Find the magnitude of the torque exerted by F on the bolt at P if |vector PQ| = 6 in. and |F| = 16 lb.

The probability of a student taking a calculus or a statistics class is found using the Addition Rule of Probability. For non-mutually exclusive events like these, the formula is P(A or B) = P(A) + P(B) - P(A and B). Substituting in the given values, we find the probability is 93%.

In probability theory, there's a rule called the Addition Rule of Probability. The rule states: the probability of the occurrence of either of two mutually exclusive events A and B is given by the sum of the probabilities of A and B.

However, if the two events aren't mutually exclusive (they can occur together), like our case here with the calculus and statistics classes, we need to adjust the formula. We subtract the probability of both of them happening. Hence, the formula becomes: P(A or B) = P(A) + P(B) - P(A and B).

If we plug in the given values: P(calculus or statistics) = P(calculus) + P(statistics) - P(calculus and statistics) = 0.10 + 0.90 - 0.07 = 0.93 or 93%.

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The probability that a student is taking either a calculus class or a statistics class is 0.93 or 93%. This uses the principle of inclusion-exclusion in probability.

To determine the probability of a student taking a calculus class or a statistics class, we use the principle of inclusion-exclusion for probabilities. This principle states:

P(A or B) = P(A) + P(B) - P(A and B)

Here, we are given the following probabilities:

Using the inclusion-exclusion principle, we get:

P(C or S) = P(C) + P(S) - P(C and S)

= 0.10 + 0.90 - 0.07

= 0.93

Thus, the probability that a student is taking either a calculus class or a statistics class is 0.93 or 93%.

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

We are given that a publisher reports that 79% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually more than the reported percentage. A random sample of 100 found that 89% of the readers owned a personal computer.

Let Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 0.79  {means that the percentage is actually less than or equal to the reported percentage}

Alternate Hypothesis, [tex]H_1[/tex] : p > 0.79  {means that the percentage is actually more than the reported percentage}

The test statics that will be used here is One-sample proportions test;

         T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = % of the readers who owned a personal computer in a sample of 100 = 89%

           n = sample size = 100

So, test statistics = [tex]\frac{0.89 -0.79}{\sqrt{\frac{0.89(1-0.89)}{100} } }[/tex]

                             = 3.196

Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.

Therefore, we conclude that percentage is actually more than the reported percentage which means we have sufficient evidence at the 0.02 level to support the executive's claim.

Answer:

Step-by-step explanation:

The detailed steps and analysis is as shown in the attached file.

To write the system of equations in matrix form, use p = [R, F] and p' = [1.5R - 0.2F + 100]. To find the vector of rabbits and foxes after time V, integrate p' twice. To find the vector of rabbits and foxes after time a, integrate p' once. To find the vector of rabbits and foxes after 2 years, substitute n=2 and use summation notation.

(a) To write the system of equations in matrix form, we have:

R' = 1.5R - 0.2F + 100

We can represent the variables R and F as a vector p, so that p = [R, F]. The derivative of p with respect to time (p') is then given by:

p' = [1.5R - 0.2F + 100]

(b) To find the vector of rabbits and foxes after time V, we can integrate the equation given in part (a) with respect to time twice:

p'' = ∫[1.5R - 0.2F + 100]dt

(c) To find the vector of rabbits and foxes after time a, we can integrate the equation given in part (a) with respect to time:

p(a) = ∫[1.5R - 0.2F + 100]dt

(d) To find the vector of rabbits and foxes after 2 years, we can substitute n=2 into the equation given in part (c) and use summation notation:

p(2) = ∫[1.5R - 0.2F + 100]dt + ∑[∫[1.5R - 0.2F + 100]dt]^n

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Answer:

Hence the values for MOE and are interval estimate of 19 yr old age drivers in 1983  are 1%(0.01) and 19±0.0196

Step-by-step explanation:

Given: Refer the paper university of Michigan transportation research  Institute website ,April,2012.

It say that 19 year old group age people ,about 87% has driving license .

And of 1200 has a random sample space ,in 1983.

To Find : MOE(margin of error ) and   Interval Estimate  in 1983.

Solution:

Given that ,there is sample space of 1200 19 yr old people  and of which 87% has driving license.

hence , we get that

87% of 1200 = sample  size.

sample size =1044 members has driving license out of 1200.

Consider 95 % of confidence level,

for that Z-score is required.

calculating the Alpha for that=1-confidence level.

=1-0.95=0.05

therefore Z-alpha=Z(0.05)=1.96

1)MOE=margin of error is given by ,

=[tex]Z-alpha*\sqrt{\frac{p(1-p)}{N} }[/tex]

here p=0.87 and N=total size =1200.

MOE=1.96*[tex]\sqrt{\frac{0.87(1-0.87)}{1200} }[/tex]  =1 %.

2) Interval estimate is given by ,

For that we should know mean,standard deviation and sample size,

we are calculating for 19 year old age of entire age of population,

Hence mean will be 19 yr-old age.

Mean=19.

Sample size=1044.

standard deviation given by,

=[tex]p*(\sqrt{(1-p)}[/tex]

=0.87*[tex]\sqrt{0.13}[/tex]=0.3136.

Hence now calculating the interval estimate ,with 95% confidence level,

μ=M±Z(Standard error )

standard error=standard deviation/sqrt(sample size)

=0.3136/32.31=0.01.

μ=19±1.96*0.01

μ=19±0.0196.

Answer: Length = x² - 6x - 16

Step-by-step explanation:

The formula for determining the area of a rectangle is expressed as. Area = length × width

Length = Area/width

The area of a rectangle is represented by the function

x³ - 2x² - 40x - 64. The width of the rectangle is x + 4. Therefore,

Length = (x³ - 2x² - 40x - 64)/(x + 4)

We would apply the method of long division. The steps are shown in the attached photo. From the photo,

Length = x² - 6x - 16

Hey love! <3

Answer:

⋆ ☄.

·˚ * I see your answer up in the stars! It's x^2 − 6x − 16 or C on your FLVS quiz!

Step-by-step explanation:

If area = length * width and our width is given as x + 4, then the length is found by dividing the area by the width; You could do that using long division, but it's easier using synthetic division. That's what we'll do here for example.

-4 |   1   -2   -40   -64  

This is only the start! The -4 inside the box comes from the factor you are dividing by. If x + 4 = 0, then x = -4.. The numbers after are the coefficients from each descending power of x. Multiply the -4 by the 1 and put that product up under the -2 and add to get:

-4 |   1   -2   -40   -64

After you finish the multiplication and simplification process you receive your given answer = x^2 − 6x − 16

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Answer:

uniformly distributed random numbers here the MATLAB code to find for any number of people and interval

X = rand

X = rand(n)

X = rand(sz1,...,szN)

X = rand(sz)

X = rand(___,typename)

X = rand(___,'like',p)

where

X = rand returns a single uniformly distributed random number in the interval (0,1).

X = rand(n) returns an n-by-n matrix of random numbers.

X = rand(sz1,...,szN) returns an sz1-by-...-by-szN array of random numbers where sz1,...,szN indicate the size of each dimension. For example, rand(3,4) returns a 3-by-4 matrix.

X = rand(sz) returns an array of random numbers where size vector sz specifies size(X). For example, rand([3 4]) returns a 3-by-4 matrix.

X = rand(___,typename) returns an array of random numbers of data type typename. The typename input can be either 'single' or 'double'. You can use any of the input arguments in the previous syntaxes.

X = rand(___,'like',p) returns an array of random numbers like p; that is, of the same object type as p. You can specify either typename or 'like', but not both.

The sum of the squared residuals of the least squares line for the given data is 8691.90.

To compute the sum of the squared residuals for the given data, we will perform the steps of calculating the least squares line manually. The least squares line represents the linear regression line that minimizes the sum of the squared differences between the observed data points and the predicted values.

Week 1: Number of Grunts = 90, Age (days) = 125

Week 2: Number of Grunts = 68, Age (days) = 141

Week 3: Number of Grunts = 39, Age (days) = 155

Week 4: Number of Grunts = 44, Age (days) = 160

Week 5: Number of Grunts = 63, Age (days) = 167

Week 6: Number of Grunts = 40, Age (days) = 174

Week 7: Number of Grunts = 62, Age (days) = 183

Week 8: Number of Grunts = 17, Age (days) = 189

Week 9: Number of Grunts = 20, Age (days) = 195

Calculate the means of the Number of Grunts (Y) and Age (X) variables:

Mean of Y (Number of Grunts) = (90 + 68 + 39 + 44 + 63 + 40 + 62 + 17 + 20) / 9 = 48.78

Mean of X (Age) = (125 + 141 + 155 + 160 + 167 + 174 + 183 + 189 + 195) / 9 = 165

Calculate the deviations from the means for each data point:

For each data point, subtract the mean of X (Age) from the specific X value and the mean of Y (Number of Grunts) from the specific Y value.

Deviation from mean for each X value:

125 - 165.67 = -40.67

141 - 165.67 = -24.67

155 - 165.67 = -10.67

160 - 165.67 = -5.67

167 - 165.67 = 1.33

174 - 165.67 = 8.33

183 - 165.67 = 17.33

189 - 165.67 = 23.33

195 - 165.67 = 29

Deviation from mean for each Y value:

90 - 48.78 = 41.22

68 - 48.78 = 19.22

39 - 48.78 = -9.78

44 - 48.78 = -4.78

63 - 48.78 = 14.22

40 - 48.78 = -8.78

62 - 48.78 = 13.22

17 - 48.78 = -31.78

20 - 48.78 = -28.78

Calculate the sum of the products of the deviations:

Sum of (Deviation from mean for X * Deviation from mean for Y)

= (-40.67 * 41.22) + (-24.67 * 19.22) + (-10.67 * -9.78) + (-5.67 * -4.78) + (1.33 * 14.22) + (8.33 * -8.78) + (17.33 * 13.22) + (23.33 * -31.78) + (29.33 * -28.78)

= -5.02 + -0.90 + 1.04 + 0.27 + 18.95 + -73.14 + 228.44 + -739.97 + -845.44

= -1407.77

Calculate the sum of the squared deviations for X:

Sum of (Deviation from mean for X)^2

= (-40.67)^2 + (-24.67)^2 + (-10.67)^2 + (-5.67)^2 + (1.33)^2 + (8.33)^2 + (17.33)^2 + (23.33)^2 + (29.33)^2

= 16572.86

Calculate the sum of the squared residuals:

Sum of squared residuals = Sum of (Deviation from mean for Y)^2

= (41.22)^2 + (19.22)^2 + (-9.78)^2 + (-4.78)^2 + (14.22)^2 + (-8.78)^2 + (13.22)^2 + (-31.78)^2 + (-28.78)^2

= 8691.90

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The sum of the squared residuals of the least squared line for the given data is 8399.

