Becca bought a shirt that was 25% off and saved $6. What was the original price of the shirt?

Step-by-step explanation:

For a community service project, the no of the students in a class bring paper towels  = [tex]\frac{3}{4}[/tex]

The students bring towels and soap = [tex]\frac{2}{5}[/tex]

The students who bring paper towels = [tex]\frac{3}{4}[/tex]

The fraction of students who bring paper towels also bring soap =  [tex]\frac{2}{5}[/tex]

Stokes' theorem says the integral of the curl of [tex]\vec F[/tex] over a surface [tex]S[/tex] with boundary [tex]C[/tex] is equal to the integral of [tex]\vec F[/tex] along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that

[tex]\displaystyle\iint_S\nabla\times\vec F\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r[/tex]

We have

[tex]\vec F(x,y,z)=x^2\,\vec\imath+4x\,\vec\jmath+z^2\,\vec k[/tex]

[tex]\implies\nabla\times\vec F(x,y,z)=4\,\vec k[/tex]

Parameterize the ellipse [tex]S[/tex] by

[tex]\vec s(u,v)=\dfrac{u\cos v}{\sqrt5}\,\vec\imath+\dfrac{u\sqrt5\sin v}4\,\vec\jmath[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex].

Take the normal vector to [tex]S[/tex] to be

[tex]\dfrac{\partial\vec s}{\partial\vec u}\times\dfrac{\partial\vec s}{\partial\vec v}=\dfrac u4\,\vec k[/tex]

Then the flux of the curl is

[tex]\displaystyle\iint_S4\,\vec k\cdot\dfrac u4\,\vec k\,\mathrm dA=\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}[/tex]

Answer:

Fraction

Step-by-step explanation:

Answer:

fraction

Step-by-step explanation:

pls mark brainliest

Given:

The given two functions are [tex]f(x)=\frac{4 x^{2}+24 x}{x^{2}-11 x+30}[/tex]  and  [tex]g(x)=\frac{x-5}{x^{2}}[/tex]

We need to determine the value of R(x)

Value of R(x):

The value of R(x) can be determined by R(x) = f(x) × g(x)

Substituting the values, we get;

[tex]R(x)=\frac{4 x^{2}+24 x}{x^{2}-11 x+30} \cdot \frac{x-5}{x^{2}}[/tex]

Multiplying the fractions, we have;

[tex]R(x)=\frac{\left(4 x^{2}+24 x\right)(x-5)}{\left(x^{2}-11 x+30\right) x^{2}}[/tex]

Let us factor out the common term 4x from the term (4x² + 24x)

Thus, we have;

[tex]R(x)=\frac{4 x(x+6)(x-5)}{\left(x^{2}-11 x+30\right) x^{2}}[/tex]

Let us factor the term (x² -11x + 30), we get;

[tex]R(x)=\frac{4(x+6)(x-5)}{x(x-5)(x-6)}[/tex]

Cancelling the common term, we have;

[tex]R(x)=\frac{4(x+6)}{x(x-6)}[/tex]

Thus, the value of R(x) is [tex]\frac{4(x+6)}{x(x-6)}[/tex]

Answer:

20%  

Step-by-step explanation:

Find the increase (subtract original number with final):

42-35= 7

Find the percent change (divide the increase by original number, then multiply by 100 to find the percentage):

[tex]\frac{7}{35}[/tex]*100=20

The percent change from 35 students to 42 students is 20%. This is calculated by using the formula for percent change: ((New Amount - Original Amount) / Original Amount) * 100.

In this case, we are trying to calculate the percent change in the number of students in Mrs. Mroc's class. The formula for percent change is:

((New Amount - Original Amount) / Original Amount) * 100

The New Amount is 42 (the number of students next year), and the Original Amount is 35 (the current number of students). So the percent change would be (((42-35)/35)*100), which equals 20%.

So, if Mrs. Mroc's class increases from 35 students to 42 students next year, the percent change will be 20%.

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The rate of increase is 1.2 as it represents b in the equations A+b power of x+c then +D.

Answer:

y=7.46496

Step-by-step explanation:

Complete Question:

For a healthy human, a body temperature follows a normal distribution with Mean of 98.2 degrees Fahrenheit and Standard Deviation of 0.26 degrees Fahrenheit. For an individual suffering with common cold, the average body temperature is 100.6 degrees Fahrenheit with Standard deviation of 0.54 degrees Fahrenheit. Simulate 10000 healthy and 10000 unhealthy individuals and answer questions 14 to 16.

