I = 3/2R - 18

Solve for R
(show your work)

The vectors are:

To find the vectors:

Use the graph to solve the problem:

We can add and subtract the vectors:

-To find b - a subtract a from b:

2a means multiply vector a by 2:

Now let's find the resulting vector 2a - b.

- Subtract b from 2a:

The resulting vector 2a - b is <8, -9>

Therefore, the vectors are:

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The complete question is given below:

The graph shows vectors a and b the resulting vector for b - a is blank, blank >, and the resulting vector for 2a -b is < blank, blank >.

The sum of two draws from a box containing 11 tickets labeled 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 is equivalent to the sum of two dice rolls, as the outcomes are not equally likely

What are dice rolls?

Dice rolls are a random event used in various games and activities. A dice typically has six sides, each side marked with a different number of , ranging from 1 to 6. To roll a dice, the player tosses it in the air, causing it to spin and land on a random side.

Dice rolls are often used in games of chance, such in role-playing games, where they are used to determine the outcome of actions or events. They are also used in probability theory, where the probability of a certain outcome can be calculated based on the number of sides on the dice and the number of times it is rolled.

The sum of two dice that are independently rolled and the sum of two draws from a box containing six tickets labeled 1, 2, 3, 4, 5, 6 are indeed equivalent. This is because there are 36 possible outcomes when rolling two dice, and each outcome is equally likely. The sum of the two dice can range from 2 (when both dice show a 1) to 12 (when both dice show a 6).

Similarly, when drawing two tickets from a box containing six tickets labeled 1, 2, 3, 4, 5, 6, there are 36 possible outcomes, with each outcome being equally likely. The sum of the two draws can range from 2 (when both draws yield a 1) to 12 (when both draws yield a 6).

However, if we use a box containing 11 tickets labeled 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, the outcomes will not be equally likely, and the sum of the two draws will not be the same as the sum of two dice rolls. For example, there is only one way to get a sum of 2 with two dice (rolling two 1s), but there are two ways to get a sum of 2 when drawing from the box (drawing a 1 and a 1 or drawing a 2 and a 0).

Therefore, we cannot say that the sum of two draws from a box containing 11 tickets labeled 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 is equivalent to the sum of two dice rolls, as the outcomes are not equally likely.

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Answer:

y = x + 7

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 1, thus

y = x + c ← is the partial equation

To find c substitute (- 3, 4) into the partial equation

4 = - 3 + c ⇒ c = 4 + 3 = 7

y = x + 7 ← equation of line

Hope this helps.

The graph of the linear equation, 5x - 9y = -7, is in the attachment below.

The linear equation that models a graph can be written in  slope-intercept equation as y = mx + b. The value of the slope, which is the rise over the run of the line is represented as m, while the value of b, which is where the line intercepts the y-axis is b.

Given the linear equation, 5x - 9y = -7, rewrite in slope-intercept form:

5x - 9y = -7

-9y = -5x - 7

-9y/-9 = -5x/-9 - 7/-9

y = 5/9x + 7/9

This means the rise over the run of the line, m is 5/9, while the y-intercept of the graph, b is 7/9.

The graph is shown below.

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Answer:

Ues math w/a/y

Step-by-step explanation:

ues it

The given system of equations can be written in matrix form as \[ \mathbf{y}^{\prime}=\left[\begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array}\right] \mathbf{y} . \] where \[ \mathbf{y}=\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right] . \]

To find the general solution, we need to find the eigenvalues and eigenvectors of the coefficient matrix. The characteristic polynomial is given by \[ \det\left(\left[\begin{array}{cc} 1-\lambda & -1 \\ 1 & 3-\lambda \end{array}\right]\right)=\lambda^{2}-4 \lambda+4=(\lambda-2)^{2} . \] Thus, the matrix has a repeated eigenvalue of λ = 2. The eigenvector corresponding to this eigenvalue is found by solving the equation \[ \left[\begin{array}{cc} -1 & -1 \\ 1 & 1 \end{array}\right] \mathbf{x}=\mathbf{0} . \] This gives us the eigenvector \[ \mathbf{x}_{1}=\left[\begin{array}{l} 1 \\ -1 \end{array}\right] . \] Since the eigenvalue is repeated, we need to find a generalized eigenvector by solving the equation \[ (\mathbf{A}-2\mathbf{I})\mathbf{x}_{2}=\mathbf{x}_{1},\] where $\mathbf{A}$ is the coefficient matrix.

This gives us the generalized eigenvector\[ \mathbf{x}_{2}=\left[\begin{array}{l} 0 \\ 1 \end{array}\right].\] The general solution to the system of differential equations is then given by\[ y(t)=c_{1}\left[\begin{array}{l} 1 \\ -1\end{array}\right]e^{2t}+c_{2}\left(\left[\begin{array}{l} 0 \\ 1\end{array}\right]+t\left[\begin{array}{l} 1 \\ -1\end{array}\right]\right)e^{2t},\] where $c_1$ and $c_2$ are constants determined by initial conditions.

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Answer:

Step-by-step explanation:

26) abs(10)=10

The absolute value of a positive number equals to that positive number. It cannot be a negative number. Therefore, the statement is false.

27)  abs(-5)<4

abs(-5)=5>4

The absolute value of a negative number equals to that number as positive, therefore, 5 is bigger/larger than 4, which makes the statement false.

Please mark me as Brainliest if you're satisfied with the answer.

The scatter plot showing Troy's data is given at the end of the answer.

A scatter plot shows a series of points (x,y), relating the input x to the output y.

For this problem, it is used because we want to graph a discrete function, as:

Both professional year and home runs are countable amounts, that cannot assume decimal values, hence we have a discrete function and use a scatter plot.

From the table presented in this problem, the points for the scatter plot are given as follows:

(1,6), (2,9), (3,13), (4,7), (5,9).

The scatter plot, labeled and with the points, is shown by the graph at the end of the answer.

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Answer:

1. ( w + 7 ) ( w + 2 )

2. ( x - 9 ) ( x - 5 )

Step-by-step explanation:

1.

w² + 9w + 14

= w² + 7w + 2w + 14

= w ( w + 7 ) + 2 ( w + 7 )

= ( w + 7 ) ( w + 2 )

Hence, factorised.

2.

x² - 14x + 45

= x² - 5x - 9x + 45

= x ( x - 5 ) - 9 ( x - 5 )

= ( x - 9 ) ( x - 5 )

Hence, factorised.

Random error is an unpredictable deviation from the true value or expected outcome in a measurement or experiment.

Random error refers to the unpredictable and uncontrollable variations that occur when making measurements or conducting experiments. It is a type of error that introduces inconsistencies and fluctuations in the data, leading to discrepancies between the observed values and the actual values.

Random errors can arise from various sources, such as equipment limitations, environmental factors, or human factors like imprecise readings or variations in technique.

Unlike systematic errors, which are consistent and biased in a particular direction, random errors are characterized by their lack of pattern or regularity. They are typically caused by factors that are difficult to quantify or account for, making it challenging to eliminate them entirely.

However, by taking repeated measurements and applying statistical analysis techniques, researchers can mitigate the impact of random errors and obtain more reliable and accurate results.

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Answer:

1..9.9

2....23.9

3...106.9

4....21.8

Step-by-step explanation:

1-9.9

2-23.9

3-106.9

4-21.8

5-95.01

6-715.8

To test the hypothesis that the mean weight of two sheets is equal (μ1 - μ2) against the alternative that it is not, and assuming equal variances, we can use a two-sample t-test. The t-statistic can be calculated using the following formula:

t = (x1 - x2) / (s_p * sqrt(1/n1 + 1/n2))

where:

x1 and x2 are the sample means of the two sheets,

s_p is the pooled standard deviation,

n1 and n2 are the sample sizes.

The pooled standard deviation (s_p) can be calculated using the following formula:

s_p = sqrt(((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2))

where:

s1 and s2 are the sample standard deviations.

To calculate the t-statistic, we need the sample means, sample standard deviations, and sample sizes.