To compute the sum of the squared residuals of the least squared line by hand, you will need to follow these steps:

1. Start by organizing the given data into two columns: one for the number of grunts and one for the age of the warthog.

Week 1: Number of Grunts = 90, Age (days) = 125

Week 2: Number of Grunts = 68, Age (days) = 141

Week 3: Number of Grunts = 39, Age (days) = 155

Week 4: Number of Grunts = 44, Age (days) = 160

Week 5: Number of Grunts = 63, Age (days) = 167

Week 6: Number of Grunts = 40, Age (days) = 174

Week 7: Number of Grunts = 62, Age (days) = 183

Week 8: Number of Grunts = 17, Age (days) = 189

Week 9: Number of Grunts = 20, Age (days) = 195

2. Calculate the mean (average) of the age and the number of grunts. To do this, add up all the values in each column and divide by the total number of data points.

Mean of the age (days):

(125 + 141 + 155 + 160 + 167 + 174 + 183 + 189 + 195) / 9 = 165

Mean of the number of grunts:

(90 + 68 + 39 + 44 + 63 + 40 + 62 + 17 + 20) / 9 = 52

3. Subtract the mean of the age from each age value to get the deviation of each data point from the mean. Similarly, subtract the mean of the number of grunts from each number of grunts value.

Deviation of the age:

125 - 165 = -40

141 - 165 = -24

155 - 165 = -10

160 - 165 = -5

167 - 165 = 2

174 - 165 = 9

183 - 165 = 18

189 - 165 = 24

195 - 165 = 30

Deviation of the number of grunts:

90 - 52 = 38

68 - 52 = 16

39 - 52 = -13

44 - 52 = -8

63 - 52 = 11

40 - 52 = -12

62 - 52 = 10

17 - 52 = -35

20 - 52 = -32

4. Square each deviation value obtained in step 3.

Squared deviation of the age:

[tex](-40)^2 = 1600[/tex]

[tex](-24)^2 = 576[/tex]

[tex](-10)^2 = 100[/tex]

[tex](-5)^2 = 25[/tex]

[tex]2^2 = 4[/tex]

[tex]9^2 = 81[/tex]

[tex]18^2 = 324[/tex]

[tex]24^2 = 576[/tex]

[tex]30^2 = 900[/tex]

Squared deviation of the number of grunts:

[tex]38^2 = 1444[/tex]

[tex]16^2 = 256[/tex]

[tex](-13)^2 = 169[/tex]

[tex](-8)^2 = 64[/tex]

[tex]11^2 = 121[/tex]

[tex](-12)^2 = 144[/tex]

[tex]10^2 = 100[/tex]

[tex](-35)^2 = 1225[/tex]

[tex](-32)^2 = 1024[/tex]

5. Sum up all the squared deviation values obtained in step 4.

Sum of squared residuals:

1600 + 576 + 100 + 25 + 4 + 81 + 324 + 576 + 900 + 1444 + 256 + 169 + 64 + 121 + 144 + 100 + 1225 + 1024 = 8399

Therefore, the sum of the squared residuals of the least squared line for the given data is 8399.

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Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: number of individuals that hold multiple jobs in a sample of 225.

And you need to calculate the probability of the proportion of individuals being between 0.14 and 0.18.

Considering that the parameter of interest is the proportion and the sample is large enough you can use the standard normal approximation to calculate this interval.

[tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]

Unfortunately, there is no given value for the population proportion of individuals that have multiple jobs. Let's say, for example, that his proportion is 10%.

You can symbolize the probabilities as:

P(0.14≤X≤0.18)= P(X≤0.18)-P(X≤0.14)

Using the approximation of the standard normal you can standardize these proportion values:

P(Z≤(0.18-0.1)/√[(0.1*0.9)/225])-P(Z≤(0.14-0.1)/√[(0.1*0.9)/225])

P(Z≤4)-P(Z≤2)

Now you have to look for the corresponding values in the table of the Z distribution (right table, positive numbers of Z)

P(Z≤4)-P(Z≤2)= 1 - 0.97725= 0.02275

I hope it helps!

Answer:

0.1276

Step-by-step explanation:

Let sample space = S

∴ n(S) = 225

Probability of multiple jobs A, that is, P(A) = 0.14

Probability of multiple jobs B, that is, P(B) = 0.18

Hence, P(A) = n(A)/n(S)

∴ 0.14 = n(A)/225

n(A) = 0.14 X 225 = 31.5%

Also, P(B) = n(B)/n(S)

∴ 0.18 = n(B)/225

n(B) = 0.18 X 225 = 40.5%

The probability that the multiple jobs, P( A2 ∩ B2) = P(A2) X P(B2)

P(A2) = 0.315 and P(B2) = 0.405

∴ P(A2 ∩ B2) =  P(A2) X P(B2) = 0.315 X 0.405 = 0.1276

Answer:

198 students and 127 adults

Step-by-step explanation:

Answer:

34.2022 < X < 39.7978

Step-by-step explanation:

For a 95% confidence interval, Z = 1.960

Sample size (n) = 33

Mean weight (X) = 37 ounces

Standard deviation (s) = 8.2 ounces

The relationship that describes a 95% confidence interval is:

[tex]X \pm 1.960*\frac{s}{\sqrt{n}}[/tex]

Applying the given data, the Lower (L) and Upper (U) limits are:

[tex]U=37 + 1.960*\frac{8.2}{\sqrt{33}} \\U=39.7978\\L=37 - 1.960*\frac{8.2}{\sqrt{33}} \\L=34.2022[/tex]

The 95% confidence interval is:

34.2022 < X < 39.7978

Elaborating on the matching of the sampling method descriptions:

To summarize, the correct matching would be:

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1: d. Convenience

2: e. Simple random

3: c. Cluster

4: a. Stratified

5: b. Systematic

1. Convenience sampling (d):

2. Simple random sampling (e):

3. Cluster sampling (c):

4. Stratified sample(s):

5. Systematic sampling (b):

Complete question:

Match the names of the given sampling method descriptions.

Situation:

1. drawing 50 names from a hat

2. ask all the students in your math class

3. randomly select two tables in the dining room and examine all the people at those two tables

4. dividing all students according to grades and selecting 10 students from each grade

5. call every 15th phone number on every 5th page of the phone book

Sampling method:

a. Stratified

b. Systematic

c. Cluster

d. Convenience

e. Simple random

Answer:

Test statistic = 3.1587

Step-by-step explanation:

We are given that the Toy-lot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A).

Also, a sample of eleven motors was tested, and it was found that the mean current was x = 1.20 A, with a sample standard deviation of s = 0.42 A.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 0.8 { claim of 0.8 A is not low}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 0.8 { claim of 0.8 A is too low}

Now, the test statistics used here will be;

           T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, X bar = sample mean = 1.20 A

                 s = sample standard deviation = 0.42 A

                 n = sample size = 11 motors

So, Test statistics = [tex]\frac{1.20 - 0.8}{\frac{0.42}{\sqrt{11} } }[/tex] ~ [tex]t_1_0[/tex]

                             = 3.1587

At 1% level of significance, t table gives a critical value of 2.764 at 10 degree of freedom. Since our test statistics is higher than the critical value so we have sufficient evidence to reject null hypothesis .

Therefore, we conclude that Toy-lot claim of 0.8 A is too low.

To determine if the Toylot company claim of 0.8 A is too low, a one-sample t-test can be used. The sample test statistic is calculated by subtracting the hypothesized mean from the sample mean, divided by the sample standard deviation divided by the square root of the sample size. This statistic can then be compared to the critical value from the t-distribution at the specified level of significance to determine if the claim is too low.

To determine if the Toylot company claim of 0.8 A is too low, we can conduct a hypothesis test using the sample data. We will use a one-sample t-test with a null hypothesis that the mean current is equal to 0.8 A. The alternative hypothesis will be that the mean current is greater than 0.8 A.

The sample test statistic is calculated by subtracting the hypothesized mean from the sample mean, and dividing by the sample standard deviation divided by the square root of the sample size. In this case, the sample test statistic is (1.20 - 0.8) / (0.42 / sqrt(11)) = 5.11.

To determine if this test statistic is statistically significant, we compare it to the critical value from the t-distribution with 10 degrees of freedom at a 1% level of significance. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the Toylot claim of 0.8 A is too low.

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Answer:

a) L(x,y,z) = (-4,0,2)+(2,8,-7)*t

b) (2-z)/7= y/8=(x+4)/2 (option B)

Step-by-step explanation:

the parametric equation of the line passing through the point P₀= (-4,0,2) and parallel to the vector v=2i + 8j - 7k is

L(x,y,z)=P₀+v*t

therefore

L(x,y,z) = (-4,0,2)+(2,8,-7)*t

or

x=x₀+vx*t = -4 + 2*t

y=y₀+vy*t = 8*t

z=z₀+vz*t = 2 -7*t

solving for t in the 3 equations we get the symmetric equation of the line:

(2-z)/7= y/8=(x+4)/2

thus the option B is correct

Answer:

a) Parametric equations

x = -4 + 2t

y = 8t

z = 2-7t

b) symmetric equations

[tex]\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}[/tex]          The answer is the option B

Step-by-step explanation:

For writing the vectorial equation of a line, we need a point in the line and its director vector, thus:

[tex]L: (x_{0},y_{0},z_{0} ) + t(a,b,c)[/tex]

                            Where   [tex](x_{0},y_{0},z_{0})[/tex] is a point in the line

                                           (a,b,c) is the director vector

Then

[tex]L: (-4,0,2) + t(2,8,-7)[/tex]

a) Parametric equations

Since the vectorial equation, we can obtain the parametric equations writing the equation for each component

x = -4 + 2t

y = 8t

z = 2-7t

b)Symmetric equations

Since the parametric equations, we isolate the parameter t

[tex]\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}[/tex]

The experimental design used in this assessment is a pretest-posttest design where the difference between pre- and posttest scores is calculated.

The experimental design used in this assessment is a pretest-posttest design. A pretest is administered before the students receive the intervention (in this case, the math course), and a posttest is administered after the intervention. The difference between the pre- and posttest scores is then calculated to determine the knowledge-gained score.

This design allows for the comparison of students' scores before and after the math course, giving insight into the effectiveness of the course in improving their knowledge. By comparing the two test scores, the professor can assess how much knowledge students have gained as a result of the course.

Example:

Before the math course, a student's pretest score is 50. After completing the course, the student's posttest score is 75. The knowledge-gained score is calculated as 75 - 50 = 25. This indicates that the student gained 25 points of knowledge between the pre- and posttests.

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Answer:

1) (f + g) (5) = 18, 2) (f + g) (1/2) = - 13/4, 3) (f - g) (-1) = 14, 4) (f * g) (7) = 456, 5) (f/g) (- 2) = - 3/4, 6) (g ○ f)(x) = 6 x - 23, (f ○ g)(x) = 6 x - 6, 7) [tex](g \circ f) (x) = \frac{9}{x} - 24[/tex], [tex](f \circ g) (x) = \frac{1}{x - 2} + 5[/tex], 8) (f ○ g)(5) = 3

Step-by-step explanation:

1) [tex](f + g) (x) = x^{2}-8\cdot x + 3 + 7 \cdot x - 5\\(f + g) (x) = x^{2} - x - 2\\(f + g) (5) = 18[/tex]

2) [tex](f + g) (x) = x^{2} - 7 \cdot x + 4 + 6 \cdot x - 7\\(f + g) (x) = x^{2} - x - 3\\(f + g) (\frac{1}{2} ) = - \frac{13}{4}[/tex]

3) [tex](f - g) (x) = x^{2} - 3 \cdot x + 4 - 4 \cdot x + 2\\(f - g) (x) = x^{2} - 7 \cdot x + 6\\(f - g) (-1) = 14[/tex]

4) [tex](f \cdot g) (x) = (x^{2}-4\cdot x + 3) \cdot (3\cdot x - 2)\\(f \cdot g) (7) = 456[/tex]

5) [tex](f / g) (x) = \frac{x^{2}-3\cdot x + 2}{4 \cdot x - 8} \\(f / g) (-2) = - \frac{3}{4}[/tex]

6) [tex](g \circ f) (x) = 3 \cdot (2 \cdot x - 8) + 1\\(g \circ f) (x) = 6 \cdot x - 23\\(f \circ g) (x) = 2 \cdot (3 \cdot x + 1) - 8\\(f \circ g) (x) = 6 \cdot x - 6[/tex]

7) [tex](g \circ f) (x) = 3 \cdot (\frac{3}{x} - 6) - 6\\(g \circ f) (x) = \frac{9}{x} - 24\\(f \circ g) (x) = \frac{3}{3 \cdot x - 6} + 5\\(f \circ g) (x) = \frac{1}{x - 2} + 5[/tex]

8) [tex](f \circ g) (x) = 2 \cdot (x^{2} - 5 \cdot x) + 3\\(f \circ g) (x) = 2 \cdot x ^{2} - 10 \cdot x + 3\\(f \circ g) (5) = 3[/tex]

1. (f + g)(5) = 18.

2. (f + g)(1/2) = -13/4.

3. (f - g)(-1) = 14.

4. (fg)(7) = 456.

5. (f/g)(-2) = -3/4.

6. (g ○ f)(x) = 6x - 23, (f ○ g)(x) = 6x - 6.

7. (g ○ f)(x) = [tex]\(\frac{9}{x+5} - 6\), (f ○ g)(x) = \(\frac{3}{3x-1}\).[/tex]

8. (f ○ g)(5) = 3.

Let's tackle each problem step by step:

1.Evaluate (f + g)(5):

  Step 1: Find f(5) and g(5).

 [tex]\(f(5) = (5)^2 - 8(5) + 3 = 25 - 40 + 3 = -12\)[/tex]

[tex]\(g(5) = 7(5) - 5 = 35 - 5 = 30\)[/tex]

  Step 2: Add f(5) and g(5).

 [tex]\((f + g)(5) = f(5) + g(5) = -12 + 30 = 18\)[/tex]

2. Evaluate (f + g)(1/2):

  Step 1: Find f(1/2) and g(1/2).

 [tex]\(f(1/2) = (1/2)^2 - 7(1/2) + 4 = 1/4 - 7/2 + 4 = 1/4 - 14/4 + 16/4 = 3/4\)[/tex]

  [tex]\(g(1/2) = 6(1/2) - 7 = 3 - 7 = -4\)[/tex]

  Step 2: Add f(1/2) and g(1/2).

[tex]\((f + g)(1/2) = f(1/2) + g(1/2) = 3/4 - 4 = -13/4\)[/tex]

3.Evaluate (f − g)(−1):

  Step 1: Find f(−1) and g(−1).

 [tex]\(f(-1) = (-1)^2 - 3(-1) + 4 = 1 + 3 + 4 = 8\)[/tex]

  [tex]\(g(-1) = 4(-1) - 2 = -4 - 2 = -6\)[/tex]

  Step 2: Subtract g(-1) from f(-1).

  [tex]\((f - g)(-1) = f(-1) - g(-1) = 8 - (-6) = 8 + 6 = 14\)[/tex]

4.Evaluate (fg)(7):

  Step 1: Find f(7) and g(7).

[tex]\(f(7) = (7)^2 - 4(7) + 3 = 49 - 28 + 3 = 24\)[/tex]

  [tex]\(g(7) = 3(7) - 2 = 21 - 2 = 19\)[/tex]

  Step 2: Multiply f(7) by g(7).

 [tex]\((fg)(7) = f(7) \times g(7) = 24 \times 19 = 456\)[/tex]

5. Evaluate (f/g)(−2):

  Step 1: Find f(-2) and g(-2).

[tex]\(f(-2) = (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12\)[/tex]

  [tex]\(g(-2) = 4(-2) - 8 = -8 - 8 = -16\)[/tex]

  Step 2: Divide f(-2) by g(-2).

  [tex]\((f/g)(-2) = \frac{f(-2)}{g(-2)} = \frac{12}{-16} = -\frac{3}{4}\)[/tex]

6. Find (g ○ f)(x) and (f ○ g)(x) for f(x) = 2x − 8, g(x) = 3x + 1:

 [tex]\((g ○ f)(x) = g(f(x)) = g(2x - 8) = 3(2x - 8) + 1 = 6x - 24 + 1 = 6x - 23\)[/tex]

 [tex]\((f ○ g)(x) = f(g(x)) = f(3x + 1) = 2(3x + 1) - 8 = 6x + 2 - 8 = 6x - 6\)[/tex]

7. Find (g ○ f)(x) and (f ○ g)(x) for f(x) = 3/x+5, g(x) = 3x − 6:

 [tex]\((g ○ f)(x) = g(f(x)) = g(\frac{3}{x+5}) = 3(\frac{3}{x+5}) - 6 = \frac{9}{x+5} - 6\)[/tex]

 [tex]\((f ○ g)(x) = f(g(x)) = f(3x - 6) = \frac{3}{3x-6+5} = \frac{3}{3x-1}\)[/tex]

8. Evaluate (f ○ g)(5):

  Step 1: Find g(5).

  [tex]\(g(5) = 5^2 - 5 \times 5 = 25 - 25 = 0\)[/tex]

  Step 2: Find f(g(5)).

 [tex]\(f(g(5)) = f(0) = 2 \times 0 + 3 = 3\)[/tex]

Answer:

y=1/2x+2

Step-by-step explanation:

Answer: y = 1/2x + 2

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the graph,

y2 = 4

y1 = 2

x2 = 4

x1 = 0

Slope,m = (4 - 2)/(4 - 0) = 2/4 = 1/2

To determine the y intercept, we would substitute x = 4, y = 4 and

m = 1/2 into y = mx + c. It becomes

4 = 1/2 × 4 + c

4 = 2 + c

c = 4 - 2 = 2

The equation becomes

y = x/2 + 2

Answer: FIRST OPTION.

Step-by-step explanation:

First, it is important to remember that the Slope-Intercept form of the equation of a line is the shown below:

[tex]y=mx+b[/tex]

Where "m" is the slope of the line and "b" is the y-intercept.

By definiton, given a System of Linear equations, if they are exactly the same line, then the System of equations have Infinely many solutions.

In this case you have the following System of Linear equations given in the exercise:

[tex]\left \{ {{y=-2x+5} \atop {y=ax+b}} \right.[/tex]

So, since you need the system has Infinite solutions, you know that the slope and the y-intercept of both lines must be equal.

Therefore, you can identify that the value of "a" and "b" must be the following:

[tex]a=-2\\\\b=5[/tex]

So the Linear System would be the shown below:

[tex]\left \{ {{y=-2x+5} \atop {y=-2x+5}} \right.[/tex]

Answer:

Probability that a randomly selected student scored less than 156 on the final exam is 0.97725 .

Step-by-step explanation:

We are given that the exam scores on a statistics final exam are normally distributed with a mean of 140 points out of 200 and standard deviation of 8.

Let X = Score of students in exam

So, X ~ N([tex]\mu = 140, \sigma^{2} =8^{2}[/tex])

The z score probability distribution is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 140

            [tex]\sigma[/tex] = standard deviation = 8

So, the probability that a randomly selected student scored less than 156 on the final exam is given by = P(X < 156)

 P(X < 156) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{156-140}{8}[/tex] ) = P(Z < 2) = 0.97725 .

Therefore, the probability that a randomly selected student scored less than 156 on the final exam is 0.97725 .

To find the probability, calculate the z-score and look it up in a table or use a calculator. The probability of scoring less than 156 is approximately 0.9772.

To find the probability that a randomly selected student scored less than 156 on the final exam, we will use the z-score formula. The z-score formula is calculated by subtracting the mean from the given score and dividing by the standard deviation. In this case, the z-score is (156 - 140) / 8 = 2.

To find the probability, we can look up the z-score in a standard normal distribution table or use a calculator like the TI-84. Using either method, we find that the probability of a student scoring less than 156 is approximately 0.9772.

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Answer:

P

Step-by-step explanation:

(all 5 correctly) = (5 chosen among 7) / (5 chosen among 10)

Answer:

(a) The 5-hour decay factor is 0.5042.

(b) The 1-hour decay factor is 0.8720.

(c) The amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

Step-by-step explanation:

The amount of caffeine in Chase's body decreases exponentially.

The 10-hour decay factor for the number of mg of caffeine is 0.2542.

The 1-hour decay factor is:

[tex]1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720[/tex]

(a)

Compute the 5-hour decay factor as follows:

[tex]5-hour\ decay\ factor=(0.8720)^{5}\\=0.504176\\\approx0.5042[/tex]

Thus, the 5-hour decay factor is 0.5042.

(b)

The 1-hour decay factor is:

[tex]1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720[/tex]

Thus, the 1-hour decay factor is 0.8720.

(c)

The equation to compute the amount of caffeine in Chase's body is:

A = Initial amount × (0.8720)ⁿ

It is provided that initially Chase had 171 mg of caffeine, 1.39 hours after consuming the drink.

Compute the amount of caffeine in Chase's body 2.39 hours after consuming the drink as follows:

[tex]A = Initial\ amount \times (0.8720)^{2.39} \\=[Initial\ amount \times (0.8720)^{1.39}] \times(0.8720)\\=171\times 0.8720\\=149.112[/tex]

Thus, the amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

The 5-hour growth/decay factor is 0.561, the 1-hour growth/decay factor is 0.869, and there are 154 mg of caffeine in Chase's body 2.39 hours after consuming the energy drink.

a. To find the 5-hour growth/decay factor, we raise the 10-hour decay factor to the power of (5/10). So, the 5-hour growth/decay factor is 0.2542^(5/10) = 0.561.

b. To find the 1-hour growth/decay factor, we raise the 10-hour decay factor to the power of (1/10). So, the 1-hour growth/decay factor is 0.2542^(1/10) = 0.869.

c. To find the number of mg of caffeine in Chase's body 2.39 hours after consuming the energy drink, we multiply the initial amount of caffeine by the 2.39-hour decay factor. So, the number of mg is 171 * 0.2542^(2.39/10) = 154.

Answer:

No , the cause and effect can be finished up through analyses as it were.  

What we have in the inquiry is only an observational investigation where we basically study 3000 grown-ups and attempt to outline the outcomes with no trial proof.  

Imagine a scenario in which individuals who had breakfast normally were inclined to maintain their weight reduction, for sure if individuals who keep up weight reduction will in general have breakfast routinely.  

henceforth the circumstances and logical results relationship cannot be built up  

So as to do so , one must lead measurable trials, for example, autonomous example t test or ANOVA examination

The student's question pertains to calculating the regression line equation that represents the relationship between rainfall and wheat yield. This can be done using a calculator's statistical functions to provide the least-squares regression line equation, which allows for the prediction of one variable based on another.

The question asks about finding the equation for a linear regression line based on the relationship between rainfall and wheat yield. To accomplish this, one must enter the given data into a calculator, generate a scatter plot, and then use the calculator's regression function to find the least-squares regression line's equation. This regression equation typically takes the form î = a + bx, where 'a' is the y-intercept and 'b' is the slope of the line.

In terms of finding the regression line manually, a commonly used method is the median-median line approach, but for this question, we are instructed to use a calculator. The process involves statistical analysis to determine the line that minimally deviates from all data points in the scatter plot. Once the equation is calculated, one can predict values such as wheat yield for a given amount of rainfall.

Answer:

58.52

Step-by-step explanation:

The predicted regression equation for predicting Exam 1 score is

y_hat=58.52+2.19x.

We have to find the predicted exam score 1 for student who did not submit homework. If the student did not submit homework 1 score then the homework 1 score will be zero. So,

y_hat=58.52+2.19(0)

y_hat=58.52.

Thus, the predicted exam score 1 for student who did not submit homework 1 score is 58.52.

Answer:

The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.

Step-by-step explanation:

The point estimate is the halfway point of the confidence interval, that is, the lower bound added to the upper bound, and then this sum is divided by 2. So

Lower bound: 62.535

Upper bound: 76.478

Point estimate:

[tex]P_{e} = \frac{62.535 + 76.478}{2} = 69.505[/tex]

The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.

The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.

Since Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478].

So, the estimation of the point is

[tex]= (62.532 + 76.478) \div 2[/tex]

= 69.505 inches

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Answer:

a) v ∈ ker(L) if only if [tex][V]_{E}[/tex] ∈ N(A)

b) w ∈ L(v) if and only if [tex][W]_{F}[/tex] is in the column space of A

See attached

Step-by-step explanation:

See attached the proof Considering the vector spaces V and W with other bases E and F respectively.

Let L be the Linear transformation form V and W and A is the matrix representing L relative to E and F

Answer:

Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.

Step-by-step explanation:

We use the central limit theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a sample size larger than 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

So

Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.

Answer:

We would have

                                    [tex]l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}[/tex]

where " l " is  length, " w"  is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

         Surface area for a box with no top = [tex]lw+2lh+2wh = 64[/tex]

where " l " is  length, " w"  is width and "h" is height.

 Step 2.

Remember as well that

                              Volume of the box = [tex]l*w*h[/tex]

Step 3

 We can now use lagrange multipliers.  Lets say,

                                    [tex]F(l,w,h) = lwh[/tex]

and

                                [tex]g(l,w,h) = lw+2lh+2wh = 64[/tex]

By the lagrange multipliers method we know that                            

                                                     [tex]\nabla F = \lambda \nabla g[/tex]

Step 4

Remember that

                          [tex]\nabla F = (wh,lh,lw)[/tex]

and

                      [tex]\nabla g = (w+2h,l+2h , 2w+2l)[/tex]

So basically you will have the system of equations

                              [tex]wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)[/tex]

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

                                   [tex]lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)[/tex]

Then you would get

                      [tex]l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)[/tex]

You can get rid of [tex]\lambda[/tex] from these equations and you would get

                         [tex]lw+2lh = lw+2wh = 2wh+2lh[/tex]

And from those equations you would get

                                             [tex]l = w =2h[/tex]

Now remember the original equation

                                    [tex]lw+2lh+2wh = 64[/tex]

If we plug in what we just got, we would have

                                 [tex]l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}[/tex]

To maximize the volume of an open-topped rectangular box with a fixed surface area, express the volume in terms of a single variable using the provided surface area equations. Solve these using calculus methods to find the maximum volume.

This question involves the field of mathematics, specifically calculus and optimization. As the task is about maximizing the volume of an open-topped rectangular box with a fixed surface area, we can look into the relation between length, width, height, surface area, and volume of the box.

Let's denote the length of the box as x, width as y, and height as z. The volume (V) of a the box is found by multiplying length, width, and height: V = xyz.

Since there is no top, the surface area (A) is calculated by adding together the areas of the bottom and the four sides, which gives us: A = xy + 2xz + 2yz = 64 m².

To maximize the volume, we need to express the volume in terms of a single variable. From the surface area equation, we can solve for z to get: z = (64 - xy) / (2x + 2y). Now substitute z into the volume equation to get V = xy * [(64 - xy) / (2x + 2y)]. Now we can take derivative of V with respect to x and y, set that equal to zero and find when the volume will be maximum. After solving, we will find the dimensions that maximize the volume of the box under given condition.

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Answer:

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

Step-by-step explanation:

For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent from other bulbs, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]

In which [tex]C_{n,x}[\tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[\tex]

And p is the probability of X happening.

Suppose 30% of the bulbs in the lot are defective.

This means that [tex]p = 0.3[/tex]

A quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.

This means that [tex]n = 16[/tex]

What is the probability that exactly 4 bulbs from the sample are defective?

This is P(X = 4).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]

[tex]P(X = 4) = C_{16,4}.(0.3)^{4}.(0.7)^{12} = 0.2040[\tex]

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

Answer:

a. 32,760 ways

b. 1365 ways

b. 1/1365

Step-by-step explanation:

Given

Number of qualified candidates= 15

Number of Vacancies = 4 ( President, CEO, COO and CFO)

a.

How many different ways can the officers be appointed?

This means that in how many ways can 4 candidates be arranged out of a total of 15.

They keyword here is arranged which means Permutation because the order of arrangement doesn't count.

So,

Number of Appointments = 15P4

Number of Appointment = 1365 ways

b. How many different ways can the committee be appointed?

Here, the order of appointments matters.

So, this means, in his many ways can a committee of 4 be selected out of a total of 15.

The keyword, selection means Combination.

So, number of Appointments = 15C4

Number of Appointments = 1365

c. What is the probability of randomly selecting the committee members and getting the five youngest of the qualified candidates?

There are 1365 ways of choosing the board members and the chance of selecting the 5 youngest members is 1, because there's only one way to select the 5 youngest.

So the probability = 1/1365

To appoint officers, there are 32,760 different ways. The committee can be appointed in 1,365 different ways. The probability of randomly selecting the four youngest candidates for the committee is approximately 0.073%.

To solve the problem of appointing corporate officers and a committee, we will go step by step. For part (a), there are 15 qualified candidates, and we need to appoint 4 different officers. Since one person cannot hold more than one officer position simultaneously, we will use permutations. The first officer can be chosen in 15 ways, the second in 14 ways, the third in 13 ways, and the fourth in 12 ways, making the total number of ways to appoint the officers:

15 × 14 × 13 × 12 = 32,760 different ways.

For part (b), after appointing the officers, 11 candidates remain (15 total candidates - 4 officers). However, since officers can also serve on the committee, we still have 15 candidates to choose from. We need to appoint 4 different committee members from these 15 candidates, which is a combination problem because the order in which they're chosen doesn't matter. The number of ways to appoint the committee is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of candidates, and k is the number of positions to fill:

C(15,4) = 15! / (4!(15-4)!) = 15! / (4! × 11!) = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 1,365 different ways.

For part (c), to find the probability of randomly selecting the five youngest candidates of the 15 qualified candidates, we should note that there are 4 positions on the committee, not 5. Assuming 'five youngest' is a typo, the probability of getting the 4 youngest candidates on the committee is:

P(selecting 4 youngest) = 1 / C(15, 4) = 1 / 1,365 ≈ 0.00073, or 0.073%.

Answer:

B

Step-by-step explanation:

The data doesn't follow any linearity. Infact, the given data set is in a range of values. It is distributed around some mean value. So it is plausible that it is normally distributed

The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible.

The question asks whether it is plausible that the compressive strength for this type of concrete is normally distributed. In order to answer this question, we can examine the normal probability plot. If the plot is acceptably linear, it suggests that a normal population distribution is plausible. On the other hand, if the plot is not acceptably linear, it suggests that a normal population distribution is not plausible.

According to the options provided, the correct answer would be C. The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible. This implies that the compressive strength for this type of concrete is not normally distributed.

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