14. If person A is healthy and person B has a cold, which of the events are the most likely? Pick the closest answer.  

a. Person B will have higher temperature than 101 degrees.

b. Person A will have temperature higher than 99 degrees

c. Person B will have temperature lower than 100 degrees

d. ​​Person A will have temperature lower than 97.5 degrees

15. What would be a range [A to B], which would contain 68% of healthy individuals? Pick the closest answer.  

a. Between 97.9 and 98.46

b. Between 99.5 and 101.6

c. Between 100.06 and 101.14

d. Between 100.1 and 102.2

16. What is the approximate probability that a randomly picked, unhealthy individual (one with the cold) would have body temperature above 101 degrees Fahrenheit? Pick the closest answer.  

a. About 10%

b. About 15%

c. About 22%

c. About 34%

Answer:

14. option a

15. option a

16. option a

explanation:

14 option (a)

person B have higher temperature than 101 degrees,  

(b) cannot be true if person a has higher temperature than 99 then he has cold most likely and as mentioned by the standard dev and mean of healthy people  

(c) AND (d) also cannot be true as they also do not satisfy the given standard dev and mean

15 option (a) because of the A to B range and not B to A range ....... as the healthy person's standard dev is 0.26. Hence the 68% data will lie in 97.94 - 98.46

but, if it was B to A, then the (c) option will be true

16 option (a) because most of the unhealthy individuals above the 0.54 standard dev will come somewhat near 90%

Answer:

218

Step-by-step explanation:

327*2/3=218

The number of people in attendance are:

Children=70

Parents=140

Grandparents=190

Let

C represent children

P represent parents

G represent grandparents

Hence:

children+parents+ grandparent=400

P=2G

C=P+50 = 2G+50

2G+50 + 2G + G =400

5G=400-50

5G = 350

G=350/5

G =70 grandparents

P=70×2

P =140 parents

C=140+50

C =190 children

Inconclusion The number of people in attendance are:

Children=70

Parents=140

Grandparents=190

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The question is a math problem requiring a system of linear equations to solve. There were 70 grandparents, 140 parents, and 190 children at the family reunion. This was determined by assigning variables to each group and using the given conditions to form equations, which were then solved sequentially.

This problem can be solved using a system of linear equations. We can represent each group at the family reunion (children, parents, and grandparents) with a variable. To create equations from the information provided:

First, replace P and C in the total equation with the equivalent as per the other equations so it reads (2G + 50 + 2G + G = 400). Simplify this equation and you get 5G + 50 = 400. Solving for G, we find that there were 70 grandparents.

Plug the value of G (70) back in the equation for P (2*70), we find that there were 140 parents.

Similarly plug P (140) in the equation for C, we find that there were 190 children.

So, in conclusion, there were 70 grandparents, 140 parents, and 190 children at the family reunion.

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Answer:

a and c

Step-by-step explanation:

Looking at the pieces of information, the pieces of information that can be gathered from these box plots are:

Olympics actually refers to the major international event which usually features different sports and normally holds once in every four years. It is also known as the Games of the Olympiad.

From the pieces of information given, we can actually conclude that the above options can be gathered from the information.

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Answer:

4 units to the right & 1 unit up.

Answer: it’s B)

The answer is the second one

Answer:

d = 11

Step-by-step explanation:

Answer:

Point of estimate of the true Proportion 'p' = 0.1941

Step-by-step explanation:

Explanation:-

Given data a random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights.

Given sample size 'n'= 340

Point of estimate of the true Proportion P  = [tex]\frac{x}{n}[/tex]

Properties of Estimation

An estimator is not expected to estimate the Population parameter without error.

An estimator should be close to the true value of un-known parameter.

Point of estimate of the true Proportion   = [tex]p = \frac{66}{340}= 0.1941[/tex]

Conclusion:-

Point of estimate of the true Proportion = 0.1941

Answer:

Three hundred fifty-five and three tenths.

Step-by-step explanation:

The word form for 355.3 kilograms is "Three hundred fifty-five point three kilograms." This representation breaks down the number into its hundreds, tens, ones, tenths, and specifies the unit of measurement, kilograms.

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Answer:

a) The null hypothesis is represented as

H₀: μ ≤ 23

The alternative hypothesis is given as

Hₐ: μ > 23

b) Check the Explanation

The conditions for a t-test to be performed are satisfied or not?

