloths. Assume that this represents a random sample.
1.52, 1.06, 0.93, 1.38, 1.47, 1.20, 1.16
a. In related, faster-moving animals, the standard deviation of the ratio of the length to the width of the anterior semicircular canals is known to be 0.09. What is the estimate of the standard deviation of this measurement in three toed sloths?
b. Based on these data, what is the most plausible range of values for the population standard deviation in the three-toed sloth? Does this range include the known value of the standard deviation in related, faster-moving species?
c. What additional assumption is required for your answer in (b)? What do you know about how sensitive the confidence interval calculation is when the assumption is not met?
Answer:
Step-by-step explanation:
Hello!
The objective is to study the semicircular canals (SC) of the inner ear of the three-toed sloth (Bradypus variegatus) to see if it is weak compared to faster animals.
The study variable is X: length to width of the anterior semicircular canal of a three-toed sloth.
A sample of 7 sloths was taken and the semicircular canal was measured:
1.52, 1.06, 0.93, 1.38, 1.47, 1.20, 1.16
∑X= 8.72
∑X²= 11.15
a.
[tex]S^2= \frac{1}{n-1} * [sumX^2-\frac{(sumX)^2}{n} ]= \frac{1}{6} *[11.15-\frac{(8.72)^2}{7} ][/tex]
S²= 0.047≅0.05
S=0.218≅ 0.22
Comparing the estimation of the variance of the length to width of the anterior semicircular canal of three-toed sloths with the known number of length to width of the anterior semicircular canal of faster animals (S=0.09), it appears that the variability os length to width of the anterior semicircular canal of sloths is greater than the length to width of the anterior semicircular canal of faster animals.
b. and c.
To estimate the most plausible range of values of the population standard deviation of the anterior semicircular canal of the sloths, you have to do an estimation per confidence interval.
To be able to make this estimation we have to assume that the variable of interest has a normal distribution. With this assumption, it is valid to use a Chi-Square statistic to estimate the population standard deviation.
[tex]X^2= \frac{(n-1)S^2}{Sigma^2} ~X^2_{n-1}[/tex]
I'll choose a confidence level of 95%
The formula for the interval is:
[tex][\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ][/tex]
[tex]X^2_{n-1;1-\alpha /2}= X^2_{6;0.975}= 14.449[/tex]
[tex]X^2_{n-1;\alpha /2}}= X^2_{6;0.025}= 1.2373[/tex]
[tex][\frac{6*0.05}{14.449} +\frac{6*0.05}{1.2373} ][/tex]
[0.2076;2.4246] ⇒This confidence interval is for the population variance, calculating the square root of each bond gives us the CI for the population standard deviation:
√[0.2076;2.4246]= [0.1441;1.5571]
The 95% CI [0.1441;1.5571] is expected to contain the true value of the population standard deviation of the length to width of the anterior semicircular canal of the three-toed sloths.
As you can see this interval does not contain the known value of the population standard deviation for faster animals, which leads to thinking there is a difference between the standard deviation of the anterior semicircular canal in both species.
I hope it helps!
In a past presidential election, it was estimated that the probability that the Republican candidate would be elected was 5/7, and therefore the probability that the Democratic candidate would be elected was 2/7 (the two Independent candidates were given no chance of being elected). It was also estimated that if the Republican candidate were elected, the probability that a conservative, moderate, or liberal judge would be appointed to the Supreme Court (one retirement was expected during the presidential term) was 1/7, 1/7, and 5/7, respectively. If the Democratic candidate were elected, the probabilities that a conservative, moderate, or liberal judge would be appointed to the Supreme Court would be 1/3, 1/6, and 1/2, respectively. A conservative judge was appointed to the Supreme Court during the presidential term. What is the probability that the Democratic candidate was elected
Answer:
The probability that the Democratic candidate was elected given that a conservative judge was appointed to the Supreme Court is [tex]\frac{14}{29}[/tex].
Step-by-step explanation:
The conditional probability of an events A given that another events B has already occurred is given by:
[tex]P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}[/tex]
The estimation of the past presidential election is provided.
