Candy can arrange in 1, 2, 3, 5, 6, 10, 15 ways
2, 3, 4, 6, 8, 12 number of people can sit at each table
Solution:
Candy brought 30 cookies to a party. List all the ways she can arrange them on a plate in equal rows.
Given that Candy brought 30 cookies to party
She has to arrange the ways she can arrange them on plate in equal rows
We have to list the factors of 30
factors of 30 = 1, 2, 3, 5, 6, 10, 15
So she can arrange in 8 ways
Possible arrangement are 1, 2, 3, 5, 6, 10, 15 rows
David is setting up tables for 24 people at the party. The same number of people will sit at each table, and no one will sit alone. How many people can sit at each table? List all possibilities.
Fcators of 24 = 1, 2, 3, 4, 6, 8, 12
Here we can eliminate 1 since given that no one will sit alone
Thus 2, 3, 4, 6, 8, 12 number of people can sit at each table
Algebra unit 5 test part 2 radical and rational expressions
The subject of the question is Mathematics, relevant to High School students. It involves simplifying and combining radicals, eliminating terms to simplify algebraic expressions, and understanding transcendental numbers and their properties.
Explanation:Add and Subtract Radicals
When working with radicals, it is essential to simplify each radical by removing perfect square roots in order to combine like radicals effectively. For instance, to add \\(\\sqrt{18} + \\sqrt{8}\\), we must first simplify. \\(\\sqrt{18}\\) becomes \\(\\sqrt{9*2}\\) or \\(\\sqrt{9}\\cdot\\sqrt{2}\\), which simplifies to \\(\\sqrt{2}\\) times 3. Similarly, \\(\\sqrt{8}\\) can be rewritten as \\(\\sqrt{4*2}\\) or \\(\\sqrt{4}\\cdot\\sqrt{2}\\), which simplifies to \\(\\sqrt{2}\\) times 2. Now we have like radicals and can combine them: 3\\(\\sqrt{2}\\) + 2\\(\\sqrt{2}\\) equals 5\\(\\sqrt{2}\\).
Simplify Algebra and Reasonableness
To simplify the algebra, we identify and eliminate terms where possible, checking for reasonableness of the answer at all times. Verify if the operations you have performed are correct and the solution looks reasonable given the original equation or expression.
Transcendental Numbers and Functions
Transcendental numbers, like \\(\\sqrt{2}\\) or \\(\\sqrt{5}\\), are those that are not the root of any non-zero polynomial equation with rational coefficients. In algebra, these play a vital role in understanding real numbers and theorems related to them.
find x from the picture
Answer: x = 120 degrees
Step-by-step explanation:
The diagram is that of a polygon with 5 sides. This means that it is a Pentagon. The sum of the interior angles in a polygon is expressed as
180(n -2)
Where n represents the number of sides that the polygon has.
Since the given polygon has 5 sides, then the sum of the interior angles would be
180(5 - 2) = 180 × 3 = 540 degrees.
Therefore,
x + x + x + 90 + 90 = 540
3x + 180 = 540
3x = 540 - 180 = 360
x = 360/3 = 120 degrees
Explain how to solve 3^(x − 4) = 6 using the change of base formula log base b of y equals log y over log b. Include the solution for x in your answer. Round your answer to the nearest thousandth.
Answer:
x = 4 + (log 6 / log 3)
x ≈ 5.631
Step-by-step explanation:
3^(x − 4) = 6
Take log base 3 of both sides.
log₃ 3^(x − 4) = log₃ 6
x − 4 = log₃ 6
Use change of base formula.
x − 4 = log 6 / log 3
Solve for x.
x = 4 + (log 6 / log 3)
x ≈ 5.631
Answer:
5.631
Step-by-step explanation:
Using the change of base formula log base b of y equals log y over log b
Log y (base b) = log y /log b
3^(x − 4) = 6
Taking the log of both sides
log 3^(x − 4) = log 6
using the logarithm law that states that
log a ^ x = x log a
x - 4 log 3 = log 6
x - 4 = log 6 / log 3
x - 4 = 1.630929754
x = 5.630929754
≈ 5.631
The probabilities that two students will not show up for class on a beautiful spring day are 0.3 and 0.4, respectively. It is a beautiful spring day. Find the following probabilities. (See Example 6. Round your answers to two decimal places.) (a) Neither will show up for class. (b) Both will show up for class. (c) Exactly one will show up for class.
Answer:
(a) 0.12
(b) 0.42
(c) 0.46
Step-by-step explanation:
Probability of student A not showing up = 0.3
Probability of student B not showing up = 0.4
(a) Neither will show up for class
[tex]P(A\ and\ B) = 0.3*0.4 = 0.12[/tex]
(b) Both will show up for class
[tex]P (A\ nor\ B) = (1-0.3)*(1-0.4) = 0.42[/tex]
(c) Exactly one will show up for class
[tex]P(A\ or\ B) = 1 -P(A\ and\ B) - P(A\ nor\ B)\\P(A\ or\ B) = 1 -0.12-0.42 = 0.46[/tex]
The weekly salary paid to employees of a small company that supplies part-time laborers averages $750 with a standard deviation of $450. (a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week. (b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week? (c) If every employee receives a 5% salary increase for the next year, how does the normal model change? (d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate? (a) If the weekly salaries are normally distributed, the fraction of employees that make more than $300 per week is approximately nothing. (Type an integer or a fraction.)
Answer:
(a) The fraction of employees is 0.84.
(b)
[tex]\mu=850\\\\\sigma=450[/tex]
(c)
[tex]\mu=787.5\\\\\sigma=472.5[/tex]
(d) No. The left part of the distribution would be truncated too much.
Step-by-step explanation:
(a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week.
We have to calculate the z-value and compute the probability
[tex]z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84[/tex]
(b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week?
The mean of the salaries grows $100.
[tex]\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850[/tex]
The standard deviation stays the same ($450)
[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2} } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2} }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2} } =\sigma[/tex]
(c) If every employee receives a 5% salary increase for the next year, how does the normal model change?
The increases means a salary X is multiplied by 1.05 (1.05X)
The mean of the salaries grows 5%, to $787.5.
[tex]\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5[/tex]
The standard deviation increases by a 5% ($472.5)
[tex]\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2} } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2} }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5[/tex]
(d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate?
No. The left part of the distribution would be truncated too much.
Normal distribution has its mean, median and mode coincident on single point. The solutions to the given problems are specified as:
a) P(X > 300) = 0.8413b) The normal model shifts to the right, 100 units, but its structure stays same.c) If salary is increased 5%, then the normal model gets scaled .d) If lowest salary = $300, and median = $525, then it isn't a normal model as median ≠ mean = $750. It is going to be negatively skewed.How to get the z scores?If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For the given case, if we take X = salary of employees weekly of the considered company, then:
[tex]X \sim N(750, 450)[/tex]
The fraction of employees that make more than $300 per week is
P(X > 300).
Using the standard normal distribution, we can estimate this fraction of employees that make more than $300 per week as:
[tex]P(X > 300 ) = 1 - P(X \leq 300) = 1 - P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{300 - 750}{450} =-1)\\ \\P(X > 300) = 1 - P(Z \leq -1)\\[/tex]
Using the z-table, the p-value for Z = -1 is: 0.1587
Thus, [tex]P(X > 300) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413[/tex]
When we add $100 to each value of X, it doesn't change the structure of the graph of normal distribution.
When we increase salary by 5%, it means, new salary [tex]Y = X + 5\% \text{of X} = \dfrac{21X}{20}[/tex]
The random variable is scaled by 21/20 = 1.05, so the graph will stay on same origin but it will be stretched a bit in thickness and height.
For the 4th case(d), the median is specified to be $525, but as it was known to us that mean is $750, so mean and median aren't coinciding, and specially, the median is in left of mean, showing that the graph is leaning on right side(negatively skewed). (we deduce it when median < mean, as median shows that mid value is reached, but mean shows that probability is still not reached, so its being late, and reaches later, showing that there is tail in the left of the graph, so being negatively skewed).
Thus, The solutions to the given problems are specified as:
a) P(X > 300) = 0.8413b) The normal model shifts to the right, 100 units, but its structure stays same.c) If salary is increased 5%, then the normal model gets scaled .d) If lowest salary = $300, and median = $525, then it isn't a normal model as median ≠ mean = $750. It is going to be negatively skewed.Learn more about standard normal distribution here:
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An analysis of the grades on the first test in History 101 revealed that they approximate a normal curve with a mean of 75 and a standard deviation of 8. The instructor wants to award the grade of A to the upper 10% of the test grades. To the nearest percent, what is the dividing point between an A and a B grade?
Select one:
a. 80
b. 85
c. 90
d. 95
Answer:
b. 85
Step-by-step explanation:
Average grade (μ) = 75
Standard deviation (σ) = 8
Assuming a normal distribution, the z-score corresponding to the upper 10% of the test grades is z = 1.28.
The minimum grade 'X' within the top 10% is given by:
[tex]z=\frac{X-\mu}{\sigma}\\1.28=\frac{X-75}{8}\\X=85.24[/tex]
Rounding to the nearest percent, the dividing point between an A and a B grade is 85.
Final answer:
The dividing point between an A and B grade for the upper 10% of the test grades is calculated using the normal distribution properties. A Z-score of 1.28 corresponds to the 90th percentile, resulting in a score of 85.24, which is rounded to 85 according to standard rounding rules.
Explanation:
To determine the dividing point between an A and B grade for the upper 10% of the test grades in History 101, we need to refer to the properties of the normal distribution. The mean score is given as 75 with a standard deviation of 8. We are looking for the score that corresponds to the 90th percentile since the instructor wants to award an A to the upper 10%. Z-scores allow us to translate percentile ranks into scores on a given normal distribution.
Using a standard normal distribution table, we find that a Z-score of approximately 1.28 corresponds to the 90th percentile. To find the actual score, we use the formula Score = Mean + (Z-score imes Standard Deviation). Plugging in the values:
Mean = 75
Standard Deviation = 8
Z-score for the 90th percentile = 1.28
Score = 75 + (1.28 imes 8) = 75 + 10.24 = 85.24
When applying the standard rules for rounding, we round 85.24 to the nearest whole number, which is 85. Therefore, the dividing point between an A and a B grade is an 85. Students need to score at this point or higher to be in the top 10% and receive an A grade.
Solve the inequality. Graph the solution. 4(n-3) -6>18
Answer:
n>9
Step-by-step explanation:
4(n-3)-6>18
4n-12-6>18
4n-18>18
4n>18+18
4n>36
n>36/4
n>9
Location is known to affect the number, of a particular item, sold by HEB Pantry. Two different locations, A and B, are selected on an experimental basis. Location A was observed for 18 days and location B was observed for 13 days. The number of the particular items sold per day was recorded for each location. On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4. Select a 99% confidence interval for the difference in the true means of items sold at location A and B. a) O [-1242,-7582]
b) O132.76, 45.24]
c)。8 1.76, 94.24]
d) 0-1 6.03,-3.97]
e)。[42.76, 55.24]
F. None of the above
Answer:
d) [-16.03,-3.97]
[tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].
Step-by-step explanation:
Notation and previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]n_A=18[/tex] represent the sample of A
[tex]n_B =13[/tex] represent the sample of B
[tex]\bar x_A =39[/tex] represent the mean sample for A
[tex]\bar x_B =49[/tex] represent the mean sample for B
[tex]s_A =8[/tex] represent the sample deviation for A
[tex]s_B =4[/tex] represent the sample deviation for B
[tex]\alpha=0.01[/tex] represent the significance level
Confidence =99% or 0.99
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex] (1)
The point of estimate for [tex]\mu_A -\mu_B[/tex] is just given by:
[tex]\bar X_A -\bar X_B =39-49=-10[/tex]
The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2=18+13-2=29[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that [tex]t_{\alpha/2}=2.756[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}[/tex]
And replacing we have:
[tex]SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03[/tex]
[tex]-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97[/tex]
So on this case the 99% confidence interval for the differences of means would be given by [tex]-16.03 \leq \mu_A -\mu_B \leq -3.97[/tex].
d) [-16.03,-3.97]
The 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], therefore the correct answer is F. None of the above.
Explanation:This question is about computing a confidence interval for the difference of two sample means. The formula for the 99% confidence interval for the difference between two means is:
(X1 - X2) ± Z * sqrt [s1^2/n1 + s2^2/n2]
Where X1 and X2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the Z-score for the desired confidence level. For a 99% confidence level, the Z-score is approximately 2.576. We plug the given values into the equation to calculate:
(39 - 49) ± 2.576 * sqrt [(8^2 / 18) + (4^2 / 13)] => -10 ± 2.576 * sqrt [3.56 + 1.23] => -10 ± 2.576 * sqrt [4.79] => -10 ± 2.576 * 2.19 => -10 ± 5.64
This means the 99% confidence interval for the difference in the true means of items sold at location A and B is [-15.64, -4.36], which is not among the given options, so the correct answer is F. None of the above.
Learn more about Confidence Interval here:https://brainly.com/question/34700241
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Which of the following is the product of complex numbers below?
(3-2i)(5+i)
Answer: 17-7i
Step-by-step explanation:
Products produced by a machine has a 3% defective rate. (a) What is the probability that the first defective occurs in the fifth item inspected? (b) What is the probability that the first defective occurs in the first five inspections? (c) What is the expected number of inspections before the first defective occurs?
Answer:
Step-by-step explanation:
Given that the Products produced by a machine has a 3% defective rate.
Each product is independent of the other with a constant prob of being defective as 0.03
X - no of defects is binomial with p =0.03
a) the probability that the first defective occurs in the fifth item inspected
=Prob for first 4 non defective and 5th defective
=[tex](0.97)^4(0.03)^1\\=0.0266[/tex]
(b) the probability that the first defective occurs in the first five inspections
=P(X=1) in binomial with n=5
= 0.9915
c) the expected number of inspections before the first defective occurs
Expected defects in n trials = np
Expected number of inspection before the first defect = 1/p
= 33.33
=34
(c) What is the expected number of inspections before the first defective occurs?
Final answer:
The probability that the first defective product occurs on the fifth item is approximately 0.0895. The probability that a defective product occurs within the first five inspections is approximately 0.1426. The expected number of inspections before a defective product occurs is around 33 items.
Explanation:
To find the probability that the first defective occurs on the fifth item inspected (part a), we consider the first four items to be non-defective and the fifth one to be defective. The probability for one non-defective item is 97%, or 0.97. The probability we are looking for is thus the product of these probabilities:
P(non-defective, non-defective, non-defective, non-defective, defective) = (0.97)⁴* 0.03 ≈ 0.0895
For part b, we want to find the probability that a defective is found at any point in the first five inspections. This is the sum of the probabilities of finding the first defective on the first, second, third, fourth, or fifth inspection:
P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th) = 0.03 + (0.97 * 0.03) + (0.97² * 0.03) + (0.97³ * 0.03) + (0.97⁴ * 0.03) ≈ 0.1426
For part c, the expected number of inspections before finding the first defective item is given by the formula E(X) = 1/p, where p is the probability of finding a defective item. Substituting the given probability:
E(X) = 1/0.03 ≈ 33.33
Therefore, we expect to inspect approximately 33 items before finding the first defective.
Which of the following definitions describe functions from the domain to the codomain given? Which functions are one-to-one? Which functions are onto? Describe the inverse function for any bijective function.
a. f: ℤ → ℕ where f is defined by f(x) = x2 + 1
b. g: ℕ → ℚ where g is defined by g(x) = 1/x
c. h: ℤ x ℕ → ℚ where h is defined by h(z,n) = z/(n+1)
f: h: ℝ2 → ℝ2 where h is defined by h(x,y) = (y+1, x+1)
Answer:
a) f is a function. It is not 1-1, it is not onto.b) g is not a function. c) h is a function. It is not 1-1, it is onto.f) h is a function. It is a bijection, and h^-1(x,y)=(y-1,x-1)Step-by-step explanation:
a) For all x∈ℤ, the number f(x)=x²+1 exists and is unique because f(x) is defined using the operations addition (+) and multiplication (·) on ℤ. Then f is a function. f is not one-to-one: consider -1,1∈ℤ. -1≠1 but f(-1)=f(1)=2- Because two different elements in the domain have the same image under f, f is not 1-1. f is not onto: x²≥0 for all x∈ℤ then f(x)=x²+1≥1>0 for all x∈ℤ. Then 0∈ℕ but for all x∈ℤ f(x)≠0, which means that one element of the codomain doen't have a preimage, so f is not onto.
b) 0∈ℕ, so 0 is an element of the domain of g, but g(0)=1/0 is undefined, therefore g is not a function.
c) Let (z,n)∈ℤ x ℕ. The number h(z,n)=z·1/(n+1) is unique and it's always defined because n+1>0, then h is a function. h is not 1-1: consider the ordered pairs (1,2), (2,5). They are different elements of the domain, but h(2,5)=2/6=1/3=h(1,2). h is onto: any rational number q∈ℚ can be written as q=a/b for some integer a and positive integer b. Then (a,b-1)∈ ℤ x ℕ and h(a,b-1)=a/b=q.
f) For all (x,y)∈ℝ², the pair h(x,y)=(y+1,x+1) is defined and is unique, because the definition of y+1 and x+1 uses the addition operation on ℝ. f is 1-1; suppose that (x,y),(u,v)∈ℝ² are elements of the domain such that h(x,y)=h(u,v). Then (y+1,x+1)=(v+1,u+1), so by equality of ordered pairs y+1=v+1 and x+1=u+1. Thus x=u and y=x, therefore (x,y)=(u,v). f is onto; let (a,b)∈ℝ² be an element of the codomain. Then (b-1,a-1)∈ℝ² is an element of the domain an h(b-1,a-1)=(a-1+1,b-1+1)=(a,b). Because h is 1-1 and onto, then h is a bijection so h has a inverse h^-1 such that for all (x,y)∈ℝ² h(h^-1(x,y))=(x,y) and h^-1(h^(x,y))=(x,y). The previous proof of the surjectivity of h (h onto) suggests that we define h^-1(x,y)=(y-1,x-1). This is the inverse, because h(h^-1(x,y))=h(y-1,x-1)=(x,y) and h^-1(h^(x,y))=h^-1(y+1,x+1)=(x,y).
The function f: ℤ → ℕ is neither one-to-one nor onto, g: ℕ → ℚ is one-to-one but not onto, h: ℤ x ℕ → ℚ is neither one-to-one nor onto, and h: ℝ² → ℝ² is a bijective function with an inverse h⁻¹(u,v) = (v-1, u-1).
Explanation:
Let's examine each function individually to determine if they define a function from the domain to the codomain, and if so, whether they are one-to-one or onto, and describe the inverse function for any bijective function.
f: ℤ → ℕ where f is defined by f(x) = x2 + 1. This is indeed a function since each element in the domain ℤ has a unique image in the codomain ℕ. It is not one-to-one because both positive and negative integers will produce the same result when squared. However, it is not onto since no element in ℕ will map to 0, which is not attained by x2 + 1 for any integer x.
g: ℕ → ℚ where g is defined by g(x) = 1/x. This represents a function since each positive integer x will have a unique reciprocal in ℚ. This function is one-to-one, as no two different positive integers have the same reciprocal, but it is not onto because certain rational numbers, like 2/3, cannot be expressed as the reciprocal of a natural number.
h: ℤ x ℕ → ℚ where h is defined by h(z,n) = z/(n+1). This defines a function where each ordered pair of integers and natural numbers corresponds to a unique rational number. However, this function is neither one-to-one nor onto. It is not one-to-one because different integer pairs could result in the same rational number (e.g., h(2,1) = h(-2,-3) = 2/2), and it's not onto as some rational numbers cannot be obtained using this formula, such as 2/3.
h: ℝ2 → ℝ2 where h is defined by h(x,y) = (y+1, x+1). This is a function that maps pairs of real numbers to pairs of real numbers, it is both one-to-one and onto (therefore, bijective), as every pair (x,y) has a unique image and every possible pair (u,v) in ℝ2 is hit. The inverse function is given by h-1(u,v) = (v-1, u-1).
Consider the equivalence relation R = {( x, y) Ix-y is an integer}.
(a) What is the equivalence class of 1 for this equivalence relation?
(b) What is the equivalence class of 1/2 for this equivalence relation?
Final answer:
The equivalence class of 1 consists of all integers plus 1, and the equivalence class of 1/2 consists of all numbers of the form 1/2 plus any integer.
Explanation:
The equivalence relation R is defined such that (x, y) is in R if and only if x - y is an integer. For any real number a, the equivalence class of a is the set of all real numbers b such that a - b is an integer.
Equivalence Class of 1
The equivalence class of 1 includes all real numbers that are an integer distance from 1. This means it contains all numbers of the form 1 + k, where k is any integer. Hence, it includes numbers like 0, 1, 2, 3, and so on, in addition to negative integers: -1, -2, -3, etc.
Equivalence Class of 1/2
Similarly, the equivalence class of 1/2 consists of all real numbers of the form 1/2 + k, where k is any integer. This set includes numbers like -1/2, 1/2, 3/2, 5/2, and so on.
Lin and Priya were working on solving this system of equations.
{1/3x+2y=4
x+y=-3
Lin's first move is to multiply the first equation by 3.
Priya's first move is to multiply the second equation by 2.
Explain why either move creates a new equation with the same solutions as the original equation.
The first equation is x + 6y = 36
the second equation is x+y = -6
(x,y) =(-6,3)
Answer:
we can conclude two things that:
If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.Either person's move will work. Lin's move eliminated the x variable, while Priya's eliminated y variable, but in the end the solution was same.Step-by-step explanation:
Why either move creates a new equation with the same solutions as the original equation?
If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.
When we multiple the two sides of any given equation by the same number, it would keep the two sides of that particular equation equal. So, whatever the the solution the first equation may get, will still work for the second equation.
Determining Lin's first move i.e. to multiply the first equation by 3.
Let us consider the equation
x/3 + 2y = 4 .....[1]
x + y = -3 .....[2]
Lin's first move is to multiply the first equation by 3.
3(x/3 + 2y) = 3(4 )
x + 6y = 12 .....[3]
Now subtract the Equation [2] from Equation [3]
x + 6y - x - y = 12 - (-3)
5y = 15
y = 3
Putting y = 3 in [2]
x + (3) = -3
x = -6
So, x = -6 and y = 3
Determining Priya's first move i.e. to multiply the Second equation by 2.
Let us consider the equation
x/3 + 2y = 4 .....[1]
x + y = -3 .....[2]
Priya's first move is to multiply the second equation by 2.
2(x + y) = 2(-3)
2x + 2y = -6 .....[3]
Now subtract the Equation [2] from [1]
x/3 + 2y - 2x - 2y= 4 - (-6)
x/3 - 2x = 10
x - 6x = 30
x = -6
Putting x = -6 in Equation [2]
x + y = -3
-6 + y = -3
y = 3
So, x = -6 and y = 3
So, from the entire analysis, we can conclude two things that:
If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.Either person's move will work. Lin's move eliminated the x variable, while Priya's eliminated y variable, but in the end the solution was same.Keywords: system of equation, solution, equation
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C be the circle of radius 7 centered at the origin oriented counterclockwise. Evaluate Contour integral Subscript Upper C Superscript Baseline Bold Upper F times d Bold r by parameterizing C.
Answer:
[tex]\oint_cF.dr=0\\[/tex]
Step-by-step explanation:
Given that a circle C of radius 7
[tex]x^{2} +y^{2} =49---(1)[/tex]
To find:
[tex]\oint_{C}F.dr[/tex]
As NO function is given so we suppose it to be:
[tex]F=<x,y>[/tex]
Parametric equations:
[tex]x=rcos\theta=7cos\theta\\y=rsin\theta=7sin\theta[/tex]
Each point on circle can be then found as
[tex]r(\theta)=<7cos\theta,7sin\theta>---(2)[/tex]
From (2) dr can be found as:
[tex]dr=<-7sin(\theta),7cos(\theta)>d\theta---(3)[/tex]
From (2) and (3)
[tex]\oint_cF.dr=\int_{0}^{2\pi}{<7cos\theta,7sin\theta><-7sin(\theta),7cos(\theta)>}\,d\theta\\\\\oint_cF.dr=\int_{0}^{2\pi}{<(-7cos\theta)(7sin\theta),(7sin(\theta))(7cos(\theta))>}\,d\theta\\\\\\\oint_cF.dr=\int_{0}^{2\pi}{-49cos\theta sin\theta+49sin(\theta)cos(\theta)}\,d\theta\\\\\oint_cF.dr=0\\[/tex]
A media research company conducted an online survey of adults in a certain country to determine their favorite sport. A friend believes professional football (with revenue of about $13 billion per year) is the favorite sport of 53% of adults in the country. Specify the null and alternative hypotheses for testing this belief. Be sure to identify the parameter of interest.
Answer:
Null hypothesis: The favorite sport of 53% of adults in the country is professional football ( with revenue of about $13 billion per year)
Alternate hypothesis: The adults in the country whose favorite sport is not professional football (with revenue of about $13 billion per year) is less than 53%
Step-by-step explanation:
The given claim is that the favorite sport of 53% of adults in the country is professional football (with revenue of about $13 per year) which is the null hypothesis. This means that the favorite sport of the remaining 47%(less than 53%) of adults in the country is not professional football (with revenue of about $13 billion per year) which is the alternate hypothesis
A random sample of 30 varieties of cereal was selected. The average number of calories per serving for these cereals is x-120. Assuming that σ = 10, find a 95% confidence interval for the mean number of calories, μ, in a serving of cereal. a. 115.30 to 124.70 b. 116.42 to 123.58 c. 118.00 to 122.00 d. 117.00 to 123.00
Answer: confidence interval = b. ( 116.42 to 123.58)
Step-by-step explanation:
Given;
Number of samples n = 30
Standard deviation r = 10
Mean x = 120
Confidence interval of 95%
Z' = t(0.025) = 1.96
Confidence interval = x +/- Z'(r/√n)
= 120 +/- 1.96(10/√30)
= 120 +/- 3.58
= ( 116.42, 123.58)
An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the age of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 95 percent confidence, what will the margin of error be?
a) About ±2.98 minutes
b) Approximately ±1.52 minute
c) Approximately ±42.14 minutes
d) z = 1.96
Answer:
Option A) About ±2.98 minutes
Step-by-step explanation:
We are given the following information in the question:
Sample mean, [tex]\bar{x}[/tex] = 191.3 minutes
Sample size, n = 200
Population standard deviation, σ = 21.5 minutes
Alpha, α = 0.05
The leaders of the organization wish to develop an interval estimate with 95 percent confidence.
[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]
Margin of error =
[tex]z_{\text{critical}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get:
[tex]\pm 1.96\times \displaystyle\frac{21.5}{\sqrt{200}} = \pm 2.9797 \approx \pm 2.98[/tex]
Option A) About ±2.98 minutes
"The Munchies Cereal Company makes a cereal from several ingredients. Two of the ingredients, oats and rice, provide vitamins A and B. The company wants to know how many ounces of oats and rice it should include in each box of cereal to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while minimizing cost. An ounce of oats contributes 8 milligrams of vitamin A and 1 milligram of vitamin B, whereas an ounce of rice contributes 6 milligrams of A and 2 milligrams of B. An ounce of oats costs $0.05, and an ounce of rice costs $0.03. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis."
Answer:
8x + 6y >/= 48 ......1
x + 2y >/= 12 .......2
The cost function is given as;
C = 0.05x + 0.03y .........3
The minimum cost is $0.24 at (0,8)
That is 0 ounces of oats and 8 ounces of rice.
Step-by-step explanation:
let x represent the number of ounces of oats
And y represent the number of ounces of rice
For Vitamin A
Minimum requirements = 48mg
x ounces of oats contribute 8mg × x
y ounces of Rice contribute 6mg × y
Therefore, we have;
8x + 6y >/= 48 ......1
For Vitamin B
Minimum requirements = 12mg
x ounces of oats contribute 1mg × x
y ounces of Rice contribute 2mg × y
Therefore, we have;
x + 2y >/= 12 .......2
The cost function is given as;
C = 0.05x + 0.03y .........3
Attached is the graphical representation.
The feasible points are (x,y) = (0,8),(2.4,4.8),(12,0)
The minimum cost is determined by substituting each point into the cost function
For (0,8)
C= 0.05(0) + 0.03(8)
C = $0.24
For (12,0)
C= 0.60
For (2.4,4.8)
C= $0.264
The minimum cost is $0.24 at (0,8)
The problem can be modelled with a system of linear inequalities to represent the constraints of the cereal company. You graph these constraints and find the feasible region. After graphing the objective function, move this line towards the origin until it just leaves the feasible region. This point gives the optimal solution.
Explanation:In this problem, we are dealing with linear equations and inequalities. The goal of the Munchies Cereal Company is to determine the amount of oats and rice, measured in ounces, to include in its cereal mix so as to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while at the same time minimizing the cost.
Let's denote the amount of oats as 'x' and the amount of rice as 'y'. The nutrition constraints can be formulated as:
8x + 6y >= 48 (to meet the requirement for vitamin A)x + 2y >= 12 (to meet the requirement for Vitamin B)And since quantities cannot be negative, we also have the constraints: x >= 0 and y >= 0. The objective is to minimize the cost, which can be expressed as C = 0.05x + 0.03y.
To solve this problem graphically, you would plot the constraint lines and see the feasible region (the area that satisfies all constraints). The cost line (C = 0.05x + 0.03y) is then drawn and moved towards the origin until the last point of the feasible region is touched. That point gives the optimal solution.
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As risk management officer at your firm, you are in charge of analyzing the data on personal injury claims filed against your firm. Some summary statistics for a random sample ofthe costs of 100 claims filed in the recent past are below.
Mean = $1,040.47 1st Quartile = $989.72
Median $1,039.71 3rd Quartile $1,088.18 .
Standard deviation = $89.50 !)o.:,
1. Which statement is correct?
The middle 50% ofthe costs are between $989.72 and $1,088.18."
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The correct statement is that the middle 50% of personal injury claim costs fall between $989.72 and $1,088.18, which represents the interquartile range. This range and the standard deviation are key in evaluating the distribution of claim costs.
The statement that the middle 50% of the costs are between $989.72 and $1,088.18 is correct in reference to the provided summary statistics of personal injury claims. This range is defined by the first and third quartiles, also known as the interquartile range (IQR). The IQR is a measure of variability and represents the span between the 25th percentile (first quartile) and the 75th percentile (third quartile), which indeed encompasses the middle 50% of data in a given sample.
In the context of personal injury claims costs at your firm, this means that half of the claim costs fall within that range, with fewer costs being less than $989.72 (the lower 25%) and fewer costs being more than $1,088.18 (the upper 25%). This can be useful information for assessing claims costs and preparing for future claims expenses. The provided standard deviation of $89.50 indicates the average amount that claim costs vary from the mean ($1,040.47).
Suppose that the average number of airline crashes in a country is 2 per month. (a) What is the probability that there will be at least 3 accidents in the next month? Probability = 0.3233 (b) What is the probability that there will be at least 6 accidents in the next two months? Probability = 1-(643/(15e^4)) (c) What is the probability that there will be at most 6 accidents in the next three months? Probability =
Answer:
a) [tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]
b) [tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]
c) [tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]
Step-by-step explanation:
Let X the random variable that represent the number of airline crashes in a country. We know that [tex]X \sim Poisson(\lambda=2)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
(a) What is the probability that there will be at least 3 accidents in the next month?
On this case we are interested on the probability of having at least three accidents in the next month, and using the complement rule we have this:
[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-2} 2^0}{0!}=0.1353[/tex]
[tex]P(X=1)=\frac{e^{-2} 2^1}{1!}=0.2707[/tex]
[tex]P(X=2)=\frac{e^{-2} 2^2}{2!}=0.2707[/tex]
And replacing we have this:
[tex]P(X\geq 3)=1-P(X<3)=1-P(X\leq 1)=1-[0.1353+0.2707+0.2707]=0.3234[/tex]
(b) What is the probability that there will be at least 6 accidents in the next two months?
For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x2= 4[/tex] accidents per 2 months.
[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-4} 4^0}{0!}=0.0183[/tex]
[tex]P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733[/tex]
[tex]P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465[/tex]
[tex]P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954[/tex]
[tex]P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954[/tex]
[tex]P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563[/tex]
Replacing we got:
[tex]P(X\geq 6)=1-P(X<6)=1-P(X\leq 5)=1-[0.0183+0.0733+0.1465+0.1954+0.1954+0.1563]=0.2148[/tex]
(c) What is the probability that there will be at most 6 accidents in the next three months?
For this case since we want the amount in the next two months the rate changes [tex]\lambda=2x3= 6[/tex] accidents per 3 months.
[tex]P(X\leq 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)[/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-6} 6^0}{0!}=0.00248[/tex]
[tex]P(X=1)=\frac{e^{-6} 6^1}{1!}=0.0149[/tex]
[tex]P(X=2)=\frac{e^{-6} 6^2}{2!}=0.0446[/tex]
[tex]P(X=3)=\frac{e^{-6} 6^3}{3!}=0.0892[/tex]
[tex]P(X=4)=\frac{e^{-6} 6^4}{4!}=0.1339[/tex]
[tex]P(X=5)=\frac{e^{-6} 6^5}{5!}=0.1606[/tex]
[tex]P(X=6)=\frac{e^{-6} 6^6}{6!}=0.1606[/tex]
[tex]P(X\leq 6)=0.00248+0.0149+0.0446+0.0892+0.1339+0.1606+0.1606=0.6063[/tex]
Devise the exponential growth function that fits the given data, then answer the accompanying question. Be sure to identify the refernce point (t=0) and units of time.Between 2003 and 2008, the average rate of inflation in a certain country was about 4% per year. If a cart of groceries cost $120 in 2003, what will it cost in 2013 assuming the rate of inflation remains constant?
Answer:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
Step-by-step explanation:
For this case we need to define some notation first.
A= population , t= represent the years after 2003, C= constant for the exponential model.
The starting point t=0 correspond to the year of 2003.
On this case we are assuming the following exponential model:
[tex]A(t) = A_o e^{Ct}[/tex]
The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:
[tex]120=A_o e^{C(0)}=A_o e^0 = A_o[/tex]
And then the model is:
[tex]A(t) =120 e^{Ct}[/tex]
Now we need to determine the value for C. Since we know that inflation increase 4% per year we have that after one year we have 1.04 times the value of the original value, and we have this equation:
[tex]1.04 A_o= A_o e^{C(1)}= A_o e^C[/tex]
And we got this:
[tex]1.04= e^C [/tex]
Applying ln on both sides we got:
[tex]ln(1.04)= C=0.0392207[/tex]
So then our model is given by:
[tex]A(t) = 120 e^{ln(1.04)t}[/tex]
For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
The cost of groceries in 2013, after applying an annual inflation rate of 4% for 10 years, will be approximately $177.63.
Explanation:To calculate the annual rate of inflation and the cost in 2013, you can use the exponential growth function, where cost =[tex]initial_{cost} * (1 + rate)^{time[/tex]. In this case, the initial cost in 2003 (t=0) is $120 and the annual rate of inflation is 4%, or 0.04. To calculate the cost in 2013, which is 10 years after 2003, apply the exponential growth function:
Cost in 2013 = $120 * (1 + 0.04)¹⁰
This calculation yields:
Cost in 2013 = $120 * (1.04)¹⁰
Cost in 2013 = $120 * 1.48024
Cost in 2013 = $177.63
Therefore, in 2013, assuming the rate of inflation remains constant, the cart of groceries will cost approximately $177.63.
Life after college. We are interested in estimating the proportion of graduates at a mid-sized university who found a job within one year of completing their undergraduate degree. Suppose we conduct a survey and find out that 348 of the 400 randomly sampled graduates found jobs. The graduating class under consideration included over 4500 students.(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?(b) Check if the conditions for constructing a confidence interval based on these data are met.(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.(d) What does "95% confidence" mean?(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data.(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider? Explain.(Please show work for all problems, thank you)
Answer:
a) The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:
[tex]\hat p=\frac{348}{400}=0.87[/tex]
b) [tex]np=400*0.87=348>10[/tex]
[tex]n(1-p)=400(1-0.87)=52>10[/tex]
So both conditions are satisifed and we can construct the confidence interval.
c) The 95% confidence interval would be given (0.837;0.903).
We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903
d) On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.
e) The 99% confidence interval would be given (0.827;0.913).
We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913
f) The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.
Step-by-step explanation:
Data given and notation
n=400 represent the random sample taken
X=348 represent the number of graduates that found jobs in the sample
[tex]\hat p=\frac{348}{400}=0.87[/tex] estimated proportion of graduates that found jobs
[tex]\alpha[/tex] represent the significance level
z would represent the statistic to calculate the confidence interval
p= population proportion of graduates that found jobs
(a) Describe the population parameter of interest. What is the value of the point estimate of this parameter?
The parameter of interest is p who represent the proportion of graduates from this university who found a job within one year after graduating, and the estimated value is:
[tex]\hat p=\frac{348}{400}=0.87[/tex]
(b) Check if the conditions for constructing a confidence interval based on these data are met.
[tex]np=400*0.87=348>10[/tex]
[tex]n(1-p)=400(1-0.87)=52>10[/tex]
So both conditions are satisifed and we can construct the confidence interval.
(c) Calculate a 95% confidence interval for the proportion of graduates who found a job within one year of completing their undergraduate degree at this university, and interpret it in the context of the data.
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.87 - 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.837[/tex]
[tex]0.87 + 1.96 \sqrt{\frac{0.87(1-0.87)}{400}}=0.903[/tex]
And the 95% confidence interval would be given (0.837;0.903).
We are confident (95%) that that the true proportion of graduates that found jobs is between 0.837 and 0.903
(d) What does "95% confidence" mean?
On this case that the 95% of the selected random samples will produce a 95% confidence interval that includes the true proportion of interest.
(e) Now calculate a 99% confidence interval for the same parameter and interpret it in the context of the data
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And replacing into the confidence interval formula we got:
[tex]0.87 - 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.827[/tex]
[tex]0.87 + 2.58 \sqrt{\frac{0.87(1-0.87)}{400}}=0.913[/tex]
And the 99% confidence interval would be given (0.827;0.913).
We are confident (99%) that that the true proportion of graduates that found jobs is between 0.827 and 0.913
(f) Compare the widths of the 95% and 99% confidence intervals. Which one is wider?
The width for the 95% interval is 0.903-0.837=0.066, and for the 99% interval 0.913-0.827=0.086. And we see that the 99% is wider since we have more confidence that the true parameter of interest would be on the range provided.
Tara wants to weigh her three stuffed animals. They will only fit on the scale two at a time. Together Addie and Missy weight 18 ounces. Missy and Corky weigh 22 ounces, and Addie and Corky weigh 12 ounces. How much does each animal weigh?
Answer: Addie weighs 4 ounces
Missy weighs 14 ounces
Corky weighs 8 ounces
Step-by-step explanation:
Let a represent the weight of Addie.
Let m represent the weight of Missy.
Let c represent the weight of Corky.
Together Addie and Missy weigh 18 ounces. This means that
a + m = 18 - - - - - - - - - 1
Missy and Corky weigh 22 ounces. This means that
m + c = 22
m = 22 - c - - - - - - - - - - 2
Addie and Corky weigh 12 ounces. This means that
a + c = 12
a = 12 - c - - - - - - - - - - - 3
Substituting equation 2 and equation 3 into equation 1, it becomes
22 - c + 12 - c = 18
34 - 2c = 18
- 2c = 18 - 34 = - 16
c = - 16/ - 2 = 8
Substituting c = 8 into equation 2, it becomes
m = 22 - 8
m = 14
Substituting c = 8 into equation 3, it becomes
a = 12 - 8
a = 4
Find at least ten solutions to the linear equation 1/2x + y = 5, and plot the points on a coordinate plane. What shape is the graph of the linear equation taking?
The linear equation 1/2x + y = 5, we can choose values for x and solve for y. When we plot the points on a coordinate plane, we get a line graph.
Explanation:To find solutions to the linear equation 1/2x + y = 5, we can arbitrarily choose values for x and solve for y.
Let's choose x = 0:
1/2(0) + y = 5
y = 5
So one solution is (0, 5).
Now let's choose x = 2:
1/2(2) + y = 5
1 + y = 5
y = 4
Another solution is (2, 4).
We can continue this process and find more solutions:
x = 4, y = 3
x = 6, y = 2
x = 8, y = 1
x = 10, y = 0
x = -2, y = 6
x = -4, y = 7
x = -6, y = 8
x = -8, y = 9
x = -10, y = 10
These are ten solutions to the equation.
If we plot these points on a coordinate plane, we will see that they all lie on a straight line.
Therefore the shape of the graph is a line. The equation represents a linear relationship between x and y.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.11 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. If the test statistic is 2.2, what is the p-value associated with this hypothesis test?(Round your answer to three decimal places.)
Answer:
At 0.05 significance level, the p-value is 0.014
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 3 minutes
Sample mean, [tex]\bar{x}[/tex] = 3.11 minutes
Sample size, n = 100
Alpha, α = 0.05
Sample standard deviation, σ = 0.5 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 3\text{ minutes}\\H_A: \mu > 3\text{ minutes}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
[tex]z_{stat} = 2.2[/tex]
Now, we calculate the p-value from the normal standard z-table.
P-value = 0.014
At 0.05 significance level, the p-value is 0.014
Teachers in a medium-sized suburban school district have an average salary of $47,500 per year, with a standard deviation of $4,600. After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. What are the new mean and standard deviation of the teacher's salaries during the year in which they recieve bonus?A. $50,125; $4,960B. $49,875; $4,830C. $49,875; $5,330D. $50,375; $4,830E. $50,375; $5,330
Answer: $49,875; $4,830C.
Step-by-step explanation:
The average salary of Teachers in a medium-sized suburban school district is $47,500 per year.
The standard deviation is $4,600
After negotiating with the school district, teachers recieve a 5% raise and a one-time $500 bonus. The bonus of $500 will not alter the mean and standard deviation because equal amount is added for each teacher.
5% increase in each teacher's salary would differ. Therefore, it will affect the mean and standard deviation by 5%. Therefore, the new mean would be
47500 + (5/100 × 47500) = $49875
The new standard deviation would be
4600 + (5/100 × 4600) = $4830
Answer:
$50,375; $4,830 (I just answered it on one of my quizzes)
Step-by-step explanation:
In your day to day, is there any occurring fact that can be interpreted as linear systems and non-linear systems?
Answer:
Step-by-step explanation:
Linear relations are those which are represented by straight line on a graph i.e. increase or decrease a variable directly effect the other in a linear way
for example
y=2x+3
increase in value of x increase the value of y linearly.
Non-linear relationship are those in which change in one entity does not correspond to change in other quantity.
Real life situation for linear relation is when we apply force on a block it accelerates and it goes on increasing as force is increasing.
For Non-linear relation
increase in temperature vs time spend
Time spent is more to increase the slight amount of temperature.
Linear and non-linear systems exist in everyday scenarios. A linear example is cost increasing by a fixed amount per item purchased, while a non-linear example might be a room's temperature: heating up quickly initially, then more slowly as it approaches the desired warmth.
Explanation:In everyday life, both linear and non-linear systems abound. A straightforward example of a linear system is calculating the total cost of goods: if each item costs the same, your total cost represents a linear relationship with the number of items - the cost increases by a set amount per item.
On the other hand, a non-linear system might relate to how the temperature of a heated room changes over time: it may start increasing rapidly, then slowly once the room approaches the desired temperature. This represents a non-linear relationship as the rate of temperature increase isn't constant.
Learn more about Linear and Non-linear Systems here:https://brainly.com/question/35354090
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Ski resorts are interested in the average age that children take their first Ski and Snowboard lessons they need thisinformation to optimally plan their ski classes match the vocabulary word with its corresponding exampleparameterpopulationvariablesamplestatisticdata1. all children who Ski and Snowboardparameter2. the 92 children who were asked when they took their first lesson3. the average age that all children take their first to lesson4. the average age then the 92 children took their first lesson5. the age that children take their first listen6. the list of the 92 ages that the children from the study took the first listen
Answer:
Matching the vocabulary word with its corresponding example:
1. All children who Ski and Snowboard = population (a group of items, units or subjects which is under reference of study e.g inhabitants of a region, numbers of cars in a city e.t.c)
2. The 92 children who were asked when they took their first lesson = Sample (a part or fraction of a population selected on some basis)
3. The average age that all children take their first to lesson = Parameter (the number that summarizes some characteristics of a population
4. The average age that the 92 children took their first lesson = statistic (a sample characteristic corresponding to the population parameter used when a sample is use to make inference about a population)
5. The age that children take their first listen = variable (anything that has attribute, quality or quantity that varies or a characteristic/attribute that describes a place, person or thing)
6. The list of the 92 ages that the children from the study took the first listen = data (facts that are collected together for analysis)
Step-by-step explanation:
1. All children who Ski and Snowboard = population (a group of items, units or subjects which is under reference of study , numbers of cars in a city e.t.c)
2. The 92 children who were asked when they took their first lesson = Sample (a part of a population selected on some basis)
3. The average age that all children take their first to lesson = Parameter (the number that summarizes some characteristics of a population
4. The average age that the 92 children took their first lesson = statistic (a sample characteristic corresponding to the population parameter used when a sample is use to make inference about a population)
5. The age that children take their first listen = variable (anything that has attribute, quality or quantity that varies or a characteristic/attribute that describes a place, person or thing)
6. The list of the 92 ages = data (facts that are collected together for analysis)
A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a population standard deviation of 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal.
What is the alternative and null hypotheses?
What is the test statistic?
What is the p-value?
What is your conclusion about the stated hypotheses at a 95% confidence level?
Answer:
[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]
[tex]p_v =P(t_{15}>2)=0.0320[/tex]
If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=25[/tex] represent the sample mean
[tex]s=2[/tex] represent the standard deviation for the sample
[tex]n=16[/tex] sample size
[tex]\mu_o =24[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
Confidence =0.95 or 95%
[tex]\alpha=0.05[/tex]
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 24[/tex]
Alternative hypothesis:[tex]\mu > 24[/tex]
We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]
Calculate the P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=16-1=15[/tex]
Since is a one-side upper test the p value would be:
[tex]p_v =P(t_{15}>2)=0.0320[/tex]
Conclusion
If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.
Consider the following function. Without finding the inverse, evaluate the derivative of the inverse at the given point. f(x)=ln(8x+e); (1,0)
We can use the inverse function derivative theorem:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=a} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(a)}}.[/tex]
In this case, we want to evaluate [tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1}[/tex], so:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}}.[/tex]
The derivative is:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x} = \dfrac{\textrm{d}}{\textrm{d}x}\left[\ln(8x + \textrm{e})\right] = \dfrac{1}{8x+\textrm{e}}\dfrac{\textrm{d}}{\textrm{d}x}\left(8x + \textrm{e}\right) = \dfrac{8}{8x+\textrm{e}}.[/tex]
The ordinate of the point is [tex]f^{-1}(1) = 0[/tex], so we evaluate:
[tex]\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0} = \dfrac{8}{8 \times 0+\textrm{e}} = \dfrac{8}{\textrm{e}}.[/tex]
Finally:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=f^{-1}(1)}} = \dfrac{1}{\dfrac{\textrm{d}f}{\textrm{d}x}\Big\vert_{x=0}} = \dfrac{1}{\dfrac{8}{\textrm{e}}} = \dfrac{\textrm{e}}{8}.[/tex]
We can check the answer by finding the inverse:
[tex]y = \ln(8x + \textrm{e}) \implies \textrm{e}^y = 8x + \textrm{e} \iff \textrm{e}^y - \textrm{e} = 8x \iff x = \dfrac{\textrm{e}^y-\textrm{e}}{8},[/tex]
so that
[tex]f^{-1}(x) = \dfrac{\textrm{e}^x-\textrm{e}}{8}.[/tex]
Therefore:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x} = \dfrac{\textrm{e}^x}{8}.[/tex]
Which finally gives the same answer as before:
[tex]\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}^1}{8} = \dfrac{\textrm{e}}{8}.[/tex]
Answer: [tex]\boxed{\dfrac{\textrm{d}f^{-1}}{\textrm{d}x}\Big\vert_{x=1} = \dfrac{\textrm{e}}{8}}.[/tex]