Let u = (1,2), v = (−3,4), and w = (5,0)

a) Draw these vectors in ℝ2 .

b) Find scalars λ1 and λ2 such that w = λ1u + λ2v.

Answer:

[tex] z = \frac{83-76}{8}=0.875[/tex]

[tex] z = \frac{79-65}{12}=1.167[/tex]

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Emilia case

Let X the random variable that represent the scores of a test, and we know that

Where [tex]\mu=76[/tex] and [tex]\sigma=8[/tex]

The z score formula is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Since Emilia made 83 points we can find the z score like this:

[tex] z = \frac{83-76}{8}=0.875[/tex]

Ferdinand case

Let X the random variable that represent the scores of a test, and we know that

Where [tex]\mu=65[/tex] and [tex]\sigma=12[/tex]

The z score formula is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Since Ferdinand made 79 points we can find the z score like this:

[tex] z = \frac{79-65}{12}=1.167[/tex]

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.

Answer:

[tex](0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401[/tex]  

[tex](0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380[/tex]  

We are confident at 95% that the difference between the two proportions is between [tex]-0.4401 \leq p_B -p_A \leq -0.1380[/tex]

1.  -.4401 ≤ p1 - p2 ≤ -.1380

4.  The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

Step-by-step explanation:

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.

1.  -.4401 ≤ p1 - p2 ≤ -.1380

2.  -.4401 ≤ p1 - p2 ≤ .1380

3.  The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%

4.  The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

5.  The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Solution to the problem

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

[tex]p_1[/tex] represent the real population proportion for San Jose

[tex]\hat p_1 =\frac{30}{73}=0.411[/tex] represent the estimated proportion for San Jos

[tex]n_1=73[/tex] is the sample size required for San Jose

[tex]p_2[/tex] represent the real population proportion for San Francisco

[tex]\hat p_2 =\frac{56}{80}=0.7[/tex] represent the estimated proportion for San Francisco

[tex]n_2=80[/tex] is the sample size required for San Francisco

[tex]z[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]  

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_1 -\hat p_1) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]  

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401[/tex]  

[tex](0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380[/tex]  

We are confident at 95% that the difference between the two proportions is between [tex]-0.4401 \leq p_B -p_A \leq -0.1380[/tex]

Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.

Based on this the correct options are:

1.  -.4401 ≤ p1 - p2 ≤ -.1380

4.  The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

In Statistics, percentiles are a representation of the relative position of a particular value within a data set. For example, if your exam score is better than k% of the rest of the class. That means your exam score is at the kth percentile.

If your test score is at the 32nd percentile it can be interpreted as follows:

-Your test score is better than only 32 percent of the other scores recorded for the test.

-32 percent of the people who took the admission test have scores which are lower than yours.

Answer:

A. 0.05kg/l

B. dy/dt = 9/1000(25 - y)

C. 20.05 kg of salt

D. 0.0025kg/l

Step-by-step explanation:

A. Concentration of salt in the tank initially,

Concentration (kg/l) = mass of salt in kg/ volume of water in liter

= 50kg/1000l

= 0.05kg/l

B. dy/dt = rate of salt in - rate of salt out

Rate of salt in = 0.025kg/l * 9l/min

= 0.225kg/min

Rate of salt out = 9y/1000

dy/dt = 0.225 - 9y/1000

dy/dt = 9/1000(25 - y)

C. Collecting like terms from the above equation,

dy/25 - y = 9/1000dt

Integrating,

-Ln(25 - y) = 9/1000t + C

Taking the exponential of both sides,

25 - y = Ce^(-9t/1000)

Calculating for c, at y = 0, t = 0;

C = 25

y(t) = 25 - 25e^(-9t/1000)

At 2.5 hours,

2.5 hours * 60 mins = 180 mins

y(180 mins) = 25 - 25e^(-9*180/1000)

= 25 - 25*(0.1979)

= 20.05kg of salt

D. As time approaches infinity, e^(Infinity) = 0,

y(t) = 25 - 25*0

Concentration (kg/l) = 25/1000

= 0.0025kg/l

The initial concentration is 0.05 kg/L. A differential equation that models the amount of salt in the tank is dy/dt = 0.225 - (9y/(1000 + 9t)). By solving this equation over an interval of 2.5 hours could find the amount of salt after this time period. Lastly, concentration of the tank's solution as time approaches infinity is 0.025 kg/L.