Denote the events as follows:
R = a Republican candidate would be elected
D = a Democratic candidate would be elected
C = a conservative judge would be appointed to the Supreme Court
M = a moderate judge would be appointed to the Supreme Court
L = a liberal judge would be appointed to the Supreme Court
The information provided is:
[tex]P(R)=\frac{5}{7},\ P(D)=\frac{2}{7}[/tex]
[tex]P(C|R)=\frac{1}{7},\ P(M|R)=\frac{1}{7},\ P(L|R)=\frac{5}{7}[/tex]
[tex]P(C|B)=\frac{1}{3},\ P(M|D)=\frac{1}{6},\ P(L|D)=\frac{1}{2}[/tex]
It is provided that a conservative judge was appointed to the Supreme Court during the presidential term.
Compute the probability that the Democratic candidate was elected given that a conservative judge was appointed to the Supreme Court as follows:
[tex]P(D|C)=\frac{P(C|D)P(D)}{P(C|D)P(D)+P(C|R)P(R)}[/tex]
[tex]=\frac{(\frac{1}{3}\times \frac{2}{7})}{(\frac{1}{3}\times \frac{2}{7})+(\frac{1}{7}\times \frac{5}{7})}[/tex]
[tex]=\frac{\frac{2}{21}}{\frac{2}{21}+\frac{5}{49}}[/tex]
[tex]=\frac{2}{21}\div \frac{29}{147}[/tex]
[tex]=\frac{14}{29}[/tex]
Thus, the probability that the Democratic candidate was elected given that a conservative judge was appointed to the Supreme Court is [tex]\frac{14}{29}[/tex].
A researcher claims that the proportion of people who are right-handed is greater than 70%. To test this claim, a random sample of 600 people is taken and its determined that 424 people are right-handed.
The following is the setup for this hypothesis test:
{H0:p=0.70
Ha:p>0.70
Find the test statistic for this hypothesis test for a proportion.
The test statistic for the given hypothesis test for a proportion is calculated to be 0.356.
Given that:
The claim is that the proportion of people who are right-handed is greater than 70%.
So, p > 0.7 is the alternative hypothesis.
From the setup of the hypothesis test, the test is a one-tailed test since the alternative hypothesis has a sign ">".
Sample size, n = 600
Number of people right-handed = 424
So, the observed sample proportion is:
[tex]\hat{\text{p}}=\frac{424}{600}[/tex]
[tex]=0.7067[/tex]
The standard deviation is:
[tex]\sigma=\sqrt{\frac{p(1-p)}{n} }[/tex]
[tex]=\sqrt{ \frac{0.7(1-0.7)}{600} }[/tex]
[tex]=0.01871[/tex]
So, the test statistic is:
[tex]\text{z}=\frac{\hat{\text{p}}-\text{p}}{\sigma}[/tex]
[tex]=\frac{0.7067-0.7}{0.01871}[/tex]
[tex]=0.356[/tex]
Hence, the test statistic is 0.356.
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(A) In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have:
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient.
(B) Now that you have one of the two variables in Part (A) isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers..