- Yes, because the students were randomly sampled.

- Yes, the sample size is larger man 30.

And the central limit theorem allows us to approximate that the random sample obtained from the population is a normal distribution.

c) Are the sampled values independent of each other?

Yes, because each student's test score does not affect other students' test scores.

d) p-value obtained = 0.004

Reject the null hypothesis because the P-value a less than the alpha = 0.10 level of significance

e) There is sufficient evidence to conclude that the population mean is greater than 23.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to check if results suggest that students who complete the core curriculum are ready for college-level mathematics.

The only condition to be ready for college is scoring above 23.

So, the null hypothesis would be that the mean of test scores of students that complete core curriculum is less than or equal to 23. That is, there isn't significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics.

And the alternative hypothesis would be that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.

Mathematically

The null hypothesis is represented as

H₀: μ ≤ 23

The alternative hypothesis is given as

Hₐ: μ > 23

b) The conditions required before performing t-test.

- The sample should be a random sample

- The dependent variable should be approximately normally distributed.

- The observations are independent of one another.

- The dependent variable should not contain any outliers

All of these conditions are satisfied for our distribution.

c) Are the sampled values independent of each other?

Yes, because each student's test score does not affect other students' test scores.

d) To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 23.6

μ₀ = 23

σₓ = standard error = (σ/√n)

where n = Sample size = 200

σ = Sample standard deviation = 3.2

σₓ = (3.2/√200) = 0.226

t = (23.6 - 23) ÷ 0.226 = 2.65

checking the tables for the p-value of this t-statistic

- Degree of freedom = df = n - 1 = 200 - 1 = 199

- Significance level = 0.10

- The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 2.65, at 0.10 significance level, df = 199, with a one tailed condition) = 0.004348 = 0.004 to 3 d.p.

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.10

p-value = 0.004

0.004 < 0.10

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis and say that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.

e) The result of the p-value obtained is that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.

Hope this Helps!!!

Answer:

[tex]t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202[/tex]    

[tex]p_v =2*P(t_{(60)}>1.202)=0.234[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from  420. So the specification is satisfied.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=424[/tex] represent the sample mean

[tex]s=26[/tex] represent the sample standard deviation

[tex]n=61[/tex] sample size  

[tex]\mu_o =420[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 420 or not, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 420[/tex]  

Alternative hypothesis:[tex]\mu \neq 420[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{424-420}{\frac{26}{\sqrt{61}}}=1.202[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=61-1=60[/tex]  

Since is a two sided test the p value would be:  

[tex]p_v =2*P(t_{(60)}>1.202)=0.234[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from  420. So the specification is satisfied.

Answer:

g = - 30

Step-by-step explanation:

[tex] - 15 - \frac{g}{3} = - 5 \\ - \frac{g}{3} = - 5 + 15 \\ - \frac{g}{3} = 10 \\ - g = 10 \times 3 \\ - g = 30 \\ \huge \red{ \boxed{g = - 30}}[/tex]

Answer:

[tex]BC=6\sqrt{2}\ units[/tex]

Step-by-step explanation:

In this problem i will assume that the side AB is the hypotenuse

The picture in the attached figure

we know that

In a right triangle, if one angle is 45 degrees, then the other angle complementary is equal to 45 degrees too

so

[tex]m\angle B=45^o[/tex]

In the right triangle ABC

[tex]cos(B)=\frac{BC}{AB}[/tex] ----> by CAH (adjacent side divided by the hypotenuse)

substitute the given values

we have

[tex]cos(45^o)=\frac{\sqrt{2}}{2}[/tex]

[tex]AB=12\ units[/tex]

substitute

[tex]BC=cos(B)(AB)[/tex]

[tex]BC=\frac{\sqrt{2}}{2}(12)=6\sqrt{2}\ units[/tex]

Answer: 6 square root 2

Step-by-step explanation:

Answer:

x = -5/4    x=1

Step-by-step explanation:

5x^2 -x -4 =0

Factor

(5x+4) (x-1)=0

Using the zero product property

5x+4 =0  x-1 =0

5x=-4       x=1

x = -5/4    x=1

Answer:

x = -0.8, 1

Step-by-step explanation:

5x² - x - 4 = 0

5x² - 5x + 4x - 4 = 0

5x(x - 1) + 4(x - 1) = 0

(5x + 4)(x - 1) = 0

x = -⅘, 1

Answer:

A. when working on an online class students have more freedom to choose an environment in which they are most comfortable

Step-by-step explanation:

some students do better at home some do better in a traditional setting but being online lets you change your setting to anywhere with online access

In the given data set, 12, 8, 6, 6, 9, 8, 7, 11, 6, the value 6 represents the mode.

We are given the following dataset: 12, 8, 6, 6, 9, 8, 7, 11, 6.

Count the frequency of each number:

12 appears 1 time.

8 appears 2 times.

6 appears 3 times.

9 appears 1 time.

7 appears 1 time.

11 appears 1 time.

The number 6 appears 3 times, more frequently than any other number.

By the definition of mode, it is the value that appears most frequently in a data set.

Since 6 occurs more times than any other number in the list, then it is the mode

The average rate of change can be calculated using the formula: (g(b) - g(a)) / (b - a), substituting -5 for a and 5 for b in our case.

In order to find the average rate of change of a function, in this case, function g, from x = -5 to x = 5, we use the formula: (g(b) - g(a)) / (b - a), where b is the ending value and a is the starting value. This gives us the slope of the tangent line that connects these two points on the graph, which reflects the average rate of change of the function over the interval. In this case, to find the average rate of change of function g from x = -5 to x = 5, we would substitute -5 for a and 5 for b in the formula to get: (g(5) - g(-5)) / (5 - (-5)). From here we can solve based on the known values of g(5) and g(-5).

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Answer:

1391000000

Step-by-step explanation:

Answer:

A. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.

B. P(Z<-4.93333) = 0.0000

C. n = 5945 samples

Step-by-step explanation:

A)

The rightnull and alternative hypotheses are given as below:

Null hypothesis H0: µ = 5.5

Alternative hypothesis Ha: µ ≠ 5.5

Conducting the one sample z test for population mean.

Test statistic formula is given as below:

Z = (Xbar - µ)/[σ/√(n)]

Given,

Xbar = 5.23

µ = 5.5

σ = 0.30

n = 16

Z = (5.23 – 5.5) / [ 0.30 /√(16)]

Z = -3.6000

P-value = 0.0003

α value = 0.05

P-value < α value

So, we reject the null hypothesis. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.

B

From the information given

Xbar = 5.23

µ = 5.6

σ = 0.30

n = 16

α value = 0.01

Z = (Xbar - µ)/[σ/√(n)]

Z = (5.23 - 5.6)/(0.30/√(16))

Z = -4.93333

P(Z<-4.93333) = 0.0000

Required probability = 0.0000

C

We are given α =0.01

Therefore, critical Z value = 2.57

E = 0.01

σ = 0.30

n = (Z*σ/E)^2

n = (2.57*0.30/0.01)^2 = 5944.41

n = 5945 samples

Answer:

θ   [tex]0.75\pi[/tex]     [tex]1.75\pi[/tex]

r   10.551  244.151

Step-by-step explanation:

The maximum value for [tex]\theta[/tex] is:

[tex]\theta_{max} = \ln r[/tex]

[tex]\theta_{max} = 2.199\pi\,rad[/tex]

The formula for the slope of the tangent line in polar coordinates is:

[tex]m = \frac{r'\cdot \sin \theta + r \cdot \cos \theta}{r' \cdot \cos \theta - r \cdot \cos \theta}[/tex]

Horizontal tangent lines have a slope of zero. So, the following relation must be satisfied:

[tex]r'\cdot \sin \theta + r \cdot \cos \theta = 0[/tex]

[tex]r'\cdot \sin \theta = - r \cdot \cos \theta[/tex]

[tex]\tan \theta = - \frac{r}{r'}[/tex]

[tex]\tan \theta = -\frac{e^{\theta}}{e^{\theta}}[/tex]

[tex]\tan \theta = -1[/tex]

[tex]\theta = \tan^{-1}(-1)[/tex]

[tex]\theta = \frac{3}{4}\pi + i\cdot \pi[/tex], for all [tex]i \in \mathbb{N}_{O}[/tex].

The maximum value of i is:

[tex]i = \frac{\theta_{max}-\frac{3}{4}\pi }{\pi}[/tex]

[tex]i = \frac{2.199-0.75}{1}[/tex]

[tex]i = 1.449[/tex] ([tex]i_{max} = 1[/tex]).

Then, values are listed below:

θ   [tex]0.75\pi[/tex]     [tex]1.75\pi[/tex]

r   10.551  244.151

Answer:

{ ( 2.193 , π / 4)   , ( 10.551 , 3π / 4) ,  ( 50.754 , 5π / 4)  , ( 244.151 , 7π / 4)  }

Step-by-step explanation:

Given:-

- The polar curve has the equation:

                           r = e^θ

- list the points starting with the smallest value of r such that:

                1 ≤ r ≤ 1000   ,  0 ≤ θ < 2π.

Find:-

List all of the points (r,θ) where the tangent line is horizontal

Solution:-

- We will first transform the polar curve to cartesian coordinate system using the parametric relations:

                  x = r*cos (θ)

                  y = r*sin (θ)

- The tangent line is horizontal when the " dy / dθ  " = 0 and " dx / dθ  " = 0, so:

                  x = e^θ*cos (θ)      ,    y = e^θ*sin (θ)

                  dx / dθ = e^θ*cos (θ) - e^θ*sin(θ)

                               = e^θ*[cos (θ) - sin(θ)]

                  dx / dθ = e^θ*[cos (θ) - sin(θ)] = 0,

                  e^θ = 0    ,      [cos (θ) - sin(θ)] = 0

                  e^θ ≠ 0 for the given interval 0 ≤ θ< 2π  

                  cos (θ) - sin(θ) = 0 , tan ( θ ) = 1 - (1st quad and 3rd quad)

                  θ = { π / 4 , 5π / 4 } , 0 ≤ θ< 2π    

- Similarly, evaluate dy/dθ = 0;

                   dy/dθ = e^θ*cos (θ) + e^θ*sin(θ)

                              = e^θ*[cos (θ) + sin(θ)]

                  dy / dθ = e^θ*[cos (θ) + sin(θ)] = 0,

                  e^θ = 0    ,      [cos (θ) + sin(θ)] = 0

                  e^θ ≠ 0 for the given interval 0 ≤ θ< 2π  

                  cos (θ) + sin(θ) = 0 , tan ( θ ) = -1 , (2nd quad and 4th quad)

                  θ = { 3π / 4 , 7π / 4 } , 0 ≤ θ< 2π

- All possibilities of " θ " over the interval satisfying the a horizontal tangent line to the given polar curve:

                  θ = { π / 4, 3π / 4 , 5π / 4 , 7π / 4 } , 0 ≤ θ < 2π  

- We will plug the evaluated list of values of "θ " in the given polar curve and determine the corresponding values of "r":

                  r = e^θ

                  θ = π / 4  , r = e^(π / 4) = 2.193        

               1: ( r , θ ) = ( 2.193 , π / 4)        

                  θ = 3π / 4  , r = e^(3π / 4) = 10.55072        

               2: ( r , θ ) = ( 10.551 , 3π / 4)  

                  θ = 5π / 4  , r = e^(5π / 4) = 50.754      

               3: ( r , θ ) = ( 50.754 , 5π / 4)  

                 θ = 7π / 4  , r = e^(7π / 4) = 244.151      

               4: ( r , θ ) = ( 244.151 , 7π / 4)          

Answer:

The randomization will be centered around 0.5

Step-by-step explanation:

n = 30

Null hypothesis, H₀ : p = 0.5

Alternative hypothesis,  : p > 0.5

The randomization will be centered around 0.5

The indication of where the randomization distribution will be centered shows that the randomization distribution is always centered at the value of the parameter given in the null hypothesis, which is p = 0.5.

Thus, the randomization distribution is always centered at the value of the parameter given in the null hypothesis, which is p = 0.5.

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Answer:

top 83 1/5

bottom 1,248

Step-by-step explanation:

i got it right

Answer:

Top 83 1/5

Bottom 1,248

Step-by-step explanation:

Just took it

a) True. Some residuals will be positive and some will be negative. b) True. Least squares minimizes the squares of the residuals. c) False. y with a caret denotes predicted values.

a) True. Some of the residuals from a least squares linear model will be positive and some will be negative. Residuals measure the difference between the observed values and the predicted values, and they can be positive if the observed value is higher than the predicted value, or negative if the observed value is lower.

b) True. Least squares regression aims to minimize the sum of squared residuals. By minimizing the squares of the residuals, we ensure that the line of best fit is as close as possible to the data points.

c) False. We write y with a caret (^) to denote the predicted values and y without a caret to denote the observed values.

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