The integral [, sin(x - 2) dx is transformed into S1, g(t)dt by applying an appropriate change of variable, then g() is: g(t) = sin g(t) = cos (5) O This option O This option g(t) = -cos (3) g(t) = sin O This option

P does not contain a line, it means that for any x in R^n, there exists a unique y in R^k such that Ax + By ≥ b. Therefore, x^ is uniquely determined by y^, and (x^, y^) is an extreme point of P.

1) The statement is true. Suppose (x^,y^) is an extreme point of P. To show that x^ is an extreme point of Q, we need to prove that for any two distinct points x_1, x_2 in Q, the line segment connecting x_1 and x_2 lies entirely in Q. Since Q is the projection of P onto x-space, it means that for any x in Q, there exists y in R^k such that Ax + By ≥ b.

Now, let's assume x_1 and x_2 are two distinct points in Q. Since they belong to Q, there exist corresponding y_1 and y_2 in R^k such that Ax_1 + By_1 ≥ b and Ax_2 + By_2 ≥ b. Since P is a polyhedron, the set of points that satisfy Ax + By ≥ b is a convex set. Therefore, the line segment connecting x_1 and x_2, denoted by [x_1, x_2], lies entirely in P. Since the projection of a convex set onto a subspace is also a convex set, [x_1, x_2] lies entirely in Q. Thus, x^ is an extreme point of Q.

2) The statement is false. Suppose x^ is an extreme point of Q. It does not necessarily imply the existence of a corresponding y^ such that (x^, y^) is an extreme point of P. This is because the projection Q onto x-space may not capture all the extreme points of P. It is possible for multiple points in P to project to the same point in Q, making it impossible to uniquely determine y^.

3) The statement is true. If x^ is an extreme point of Q and P does not contain a line, then there exists a corresponding y^ such that (x^, y^) is an extreme point of P. Since P does not contain a line, it means that for any x in R^n, there exists a unique y in R^k such that Ax + By ≥ b. Therefore, x^ is uniquely determined by y^, and (x^, y^) is an extreme point of P.

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a)From this 230 is divisible by 10.Then 995 and 526 are not divisible by 10.b)From this 230 and 995 are divisible by 5 and 526 is not divisible by 5.c) From this 230,995,526 are divisible by 2.

Divisibility by 10: A number is divisible by 10 if its last digit is 0. Therefore, 230 is divisible by 10 since its last digit is 0. 995 and 526 are not divisible by 10 since their last digits are not 0. The mathematical formula for divisibility by 10 is: (10 | N) = 0,Where N is the number that is being tested for divisibility. Calculation: For example, let's take the number 230. (10 | 230) = 0230/10 = 23 Therefore, 230 is divisible by 10. Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5. Therefore, 230 and 995 are divisible by 5 since their last digits are 0 and 5 respectively. 526 is not divisible by 5 since its last digit is not 0 or 5. The mathematical formula for divisibility by 5 is: (5 | N) = 0Where N is the number that is being tested for divisibility. Calculation: For example, let's take the number 995. (5 | 995) = 0995/5 = 199 .Therefore, 995 is divisible by 5. Divisibility by 2: A number is divisible by 2 if its last digit is 0, 2, 4, 6 or 8. Therefore, 230, 995, and 526 are all divisible by 2 since their last digits are 0, 5 and 6 respectively. The mathematical formula for divisibility by 2 is: (2 | N) = 0Where N is the number that is being tested for divisibility. Calculation: For example, let's take the number 526. (2 | 526) = 0.526/2 = 263 .Therefore, 526 is divisible by 2.

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Answer: 98.6

Step-by-step explanation:

5.8 can be broken down into 5 & 0.8

17 timed 5 = 85

17 timed 0.8 = 13.6

85+13.6 = 98.6

In this case, the mean is 5.5 cm and the standard deviation is 0.1 cm. So, one standard deviation below the mean is 5.4 cm and one standard deviation above the mean is 5.6 cm. Therefore, about 68% of the data falls between 5.4 cm and 5.6 cm, which includes the range of 5.3 cm to 5.7 cm.

Your question involves a normal distribution with a mean (µ) of 5.5 cm and a standard deviation (σ) of 0.1 cm. You want to find the percentage of data between 5.3 cm and 5.7 cm.

First, we need to standardize the scores using the z-score formula: z = (x - µ) / σ

For 5.3 cm: z1 = (5.3 - 5.5) / 0.1 = -2
For 5.7 cm: z2 = (5.7 - 5.5) / 0.1 = 2

Now, referring to a standard normal distribution table or using a calculator, we find the area between these z-scores:

This can be determined using the empirical rule, also known as the 68-95-99.7 rule, which states that for a normal distribution:
- About 68% of the data falls within one standard deviation of the mean
- About 95% of the data falls within two standard deviations of the mean
- About 99.7% of the data falls within three standard deviations of the mean

P(-2 < z < 2) = P(z < 2) - P(z < -2) = 0.9772 - 0.0228 = 0.9544

Converting it to a percentage, we get 95.44%, which is approximately 95%.