To solve the problem, we need to know the basics of differential equations and concepts related to concentration calculations. Starting with the initial conditions, we see that:

(a) The initial concentration is calculated by dividing the amount of salt by the volume of the solution. Hence, concentration = 50 kg/1000 L = 0.05 kg/L.

(b) The differential equation that models the salt in the tank can be derived using the inflow and outflow rates. The amount of salt entering the tank per minute is 9 L/min * 0.025 kg/L = 0.225 kg/min. The amount of salt leaving the tank per minute is 9 L/min * y kg/L; where y is the current amount of salt in the tank divided by the current volume of the solution in the tank (1000 L + 9t min). Hence, the differential equation is dy/dt = 0.225 - 9y/(1000 + 9t).

(c) To find the amount of salt after 2.5 hours, we would need to solve the differential equation given above with the initial condition y(0) = 50 kg, over the interval from t=0 to t=2.5 hours (or 150 minutes). This requires calculus skills specifically related to the solution of differential equations.

(d) At the limit as time approaches infinity, the volume in the tank approaches a constant (because inflow equals outflow), and so does the amount of salt in the tank (because inflow equals outflow). Hence, the differential equation becomes dy/dt = 0, yielding y = constant. This constant is the inflow rate divided by the outflow rate, or 0.225 kg/min / 9 L/min = 0.025 kg/L. This is the concentration of the solution as time approaches infinity.

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Answer:

dC = 2.44

Relative error = 19%

Step-by-step explanation:

[tex]C = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)[/tex]

Δv = 3 ; ΔT = 1 , v = 23, T = 8

Use differential Calculus

[tex]dC = Cv.dv + Ct.dt[/tex]

[tex]Cv = dC/dv = -5.72*v^(-0.84) + 0.0684*T*v^(-0.84)\\Ct = dC/dT = 0.6215 + 0.4275*v^(0.16)\\dC = modulus (Cv.dv) + modulus (CT.dT)\\dC = (-5.72*v^(-0.84) + 0.0684*T*v^(-0.84))* 3 + (0.6215 + 0.4275*v^(0.16))*1\\dC = 2.44[/tex]

Relative Error

dC / C @(T = 8, v=23)  * 100 = 19 %

The maximum possible propagated error in calculating wind chill is approximately 7.6425°F, and the relative error is approximately 3.06.

The formula for wind chill (C) is given by:

⇒ C = 35.74 + 0.6215T − 35.75[tex]v^{0.16}[/tex] + 0.4275T [tex]v^{0.16}[/tex]

Here, T is the temperature in degrees Fahrenheit, and v is the wind speed in miles per hour. We need to estimate the maximum possible propagated error in calculating the wind chill based on the given ranges for wind speed (v = 23 ± 3 mph) and temperature (T = 8° ± 3°).

⇒ ∂C/∂T = 0.6215 + 0.4275v0.16

⇒ ∂C/∂v = -5.72v-0.84 + 0.0684Tv-0.84

⇒ ∂C/∂T | (8,23) = 0.6215 + 0.4275(230.16)

≈ 0.6215 + 0.4275(1.8838)

≈ 1.4269

⇒ ∂C/∂v | (8,23) = -5.72(23-0.84) + 0.0684(8)(23-0.84)

≈ -5.72(0.1957) + 0.5472(0.1957)

≈ -1.1206

⇒ dC ≈ |∂C/∂T| × dT + |∂C/∂v| × dv

⇒ dC ≈ |1.4269| × 3 + |-1.1206| × 3

≈ 4.2807 + 3.3618

≈ 7.6425

Therefore, the maximum possible propagated error is approximately 7.6425°F.

⇒ Relative Error = 7.6425 ÷ |-2.5|

≈ 3.06 (rounded to two decimal places).

Hence, the maximum possible propagated error is approximately 7.6425°F and the relative error is approximately 3.06.

Complete question:

The formula for wind chill C (in degrees Fahrenheit) is given by C = 35.74 + 0.6215T − 35.75[tex]v^{0.16}[/tex] + 0.4275T[tex]v^{0.16}[/tex]

where v is the wind speed in miles per hour and T is the temperature in degrees Fahrenheit. The wind speed is 23 ± 3 miles per hour and the temperature is 8° ± 3°. Use dC to estimate the maximum possible propagated error (round your answer to four decimal places) and relative error in calculating the wind chill (round your answer to two decimal places).

Answer:

matrices can be simple added by ading the adjacent elements of the two matrices

The matrix multiplication is done by multiplying row of first matrix to columns of the other matrix.

each multiplication for each element of the resultant matrix.

Step-by-step explanation:

Answer:

[tex]A)\,\,det(A)=1[/tex]

[tex]B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] [/tex]

Step-by-step explanation:

[tex]det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|[/tex]

Expanding with first row

[tex]det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1[/tex]

To find inverse we first find cofactor matrix

[tex]C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\[/tex]

[tex]C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\[/tex]

Cofactor matrix is

[tex]C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right][/tex]

Answer:

True

Step-by-step explanation:

A linear system of equations will have a unique solution if and only if the number of variables equal the number of independent equations.

By independent equations we mean the same equation not repeated by multiplying by any constant.

Suppose number of variables are n, we must have the determinant formed by the coefficients non zero to have a unique solutions.

Here the no of equations are less than the number of variables.  So we cannot have a unique solution but can have a parametric solution using the number of parameters as n-m where n = number of variables and m = the number of independent equations given.

So the given statement is true.

Answer:

The given statement is false.

Step-by-step explanation:

We are given the following statement:

"If a linear system has the same number of equations and variables, then it must have a unique solution."

The given statement is false because If a linear system has the same number of equations and variables, then it may have unique solution, no solution or infinitely many solution.

There could be three types of solution

depending on the augmented matrix and coefficient matrix of the linear system.

Answer:

Skew lines

Step-by-step explanation:

Two lines are given and we have to find out whether they are parallel, skew, or intersecting

[tex]x =t , y = 1 + 2t, z = 2 + 3t, l2 : \\ x = 3 -4s,, y = 2 -3s, z = 1 + 2s[/tex]

direction ratios of these two lines are

(1,2,3) and (-4, -3,2) (coefficient of parameters)

Obviously these two are neither equal nor proportional

Hence we get not parallel lines.  If these intersect there must be a common point making

[tex]t= 3-4s:   1+2t =2-3s and 2 + 3t =1+2s[/tex]

Consider first two equation

[tex]t+4s =3\\2t+3s = 1\\2t+8s = 6\\-5s =-5\\s=1[/tex]

when s=1, t = -1

Let us check whether these two values satisfy the third equation

2+3t = -1 and 1+2s = 1+2 =3

Not equal.  so there is no common point between them

These two are skew lines

The given lines l1 and l2 intersect at the point (7, -10, -4).

The given lines, l1 and l2, can be analyzed to determine whether they are parallel, skew, or intersecting. By comparing the equations of the two lines, we can see that they intersect at a single point. To find the point of intersection, we can set the x, y, and z coordinates of l1 equal to the x, y, and z coordinates of l2, and solve for the parameter t. By substituting the value of t back into the equations of l1, we can find the point of intersection to be (7, -10, -4).

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Answer:

a) [tex]H_{0}: p = 0.66\\H_A: p > 0.66[/tex]

b) P-value = 0.2650

c) No, this programme will not be recommended as there is no real improvement over the national average.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 240

p = 66% = 0.66

Alpha, α = 0.05

Number of students admitted to law school , x = 163

a) First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.66\\H_A: p > 0.66[/tex]

This is a one-tailed(right) test.  

Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{163}{240} = 0.6792[/tex]

[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

Putting the values, we get,

[tex]z = \displaystyle\frac{0.6792-0.66}{\sqrt{\frac{0.66(1-0.66)}{240}}} = 0.6279[/tex]

b) Now, we calculate the p-value from the table.

P-value = 0.2650

c) Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is no real improvement over the national average.

No, this programme will not be recommended as there is no real improvement over the national average.

Answer:

Since the equation was missing, I solved it with another equation and got an answer of T(0) = <3j / 5 + 4k / 5>.

Please see my explanation. I hope this helps

Step-by-step explanation:

The question asked us to find out unit tangent vector.

Recall unit vector = vector / magnitude of vector

Since the question is missing with an equation. I suppose an equation.

r(t)=Cost i, 3t j, 2Sin(2t) k at t=0

Lets take out differentiation

r'(t) = <(-Sint), 3, 2(Cos(2t)(2))>

r'(t)= <-Sint, 3, 4Cos(2t)>

Now substitute t=0 in the differentiate found above.

r'(0)= <-Sin(0), 3, 4Cos(2*0)>

r'(0)= <0, 3, 4(1)>

r'(0)= <0,3,4>

vector r'(0)=<0i, 3j, 4k>

Now lets find out magnitude of vector

|r'(0)| = [tex]\sqrt{0^{2}+3^{2}+4^{2} }[/tex]

|r'(0)| = [tex]\sqrt{0+9+16}[/tex]

|r'(0)| = [tex]\sqrt{25}[/tex]

|r'(0)| = 5

Unit Tangent Vector

T(0) = <0, 3, 4> / 5

T(0) = <3j / 5  +  4K / 5>

To find the unit tangent vector, we first need to find the velocity vector. Given that the position vector is r(t) = Acos(wt)i + Asin(wt)j, we can find the derivative of this vector to get the velocity vector. To find the unit tangent vector, we divide the velocity vector by its magnitude.

To find the unit tangent vector, we first need to find the velocity vector. Given that the position vector is r(t) = Acos(wt)i + Asin(wt)j, we can find the derivative of this vector to get the velocity vector:

v(t) = -Aw*sin(wt)i + Aw*cos(wt)j

To find the unit tangent vector, we divide the velocity vector by its magnitude:

T(t) = (v(t))/(|v(t)|) = (-Aw*sin(wt)i + Aw*cos(wt)j)/(sqrt((Aw*sin(wt))^2 + (Aw*cos(wt))^2))

So, the unit tangent vector at any point is T(t).

Answer:

1. x = (199/130)e^(-2t) - (33/65)e^(-4t) - (18/65)cos3t - (1/65)sin3t

2. x = (191/128) - (63/128)e^(-4t) + (t³/12) - (t²/16) + (t/32)

Step-by-step explanation:

Steps are shown in the attachment.

Answer:

(0.185, 0.413)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 87, x = 26, p = \frac{x}{n} = \frac{26}{87} = 0.2989[/tex]

98% confidence interval

So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.325[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 - 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.185[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 + 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.413[/tex]

So the correct answer is:

(0.185, 0.413)

Answer: the cost of a cheeseburger is $4

Step-by-step explanation:

Let x represent the cost of one cheeseburgers.

Let y represent the cost of one Soda.

The patriot diner sells 2 cheeseburgers and one soda for $11.00. It means that

2x + y = 11 - - - - - - - - - - - 1

She also sells 3 cheeseburgers and 2 sodas for $18.00. It means that

3x + 2y = 18 - - - - - - - - - - -2

Multiplying equation 1 by 3 and equation 2 by 2, it becomes

6x + 3y = 33

6x + 4y = 36

Subtracting, it becomes.

- y = - 3

y = 3

Substituting y = 3 into equation 1, it becomes

2x + 3 = 11

2x = 11 - 3 = 8

x = 8/2 = 4

Answer:

The answer to your question is letter C

Step-by-step explanation:

Equations

                             3x  - y  = 14     ----------- (I)

                             4x + y =   21    ----------- (II)

Solve the system of equations by the elimination method.

                            3x -  y  = 14

                            4x + y  = 21

                            7x   0   = 35

Solve for x, divide both sides by 7

                                  7x/ 7 = 35/7                        

                                         x = 5

Substitute x in equation I

                            3(5) - y = 14

                            15 - y = 14

Subtract 15 in both sides

                            15 - 15 - y = 14 - 15

Simplify

                                      - y = -1

Multiply by -1

                                        y = 1

Solution

                                      (5, 1)                          

Answer:

140·80=11200

Step-by-step explanation:

From exercise we have example for how we find  half the sum of 246 and 388, we can enter "(246+388)/2".  

Based on this example, we will calculate what is required in the task using a calculator. So we use a calculator to find the product of the following numbers, 140 and 80.

We calculate, and we get

140·80=11200

Answer: The mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.

Step-by-step explanation:

if there is a linear relationship between two variables x and y , then we represent this relation in the form of equation as :[tex]y= mx+c[/tex]   (*)

, where m = rate of change of y with respect to x

c= Constant or the value of y when x=0.

We assume there is a linear relationship between the curb weight of a car (in thousands of kilograms) and its combined mileage (in miles per gallon).

Let y= combined mileage (in miles per gallon)

x= curb weight of a car (in thousands of kilograms)

When x= 3 thousand kg  , y = 9 mpg

⇒ [tex]9= m(3)+c[/tex]                   [Put values in (*)]

⇒ [tex]9= 3m+c--------(1)[/tex]

When x= 1.9 thousand kg  , y = 21 mpg

⇒ [tex]21= m(1.9)+c[/tex]                   [Put values in (*)]

⇒ [tex]21= 1.9m+c--------(2)[/tex]

Eliminate the equation (1) from (2) , we get

[tex]1.9m-3m=21-9[/tex]

[tex]-1.1m=12\\\Rightarrow\ m=\dfrac{12}{-1.1}=\dfrac{-120}{11}[/tex]

Put value of m in (1) , we gte

[tex]9= 3(\dfrac{-120}{11})+c[/tex]

[tex]9= \dfrac{-360}{11}+c\\\Rightarrow\ c=9+\dfrac{360}{11}=\dfrac{459}{11}[/tex]

Substitute the value of m and c in (*) , we get

[tex]y= \dfrac{-120}{11}x+\dfrac{459}{11}[/tex]

When x= 1.1

[tex]y= \dfrac{-120}{11}(1.1)+\dfrac{459}{11}[/tex]

[tex]y= \dfrac{-120}{11}(\dfrac{11}{10})+\dfrac{459}{11}[/tex]

[tex]y=-12+\dfrac{459}{11}=\dfrac{327}{11}\approx29.72[/tex]

Hence, the mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.

Answer:

(a) 142 km trip requires 15.16 Liters of Gasoline

(b) The money spent is 69.34 euros or $87.37.

(c) The average speed of sprinter is 34.45 km/h.

(d) 4 lb of candy contains 0.559 kg of dietary fat.

Step-by-step explanation:

(a)

Given that;

Mileage = 22 mi/gal

Converting it to km/L

Mileage = (22 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)

Mileage = 9.36 km/L

Now, the gasoline required for 142 km trip:

Gasoline Required = (Length of trip)/(Mileage)

Gasoline Required = (142 km)/(9.36 km/L)

Gasoline Required = 15.16 L

(b)

Given that;

Mileage = 30 mi/gal

Converting it to km/L

Mileage = (30 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)

Mileage = 12.77 km/L

Now, the gasoline required for 142 km trip:

Gasoline Required = (Length of trip)/(Mileage)

Gasoline Required = (115 km/day)(7 days/week)/(12.77 km/L)

Gasoline Required = 63 L/week

Now, we find the cost:

Weekly Cost = (Gasoline Required)(Unit Cost)

Weekly Cost = (63 L/week)(1.1 euros/L)

Weekly Cost = 69.34 euros/week = $87.37

since, 1 euro = $1.26

(c)

Average Speed = (Distance Travelled)/(Time Taken)

Average Speed = 200 m/20.9 s

Average Speed = (9.57 m/s)(3600 s/1 h)(1  km/ 1000 m)

Average Speed = 34.45 km/h

(d)

22.7 g piece contains = 7 g dietary fat

(22.7 g)(1 lb/453.592 g) piece contains = (7 g)(1 kg/1000 g) dietary fat

0.05 lb piece contains = 0.007 kg dietary fat

1 lb piece contains = (0.007/0.05) kg dietary fat

4 lb piece contains = 4(0.007/0.05) kg dietary fat

4 lb piece contains = 0.559 kg dietary fat

The expression  [tex]Cosec\theta*Tan\theta[/tex] in terms of [tex]Sin\theta[/tex] and  [tex]Cos\theta[/tex] in simplified form is  [tex]Cosec\theta*Tan\theta = Sin\theta[/tex]

Given expression:

[tex]Cosec\theta*Tan\theta[/tex]

Express [tex]cosec\theta[/tex]  and  [tex]Tan\theta[/tex]  in terms of [tex]Sin\theta[/tex]  and [tex]Cos\theta[/tex]

Now,

[tex]Cosec\theta = \dfrac{1}{Sin\theta}[/tex]

[tex]Tan\theta = \dfrac{Sin\theta}{Cos\theta}[/tex]

Substitute the expression of  [tex]Cosec\theta[/tex]  and  [tex]Tan\theta[/tex] in original expression:

[tex]Cosec\theta*Tan\theta = \dfrac{1}{Sin\theta} * \dfrac{Sin\theta}{Cos\theta}[/tex]

[tex]Sin\theta[/tex]  is cancel out in the right-hand side of the equation:

[tex]Cos\theta[/tex] can be written as:

[tex]\dfrac{1}{Cos\theta} = \dfrac{1}{\dfrac{1}{Sin\theta} }[/tex]

[tex]= Sin\theta[/tex]

For the given expression:

[tex]Cosec\theta*Tan\theta = Sin\theta[/tex]

The simplified expression is [tex]Cosec\theta*Tan\theta = Sin\theta[/tex]

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The expression csc θ tan θ simplifies to sec θ which further simplifies to [tex]\frac{1}{cos\theta}[/tex].

To express csc θ tan θ in terms of sin θ and cos θ, we start by rewriting each trigonometric function:

Answer:

[tex] T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)[/tex]

[tex] P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z<0) = 0.5[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(12,0.3)[/tex]  

Where [tex]\mu=12[/tex] and [tex]\sigma=0.3[/tex]

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}[/tex]

If we solve for the total T we got:

[tex] T= n \bar X[/tex]

For this case then the expected value and variance are given by:

[tex] E(T) = n E(\bar X) =n \mu[/tex]

[tex] Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2[/tex]

And the deviation is just:

[tex] Sd(T) = \sqrt{n} \sigma[/tex]

So then the distribution for the total would be also normal and given by:

[tex] T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)[/tex]

And we want this probability:

[tex] P(T\leq 72)[/tex]

And we can use the z score formula given by:

[tex] z = \frac{x-\mu}{\sigma}[/tex]

[tex] P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z<0) = 0.5[/tex]

Answer:

Dependent: fee

Independent: time

Step-by-step explanation:

The scenario described in question indicates that fee is the dependent variable and time is independent variable. The reasoning is that fee depends on the time(hour). The fee will increase as the time increases. Whereas change in time cause change in fee. So, fee is the dependent variable and time is independent variable.  

Final answer:

In the SCUBA equipment rental situation, the independent variable is the number of hours the equipment is rented, and the dependent variable is the total fee. The equation for the total fee is y = 25 + 12.50x, with $25 representing the y-intercept and $12.50 the slope of the hourly rental cost.

Explanation:

For the SCUBA equipment rental situation at the vacation resort, the independent variable is the number of hours the equipment is rented. Since the renter can decide how many hours to rent the equipment, it is considered independent. On the other hand, the dependent variable is the total fee for renting the equipment because it depends on the number of hours the equipment is rented for.

In this context, the equation expressing the total fee (y) in terms of the number of hours (x) the equipment is rented is:
y = 25 + 12.50x.

Here, $25 is the up-front fee which is also the y-intercept of the equation, while $12.50 per hour is the slope. The y-intercept represents the initial cost of renting the equipment before any hourly charges apply, and the slope represents the rate at which the total cost increases for every additional hour the equipment is rented.

Answer:

We conclude that the valve performs above the specifications.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 6.3 pounds per square inch

Sample mean, [tex]\bar{x}[/tex] = 6.5 pounds per square inch

Sample size, n = 130

Alpha, α = 0.02

Population standard deviation, σ = 0.8

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 6.3\text{ pounds per square inch}\\H_A: \mu > 6.3\text{ pounds per square inch}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{6.5 - 6.3}{\frac{0.8}{\sqrt{130}} } = 2.85[/tex]

Now, [tex]z_{critical} \text{ at 0.02 level of significance } = 2.05[/tex]

Decision rule:

If the calculated statistic is greater than the the critical value, we reject the null hypothesis and if the calculated statistic is lower than the the critical value, we accept the null hypothesis

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.

Thus, we conclude that the valve performs above the specifications.

Answer:

Step-by-step explanation:

You expect around K kids to come to your house, and each kid is to be given three pieces of candy. This means that the total number of candies that you would buy is

3 × K = 3K

Each bag of candy you can buy contains N pieces of candy for P dollars.

Therefore,

If N pieces of candy cost $P, then

3K pieces of candy would cost $M

Therefore, the algebraic expression

to tell you how much should you expect to have to pay (M) would be

M = 3KP/N

The algebraic expression that tells the amount you expect to pay is [tex]\frac{3KP}N[/tex]

The given parameters are:

The total amount paid for N pieces of candy is calculated as:

[tex]Total = Kids \times 3 \times Unit\ Rate[/tex]

So, we have:

[tex]Total = K \times 3 \times \frac PN[/tex]

Evaluate the product

[tex]Total = \frac{3KP}N[/tex]

Rewrite the above equation as:

[tex]M= \frac{3KP}N[/tex]

Hence, the algebraic expression that tells the amount you expect to pay is [tex]\frac{3KP}N[/tex]

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Answer:

False

Step-by-step explanation:

Consider the equations with the same number of equations and variables as shown below,

Case 1

                        [tex]x_{1} + x_{2} = 0\\x_{1} + x_{2} = 1[/tex]

This equation has no solution because it is not possible to have two numbers that give a sum of 0 and 1 simultaneously.

Case 2

                       [tex]x_{1} + x_{2} = 1\\2x_{1} + 2x_{2} = 2[/tex]

This equation has infinitely many possible solutions.

Therefore it is FALSE to say a linear system with the same number of equations and variables, must have a unique solution.

The statement that a linear system with the same number of equations and variables must have a unique solution is false. Other considerations, such as whether the system is consistent or inconsistent and dependent or independent, can impact the amount of solutions a system has.

The statement is false. Even if a linear system has the same number of equations and variables, it does not necessarily mean that it will have a unique solution. Rather, whether a system has a unique solution, no solutions, or infinitely many solutions depends on whether the system is consistent or inconsistent and dependent or independent.

For instance, consider two linear equations: x + y = 5 and 2x + 2y = 10. Even though these two equations have the same number of variables and equations, they represent the same line and thus have infinitely many solutions. Similarly, consider the system x + y = 5 and x + y = 6. These two equations have also the same number of equations and variables but they are parallel lines and do not intersect, so this system does not have any solution.

So, the number of variables and equations is not always enough to determine the number of solutions to a linear system.

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Answer:

10⁵ = 100,000

Step-by-step explanation:

Data provided in the question:

Number of available choices = 10

for the sample of 5 i.e n = 5

Repetition is allowed

Thus,

Total samples of 5 that are possible = ( Number of available choices )ⁿ

thus,

Total samples of 5 that are possible = 10⁵

Hence,

The Total samples of 5 that are possible = 100,000

Final answer:

The probability of drawing a 7 followed by a king from a standard 52-card deck without replacing the first card is 4/663, which rounds to approximately 0.006 when rounded to the nearest thousandth.

Explanation:

The student is asking to find the probability of drawing a 7 for the first card and a king for the second card from a standard 52-card deck, without replacing the first card. To solve this, we need to calculate the probability of each event occurring consecutively.

First, the probability of drawing a 7 from a deck of 52 cards is 4/52 or 1/13, because there are 4 sevens in the deck. After drawing a seven, there are now 51 cards left in the deck. Next, the probability of drawing a king from the remaining 51 cards is 4/51, since there are still 4 kings in the deck.

To find the combined probability of both events happening, we multiply these probabilities together:

Probability of drawing a 7 and then a king = (1/13) × (4/51)

Which simplifies to:

Probability = 4/663

Rounded to the nearest thousandth, the probability is approximately 0.006.

Exit polls commonly use simple random sampling, stratified sampling, and systematic sampling to ensure a representative sample of voters. Cluster sampling may be used, but convenience sampling is typically avoided to prevent bias.

Exit polls typically employ a variety of sampling methods to ensure a representative sample of voters. These methods can include simple random sampling, where each person has an equal chance of being selected, and stratified sampling, which involves dividing the population into subgroups and sampling from each. Systematic sampling can also be used by selecting every nth person to participate in the poll.

Cluster sampling could be utilized if the population is divided into clusters, and a random selection of clusters is made. Convenience sampling, however, is generally not used in exit polling because it can introduce bias. Instead, methods that provide each individual an equal chance of selection are preferred to obtain a representative sample.

Answer:

The correct number of servings = 6.

By saying there were [tex]3\frac{3}{5}[/tex] servings of soup, Hector must have intended to say that there were [tex]3\frac{3}{5}[/tex] quarts of soup for serving.

Step-by-step explanation:

Given:

Total quarts of soup made by students = [tex]4\frac{1}{2}[/tex]

Fraction of the soup served to  friends = [tex]\frac{4}{5}[/tex]

Each serving = [tex]\frac{3}{5}[/tex] quarts

Hector incorrectly says that that there were [tex]3\frac{3}{5}[/tex] servings of soup.

To find the correct number of servings and to identify Hector's mistake.

Solution:

Total quarts of soup made = [tex]4\frac{1}{2}=\frac{9}{2}[/tex]

Fraction of total quarts served = [tex]\frac{4}{5}[/tex]

Thus, total quarts of soup served = [tex]\frac{9}{2}\times \frac{4}{5}[/tex]

⇒ [tex]\frac{18}{5}=3\frac{3}{5}[/tex]

Thus, total quarts of soup for serving  =[tex]\frac{18}{5}[/tex] or [tex]3\frac{3}{5}[/tex]

Each serving  = [tex]\frac{3}{5}[/tex] quarts

Total number of servings can be given as:

⇒ [tex]\frac{Total\ quarts\ of\ serving}{Size\ of\ each\ serving}[/tex]

⇒ [tex]\dfrac{\frac{18}{5}}{\frac{3}{5}}[/tex]

To divide fractions, we take reciprocal of the divisor and replace divsion with multiplication.

⇒ [tex]\frac{18}{5}\times \frac{5}{3}[/tex]

⇒ [tex]\frac{18}{3}[/tex]

⇒ 6  servings

Thus, the correct number of servings = 6.

By saying there were [tex]3\frac{3}{5}[/tex] servings of soup, Hector must have intended to say that there were [tex]3\frac{3}{5}[/tex] quarts of soup for serving.

Answer:

The answer is "False."

Explanation:

An area is considered the amount of space that an object occupies. A filled-area subjects the object to be made  up of lines. These lines connect with each other to form an "edge."

The connection of the lines in order to define the object's shape or area are considered "snap points." Remember that "polygons" are made of line segments, where their endpoints meet with each other in order to define its closed shaped. Thus, it needs to conform to "snap points."

This explains the answer.

The given statement "Like the edges of a filled-in area, the endpoints of a polygon do not need to conform to snap points" is false.

The edge of a filled-in area is the boundary of the surface. Area is a two-dimensional space covered by any surface. The boundary of the surface is generally the edges of the area. The edges or the boundary should connect in order to make a closed area or surface.

Similarly, snap points are the end points of the sides of a polygon. These endpoints need to connect with each other in order to make a closed figure. Polygon is a closed figure with n number of sides.

Therefore, the given statement "Like the edges of a filled-in area, the endpoints of a polygon do not need to conform to snap points" is false.

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