Answer:
C = 17/7 or 2 3/7
D = -4/7
Step-by-step explanation:
3C+4D=5
2C+5D=2
2C = 2 - 5D
C = 1 - 2.5D
3(1 - 2.5D) + 4D = 5
3 - 7.5D + 4D = 5
-3.5D = 2
D = -4/7
C = 1 - 2.5(-4/7)
C = 1 + 10/7 = 17/7
In a large, randomly mating population with no forces acting to change gene frequencies, the frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000, or 0.09. What percentage of the population carries this trait but displays the dominant phenotype, short eyelashes
Answer:
16.38%
Step-by-step explanation:
Given
Frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000
Let p = Probability of homozygous recessive individuals having character extra-long eyelashes = 0.09
p + q = 1 where q = Probability of homozygous recessive individuals not having character extra-long eyelashes
0.09 + p = 1
p = 1 - 0.09
p = 0.91
The frequency of the carrier individual is calculated by npq
Where n = mating partners = 2
Frequency = 2 * 0.09 * 0.91
Frequency = 0.1638 or 16.38%
Answer:
The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals which is
0.42 or 42 %
Step-by-step explanation:
To solve the question, we note that
p²+2pq+q²=1 and p + q =1
where
p = frequency of the dominant allele
q= frequency of the recessive allele
p² = frequency of homozygous dominant allele
q² = frequency of homozygous recessive allele
2·p·q = frequency of hetrozygous individuals
The frequency of the homozygous recessive individuals q² is 0.09
Therefore the frequency of q = √(0.09) = 0.3, therefore p = 1 - 0.3 =0.7
The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals or 2×p×q = Ae = 2*0.3*0.7 = 0.42
→ 0.42× 100 = 42 %
(1 point) On a piece of paper, sketch each of the following surfaces: (i) z = x2 +y2 +6 (ii) z = 3x2 Use your graphs to fill in the following descriptions of cross-sections of the surfaces. (a) For (i) (z = x2 2 + 6): Cross sections with x fixed give a downward opening parabola in a plane parallel to the yz-plane a downward opening parabola in a plane parallel to the xz-plane a downward opening parabola in a plane parallel to the xy-plane Cross sections with y fixed give Cross sections with z fixed give (b) For (ii) (z = 3x2): Cross sections with x fixed give Cross sections with y fixed give Cross sections with z fixed give an empty set, or one or two vertical lines in a plane parallel to the yz-plane> an empty set, or one or two vertical lines in a plane parallel to the xz-plane a downward opening parabola in a plane parallel to the xy-plane
Answer:
ht j5yh bgmn gmjdt ky
Step-by-step explanation:
w245358697809908
The cross sections of the given mathematical functions provide a range of shapes, from parabolas to circles or lines, depending on whether x, y or z are fixed.
Explanation:Interpreting and sketching the three-dimensional functions in question gives us an understanding of the cross-sections. For the first function i) z = x2 + y2 + 6, when x is fixed, the cross section gives us a parabola facing upwards in yz-plane. Moving ahead, when y is fixed, it gives us a similar parabola in the xz-plane. When z is fixed, we end up with a circle in xy-plane.
Going to the second function ii) z = 3x2, cross sections with x fixed results in a vertical line in yz-plane as z is not a function of y here. For y being fixed, it will again give us a vertical line but in the xz-plane because z and y are again unrelated. If z is fixed, we obtain a parabola opening upwards (or downwards depending on the sign) parallel to xy-plane.
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A small social network contains seven people who are network friends with six other people in the network, one person who is network friends with five other people in the network, and five people who are network friends with four other people in the network. The rest are network friends with three other people in the network. The network contains 50 pairs of network friends.
(a) How many people are network friends with three other people in the network?
(b) How many people are in the network?
Answer:
(a) 11 people
(b) 24 people
Step-by-step explanation:
Part (a)
Let X be the people who are network friends with three other people in the network.
The total pairings = 50
Since [tex]v[/tex]∈[tex]V[/tex],
[tex](7)(6) + (1)(5) + (5)(4) + (X)(3) = (2)(50)[/tex]
Solve for X to find the people who are network friends with three other people,
[tex]67 + 3X = 100[/tex]
[tex]X = 11[/tex]
Answer: There are 11 people who are network friends with three other people in the network.
Part (b)
Total people in the network,
[tex]= 7 + 1 + 5 + X\\ = 13 + 11\\ = 24[/tex]
Answer: There are a total of 24 people in the network.
There are 6 people who are network friends with three other people in the network. The total number of people in the network is 19.
Explanation:This problem involves understanding the concept of pairs in network friends. Each friendship forms a pair. So, if one person has 6 friends, this would contribute 6 pairs. Here, there are 7 people with 6 friends, 1 person with 5 friends, and 5 people with 4 friends. Therefore, these people contribute (7x6) + (1x5) + (5x4) = 42 + 5 + 20 = 67 pairs. But, we're told there are only 50 pairs in total. The reason for this discrepancy is that each pair is being counted twice. When we divide the total pairs (67) by 2, we get 33.5 but since the number of pairs must be an integer, we subtract this half pair to get 33 pairs for the listed people. That means, there must be 50 - 33 = 17 pairs remaining for the people who are network friends with three other people. Given that each of these people contributes 3 pairs, there must be 17 / 3 = 5.67 people, but since we're talking about whole people, we need to round this up to 6. Therefore, there are 6 people who are network friends with three other people in the network. Now, adding up all the people in the network, we have 7 (people with 6 friends) + 1 (person with 5 friends) + 5 (people with 4 friends) + 6 (people with 3 friends) = 19 people in total.
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3. It is known that 80% of all college professors have doctoral degrees. If 10 professors are randomly selected, find the probability that a. five have a doctoral degree b. fewer than 4 have doctoral degrees. c. At least 6 have doctoral degrees. d. Between 5 and 7 (inclusive) have doctoral degrees. e. What is the expected number of college professors with doctoral degrees
Answer:
a) [tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
b) [tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
c) [tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
d) [tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
e) [tex] E(X) = n*p = 10*0.8 =8[/tex]
Step-by-step explanation:
Previous concepts
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".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=10, p=0.8)[/tex]
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]
Part a
We want to find this probability:
[tex] P(X=5)[/tex]
And using the pmf we got:
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
Part b
[tex] P(X<4) = P(X \leq 3) = P(X=0) +P(X=1) +P(X=2) +P(X=3)[/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
And adding we got:
[tex] P(X<4) =0.000864[/tex]
Part c
For this case we can use the complement rule like this:
[tex] P(X \geq 6) = 1-P(X<6) = 1-P(X\leq 5) =1-[P(X=0) +....+P(X=5)][/tex]
[tex]P(X=0)=(10C0)(0.8)^0 (1-0.8)^{10-0}=1.024x10^{-7}[/tex]
[tex]P(X=1)=(10C1)(0.8)^1 (1-0.8)^{10-1}=4.096x10^{-6}[/tex]
[tex]P(X=2)=(10C2)(0.8)^2 (1-0.8)^{10-2}=7.37x10^{-5}[/tex]
[tex]P(X=3)=(10C3)(0.8)^3 (1-0.8)^{10-3}=0.000786[/tex]
[tex]P(X=4)=(10C4)(0.8)^4 (1-0.8)^{10-4}=0.0055[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
And replacing we got:
[tex] P(X \geq 6) = 1- 0.0328= 0.967[/tex]
Part d
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)[/tex]
[tex]P(X=5)=(10C5)(0.8)^5 (1-0.8)^{10-5}=0.0264[/tex]
[tex]P(X=6)=(10C6)(0.8)^6 (1-0.8)^{10-6}=0.088[/tex]
[tex]P(X=7)=(10C7)(0.8)^7 (1-0.8)^{10-7}=0.2013[/tex]
And replacing we got:
[tex] P(5 \leq X \leq 7) = P(X=5) +P(X=6) +P(X=7)=0.0264 +0.088+0.2013=0.316 [/tex]
Part e
The expected number is given by:
[tex] E(X) = n*p = 10*0.8 =8[/tex]
This answer provides step-by-step explanations for finding probabilities of professors with doctoral degrees and the expected number of professors with doctoral degrees.
Expected number of college professors with a doctoral degree:
Find the probability that five have a doctoral degree: Using the binomial probability formula, [tex]P(X = 5) = 10C5 * 0.8^5 * 0.2^5[/tex]
Find the probability that fewer than 4 have a doctoral degree: Calculate P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
Find the probability that at least 6 have a doctoral degree: Calculate P(X ≥ 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)).
Find the probability that between 5 and 7 inclusive have a doctoral degree: Calculate P(5 ≤ X ≤ 7) = P(X = 5) + P(X = 6) + P(X = 7).
Expected number of professors with a doctoral degree: Multiply the total number of professors by the probability (0.8).
The cost of flying a passenger plane from point A to point B is $70 comma 00070,000. The airline flies this route four times per day at 7 AM, 10 AM, 1 PM, and 4 PM. The first and last flights are filled to capacity with 240240 people. The second and third flights are only half full. Find the average cost per passenger for each flight.
Answer:
I. $291.67 per passenger
II. $583.33 per passenger
Step-by-step explanation:
The cost of flying the passenger plane from A to B is $70,000.
If the Capacity of the plane is 240 people.
(I)At 7A.M and 4P.M., the average cost per passenger is given as:
Average Cost=Total Cost/Number of Passengers
=70000/240=$291.67 per passenger
(II)At 10AM and 1PM, the flight is only half full, it has 120 passengers.
Average Cost = 70000/120=$583.33 per passenger
Answer: The cost will be $291.70 and $583.40 for the first and the last and for the second and third flights, respectively.
Step-by-step explanation: The cost is $70,000, the full capacity of seats is 240 and half capacity is 120.
In the first and last flights, which have the full capacity, the average cost per passenger is:
C = [tex]\frac{70,000}{240}[/tex] = 291.70
In the second and third ones, the average cost is:
C = [tex]\frac{70,000}{120}[/tex] = 583.4
The average cost per passenger for the first and the last flight are $291.70, while for the second and third are $583.4
Consider a with 3 × 3 grid where each cell contains a number of coins; for example, the following represents a possible configuration of coins on the grid (the integer in each cell is the number of coins in that cell):12 3 11 8 42 13 0This configuration is transformed in stages as follows: in each step, every cell sends a coin to all of its neighbors (horizontally or vertically, not diagonally), but if there aren’t enough coins in a cell to send one to each of its neighbors, it sends no coins at all. For example, the above would result in the following after one step:11 2 34 7 21 12 2a) Show that every staring configuration results in stable configuration (one that no longer changes in this process), or repeatedly cycles through ???? configurations for some positive integer ???? (i.e., those same ???? configurations appear repeatedly in the sequence over and over as the transformation is applied).b) In the case that the initial configuration eventually cycles through ???? configurations, what are the possible values of ?????c) Either prove that for some positive integer ????, every configuration will reach a stable configuration or a repetition of a ????-cycle in ???? or fewer steps, or prove there is no such B.
Answer:
Step-by-step explanation:
Check attachment for solution
What is slope of line between each pair of points?
(1, -19) and (-2, -7)
The slope of the line is [tex]m=-4[/tex]
Explanation:
Given that the pair of points are [tex](1,-19)[/tex] and [tex](-2,-7)[/tex]
We need to determine the slope of the line.
The slope of the line can be determined using the formula,
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Substituting the points [tex](1,-19)[/tex] and [tex](-2,-7)[/tex] for the coordinates [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] in the above formula, we have,
[tex]m=\frac{-7-(-19)}{-2-1}[/tex]
Simplifying the terms, we get,
[tex]m=\frac{-7+19}{-2-1}[/tex]
Adding the terms, we have,
[tex]m=\frac{12}{-3}[/tex]
Dividing, we get,
[tex]m=-4[/tex]
Thus, the slope of the line between the pair of points is [tex]m=-4[/tex]
An ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken from the four populations. The numerator and denominator (respectively) degrees of freedom for the critical value of F are _____.
Answer:
Step-by-step explanation:
Given that an ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken from the four populations.
Hence total observations are 30*4 =120
No of groups = 3
Hence numerator df = 3-1 =2
Now total degrees of freedom = 120-1 =119
So denominator degrees of freedom = 119-2 = 117
Thus F statistic will have numerator as 2 degrees of freedom and denominator as 117 degrees of freedom.
i really need help:(
Answer: the height of the building is 40ft
Step-by-step explanation:
Looking at the right angle triangle formed,
With angle P as the reference angle, the length shadow of the building on ground represents the adjacent side of the right angle triangle.
The height of the building represents the opposite side of the right angle triangle.
a) to determine the height of the building, x, we would apply the tangent trigonometric ratio which is expressed as
Tan θ, = opposite side/adjacent side. Therefore, the equation becomes
Tan P = x/50
b) 0.8 = x/50
x = 50 × 0.8
x = 40 ft
A restaurant purchased kitchen equipment on January 1, 2017. On January 1, 2019, the value of the equipment was $14 comma 550. The value after that date was modeled as follows. V(t)equals 14 comma 550 e Superscript negative 0.158 t a) What is the rate of change in the value of the equipment on January 1, 2019
Answer:
[tex]\frac{dV(t)}{dt} =[/tex] - 1675.38
Step-by-step explanation:
In 2017, the vakue of the kitchen equipment was $14550
V(0)=$14550
Its value after then was modelled by [tex]V(t)=14550e^{-0.158t[/tex]
We are required to find the rate of change in value on January 1, 2019
[tex]V(t)=14550e^{-0.158t[/tex]
[tex]\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t[/tex]
[tex]\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t[/tex]
[tex]\\Let u= -0.158t,\frac{du}{dt}=-0.158[/tex]
[tex]\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}[/tex]
[tex]\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}[/tex]
In 2019, i.e. 2 years after, t=2
The rate of change of the value
[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}[/tex]
=[tex]\frac{dV(t)}{dt} =-2298.9e^{-0.316}[/tex]= - 1675.38
The rate of change in the value of the equipment on January 1, 2019, is -2,302.9e^(-0.158t).
Explanation:The rate of change in the value of the equipment on January 1, 2019, can be found by taking the derivative of the given model equation V(t) = 14,550e^(-0.158t).
The derivative of V(t) with respect to t is dV/dt = -0.158(14,550)e^(-0.158t), which simplifies to dV/dt = -2,302.9e^(-0.158t).
Therefore, the rate of change in the value of the equipment on January 1, 2019, is -2,302.9e^(-0.158t).
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Use mathematical induction to prove that if L is a linear transformation from V to W, then L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) g
Answer:
The proof is shown in the explanation below.
Step-by-step explanation:
Analysis:
The proof by induction focuses on n. In this case, let n = 1, and [tex]L^{1}[/tex] will be a linear operator since [tex]L^{1} = L[/tex]
The exercise will show that [tex]L^{n}[/tex] is a linear operator on V and that [tex]L^{n+1}[/tex] is also a linear operator on V. This, follows that:
[tex]L^{n+1} (av) = L(L^{m}(v_{1}+v_{2})\\ = L(L^{m} (v_{1} + L^{m}v_{2})\\ = L(L^{m}v_{1} + L(L^{m}v_{2})\\ = L^{m+1}(v_{1}) + L^{m+1}(v_{2})[/tex]
Answer/Step-by-step explanation:
For the mathematical induction,
We show that the equation
L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn) is true for
L = 1,
Assume it is true for L = n and show that it is true for L = n + 1.
If L = 1, the equation become
(α1v1 + α2v2 +· · ·+αnvn)= α1(v1) + α2 (v2)+· · ·+αn (vn). Therefore, the Right Hand side(RHS) = Left Hand side(LHS)
When L = n, we assume the following is true
(α1nv1 + α2nv2 +· · ·+αnvn)= α1n(v1) + α2n (v2)+· · ·+αn (vn)
Then, when L = n + 1,
n +1 (α1v1 + α2v2 +· · ·+αvn)= α1(n +1) (v1) + α2(n +1) (v2)+· · ·+αn(n + 1)(vn).
Open the bracket,
n(α1v1 + α2v2 +· · ·+αvn) + α1v1 + α2v2 +· · ·+αnvn = α1n (v1) + α2v2 +· · ·+αvn ) + α1(v1) + α2v2+· · ·+αn(vn)
Since we assume the the equation is true for L = n, and eliminating some terms, then
L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1) + α2L (v2)+· · ·+αnL (vn)
PLEASE HELP ME
select the most reasonable metric unit for a glass containing 350 units of liquid. Choose from L, mL, kg, g, and mg.
Answer:
mL
Step-by-step explanation:
Ml
The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories.
a. Draw a Venn diagram summarizing the variables and their associated probabilities
b. What percent of Americans live below the poverty line and only speak English at home?
c. What percent of Americans live below the poverty line or speak a foreign language at home?
Answer:
b. 10.4
c. 26.9
Step-by-step explanation:
Let the universal set U = 100% which is the total no of people in the American community
Let A = 14.6% which is the total no of people living below poverty line
Let B = 20.7% which is the total no of people speaking foreign Language
C = 4.2% no of people who both speak foreign language and live below poverty line
X = no of people who neither live below poverty line nor speak foreign language
P (A) = 14.6%
P (B) = 20.7%
P (C) = P (A ∩ B) = 4.2%
P (A – C) = P (A ∩ U) = 14.6 – 4.2 = 10.4%
P (B – C) = P (B ∩ U) = 20.7 – 4.2 = 16.5%
P (X) = P (A ᴜ B) c =100 – (10.4 + 4.2 + 16.5) = 68.9%
a. The venn diagram is as shown above
b. Percent of Americans who live below poverty line and Speak English at home(minus foreign lang speakers living below poverty line) that is A only
= A – C
= 14.6 – 4.2
= 10.4%
c. Percentage of Americans Living below poverty line or Speaking foreign language
= A only + B only
A only = A – C ( People living below poverty line only)
= 14.6 -4.2
= 10.4%
B only = B – C ( people speaking foreign languages only)
= 20.7 – 4.2
= 16.5%
Hence
A only + B only = 10.4 + 16.5 = 26.9%
A movie theater offers 6 showings of a movie each day. A total of 1000 people come to see the movie on a particular day. The theater is interested in the number of people who attended each of the six showings. How many possibilities are there for the tallies for each showing for that day?
Answer:
2.03 × 10¹⁴ different possibilities.
Step-by-step explanation:
We want to know the different ways 1000 people can come out to watch a movie at 6 different times of the day.
This is a permutation and combination problem.
Dividing 1000 people amongst 6 movie showings = 1005C5 = 1005!/(1005-5)!5!
= (1005×1004×1003×1002×1001×1000!)/(1000!5!) = (1005×1004×1003×1002×1001)/5! = 2.03 × 10¹⁴ different possibilities.
According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health sciences is $51,541. The average starting salary for new college graduates in business is $53,901 (National Association of Colleges and Employers website, January 2015). Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates in health sciences is $11,000. Assume that the standard deviation for starting salaries for new college graduates in business is $15,000.
a. What is the probability that a new college graduate in business will earn a starting salary of at least $65,000?
b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000?
c. What is the probability that a new college graduate in health sciences will earn a starting salary of less than $40,000?
d. How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences?
Answer:
Part (a) : 0.2297
Part (b) : 0.1112
Part (c) : 0.1469
Part (d) : 77,171
Step-by-step explanation:
Given info on Health Sciences:
Mean = $51,541
Standard Deviation = $11,000
Given info on Business:
Mean = $53,901
Standard Deviation = $15,000
Part (a)
Let X represents the new college graduate in business,
P (X ≥ 65,000) = 1 - P (X < 65,000)
= 1 - P ( z < [tex]\frac{65,000 - 53,901}{15,000}[/tex] )
= 1 - P ( z < 0.74)
= 1 - 0.77035
= 0.2297
Part (b)
Let Y represents the new college graduate in Health Sciences,
P (Y ≥ 65,000) = 1 - P (Y < 65,000)
= 1 - P ( z < [tex]\frac{65,000 - 51,541}{11,000}[/tex] )
= 1 - P ( z < 1.22)
= 1 - 0.88877
= 0.1112
Part (c)
Let Y represents the new college graduate in Health Sciences,
P (Y < 40,000) = P (Y < [tex]\frac{40,000-51,541}{11,000}[/tex])
= P ( z < -1.05 )
= 0.1469
Part (d)
To have a starting salary higher than 99%, the z-score = 2.33. Let A represents the salary of a new college graduate in health sciences higher than 99% of all starting salaries.
[tex]2.33 = \frac{A - 51,541}{11,000}[/tex]
[tex]A = 77,171[/tex]
77,171 new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in health sciences.
Oslo Company's target quality characteristic, T, for one of its key components is set at 82. Using the Taguchi Quality Loss Function (QLF) the company has determined the cost coefficient, k, to be $6,000. What is the estimated loss, L(x), if the value of the quality characteristic, x, is 85
Answer:
The estimated loss is $5,400 if the value of the quality characteristic,x is 85.
Step-by-step explanation:
We are given the following in the question:
Target value,T = 82
Cost coefficient, k = $6,000
x = 85
The Taguchi Quality Loss Function (QLF) is given by:
[tex]L(x) = k(x -T)^2[/tex]
where L(x) is the loss, k is the cost coefficient, T is the target value or the mean value.
We have to estimate the loss if the value of the quality characteristic is 85.
We put x = 85
[tex]L(85) = 6000(85 -82)^2\\L(85) =54000[/tex]
Thus, the estimated loss is $5,400 if the value of the quality characteristic,x is 85.
Sean, a high school wrestler, has agreed to participate in a study of cardiovascular conditioning. He is left somewhat confused when, at the first research session, he is asked to complete a questionnaire about commonly purchased grocery items. Sean's confusion indicates a lack of ________ regarding the task. Question 2
Answer:
Face validity.
Step-by-step explanation:
Face validity is considered the weakest form of validity as it does a superficial, subjective assessment which does not involve objective approach, revealing the deeper intent of the test being used. It involved a process similar to skimming the surface of an item or a book to make opinions. Though it is a weak form of validity, it is the easiest to apply to research. Face validity is an informal approach to identifying how suitable the content of a test is to the research.
It generally asks the question:
Is the content of the test suitable to its aims?
a rectangular box with two sqaure opposite ends is to hold 8000 cubic inches. find the dimensions of the cheapest box if tge recangular sides cost 15 times more as much per square inche as the top, bottom, and square ends
Answer:
the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in
Step-by-step explanation:
since the box has a volume V
V= x*y*z = b=8000 in³
since y=z (square face)
V= x*y² = b=8000 in³
and the cost function is
cost = cost of the square faces * area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides
C = a* 2*y² + a* 2*x*y + 15*a* 2*x*y = 2*a* y² + 32*a*x*y
to find the optimum we can use Lagrange multipliers , then we have 3 simultaneous equations:
x*y*z = b
Cx - λ*Vx = 0 → 32*a*y - λ*y² = 0 → y*( 32*a-λ*y) = 0 → y=32*a/λ
Cy - λ*Vy = 0 → (4*a*y + 32*a*x) - λ*2*x*y = 0
4*a*32/λ + 32*a*x - λ*2*x*32*a/λ = 0
128*a² /λ + 32*a*x - 64*a*x = 0
32*a*x = 128*a² /λ
x = 4*a/λ
x*y² = b
4*a/λ * (32*a/λ)² = b
(a/λ)³ *4096 = 8000 m³
(a/λ) = ∛ ( 8000 m³/4096 ) = 5/4 in
then
x = 4*a/λ = 4*5/4 in = 5 in
y=32*a/λ = 32*5/4 in = 40 in
then the box has dimensions 5 in x 40 in x 40 in
Determine whether the argument is valid or invalid.
Loretta's hobby is stamp collecting. If her husband likes to fish, then Loretta's hobby is not stamp collecting. If her husband does not like to fish, then Nathan likes to read. Therefore, Nathan likes to read.
Answer:
The argument is valid.
Step-by-step explanation:
We are told "Loretta's hobby is stamp collecting."
"If her husband likes to fish, then Loretta's hobby is not stamp collecting."
Her hobby is stamp collecting, so her husband does not like to fish.
"If her husband does not like to fish, then Nathan likes to read."
Her husband does not like to fish; therefore, Nathan likes to read.
The argument is valid.
The argument is invalid as it fails the informal test of validity; the premises about Loretta and her husband's hobbies do not logically lead to the conclusion about Nathan's reading habits.
Explanation:The task involves assessing the validity of an argument by applying the informal test of validity. Let's deconstruct the argument provided. It states that Loretta's hobby is stamp collecting, and presents two conditional statements involving her husband's hobbies and Nathan's interest. Finally, it concludes that Nathan likes to read.
To apply the informal test of validity, we must determine if it's possible for all premises to be true while the conclusion could still be false. In this argument, even if all the stated conditions about Loretta and her husband's hobbies are true, they don't logically necessitate the conclusion about Nathan's reading habits. Thus, the argument is invalid because it's possible to imagine a scenario where all premises are true, but Nathan does not like to read. The premises about Loretta and her husband have no logical connection to Nathan's interests, making it invalid.
A valid argument ensures that if the premises are true, the conclusion must also be true. However, this argument fails to meet that criterion, indicating its invalidity based on the informal test.