So, the answer is A. 95%.

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Using the two parallel line theorems we proved that ∠8 ≅ ∠4.

In the given question,

Given: f || g

Prove: ∠8 ≅ ∠4

We using given diagram in proving that ∠8 ≅ ∠4

Since f || g, by the Corresponding Angles Postulate which states that "When a transversal divides two parallel lines, the resulting angles are congruent." So

∠8≅∠6

Then by the Vertical Angles Theorem which states that "When two straight lines collide, two sets of linear pairs with identical angles are created."

∠6≅∠4

Then, by the Transitive Property of Congruence which states that "All shapes are congruent to one another if two shapes are congruent to the third shape."

∠8 ≅ ∠4

Hence, we proved that ∠8 ≅ ∠4.

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The reason for ∠8 ≅ ∠4 is alternate exterior angles

The question has to do with finding the proven with the given.

We have been given line f is parallel to line g.

We need to prove angle 8 is congruent to angle 4.

From given figure we observe that line f || line g and these parallel lines are intersected by a transversal.

And the angle 8 are angle 4 are alternate exterior angles.

We know that alternate exterior angles are congruent.

so, ∠8 ≅ ∠4

Therefore, the reason for ∠8 ≅ ∠4 is alternate exterior angles

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Answer:

EFGH ~ IJKL

Step-by-step explanation:

There sides are both parallel.

Hope this helped you :)

Molly covered 1/2 of the total distance in the first two days.

To find the fraction of the distance Molly covered in the first two days, we need to add the fractions for each day.

On the first day, Molly covered 2/10 of the distance.

On the second day, Molly covered 3/10 of the distance.

To find the total distance covered in the first two days, we add these fractions together:

2/10 + 3/10 = 5/10

Simplifying the fraction, we get:

5/10 = 1/2

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Answer:

$5,720.40

Step-by-step explanation:

Answer:

36

Step-by-step explanation:

Since Volume = Width x Length x Height.

We know everything but volume.

So you just do 2x3x6.

You would get 36.

Answer:

V = 36 ft³

General Formulas and Concepts:

Pre-Alg

Geometry

Step-by-step explanation:

Step 1: Define variables

w = 3 ft

l = 6 ft

h = 2 ft

Step 2: Find Volume

Answer:

B. 2.2 kilometers

Step-by-step explanation:

⭐What is the change in the cyclist's elevation?

Thus, we need to solve for the value of the height (x) of the hill.

Notice that the hill is in the shape of a right triangle. In order to find an unknown side length for right triangles, we need to use a trigonometric function.

⭐What is a trigonometric function?

First, let's see what kind of side lengths we are given (opposite, hypotenuse, or adjacent).

We are given a side length of 10km, which is the hypotenuse of the triangle, and we are given a side length of x, which is the opposite of θ (13°).

Therefore, we will use the trigonometric function sin(θ) to solve for x.

Substitute the side lengths and angle measures we are given:

\(sin(13) = \frac{x}{10}\)

Set the equation equal to x:

\(10sin(13) = x\)

Solve for x. I recommend using a scientific calculator (e.g. Desmos Scientific Calculator):

\(2.2 = x\)

∴ The cyclist's change in elevation is 2.2 kilometers.

Answer:

16.2

Step-by-step explanation:

calculator :0

In 2012, 8 percentage of high school seniors sampled had used lad sometime during their life.

LSD use among secondary school understudies is a specific concern. In excess of 8% of secondary school seniors in the United States utilized the medication something like once in the course of their life, and almost 4% involved the medication in the previous year, as per the University of Michigan's Monitoring the Future Survey. In spite of the fact that by the 1980s LSD use had extraordinarily declined, it made a resurgence during the 1990s, basically among teens. An overview led by Monitoring the Future Study discovered that 13.6 percent of 1997's secondary school seniors had tried different things with LSD no less than once contrasted with just 7.2 percent in 1986.

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The scale factor is equal to 4 which is calculated by determining the coordinates.

Let's begin by determining the coordinates.

These are the ones we begin with!

D (1,1)

E (2,2)

F (4,2)

G (3,1)

Now we must locate the best image cords.

D' (4,4)

E' (8,8)

F' (16,8)

G' (12,4)

We're looking for the scale factor, so we'll see what number is multiplied by those coordinates to get the prime coordinates.

D is (1,1) and D' is (4,4), but we know that 1x4 equals 4, so 4 is the scale factor by which we multiply all of the numbers.

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Answer:

37

Step-by-step explanation:

i just know

Answer:

10/9

Step-by-step explanation:

average rate of change in 0≤x≤9

=(3-(-7))/(9-0)=10/9

Answer:

The solution is 10/9

Step-by-step explanation:

Step 1: identify the right approach:

Given a function g(x) over a period of time, to find the average rate of change of g(x) over this period of time, we divide the change in g(x) by the change in x.

Step 2: Perform the calculation:

As shown in picture,

x = 0 => g(x) = -7

x = 9 => g(x) = 3

=> The change of x over interval [0, 9] is:

x_change = 9 - 0 = 9

=> The change of g(x) over interval [0, 9] is:

g(x)_change = 3 - (-7) = 10

=> The average rate (r) of change of g(x) over interval [0, 9] is:

r = g(x)_change/x_change = 10/9

Hop this helps!

:)

A numeric variable is used with the VAR statement.

Given statement,

What type of variable is analyzed using a PROC TTEST? That is, what type of variable do we use with the VAR statement,

Here, we use;

→ Numeric Variable

Hence, a Numeric Variable type of variable is analyzed using a PROC TTEST.

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The formula is weight of object = mass x gravity

if you punch the numbers in the formula correctly the mass in kg is 875.634 kg x 9.8 m/s =  8,581.2132

Answer:

8

Step-by-step explanation:

common Sense hope it helps

The integral ∫[0 to 1] x² dx evaluates to 1/3.

To evaluate this integral, we can use the power rule for integration. Applying the power rule, we increase the power of x by 1 and divide by the new power. Thus, integrating x² gives us (1/3)x³.

To evaluate the definite integral from x = 0 to x = 1, we substitute the upper limit (1) into the antiderivative and subtract the result when the lower limit (0) is substituted.

Using the Fundamental Theorem of Calculus, the area under the curve is given by the expression A = ∫[0 to 1] f(x) dx. For this case, f(x) = x².

To approximate the area using right endpoints, we can use a Riemann sum. Dividing the interval [0, 1] into subintervals and taking the right endpoint of each subinterval, the Riemann sum can be expressed as lim[n→∞] Σ[i=1 to n] f(xᵢ*)Δx, where f(xᵢ*) is the value of the function at the right endpoint of the i-th subinterval and Δx is the width of each subinterval.

In this specific case, since the function f(x) = x² is an increasing function on the interval [0, 1], the right endpoints of the subintervals will be f(x) values.

Therefore, the area under the curve from x = 0 to x = 1 can be expressed as lim[n→∞] Σ[i=1 to n] (xi*)²Δx, where Δx is the width of each subinterval and xi* represents the right endpoint of each subinterval.

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The resulting value on expansion of \((3+\sqrt{-16} )(6-\sqrt{-64} )\) is 50

Given the surd product;

\((3+\sqrt{-16} )(6-\sqrt{-64} )\)

Since the square root of a negative number will give a complex number that is:

\(\sqrt{-1} = i\)

The given expression above then becomes:

\((3+4i )(6-8i)\)

Expand

\(=(3+4i )(6-8i)\\=18-24i+24i-32i^2\\=18-32i^2\\=18-32(-1)\\=18+32\\= 50\)

Hence the resulting value on expansion is 50.

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The conclusion about the relationship between age and income is that there is no specific or direct relationship between age and income. Income can vary greatly across different age groups, and age alone is not a reliable predictor of income.

When considering the relationship between age and income, it is important to understand that age alone does not determine a person's income level. While it is true that individuals generally accumulate more work experience and potentially higher income as they grow older, there are many other factors at play. Income is influenced by various factors such as education, occupation, skills, industry, location, and individual choices.

For example, two individuals of the same age may have vastly different incomes based on their education levels. Education has a stronger correlation with income than age alone. Higher education typically leads to increased earning potential and access to higher-paying job opportunities. However, it is important to note that even within similar education levels, income can still vary based on other factors mentioned earlier.

Therefore, it is crucial to consider a multitude of factors when examining the relationship between age and income. Age alone does not provide a reliable basis for predicting or determining someone's income level, as it is influenced by a wide range of individual circumstances and external factors.

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Answer:

1/6

Step-by-step explanation:

I would say 1/6 because there is 3 white chocolate and 3/18 is 1/6 simplified :)

Answer:

B. 1/6

Step-by-step explanation:

This is because there are 3 white chocolates and 18 total chocolates so you have a 3/18 chance, however 3/18 reduced is 1/6 becuqse you divide by 3 on both sides to get 1/6.

The null hypothesis is rejected as the hypothesized value of 65 falls outside the calculated confidence interval of (53.8832, 67.1168).

To test the hypothesis at the 5% level of significance, we need to calculate a confidence interval and check if the hypothesized value falls within it. In this case, the student believes the average grade on the statistics final examination is 65, and a sample of 25 final examinations is taken, with an average grade of 60.5 and a standard deviation of 16.

Step 1: Calculate the standard error of the mean.

The standard error of the mean (SE) is the standard deviation divided by the square root of the sample size. In this case, the sample size is 25 and the standard deviation is 16. Therefore, the standard error of the mean is calculated as follows:

SE = 16 / √25 = 16 / 5 = 3.2

Step 2: Calculate the confidence interval.

Using the sample mean (60.5) and the standard error (3.2), we can calculate the confidence interval. Assuming a normal distribution, we use a t-distribution with degrees of freedom (df) equal to the sample size minus 1 (25 - 1 = 24) and a desired confidence level of 95% (1 - α = 0.95). The critical value for a 95% confidence interval with 24 degrees of freedom is approximately 2.064 (obtained from statistical tables or software). The confidence interval can be calculated as follows:

CI = sample mean ± (critical value * SE)

CI = 60.5 ± (2.064 * 3.2)

CI = 60.5 ± 6.6168

CI = (53.8832, 67.1168)

Step 3: Test the hypothesis.

The hypothesized value of 65 falls outside the confidence interval (53.8832, 67.1168). Therefore, we reject the null hypothesis, which suggests that the average grade on the statistics final examination is 65. This indicates that the sample data provides evidence that the true population average is different from 65.

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The value of the measure of each sides indicated are:

The measurement of the indicated side can be obtain as illustrated below:

1. The value of x can be obtain as follow:

Tan θ = Opposite / Adjacent

Tan 20 = x / 11

Cross multiply

x = 11 × Tan 20

Value of x = 4

2. The value of x can be obtain as follow:

Sine θ = Opposite / Hypotenuse

Sine 27 = 5 / x

Cross multiply

x × Sine 27 = 5

Divide both sides by Sine 27

x = 5 / Sine 27

Value of x = 11

3. The value of x can be obtain as follow:

Sine θ = Opposite / Hypotenuse

Sine 27.2 = 10 / x

Cross multiply

x × Sine 27.2 = 10

Divide both sides by Sine 27.2

x = 10 / Sine 27.2

Value of x = 21.9

4. The value of x can be obtain as follow:

Tan θ = Opposite / Adjacent

Tan 25 = 15 / x

Cross multiply

x × Tan 25 = 15

Divide both sides by Tan 25

x = 15 / Tan 25

Value of x = 32.2

5. The value of x can be obtain as follow:

Cos θ = Adjacent / Hypotenuse

Cos 64 = 4 / x

Cross multiply

x × Cos 64 = 4

Divide both sides by Cos 64

x = 4 / Cos 64

Value of x = 9.1

6. The value of x can be obtain as follow:

Tan θ = Opposite / Adjacent

Tan 33 = x / 14

Cross multiply

x = 14 × Tan 33

Value of x = 9.1

7. The value of x can be obtain as follow:

Sine θ = Opposite / Hypotenuse

Sine 49.6 = 5.6 / x

Cross multiply

x × Sine 49.6 = 5.6

Divide both sides by Sine 49.6

x = 5.6 / Sine 49.6

Value of x = 7.4

8. The value of x can be obtain as follow:

Cos θ = Adjacent / Hypotenuse

Cos 40 = x / 10.2

Cross multiply

x = 10.2 × Cos 40

Value of x = 7.8

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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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Answer:

23

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

The domain of the function is 0 ≤ t ≤ 3 and the greatest concentration of the medication is 2mg/L

The equation of the function is given as:

\(C(t) = \frac{-18t^2 + 54t}{t^2 + 7t + 10}\)

Time can be 0 or greater but cannot be negative.

So, the domain of the function includes t ≥ 0

Set the function to 0 to determine the maximum value of t.

\(\frac{-18t^2 + 54t}{t^2 + 7t + 10} = 0\)

Cross multiply

\(-18t^2 + 54t = 0\)

Factor out -18t

-18t(t - 3) = 0

Divide both sides by -18t

t - 3 = 0

Add 3 to both sides

t = 3

This means that the maximum value of t is 3

Hence, the domain of the function is 0 ≤ t ≤ 3

From the graph of the function (see attachment), we have the maximum value to be;

C(t) = 2

Hence, the greatest concentration of the medication is 2mg/L

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Answer:

15

Step-by-step explanation: