the difference between the percentage of Calories that come from fat in cheeseburger A and the percentage of Calories that come from fat in cheeseburger B

Given that x is in quadrant 3, its value for tan x=1/2 will be 206.565°, and its value for cos(206.565°) will be -0.894.

Trigonometry is the study of angles and the angular relationships between planar and three-dimensional shapes. The cosecant, cosine, cotangent, secant, sine, and tangent are among the trigonometric functions (also known as the circle functions) that make up trigonometry. The sine, cosine, and tangent are the three fundamental trigonometric operations. The cotangent, secant, and cosecant functions are derived from these three basic functions. On these functions, all trigonometrical concepts are built.

Here,

Tan x=1/2

π<x<3π/2

x=tan⁻¹/²(1/2)

x=26.565°

As the x is in quadrant 3,

=180+26.565°

=206.565°

cos(206.565°)=-0.894

The value of x for tan x=1/2 will be 206.565° as x is in quadrant 3 and for the same value of x, cos(206.565°) will be -0.894.

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Answer:

-sqrt 5 - 2 sqrt 5/10

Step-by-step explanation:

Edge 2023

From the given data, the marginal cost and the actual cost of manufacturing is $0.097 and $218.803 respectively.

a. To find the marginal cost function, we need to take the derivative of the cost function with respect to x:

c'(x) = -9/x² + 2x/9

b. To find the marginal cost of manufacturing 12 food processors, we substitute x = 12 into the marginal cost function:

c'(12) = -9/12² + 2(12)/9 = -0.125 + 0.222 = 0.097

Therefore, the marginal cost of manufacturing 12 food processors is $0.097.

c. To find the actual cost of manufacturing the thirteenth food processor, we need to evaluate the cost function at x = 13:

c(13) = 200 + 9/13 + 13²/9 = 200 + 0.692 + 18.111 = $218.803

Therefore, the actual cost of manufacturing the thirteenth food processor is $218.803.

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Exponential form is a way of representing repeated multiplications of the same number by writing the number as a base with the number of repeats written as a small number to its upper right.

In the exponential form, the exponent indicates the number of times the base is used as a factor.

For example, in the case of 16 it can be written as 2 × 2 × 2 × 2  = \(2^{4}\), where 2 is the “base” and 4 is the “exponent.

A product in which the factors are identical is called a power of that factor.  The number that is repeated is called the base, and the number of times it repeats is called the exponent, power or degree. And the power that is written on the right upper side are called exponents. when multiplying the numbers having same base and different exponents then the base is kept same and the exponents are added.

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The width of the house driveway in the scale drawing is 4 inches.

Scale drawing is drawing that shows a real object with accurate sizes reduced or enlarged by a certain amount (called the scale).

Therefore, he measured a house and its lot and made a scale drawing.

The scale of the drawing was 1 inch : 5 feet.

The houses driveway is 20 feet wide in real life.

Therefore,

5 feet = 1 inch

20 feet = ?

cross multiply

Therefore,

width of the driveway on the scale drawing =  20 × 1 / 5

width of the driveway on the scale drawing =  20 / 5

width of the driveway on the scale drawing =  4 inches.

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The answer are: a. Total outflows: $2,007,901, b. Total inflows: $827,080, c. Net present value: $824,179, d. Should the old issue be refunded with new debt? Yes

To determine whether the old bond issue should be refunded with new debt, we need to calculate the present value of total outflows, the present value of total inflows, and the net present value (NPV). Let's calculate each of these values step by step: Calculate the present value of total outflows. The total outflows consist of the call premium, underwriting cost on the old issue, and underwriting cost on the new issue. Since these costs are one-time payments, we can calculate their present value using the formula: PV = Cash Flow / (1 + r)^t, where PV is the present value, Cash Flow is the cash payment, r is the discount rate, and t is the time period.

Call premium on the old issue: PV_call = (7% of $22 million) / (1 + 0.1)^16, Underwriting cost on the old issue: PV_underwriting_old = $530,000 / (1 + 0.1)^16, Underwriting cost on the new issue: PV_underwriting_new = $680,000 / (1 + 0.1)^16. Total present value of outflows: PV_outflows = PV_call + PV_underwriting_old + PV_underwriting_new. Calculate the present value of total inflows. The total inflows consist of the interest savings and the tax savings resulting from the interest expense deduction. Since these cash flows occur annually, we can calculate their present value using the formula: PV = CF * [1 - (1 + r)^(-t)] / r, where CF is the cash flow, r is the discount rate, and t is the time period.

Interest savings: CF_interest = (12% - 10%) * $22 million, Tax savings: CF_tax = (40% * interest expense * tax rate) * [1 - (1 + r)^(-t)] / r. Total present value of inflows: PV_inflows = CF_interest + CF_tax.  Calculate the net present value (NPV). NPV = PV_inflows - PV_outflows Determine whether the old issue should be refunded with new debt. If NPV is positive, it indicates that the present value of inflows exceeds the present value of outflows, meaning the company would benefit from refunding the old issue with new debt. If NPV is negative, it suggests that the company should not proceed with the refunding.

Now let's calculate these values: PV_call = (0.07 * $22,000,000) / (1 + 0.1)^16, PV_underwriting_old = $530,000 / (1 + 0.1)^16, PV_underwriting_new = $680,000 / (1 + 0.1)^16, PV_outflows = PV_call + PV_underwriting_old + PV_underwriting_new. CF_interest = (0.12 - 0.1) * $22,000,000, CF_tax = (0.4 * interest expense * 0.4) * [1 - (1 + 0.1)^(-16)] / 0.1, PV_inflows = CF_interest + CF_tax. NPV = PV_inflows - PV_outflows.  If NPV is positive, the old issue should be refunded with new debt. If NPV is negative, it should not.

Performing the calculations (rounded to the nearest whole dollar): PV_call ≈ $1,708,085, PV_underwriting_old ≈ $130,892, PV_underwriting_new ≈ $168,924, PV_outflows ≈ $2,007,901,

CF_interest ≈ $440,000, CF_tax ≈ $387,080, PV_inflows ≈ $827,080.  NPV ≈ $824,179.  Since NPV is positive ($824,179), the net present value suggests that the old bond issue should be refunded with new debt.

Therefore, the answers are:

a. Total outflows: $2,007,901

b. Total inflows: $827,080

c. Net present value: $824,179

d. Should the old issue be refunded with new debt? Yes

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The present value of an amount of $3,000 due to be paid in 3 years is equal to an amount, that with accumulated desired interest would grow to be $3,000 three years from now. This is because the present value is the value of the future payment today, after taking into account the time value of money and the interest rate. The answer to this question is C.

To calculate the present value of $3,000 due in 3 years, we need to discount the future payment back to its present value using the interest rate. This means that we need to find an amount that, when invested today at the given interest rate, will grow to be $3,000 in 3 years.

For example, if the interest rate is 5%, the present value of $3,000 due in 3 years would be approximately $2,530. This means that if you invest $2,530 today at 5% interest, it will grow to be $3,000 in 3 years.

Therefore, the correct answer is C, and we need to know the interest rate to calculate the present value accurately. Answer A is incorrect because the expected value of $3,000 does not take into account the time value of money and the interest rate. Answer B is incorrect because the present value should always be less than the future value if the interest rate is greater than zero. Answer D is incorrect because the expected value and the present value are not the same.

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NOVA's (Analysis of Variance) are used when a study involves comparing multiple groups. This statistical method is applied when there are five or more groups to compare, three or more groups to compare, one or more groups to compare, or four or more groups to compare.

NOVA allows researchers to determine if there are significant differences between the means of these groups, helping to assess the impact of different factors or interventions. NOVA (Analysis of Variance) is a statistical technique used to compare the means of multiple groups. It is employed when a study involves the comparison of different groups to examine the effects of various factors or interventions. NOVA is particularly useful when there are five or more groups to compare, as it allows researchers to determine if there are significant differences between these groups. By analyzing the variability within and between the groups, NOVA helps assess the impact of different factors on the outcome being studied. Furthermore, NOVA is also employed when there are three or more groups to compare. In this case, the analysis allows researchers to evaluate whether there are statistically significant differences among the means of the groups. It provides a comprehensive understanding of the variations between the groups and helps determine if the observed differences are due to chance or if they can be attributed to the variables being examined. Additionally, NOVA is applicable when researchers want to compare one or more groups. By using this statistical method, they can assess if there are significant differences in the means of these groups. This allows for a more comprehensive understanding of the data and helps draw meaningful conclusions about the impact of different factors on the outcome of interest. Lastly, NOVA is used when there are four or more groups to compare. By employing this statistical technique, researchers can analyze the differences between the means of these groups and determine if these differences are statistically significant. NOVA enables them to identify the presence of meaningful variations between the groups, aiding in the interpretation of the study results and supporting evidence-based decision-making. In summary, NOVA's (Analysis of Variance) are employed when a study involves comparing multiple groups. Whether there are five or more groups, three or more groups, one or more groups, or four or more groups to compare, NOVA allows researchers to determine if there are significant differences between the means of these groups. By utilizing this statistical method, researchers gain valuable insights into the impact of different factors or interventions and can draw meaningful conclusions from their data.

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Answer:

5(2x - 4) - 2y

if x= -2

5( (2×-2) -4)-2y

5(-4 -4) -2y

if y=6

5(-8) -2(6)

5(-8) -12

-40 - 12

= -52

We are told that there are the names of 12 girls and 14 boys in a hat. we are asked about the probability of drawing the name of a boy. To do that, we need to divide the number of boy's names over the total amount of names in the bag, like this:

Therefore, the probability is 0.538 or 53.8%

(a) ( x_{t+1}=\frac{1}{2} x_{t}+3 ), the solution to this difference equation is x_t = 2^t + 3, The difference equations in this problem are both linear difference equations with constant coefficients.

This can be found by solving the equation recursively. For example, the first few terms of the solution are

t | x_t

--- | ---

0 | 3

1 | 7

2 | 15

3 | 31

The general term of the solution can be found by noting that

x_{t+1} = \frac{1}{2} x_t + 3 = \frac{1}{2} (2^t + 3) + 3 = 2^t + 3

(b) ( x_{t+1}=-3 x_{t}+4 )

The solution to this difference equation is

x_t = 4 \cdot \left( \frac{1}{3} \right)^t + 4

This can be found by solving the equation recursively. For example, the first few terms of the solution are

t | x_t

--- | ---

0 | 4

1 | 5

2 | 2

3 | 1

The general term of the solution can be found by noting that

x_{t+1} = -3 x_t + 4 = -3 \left( 4 \cdot \left( \frac{1}{3} \right)^t + 4 \right) + 4 = 4 \cdot \left( \frac{1}{3} \right)^t + 4

The difference equations in this problem are both linear difference equations with constant coefficients. This means that they can be solved using a technique called back substitution.

Back substitution involves solving the equation recursively, starting with the last term and working backwards to the first term.

In the first problem, the equation can be solved recursively as follows:

x_{t+1} = \frac{1}{2} x_t + 3

x_t = \frac{1}{2} x_{t-1} + 3

x_{t-1} = \frac{1}{2} x_{t-2} + 3

...

x_0 = \frac{1}{2} x_{-1} + 3

The general term of the solution can be found by noting that

x_{t+1} = \frac{1}{2} x_t + 3 = \frac{1}{2} (2^t + 3) + 3 = 2^t + 3

The second problem can be solved recursively as follows:

x_{t+1} = -3 x_t + 4

x_t = -3 x_{t-1} + 4

x_{t-1} = -3 x_{t-2} + 4

...

x_0 = -3 x_{-1} + 4

The general term of the solution can be found by noting that

x_{t+1} = -3 x_t + 4 = -3 \left( 4 \cdot \left( \frac{1}{3} \right)^t + 4 \right) + 4 = 4 \cdot \left( \frac{1}{3} \right)^t + 4

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The probability there is at least one pair is therefore 1−p.

acc to our question-

hence,The probability there is at least one pair is therefore 1−p.

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The applicants who are granted admission is 0.8413, The applicants who will score 860 or higher is 0.0228, and there will be 2023 applicants that are expected to be qualified to the university.

a) The scores in this problem are known to be normally distributed with a mean of 700 and a standard deviation of 80. To find the proportion of all applicants taking the exam who are granted admission, we must compute the Z-score of 620.

We then need to find the area under the normal curve to the right of this Z-score.1. Z-score of 620: (620 - 700)/80 = -1.00Therefore, P(X ≥ 620) = P(Z ≥ -1.00) = 0.8413 (using a standard normal table)

So, the proportion of all applicants taking the exam who are granted admission is approximately 0.8413.

b) To find the proportion of all applicants who will score 860 or higher on the exam, we must compute the Z-score of 860.

We then need to find the area under the normal curve to the right of this Z-score.2. Z-score of 860: (860 - 700)/80 = 2.00

Therefore, P(X ≥ 860) = P(Z ≥ 2.00) = 0.0228 (using a standard normal table)So, the proportion of all applicants who will score 860 or higher on the exam is approximately 0.0228.

c) Using the proportions calculated in parts (a) and (c), we can expect the following number of applicants to be qualified for admission to the university.

Qualified applicants = (2400)(0.8413) = 2023 (rounded to the nearest whole number)

Therefore, we expect 2023 applicants to be qualified for admission to the university.

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The weight difference between the lightest and heaviest newborns delivered in a hospital will be 5.5 pounds.

Based on the given data we can say that the weight of the lightest baby is 4.5 and the weight of the heaviest baby is 10 pounds.

To find the difference we can subtract the weight from each other =

Weight of heaviest - the weight of lightest = 10 pounds -4.5 pounds  = 5.5 pounds.

Therefore based on the given histogram data we can find that the difference between the heaviest baby and the lightest baby is 5.5 pounds.

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True. The null hypothesis (denoted as H0) is a statement about a population parameter, typically a population mean or proportion, and it is formulated based on the assumption that there is no effect, relationship, or difference in the population.

However, in statistical hypothesis testing, data is often collected from a sample, not the entire population, due to practical limitations.

The sample data is then used to assess the evidence against the null hypothesis.

The null hypothesis is always stated in terms of the population, even though the data being analyzed comes from a sample.

Based on the provided scenario, the statement can be classified as: b. This is a nonsampling error that causes bias.

Nonsampling errors refer to errors that occur in data collection, processing, or analysis that are not related to the sampling method.

The scenario described, where individuals may not provide truthful answers due to fear of their responses being shared with law enforcement officials, introduces a bias in the collected data.

This bias occurs because the reported data does not accurately represent the true prevalence of illicit drug use in the population.

The fear of legal consequences leads to a systematic deviation from the true values, which is a form of bias.

Therefore, option b is the correct classification for this scenario.

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Given: An expression

Required: To write an equation using the expression.

Explanation: An equation is a statement that equates two quantities. It contains an = sign and some constants or variables.

We can create an equation with the given expression using constants only or by introducing some variables.

Equation using constants-

Equation using variables-

Final Answer: The equation using the expression given is-

Answer:

3.5

Step-by-step explanation:

28 days = 4 weeks

28 divided by 4 = 1 week

14 gallons divided by 4 = 3.5

Answer:

See below

Step-by-step explanation:

2/3 of the height is brick     height is   2  3/4 feet

 Brick portion =   2/3  *  2  3/4   ft

Answer:

\(\huge\boxed{(2x)^\frac{1}{2}\times(2x^3)^\frac{3}{2}=4x^5}\)

Step-by-step explanation:

\((2x)^\frac{1}{2}\times(2x^3)^\frac{3}{2}\qquad\text{use}\ (ab)^n=a^nb^m\\\\=2^\frac{1}{2}x^\frac{1}{2}\times2^\frac{3}{2}(x^3)^\frac{3}{2}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^\frac{1}{2}x^\frac{1}{2}\times2^\frac{2}{3}x^{(3)(\frac{3}{2})}=2^\frac{1}{2}x^\frac{1}{2}\times2^\frac{2}{3}x^\frac{9}{2}\\\\\text{use the commutative and associative property}\\\\=\left(2^\frac{1}{2}\times2^\frac{3}{2}\right)\left(x^\frac{1}{2}\times x^\frac{9}{2}\right)\qquad\text{use}\ a^n\times a^m=a^{n+m}\)

\(=2^{\frac{1}{2}+\frac{3}{2}}x^{\frac{1}{2}+\frac{9}{2}}=2^\frac{1+3}{2}x^{\frac{1+9}{2}}=2^\frac{4}{2}x^\frac{10}{2}=2^2x^5=4x^5\)

Answer:

\( 4x^5 \)

Step-by-step explanation:

\( (2x)^\frac{1}{2} \times (2x^3)^\frac{3}{2} = \)

\(= (2x)^\frac{1}{2} \times (2x \times x^2)^\frac{3}{2}\)

\(= [(2x)^\frac{1}{2} \times (2x)^\frac{3}{2}] \times (x^2)^\frac{3}{2}\)

\(= (2x)^{\frac{1}{2} + \frac{3}{2}} \times x^{{2} \times \frac{3}{2}}\)

\(= (2x)^{\frac{4}{2}} \times x^{\frac{6}{2}}\)

\( = 2^2x^2 \times x^3 \)

\( = 4x^5 \)

The maximum likelihood estimate of the mean and variance of the normal distribution are the sample mean and sample variance, respectively. This is because the normal distribution is a parametric distribution, and the parameters can be estimated from the data using the likelihood function.

The maximum likelihood estimate of the mean and variance of the normal distribution are given by the sample mean and sample variance, respectively. The normal distribution is a continuous probability distribution that is symmetrical and bell-shaped. It is often used to model data that follows a normal distribution, such as the height of individuals in a population.
When we have a random sample from a normal distribution, we can estimate the mean and variance of the population using the sample mean and sample variance, respectively. The maximum likelihood estimate (MLE) of the mean is the sample mean, and the MLE of the variance is the sample variance.
To find the MLE of the mean and variance of the normal distribution, we use the likelihood function. The likelihood function is the probability of observing the data given the parameter values. For the normal distribution, the likelihood function is given by:
L(μ, σ² | x₁, x₂, ..., xn) = (2πσ²)-n/2 * e^[-1/(2σ²) * Σ(xi - μ)²]
where μ is the mean, σ² is the variance, and x₁, x₂, ..., xn are the observed values.
To find the MLE of the mean, we maximize the likelihood function with respect to μ. This is equivalent to setting the derivative of the likelihood function with respect to μ equal to zero:
d/dμ L(μ, σ² | x₁, x₂, ..., xn) = 1/σ² * Σ(xi - μ) =
Solving for μ, we get:
μ = (x₁ + x₂ + ... + xn) / n
This is the sample mean, which is the MLE of the mean.
To find the MLE of the variance, we maximize the likelihood function with respect to σ². This is equivalent to setting the derivative of the likelihood function with respect to σ² equal to zero:
d/d(σ²) L(μ, σ² | x₁, x₂, ..., xn) = -n/2σ² + 1/(2σ⁴) * Σ(xi - μ)² = 0
Solving for σ², we get:
σ² = Σ(xi - μ)² / n
This is the sample variance, which is the MLE of the variance.
In conclusion, the maximum likelihood estimate of the mean and variance of the normal distribution are the sample mean and sample variance, respectively. This is because the normal distribution is a parametric distribution, and the parameters can be estimated from the data using the likelihood function.

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We can drag the particles in mass/grams measurement to the corresponding descriptions as follows:

1.  1.6744 × 10⁻²⁴: Two electrons and 0ne proton

2. 5.021 × 10⁻²⁴: Two protons and one neutron

3. 5.0224 × 10⁻²⁴: One proton and two neutrons

4. 3.3484  × 10⁻²⁴: One electron, one proton, and one neutron

To match the measurements to the descriptions first note that one neutron is 1.6749 × 10⁻²⁴. One proton is equal to  1.6726 × 10⁻²⁴ and one electron is equal to  9.108 × 10⁻²⁸.

To obtain the right combinations, we have to add up the particles to arrive at the constituents. So, for the figure;

1.6744 × 10⁻²⁴, we would

Add 2 electrons and one proton

= 2(9.108 × 10⁻²⁸) + 1.6726 × 10⁻²⁴

= 1.6744 × 10⁻²⁴

The same applies to the other combinations.

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Answer:

The speed V = 194.03 mph

Direction = 3.6° northeast

Step-by-step explanation:

The distance of the trip = 291 miles

The direction of flight = 9.1 degrees northeast

Speed of the prevailing wind = 25 mph

Direction of wing = southeast = 45 degrees South of East

The speed heading to Ivy Cliffs = V₁

V×(sin(9.1) + cos(9.1)) + 25×(-sin(45) + cos(45))  × t₁ = 291 miles

V×(sin(9.1) + cos(9.1)) + 25×(sin(45) + -cos(45))  × t₂ = 291 miles

t₁ + t₂ = 3 hours

(V×(sin(9.1)-25×(sin(45))j + (V×cos(9.1) + 25×cos(45))i

The magnitude V² = V²+29.32·V +625= 291²/t₁²......(1)

Also on the return trip we have;

V²-29.32·V +625= 291²/(3-t₁)²..........................................(2)

Subtracting equation (2) from (1) gives;

58.64·V = 291²/t₁² - 291²/(3-t₁)² = 291²×(6·t-9)/(t²·(t-3)²)

V = 291²×(6·t-9)/(t²·(t-3)²)/58.64

Substituting the value of V in (2) with a graphing calculator gives;

t₁= 1.612 or 1.387

Given that magnitude of the speed going > return = V² for  t₁ < t₂

t₁ = 1.387, t₂ = 1.612

From  V²+29.32·V +625= 291²/t₁², we have

V²+29.32·V +625= 291²/1.387²

Which gives

V²+29.32·V -43336.5 = 0

(V + 233.35)(V-194.03) = 0

V = -233.35 mph or V = 194.03 mph

Given that V is a natural number, we have, V = 194.03 mph.

The direction is given by the relation;

V×(sin(9.1) + cos(9.1)) + 25×(-sin(45) + cos(45))  × t₁ = 291 miles

V×(sin(9.1) + cos(9.1)) + 25×(-sin(45) + cos(45))  × t₁ = 291 miles

194.03×sin(9.1 degrees)-25×sin(45 degrees)j + 194.03×cos(9.1 degrees) + 25×cos(45 degrees)i = 291/1.387

13j + 209.27i = 208.81 mph

The angle tan θ = 13/209 = 0.00622

θ = tan⁻¹(13/209.27) = 3.6°.

Answer: 80 ways

Step-by-step explanation:

If 8 tee shirts can be worn in 8 ways  

similarly for 2 belts = 2 ways

five pair of jeans= 5 ways  

so total number of ways=5*2*8=80 ways

The answers are:

(a) The x component of the vector is 41.568 m.
(b) The y component of the vector is 24.0 m.

(a) The displacement vector \(\( \vec{r} \)\) in the xy plane has a magnitude of 48.0 m and is directed at an angle of \(\( \theta = 30.0^\circ \)\) in the figure.
To determine the x component of the vector, we can use the trigonometric identity \(\( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \).\)
In this case, the adjacent side represents the x component, and the hypotenuse is the magnitude of the vector.

So, the x component can be calculated as:
\(\( \text{x component} = 48.0 \, \mathrm{m} \times \cos(30.0^\circ) \)\( \text{x component} = 48.0 \, \mathrm{m} \times 0.866 \)\( \text{x component} = 41.568 \, \mathrm{m} \)\)
Therefore, the x component of the vector is 41.568 m.

(b) To determine the y component of the vector, we can use the trigonometric identity\(\( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \).\)
In this case, the opposite side represents the y component, and the hypotenuse is the magnitude of the vector.

So, the y component can be calculated as:
\(\( \text{y component} = 48.0 \, \mathrm{m} \times \sin(30.0^\circ) \)\( \text{y component} = 48.0 \, \mathrm{m} \times 0.5 \)\( \text{y component} = 24.0 \, \mathrm{m} \)\)
Therefore, the y component of the vector is 24.0 m.

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Answer:

6/4 and can be simplified as 3/2

Step-by-step explanation:

The vector in the direction is [1/sqrt(46), 3/sqrt(46), 2/sqrt(46), 0]

A unit vector in the direction of u is u/|u| where |u| is the magnitude of u.

To find the magnitude of u, we use the formula:

|u| = sqrt(1^2 + 6^2 + 3^2 + 0^2) = sqrt(46)

So, a unit vector in the direction of u is:

u/|u| = [1/sqrt(46), 6/sqrt(46), 3/sqrt(46), 0/sqrt(46)]

Simplifying the vector, we get:

[1/sqrt(46), 3/sqrt(46), 2/sqrt(46), 0]

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No a subspace of a finite dimensional vector space cannot be an infinite dimensional vector space

This is due to the fact that a subspace is a subset of a larger vector space and must have the same dimension as its parent vector space. In other words we can say that, if the parent vector space is finite dimensions,then the subspace must be as well.

To put it in another way, a vector space is finite dimensional if it can be traversed by a limited number of vectors. A subspace can only be spanned by a subset of the vectors that span the parent vector space since it is a subset of a larger vector space. As a result, if the parent vector space has finite dimensions, the subspace must be as well.

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