Answer:
a) Continuous since the weight can take decimals for example 2.3 pounds.
b) Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters
c) Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous
d) Discrete since that's a binary variable and only take integers.
e) Discrete since the number of errors can't be 4.3 for example, and can't take decimals.
Step-by-step explanation:
Previous concepts
A continuous random variable by definition is a "random variable where the data can take infinitely many values" defined on a interval.
And a discrete random variable is a random variable that only can takes integers and is defined over a domain
Solution to the problem
a. Weight of the book (e.g., 2.3 pounds)
Continuous since the weight can take decimals for example 2.3 pounds.
b. Number of chapters in the book (e.g., 10 chapters)
Discrete since the number of chapters in a book represent an integer, we can't say that abook have 4.3 chapters because not makes sense, we say the book have 4 or 5 chapters
c. Width of the book (e.g., 8 inches)
Continuous since we can measure the width and we can get 8.3 inches so then the variable can takes decimals and for this reason is Continuous
d. Type of book (0=hardback or 1=paperback)
Discrete since that's a binary variable and only take integers.
e. Number of typographical errors in the book (e.g., 4 errors)
Discrete since the number of errors can't be 4.3 for example, and can't take decimals.
For a random sample of 90 overweight men, the mean number of pounds that they were overweight was 35. The standard deviation of the population was 3.5 pounds. Find the best point estimate of the number of excess pounds that they weighed.
The best point estimate for the number of excess pounds that the overweight men weighed is the mean value given, which is 35 pounds. Furthermore, the standard deviation of 3.5 pounds indicates that majority of these men's excess weight is within 3.5 pounds of this mean value.
Explanation:In this problem, you have been given a random sample of 90 overweight men with a mean weight being overweight by 35 pounds. This mean (35 pounds) represents the best point estimate of the amount of excessive weight that the men of your study hold. What this means is that our best guess, considering this sample data, of an overweight man's excess weight is 35 pounds. We compute this mean value by adding up all the data values and dividing by the total number, in this case, 90. The standard deviation tells us somewhat about how much variation exists in the population's weights from this mean value. In this scenario, the standard deviation is 3.5 pounds, which means that the majority of the men in this study were within 3.5 pounds of the mean excess weight of 35 pounds.
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The best point estimate of the number of excess pounds that the sample of 90 overweight men weighed is the mean of their weights, which is given as 35 pounds.
Explanation:In this problem, the best point estimate of the number of excess pounds that the men weighed would be the mean of their weights. Since we are given that the mean number of excess pounds that they were overweight was 35, the best point estimate would also be 35. This is because in statistics, the mean of a sample is used as the best point estimate of the population mean. The fact that they mentioned the standard deviation of 3.5 pounds does not affect the best point estimate, as that is related more to the dispersion or spread of the data around the mean, and not the central point (which is the mean).
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Cooley et al. (2009) randomly assigned either of two treatments, naturopathic care (diet counseling, breathing techniques, vitamins, and a herbal medicine) or stan- dardized psychotherapy (psychotherapy with breathing techniques and a placebo added), to 81 individuals having moderate to severe anxiety. Anxiety scores decreased an average of 57% in the naturopathic group and 31% in the psychotherapy group. Is this an experimental or observational study?
Answer: Experimental study
Step-by-step explanation:
This is an experimental study because it involves determination of the effect of each treatment on separate individuals. It involves selecting the 81 individuals randomly to be treated with a particular treatment (either of the two treatment). In experimental study the researcher apply separate treatments on a different groups and measure its effect on each of the groups.
This is an experimental study where participants were randomly assigned to either naturopathic care or standardized psychotherapy to assess their effects on anxiety scores.
Explanation:This is an example of an experimental study. In an experimental study, researchers assign participants to different groups and manipulate variables to determine cause-and-effect relationships. In this case, Cooley et al. randomly assigned individuals to either naturopathic care or standardized psychotherapy, which are the two treatments being compared. They then measured the effects of these treatments on anxiety scores.
The fact that participants were randomly assigned to the two treatments suggests that this study was experimental rather than observational.
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In a sequence of numbers, (look at image below)
Answer:
aₙ = 6n - 6
Step-by-step explanation:
a1 = 6 x 1 - 6 = 0
a₂ = 6 x 2 - 6 = 6
Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls numbered 1, . . . , 12. Finally, box C contains four balls numbered 1, . . . , 4. One ball is selected from each urn uniformly at random.
(a) What is the probability that the ball chosen from box A is labeled 1 if exactly two balls numbered 1 were selected
(b) What is the probability that the ball chosen from box B is 12 if the arithmetic mean of the three balls selected is exactly 7?
Answer:
a) 0.73684
b) 2/3
Step-by-step explanation:
part a)
[tex]P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}[/tex]
Using conditional probability as above:
(A,B,C)
Cases for numerator when:
P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)
= [tex](\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111[/tex]
Cases for denominator when:
P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)
[tex]= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4}) + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222[/tex]
Hence,
[tex]P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684[/tex]
Part b
[tex]P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}[/tex]
Cases for denominator when:
P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)
[tex]= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}[/tex]
Cases for numerator when:
P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)
[tex]= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}[/tex]
Hence,
[tex]P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}[/tex]
Final answer:
The question involves calculating probabilities of selecting specific numbered balls from different boxes under given conditions. Part (a) addresses the probability of selecting a ball numbered 1 from box A when exactly two balls numbered 1 are chosen in total, which is generically 1/6. Part (b) asks about the probability of choosing a ball numbered 12 from box B when the arithmetic mean of numbers on the selected balls is 7, which needs combinatorial analysis of possible valid combinations.
Explanation:
The student's question involves probability and combinatorics, focusing on selecting balls from different boxes. There are two parts to the question:
(a) Probability that the ball chosen from box A is labeled 1 if exactly two balls numbered 1 were selected
For exactly two balls numbered 1 to be selected across the three boxes, one must come from box A and one must either come from box B or C. The probability of choosing the number 1 ball from box A is 1/6, assuming a uniform random selection. However, since the condition specifies exactly two balls numbered 1 need to be selected, and there's no way to determine how this condition impacts the selection without further context (like an overall occurrence rate), this part can only provide the probability of selecting a number 1 from box A in general, which is 1/6.
(b) Probability that the ball chosen from box B is 12 if the arithmetic mean of the three balls selected is exactly 7
To find the probability that the ball chosen from box B is 12 given that the arithmetic mean of the three balls is exactly 7, we must consider that the total sum of the numbers on the balls must be 21 (since 7*3=21). If the ball from box B is 12, the sum of the numbers from the other two balls must be 9. With these criteria, we can explore the combinations that result in a sum of 9 from boxes A and C. However, without specific combinations provided, detailed calculation is not feasible in this response. Generally, this would involve enumerating all valid combinations from boxes A and C that sum to 9, and then dividing that by the total number of possible draws from all boxes, taking into account the condition specified.
Prove the following theorems algebraically:
(a) X(X′ + Y) = XY
(b) X + XY = X
(c) XY + XY′ = X
(d) (A + B)(A + B′) = A
Answer:
Algebraic Theorem Proofs
Step-by-step explanation:
We have to prove the following in the question:
Theorems used:
[tex]XX' = 0\\X + 1 = X\\X + 0 = X\\X+X'=1\\XX = X\\X+X=X[/tex]
(a) X(X′ + Y) = XY
[tex]X(X'+ Y) \\=XX'+XY\\=0+XY\\=XY[/tex]
(b) X + XY = X
[tex]X + XY \\=X(1+Y)\\=XY[/tex]
(c) XY + XY′ = X
[tex]XY + XY'\\=X(Y+Y')\\=X(1)\\=X[/tex]
(d) (A + B)(A + B′) = A
[tex](A + B)(A + B')\\=AA + AB' +BA + BB'\\=A + A(B+B')+0\\=A+A(1)\\=A[/tex]
I would be happy to provide step-by-step solutions to each of these theorems.
(a) Suppose we have X(X′ + Y) = XY. If we were to simplify using the law of distribution, we would have X(X′) + XY. However, since X′ is the complement of X or the negation of X, X(X′) becomes zero and the equation simplifies to 0 + XY which is XY. If we equate the two expressions X(X′ + Y) and XY, they are not equivalent.
(b) Now we have X + XY = X. On the left side, the expression can be rewritten as X(1 + Y). But observing it closely, the factor of X is multiplied by the bracket. However, bringing X out in front results in an equation that isn't equivalent to X.
(c) Then we move on to XY + XY′ = X. First, we attempt to simplify the left side of the equation using the law of distribution. It becomes X(Y + Y′). It is important to remember that Y + Y′ equates to 1 based on Boolean algebra. Hence, our equation simplifies to X(1), or simply "X". But considering the fact that Y' means the complement or the negation of Y, XY' would be zero (because anything multiplied by zero is zero). Therefore, the equation is not equivalent to X.
(d) Finally, (A + B)(A + B′) = A. The left-hand side of the equation can be expanded using distributive law into AA′ + AB + AB' + BB′. Looking at it closely, AA′ and BB′ become zero because of the property that states an element and its complement's product is always zero. Coming to AB and AB', since A' is the complement of A, AB + AB' simplifies to 0 + 0 which is equal to 0. But 0 doesn't equate to A making the equation non-equivalent.
In summary, all four theorems failed to prove equivalent under Boolean algebra rules.
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How many anagrams can be created from the word 'masslessness' if the new words do not need to be meaningful?
Answer: 332640 anagrams
Step-by-step explanation:
Given the word:
'masslessness'
Total number of letters= 12
Number of S = 6
E =2
The rest are one each.
The number of possible arrangements can be written as;
N = 12!/(6!2!)
N = 332640
The number of anagrams that can be created from the word 'masslessness' is 239500800, calculated using the formula for permutations with repetition.
Explanation:This question is an application of the concept of permutations from combinatorics in mathematics. In this case, the word 'masslessness' consists of 12 characters including 3 's', 2 's', and 2 'l'. The formula for calculating permutations when there are repeating members within a set is n!/(r1! * r2! * ... * rn!), where n is the total number of members and r is the number of repeating members. Applying this formula, the number of anagrams of 'masslessness' would be 12!/(3! * 2! * 2!) = 239500800.
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If the sun is 68° above the horizon, find the length of the shadow cast by a building 88 ft tall. Round your answer to the nearest tenth.
Answer:
Length of shadow = 35.56 ft
Step-by-step explanation:
∠ BAC = 68°
Height of building = BC = 88 ft
Length of shadow = AC = ?
[tex]tan(68^{o})=\frac{BC}{AC}\\\\tan(68^{o})=\frac{88}{b}\\\\b=\frac{88}{tan(68^{o})}\\\\b=\frac{88}{2.475}\\\\b=35.56 \,ft[/tex]
The Length of shadow is 35.56 ft.
Given that,
If the sun is 68° above the horizon.
Based on the above information, the calculation is as follows:
[tex]tan 68 ^{\circ} = BC \div AC\\\\tan 68 ^{\circ} = 88 \div b\\\\b = 88 \div tan 68 ^{\circ} \\\\= 88 \div 2.475[/tex]
= 35.56 ft.
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Which level of measurement consists of categories only where data cannot be arranged in an ordering scheme?
Answer:
Nominal level of measurement is the one where data cannot be arranged in an ordering scheme.
Explanation:
This Nominal measurement level is generally characterized by various data that consist of categories of various types,labels of various kinds and also names.
The data that are involved in this type of measurement usually cannot be arranged in an orderly manner or in an ordering scheme.
One of the best example of these type of measurement includes survey responses such as choosing between Yes or No or sometimes undecided.
Thus we can consider nominal level measurement as the answer.
Answer:
nominal
Step-by-step explanation:
Convert 7 kl 345 L to liters. (Enter a number as an exact integer or decimal.)
Answer:
The answer to your question is 7345 l
Step-by-step explanation:
Data
Convert 7kl to 345 L
Process
- To solve this question we must know the equivalence of kl to L.
1 kl ----------------- 1000 l
- Now, use proportions and cross multiplication to solve it
1 kl ---------------- 1000 l
7 kl---------------- x l
x = (7 x 1000) / 1
x = 7000 / 1
x = 7000 l
- Calculate the result
7000 + 345 = 7345 l
Based on a poll, a newspaper reported that between 52% and 68% of voters would be likely to vote for a school bond issue. What is the margin of error of the poll? A) 8% B) 1096 C) 34% D) 26%
Answer:
A) 8%
Step-by-step explanation:
A confidence interval of proportions has an upper end and a lower end.
The margin of error is the difference between these points divided by two. Also, it is the upper end subtracted by the estimated proportion, or the estimated proportion subtracted by the lower end.
In this problem, we have that:
Upper end: 0.68
Lower end: 0.52
Margin of error
[tex]M = \frac{0.68 - 0.52}{2} = 0.08[/tex]
The margin of error is 8%.
So the correct answer is:
A) 8%
The histogram below displays the distribution of 50 ages at death due to trauma (unnatural accidents and homicides) that were observed in a certain hospital during a week. Which of the following are the appropriate numerical measures to describe the center and spread of the above distribution?
a. The mean and the median
b. The IQR and the standard deviation
c. The mean and the standard deviation
d. The median and the IQR
Answer:
d. The median and the IQR
Step-by-step explanation:
The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median. This means that the median is the appropriate numerical measure to describe the center.
The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.
The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.
The quartiles are used to measure the spread of a data-set.
The interquartile range(IQR), is Q3-Q1.
So the correct answer is:
d. The median and the IQR
Appropriate numerical measures of centre and spread of the distribution are : D) Median and IQR respectively.
Median is the positional average showing the mid point of a data. Median divides the data into two equal halves, both containing 50% of the data.
Quartiles are also a positional measure, dividing the data into 4 equal quarters, Q1 & Q3 covering 25% & 75% of the data respectively.
Interquartile Range shows the difference between Quartile 1 & Quartile 3. It is a measure of dispersion of data, denoting the spread & distribution of data.
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Let t^2y''+10ty'+8y=0Find all values of r such that y = tr satisfies the differential equation for t > tr0. If there is more than one correct answer, enter your answers as a comma separated list.
Answer:
The correct question is
Let [tex] t^2y''+10ty'+8y=0 [/tex]. Find all values of r such that [tex] y = t^r [/tex] satisfies the differential equation for t > 0. If there is more than one correct answer, enter your answers as a comma separated list.
The answer is r = 8, r = 1
Step-by-step explanation:
By substituting [tex] y = t^r [/tex] into the given equation [tex] t^2y''+10ty'+8y=0 [/tex].
That is substituting [tex] y = t^r [/tex] , [tex] y' = rt^{r-1} [/tex], [tex] y'' = r(r-1)t^{r-2} [/tex] into the given equation, we have
[tex] t^2 r(r-1)t^{r-2} +10t rt^{r-1}+8 t^r =0 [/tex]
Implies that [tex] r(r-1)t^r +10rt^r + 8 t^r =0 [/tex]
Implies that [tex] [r(r-1) +10r + 8]t^r = 0[/tex]
Implies that [tex] [r^2 - r +10r + 8]t^r = 0[/tex]
Implies that [tex] [r^2 - 9r + 8]t^r = 0[/tex]
Implies that [tex] r^2 - 9r + 8 = 0[/tex], assuming [tex] t^r \ne 0[/tex] being the assumed non- trivial solution.
The by solving this quadratic equation [tex] r^2 - 9r + 8 = 0[/tex] using factorization method, we have
[tex] r^2[/tex] – r – 8r + 8 = 0
Implies that r(r – 1) – 8(r – 1) = 0
Implies that (r – 8)(r – 1) = 0
Implies that r – 8 = 0 or r – 1 = 0
Implies that r = 8 or r = 1
Therefore, the value of r that satisfies the given equation is r = 8, r = 1.
Final answer:
The values of r such that y = t^r satisfies the given differential equation are r = -8 and r = -1.
Explanation:
The student is tasked with finding values of r such that y = t^r satisfies the given differential equation t^2y'' + 10ty' + 8y = 0 for t > 0. To solve this, we will substitute y = tr into the equation and find the characteristic equation that the values of r must satisfy.
Using the proposed solution y = t^r, we calculate the first and second derivatives of y:
y' = rt^{r-1}
y'' = r(r-1)t^{r-2}
Substituting these derivatives back into the original differential equation, we get:
t^2(r(r-1)t^{r-2}) + 10t(rt^{r-1}) + 8t^r = 0
Simplifying, we find the characteristic equation:
r(r-1) + 10r + 8 = 0
Which factors to:
(r + 8)(r + 1) = 0
Thus, the possible values of r are:
r = -8
r = -1
Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 μg/l (micrograms per liter). The untrained noses of consumers may have a higher threshold, however.
Here are the DMS odor thresholds for 10 untrained students:
30 30 42 35 22 33 31 29 19 23
Assume that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 μg/l.
(a) Give a 95% CI for the mean DMS odor threshold among all students.
(b) Are you convinced that the mean odor threshold for students is higher than the published threshold, 25 μg/l? Carry out a significance test with an α = 0.05 significance level to justify your answer.
Answer:
a) [tex]29.4-1.96\frac{7}{\sqrt{10}}=25.06[/tex]
[tex]29.4+1.96\frac{7}{\sqrt{10}}=33.74[/tex]
So on this case the 95% confidence interval would be given by (25.06;33.74)
b) [tex]z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988[/tex]
[tex]p_v =P(z>1.988)=0.0234[/tex]
If we compare the p value and the significance level given [tex]\alpha=
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".
Part a
Data given: 30 30 42 35 22 33 31 29 19 23
We can calculate the sample mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}= 29.4[/tex] the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=7[/tex] represent the population standard deviation
n=10 represent the sample size
95% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]29.4-1.96\frac{7}{\sqrt{10}}=25.06[/tex]
[tex]29.4+1.96\frac{7}{\sqrt{10}}=33.74[/tex]
So on this case the 95% confidence interval would be given by (25.06;33.74)
Part b
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu \leq 25[/tex]
Alternative hypothesis :[tex]\mu>25[/tex]
Compute the test statistic
The statistic for this case is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-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]z=\frac{29,4-25}{\frac{7}{\sqrt{10}}}=1.988[/tex]
Give the appropriate conclusion for the test
Since is a one side right tailed test the p value would be:
[tex]p_v =P(z>1.988)=0.0234[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
The mean DMS odor threshold among all students is estimated to be between 29.959 μg/l and 34.441 μg/l with 95% confidence. A significance test shows that the mean odor threshold for students is higher than the published threshold of 25 μg/l.
Explanation:The mean DMS odor threshold among all students can be estimated using a confidence interval. In this case, since the sample size is large and the population standard deviation is known, we can use a normal distribution to construct the confidence interval. Calculating the 95% confidence interval, we find that the range is (29.959, 34.441) μg/l.
To determine if the mean odor threshold for students is higher than the published threshold of 25 μg/l, we can conduct a significance test. Using a one-sample t-test with a significance level of 0.05, we compare the sample mean (mean of the untrained students' thresholds) to the hypothesized mean of 25 μg/l. Calculating the test statistic and comparing it to the critical value, we find that the test statistic falls in the rejection region. Therefore, we can conclude that the mean odor threshold for students is indeed higher than the published threshold of 25 μg/l.
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A sample is a subset of all possible data values for a given subject under consideration.
(A) True
(B) False
Answer: True
Step-by-step explanation:
A subset is a term used in the sets topic In mathematics to describe a Part of A larger values being considered.
Sample In experimental procedures is known as Part of what is to be used to determine the reaction of a larger mass of specimen.
Taking these two definitions by side, it is quite evident that both of them represent taking part of a larger picture to form a smaller picture which does not totally deviate from the larger picture being considered.
Find the mean of the data summarized in the given frequency distribution. A company had 110 employees whose salaries are summarized in the frequency distribution below. Find the mean salary. Round to the nearest hundredth. Salery ($) Employees 5,001-10,000 27 10,001-15,000 23 15,001-20,000 12 20,001-25,000 18 25,001-30,000 30
Answer:
[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]
Step-by-step explanation:
For this case we can construct the following table in order to find the mean for the grouped data:
Interval Frequency (fi) Midpoint (Xi) Xi *fi
5001-10000 27 7500.5 202513.5
10001-15000 23 12500.5 287511.5
15001-20000 12 17500.5 210006
20001-25000 18 22500.5 405009
25001-30000 30 27500.5 825015
Total 110 1930055
And the mean is calculated with the following formula:
[tex] \bar X= \frac{\sum_{i=1}^n x_i f_i}{n}[/tex]
Where [tex] n = \sum_{i=1}^n f_i = 110[/tex]
So then if we replace into the formula we got:
[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]
Solve the separable differential equation:dx/dt= x^2+ (1/9) and find the particular solution satisfying the initial condition: x(0)=6
Final answer:
To solve the given separable differential equation, we separate the variables, integrate both sides, and then apply the initial condition to find the particular solution that fits x(0) = 6.
Explanation:
The question asks to solve the separable differential equation dx/dt = x^2 + (1/9) and find the particular solution that satisfies the initial condition x(0) = 6. To solve this, first, we separate the variables and integrate both sides. The separated form of the equation is dx / (x^2 + (1/9)) = dt. Integrating both sides, we utilize the integral form that applies to the left side, leading us to an arctan function after simplification. This results in the solution to the differential equation. Subsequently, to find the particular solution satisfying the initial condition, we substitute x(0) = 6 into the obtained general solution and solve for the integration constant.
With x(0) = 6, we determine the value of the constant that makes the solution particular to the initial condition provided. By inserting this constant back into the general solution, we acquire the exact expression for x(t) that meets the initial condition specified.
Find the sumofthe geometrical progression of five terms, of which the first term is 7 and the multiplier is 7.Verify that the sum isthe product of 2801 and 7.
Answer:
The sum of first five term of GP is 19607.
Step-by-step explanation:
We are given the following in the question:
A geometric progression with 7 as the first term and 7 as the common ration.
[tex]a, ar, ar^2,...\\a = 7\\r = 7[/tex]
[tex]7, 7^2, 7^3, 7^4...[/tex]
Sum of n terms in a geometric progression:
[tex]S_n = \displaystyle\frac{a(r^n - 1)}{(r-1)}[/tex]
For sum of five terms, we put n= 5, a = 7, r = 7
[tex]S_5 = \displaystyle\frac{7(7^5 - 1)}{(7-1)}\\\\S_5 = 19607[/tex]
The sum of first five term of GP is 19607.
Verification:
[tex]2801\times 7 = 19607[/tex]
Thus, the sum is equal to product of 2801 and 7.
To find the sum of a geometric progression with a common ratio of 7, we use the formula Sum = (first term) * (1 - (common ratio)^n) / (1 - common ratio). When we substitute the values given in the question, the sum comes out to be the product of 2801 and 7.
Explanation:A geometric progression is a sequence in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In this case, the common ratio is 7.
The sum of a geometric progression can be found using the formula:
Sum = (first term) * (1 - (common ratio)^n) / (1 - common ratio)
Plugging in the values for this problem, we have:
Sum = 7 * (1 - 7^5) / (1 - 7) = 7 * (-39304) / (-6) = 2801 * 7
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The sampling distribution for the variance estimator in ANOVA is chi-square regarless of the assumptions of the data
Answer:
False
Step-by-step explanation:
The mentioned statement is not true because the F sampling distribution is used for variance estimator in the Analysis of the variance. The Analysis of variance technique used to compare two different variance estimate to check the equality of population mean. The chi-square statistic involve the comparison of variance with specified variance whereas F-statistic is calculated by taking the ratio of two variances.
The sampling distribution for the variance estimator in Analysis of Variance (ANOVA) is Chi-square if key assumptions such as random selection, equal standard deviations, and normally distributed populations are met. If not, further statistical analysis is required.
Explanation:In Analysis of Variance (ANOVA), the sampling distribution for the variance estimator is indeed Chi-square, but only if the data meets certain assumptions. These assumptions are:
If these assumptions are met, we can proceed to calculate the F ratio, which is our test statistic for Analysis of Variance (ANOVA). However, if any of these assumptions aren't met, further statistical testing is required. Even in such cases, it's important to know the distribution of sample means, the sums, and the F Distribution to make accurate statistical conclusions.
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A parcel of land is 8 ft longer than it is wide. Each diagonal from one corner to the opposite corner is 232 ft long.
What are the dimensions of the parcel?
width?
length?
The dimensions of the parcel are 80 ft (width) and 88 ft (length).
Explanation:To find the dimensions of the parcel, we can set up a system of equations using the given information. Let's denote the width of the parcel as x ft. Since the parcel is 8 ft longer than it is wide, the length can be represented as x + 8 ft. We can use the Pythagorean theorem to find the length of the diagonal: x^2 + (x + 8)^2 = 232^2. Solving this equation will give us the width and length of the parcel.
Expanding the equation, we get: x^2 + (x^2 + 16x + 64) = 53824. Combining like terms, we have: 2x^2 + 16x + 64 = 53824. Rearranging the equation, we have a quadratic equation: 2x^2 + 16x + (-53760) = 0. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.
By factoring, we find that the dimensions of the parcel are 80 ft (width) and 88 ft (length).
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A standard number cube numbered one through six on each side is rolled five times. What is the probability of landing on a three or four on all five rules?
Answer:
1/243 ≈ 0.004 (which is a low probability)Explanation:
The event landing on a three or four on a rolling is independent of the event of landing on a three or four on any other rolling.
Thus, for independent events you calculate the joint probability as the product of the probabilities of each event:
P(A and B and C and D and E) = P(A) × P(B) × P(C) × P(D) × P(E)Here, P(A) = P(B) = P(C) = P(D) = P(E) = probability of landing on a three or four.
Now to calculate the probability of landing on a three or four, you follow this reasoning:
Positive outcomes = three or fourNumber of positive outcomes = 2 Possible outcomes = one, two, three, four, five, or sixNumber of possible outcomes = 6Probability = numer of positive outcomes / numer of possible outcomesProbability = 2/6 = 1/3Finally, probability of landing on a three or four on all five rolls: (1/3)⁵ = 1/243 ≈ 0.004 ← answer
The sales of digital cameras (in millions of units) in year t is given by the function
f(t) = 3.06t + 6.84 (0 ≤ t ≤ 3)
where t = 0 corresponds to the end of 2001. Over that same period, the sales of film cameras (in millions of units) is given by
g(t) = −1.85t + 16.48 (0 ≤ t ≤ 3).
(a) Show that more film cameras than digital cameras were sold in 2001.
digital cameras
million
film cameras
million
(b) When did the sales of digital cameras first exceed those of film cameras?
Explanation on determining which type of cameras sold more in 2001 and when digital camera sales surpassed film camera sales.
Explanation:(a) To show that more film cameras than digital cameras were sold in 2001, we need to substitute t = 0 into both functions:
For digital cameras: f(0) = 3.06(0) + 6.84 = 6.84 million units
For film cameras: g(0) = -1.85(0) + 16.48 = 16.48 million units
Therefore, more film cameras (16.48 million units) were sold in 2001 than digital cameras (6.84 million units).
(b) To find when digital camera sales first exceeded film camera sales, we need to set the two functions equal to each other and solve for t:
3.06t + 6.84 = -1.85t + 16.48
4.91t = 9.64
t ≈ 1.96 years, which is approximately at the end of 2003.
A dog groomer is offering 20% off a full groom for one dog, and 15% off for each additional dog. You decide to take your two furry friends to be groomed. Normally, a full groom costs $65 for each dog. Assuming there is no sales tax, what is the total discounted price for two dogs to be groomed? Report your answer to two decimal places. Do not include the dollar sign, $, in the answer box below.
Answer:the total discounted price for two dogs to be groomed is 22.75
Step-by-step explanation:
A dog groomer is offering 20% off a full groom for one dog, and 15% off for each additional dog. Normally, a full groom costs $65 for each dog.
If you decide to take your two furry friends to be groomed,
the discount on the first one would be
20/100 × 65 = 0.2 × 65 = 13
the discount on the second one would be
10/100 × 65 = 0.15 × 65 = 9.75
the total discounted price for two dogs to be groomed would be
13 + 9.75 = 22.75
Answer:
First, calculate the discounted price for the first dog,
Discount=(20/100) $65=$13.
The discounted price is $65−$13=$52. Now, we can calculate the discounted price for the second dog,
Discount=(15/100) $65=$9.75.
The discounted price for the second dog is $65−$9.75=$55.25. The total discounted price for both dogs is,
$52+$55.25=$107.25
Step-by-step explanation:
The function f is differentiable and ∫x0(3f(t)+5t)dt=sin(x). Determine the value of f′(1π).
Answer:
f'(π) = (-5/3)
Step-by-step explanation:
for the equation
∫₀ˣ (3f(t)+5t)dt=sin(x)
3*∫₀ˣ f(t) dt+5*∫₀ˣ t*dt=sin(x)
then
∫₀ˣ t*dt = (t²/2) |₀ˣ = x²/2 - 0²/2 = x²/2
thus
3*∫₀ˣ f(t) dt + 5* x²/2 = sin(x)
∫₀ˣ f(t) dt = 1/3*sin(x) - 5/6*x²
then applying differentiation
d/dx ( ∫₀ˣ f(t) dt) = f(x) - f(0) ( from the fundamental theorem of calculus)
d/dx (1/3*sin(x) - 5/6*x²) = 1/3*cos(x) - 5/3*x
therefore
f(x) - f(0) = 1/3*cos(x) - 5/3*x
f(x) = 1/3*cos(x) - 5/3*x + f(0)
applying differentiation again (f'(x) =df(x)/dx)
f'(x) = -1/3*sin(x) - 5/3
then
f'(π) = -1/3*sin(π) - 5/3 = 0 - 5/3 = -5/3
f'(π) = (-5/3)
Given expression is:
→ [tex]\int x_0 (3f(t)+5t) dt = sin(x)[/tex]
or,
→ [tex]3\times \int x_0 f(t) dt+5\times \intx_0 t\times dt = sin(x)[/tex]
then,
→ [tex]\int x_0 t\times dt = (\frac{t^2}{2} ) x_0[/tex]
[tex]\frac{x^2}{2} -\frac{0^2}{2} = \frac{x^2}{2}[/tex]
thus,
→ [tex]3\times \int x_0 f(t) +5\times \frac{x^2}{2} = sin(x)[/tex]
[tex]\int x_0 f(t) dt = \frac{1}{3}\times sin(x) - \frac{5}{6}\times x^2[/tex]
By applying differentiation, we get
→ [tex]\frac{d}{dx} (\int x_0 f(t) dt) = f(x) - f(0)[/tex]
[tex]\frac{d}{dx}(\frac{1}{3} sin(x)) - \frac{5}{6}x^2= \frac{1}{3} cos(x) - \frac{5}{3} x[/tex]
Therefore,
→ [tex]f(x) -f(0) = \frac{1}{3}cos (x) - \frac{5}{3} x[/tex]
[tex]f(x) = \frac{1}{3}cos (x) -\frac{5}{3} x+f(0)[/tex]
By differentiating again, we get
→ [tex]f'(x) = -\frac{1}{3}sin(\pi) -\frac{5}{3}[/tex]
[tex]= 0-\frac{5}{3}[/tex]
[tex]= -\frac{5}{3}[/tex]
Thus the response above is correct.
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(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0.1 ≤ P(A ∩ B) ≤ 0.
Answer with Step-by-step explanation:
We are given that
P(A)=0.4 and P(B)=0.7
We know that
[tex]P(A)+P(B)+P(A\cap B)=P(A\cup B)[/tex]
We know that
Maximum value of [tex]P(A\cup B)[/tex]=1 and minimum value of [tex]P(A\cup B)[/tex]=0
[tex]0\leq P(A\cup B )\leq 1[/tex]
[tex]0\leq P(A)+P(B)-P(A\cap B)\leq 1[/tex]
[tex]0\leq 0.4+0.7-P(A\cap B)\leq 1[/tex]
[tex]0\leq 1.1-P(A\cap B)\leq 1[/tex]
[tex]0\leq 1.1-P(A\cap B)[/tex]
[tex]P(A\cap B)\leq 1.1[/tex]
It is not possible that [tex]P(A\cap B)[/tex] is equal to 1.1
[tex]1.1-P(A\cap B)\leq 1[/tex]
[tex]-P(A\cap B)\leq 1-1.1=-0.1[/tex]
Multiply by (-1) on both sides
[tex]P(A\cap B)\geq 0.1[/tex]
Again, [tex]P(A\cup B)\geq P(B)[/tex]
[tex]0.4+0.7-P(A\cap B)\geq 0.7[/tex]
[tex]1.1-P(A\cap B)\geq 0.7[/tex]
[tex]-P(A\cap B)\geq -1.1+0.7=-0.4[/tex]
Multiply by (-1) on both sides
[tex]P(A\cap B)\leq 0.4[/tex]
Hence, [tex]0.1\leq P(A\cap B)\leq 0.4[/tex]
Find values for θ that make each statement true (show steps)
1) sinθ = cos(25)
θ =
2) sin(θ/3 + 10 ) = cos θ
θ =
Answer:
Step-by-step explanation:
1. sinФ = cos 25
25 ° is in the between 0 and 90°
therefore it can simply represent
cosФ= sin (90-Ф) = sin (90 - 25) = sin 65
2. sin(Ф/3 + 10) = cos Ф
cos Ф = sin (90 -Ф)
sin(Ф/3 + 10) = cos Ф
sin(Ф/3 + 10) = cos Ф = sin(90-Ф)
Ф/3+10=90-Ф
10Ф+3/30 = 90-Ф
10Ф+3 = 30(90-Ф)
10Ф+3 = 2700-30Ф
10Ф+30Ф=2700-3
40Ф = 2697
Ф = 2697 / 40 = 67.425 ≅ 67.4°
A headline in a major newspaper read, "Breast-fed youth found to do better in school."a. Do you thinkthis statement was based on an observational study or a randomized experiment? Explain.b. Given your answer in part (a), which of these two alternative headlines do you think would be preferable: "Breast-feeding leads to better school performance" or "Link found between breast-feeding and school performance"? Explain.
Answer:
Randomized experiment
Step-by-step explanation:
Such a statement is said to be Randomized experiment as the kids in the school have been selected at random and their performances are observed and recorded. Later it is inquired if they are breast fed or not. This information is later assessed and a conclusion is reached. This type of research or experiment is Randomized experiment.
Considering that the experiment is Randomized, the statement should be "Link found between breast-feeding and school performances". This statement does not restrict the performances of the student specifically on breast-feeding but shows a probable affect of it.
A wireless service provider has a support center customers can call to get questions answered about their cell phone accounts. The manager in charge of the support center has recently conducted a study in which she surveyed 2,300 customers. The Customers who called the support center were transferred to a third party who asked the customers a series of questions.
a. Indicate whether the data generated from this study will be considered cross-sectional or time series. Explain Why.
b. One of the questions asked customers was approximately how many minutes they had been on hold waiting to get through to a support person. What level of data measurement is obtained from this question? Explain.
c. Another question asked the customer to rate the service on a scale of 1-7, with 1 being the worst possible service and 7 being the best possible service. Which level of data measurement is achieved from this question? Will the data be quantitative or qualitative? Explain.
Answer:
A) The data generated is a cross-sectional data because time is not a factor .Secondly. 2,300(variables) people are selected at a particular time.
B) It is a ratio level of measurement because zero has meaning and it is clearly defined that is if the customer did not wait and was attended to immediately then the value is zero. Secondly there is a greater than and less than relationship that exits between the data for example if a customer wait for 20 minutes and another customer waits for 25 minutes, 25 is greater than 20. Hence the greater than and less than relation between data.
C)The level of measurement achieve from the data collected from this question is ordinal and the data is a qualitative. It’s ordinal because the relationship between the data set is focused on the order at which the data value are arranged. Secondly it’s a qualitative data set in the data value represents the difference in quality and not necessarily how much the difference is.
Step-by-step explanation:
Step One: Definition Of Cross-sectional data,
This can be defined as a type of data that is been collected by noting or observing many variables at a given period of time
Step Two:Definition Of Time series data
This can be defined as a type of that that is been collected by noting or observing a particular variable at different points of time. They are usually collected at fixed intervals
Step Three: Difference between cross-sectional data and time series data
The main difference between time series data and cross-sectional data is that for time series data the focus of observation for collecting the data s on the same variable over a period of time while for cross sectional data the focus is on the different variable at the same point in time
Step Four: Definition Of Level of measurement
This defines the relationship that exits among data values. The of level of measurement includes nominal, ordinal, interval, ratio
Step Five:Definition Of Ratio level of measurement
The ratio level of measurement or the ratio scale gives a description of the order and the exact difference that exit between data values, it also has a clear definition for zero value.
Step Six:Definition Of Ordinal level of measurement
For this level of measurement the main focus is the order of the data values, but the exact difference between each data values is not really specified looking at our question part c we know that a #7 is better than a #6 but we cannot say how much better it is.
The data from the wireless service provider's survey is cross-sectional, as it captures data at a single point in time. The question about hold time collects quantitative continuous data, while the service rating question provides ordinal data, which is qualitative.
The data generated from the study conducted by the wireless service provider's support center would be considered cross-sectional because it captures data at a single point in time rather than over multiple periods. Cross-sectional data is collected by observing many subjects at the same point in time or without regard to differences in time.
The level of data measurement obtained from the question asking customers about how long they had been on hold is quantitatively continuous. This is because the time spent on hold can be measured in units (such as minutes) that can take on any value within a given range, including fractions of a minute.
Regarding the customer service rating on a scale of 1-7, the level of data measurement is ordinal. This type of data is qualitative, as it categorizes data into different levels with a meaningful order, but the differences between the levels are not mathematically uniform or meaningful. It captures the order of the ratings but does not quantify the difference between each level.
Which of the following statements properly define a variable? Select all that apply. a. Let w represent Enrique's weight b. Let d represent Enrique's distance (in miles). c. Let d represent Enrique's distance from home (in miles). d. Let d represent Enrique's distance from home. e. Let w represent Enrique's weight in pounds
Answer: c. Let d represent Enrique's distance from home (in miles).
e. Let w represent Enrique's weight in pounds
Step-by-step explanation:
A variable is a particular type of value that denotes an unknown quantity.
It is denotes by using alphabets.It can change.To define a variable , we need define a amount or quantity with units and in case of length an exact position is required.
a. Let w represent Enrique's weight
It is not a variable because no units is defined.
b. Let d represent Enrique's distance (in miles).
It is not a variable because the points to establish distance traveled by Enrique is not given .
c. Let d represent Enrique's distance from home (in miles).
It is a variable as both units and position as mentioned.
d. Let d represent Enrique's distance from home.
It is not a variable because no units is defined.
e. Let w represent Enrique's weight in pounds.
It is a variable because unit of weight is mentioned.
Hence, the correct statements that properly define a variable are :
c. Let d represent Enrique's distance from home (in miles).
e. Let w represent Enrique's weight in pounds
Suppose that the probability that a person is killed by lightning, in a year is, independently of other people, 1/(500 million). Assume that the U.S. population is 300 million. Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places. Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.
Answer:
a) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
The individual pprobabilities are:
[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]
[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]
[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]
And if we replace we got:
[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]
b) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
And we can calculate this with the following excel formula:
[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]
And we got:
[tex] P(X\geq 3) = 0.023115[/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=300000000, p=\frac{1}{500000000})[/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]
(a) Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places.
For this case we want this probability, we are going to use the complement rule:
[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
The individual probabilities are:
[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]
[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]
[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]
And if we replace we got:
[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]
(b) Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.
For this case we want this probability, we are going to use the complement rule:
[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
And we can calculate this with the following excel formula:
[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]
And we got:
[tex] P(X\geq 3) = 0.023115[/tex]
The standard deviation of the sampling distribution of the mean is called the:
a. standard error of the mean
b. standard error of variability
c. standard error of the estimate
d. standard error of the sample
Answer:
a. Standard error of mean
Step-by-step explanation:
The standard error of mean is computed by dividing the standard deviation to the square root of sample size. It can be represented as
[tex]Standard error=\frac{standard deviation}{\sqrt{n}}[/tex]
The standard deviation of the sampling distribution is known as the standard error of mean because it measures the accuracy of sample mean as compared to population mean
The standard deviation of the sampling distribution of the mean is referred to as the Standard Error of the Mean, which measures the dispersion of sample means around the true population mean.
Explanation:The standard deviation of the sampling distribution of the mean is known as the Standard Error of the Mean. When we say sampling distribution of the mean, we're referring to the distribution of means for all possible samples of a given size from the same population. The Standard Error of the Mean measures the dispersion of these sample means around the true population mean.
For example, if you have a population of test scores with a mean of 70 and a standard deviation of 10, and you were to take many samples of 30 scores from this population, the standard error of the mean would be the standard deviation of all those sample means.
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