Once you provide the specific values for these variables, I can assist you in calculating the t-statistic to 3 decimal places.

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To test the hypothesis that the mean weight of the two sheets is equal (μ1 - μ2) against the alternative that it is not, we can use a paired t-test assuming equal variances. The paired t-test is used when we have paired data or measurements on the same subjects or objects.

The t-statistic for a paired t-test is calculated as follows:

t = (X1 - X2) / (s / √n)

where X1 and X2 are the sample means of the two samples, s is the pooled standard deviation, and n is the number of pairs.

Please provide the sample means, standard deviation, and sample size for each sheet so that we can calculate the t-statistic to 3 decimal places.

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The probability that a single randomly selected value is between -0.2 and 32.7 is 0.0039.

We have to find the required probability:

we have the normal distribution with u = 0.2 and 32.7

sample size = n = 108

the values have normal distribution,

let x be the values, so x follows normal distribution,

so x ~ n(u = 0.2, 32.7)

If x~ n then z = x-n ~(0,1)

let s = sample mean

a) to find the probability distribution of randomly selected value =

P (-0.2 < N (0,1) < 32.7)

when we solve for n, the probability that a single randomly selected value is between -0.2 and 32.7 is 0.0039.

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Answer:

B - 55

Step-by-step explanation:

Answer:

25.2X + 3.4 centimeters.

Step-by-step explanation:

The formula for the perimeter of a rectangle

= 2L + 2W

= 2(L + W)

From the question:

Width of a rectangle (W) = 9.7 X - 4.7 centimeters

Length of a rectangle (L) = 2.9 X + 6.4 centimeters

Imputing it into the formula

2(L + W)

2( 9.7X - 4.7 + 2.9X + 6.4)

2 (9.7X + 2.9X - 4.7 + 6.4)

2( 12.6X + 1.7)

= 25.2X + 3.4 centimeters.

Therefore, the perimeter of the rectangle = 25.2X + 3.4 centimeters.

Answer:The answer is 18.1

Step-by-step explanation:

Answer:

35 Degrees

Step-by-step explanation:

Given in figure.

Happy to help :)

The solution is, the value of the angle ADF = 35° .

Here, we have,

from the given diagram, we get,

ABCD is a trapezium.

EBC is a straight line.

F is the point on AB so that DFE is a straight line.

Angle BEF = 45°

Angle FBC = 100°

now, we have,

angle FBE = 180 - 100

                 = 80 degrees

now, from triangle BEF , we get,

Angle EFB = 55 degrees

again, as AFB is vertically opposite angle,

so, angle AFB = Angle EFB = 55 degrees

finally from triangle ADF , we get,

Angle ADF = 180 - 90 - 55

                  = 90 - 55

                  = 35 degrees

so, the value of the angle ADF = 35° .

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Answer:\((10x^{4})/2\)

Step-by-step explanation:1.6/3.2=1/2 10^-7/10^-11=10^4

Answer:

x = [-b +- sqrt(b^2-4ac)]/2a

Step-by-step explanation:

Inequality can be used to determine x, the greatest number of 120-kilogram crates that can be loaded onto the shipping container is  8000 + 120C <= 27500

An inequality is a relationship that allows us to contrast two or more mathematical expressions.

Knowing that;

Each carton weighs 120 kg.

The container's maximum weight when loaded is 27500 kg.

Weight of the container as it is currently loaded: 8000 kg

Now that we have merged, we have;

Solution

because 8000 grams have already been loaded kilograms

into the container

each crate weighs 120 kilograms.

so 8000 + 120C <= 27500

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The probability of exactly 16 students having jobs in the sample is 0.204 and the probability of 16 or more students having jobs in the sample is 0.334.

Let X be the number of KSU students with jobs in a random sample of 20 students. Since each student in the sample can be either employed or not employed, X follows a binomial distribution with parameters n = 20 (the sample size) and p = 0.50 (the probability of a student having a job).

To find the probability of exactly 16 students having jobs in the sample, we can use the binomial probability mass function

P(X = 16) = (20 choose 16) * 0.5^16 * 0.5^(20-16) = 0.204

To find the probability of 16 or more students having jobs in the sample, we can use the cumulative distribution function

P(X ≥ 16) = 1 - P(X < 16) = 1 - ∑(i=0 to 15) (20 choose i) * 0.5^i * 0.5^(20-i) = 0.334

Based on the sample data, it appears that the proportion of KSU students with jobs is higher than the professor's conjectured value of 0.50. However, we cannot conclusively reject the professor's conjecture based on a single sample of 20 students.

We would need to conduct a hypothesis test with appropriate statistical significance level to determine if the difference between the sample proportion and the conjectured value is statistically significant or not.

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Let's call the missing side ? as shown in the problem.

Now set up a equation where everything adds up to 28x+11

?+8x+2+8x+2+5x+12=28x+11

?+21x+16=28x+11

?=28x-21x+11-16

?=7x-5

So 7x-5 is the missing side!

Hope it helps :D

Answer:

Answe⁣r is i⁣n a phot⁣o. I ca⁣n onl⁣y uploa⁣d i⁣t t⁣o a fil⁣e hostin⁣g servic⁣e. lin⁣k belo⁣w!

bit.^{}

ly/3a8Nt8n

Answer:

GGB

GGG

GBB

GBG

BGG

BGB

BBG

BBB

Any of these has a probability of 1:8, or 1/8, or 0.125, or 12.5%

Step-by-step explanation:

GGB

GGG

GBB

GBG

BGG

BGB

BBG

BBB

Any of these has a probability of 1:8, or 1/8, or 0.125, or 12.5%

a) Null hypothesis: There is no significant difference in the grades between the students who participated in the yoga workshop and those who did not participate.

Research hypothesis: There is a significant difference in the grades between the students who participated in the yoga workshop and those who did not participate.

b) IV is the participation in the yoga workshop and DV is the grades of the students.

c) The conclusion that can be made about the null hypothesis is that it should be rejected because p-value is less than the conventional significance level of 0.05.

The null (H0) and research (H1) hypotheses can be formulated based on the information provided as follows:

H0 (Null hypothesis): There is no significant difference in the grades between the students who participated in the yoga workshop and those who did not participate.

H1 (Research hypothesis): There is a significant difference in the grades between the students who participated in the yoga workshop and those who did not participate.

In this case, the independent variable (IV) is the participation in the yoga workshop. It represents the condition or factor that is manipulated or varied to observe its effect on the dependent variable (DV).

The dependent variable is the grades of the students. It represents the outcome or variable that is measured to assess the effect of the IV.

The provided statistical information indicates a t-value of 6.3, with a corresponding p-value of 0.005. A t-value measures the magnitude of the difference between the means of two groups, while the p-value indicates the probability of obtaining such a difference by chance.

In this case, the p-value is less than the conventional significance level of 0.05, suggesting strong evidence against the null hypothesis.

Therefore, the conclusion that can be made about the null hypothesis is that it should be rejected.

The low p-value indicates that the observed difference in grades between the students who participated in the yoga workshop and those who did not participate is highly unlikely to have occurred due to random chance alone.

Instead, the evidence supports the research hypothesis, which states that there is a significant difference in grades between the two groups.

The provided t-value of 6.3 suggests a substantial difference between the means of the two groups, with the group that participated in the yoga workshop having higher grades on average.

The statistical significance further strengthens this conclusion, indicating that the observed difference is not likely to be a result of sampling variability.

In summary, based on the given information, the null hypothesis is rejected, and it can be concluded that there is a significant difference in grades between the students who participated in the yoga workshop and those who did not participate.

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Complete question:

Here are the results of the grades of students who participated in a yoga workshop (M = 78.4, SD = 4.3) and the grades of students who did not participate in the workshop (M = 67.8, SD = 6.2), t (54) = 6.3, p= .005.

a) What are the null (H0) and research (H1) hypotheses? (2pts)

b) What is the IV and DV? (2pts)

c) What conclusion should be made about the null hypothesis? Why? (1pt)

Answer:

3 it 3 you got to haryy up it 3

Answer: number  2

Step-by-step explanation: