Two widgets and five gadgets cost $57. One widget and three gadgets cost $32.70 How much does one gadget cost? ​

To diagonalize C, we first need to find its eigenvalues and eigenvectors. The characteristic equation for C is det(C -

                                λI)  =  0, which gives us (1 - λ)(7 - λ)  -  6  =                                

                                   0. Solving for λ, we get λ1  =  1 and λ2  =                                    

               7. To find the eigenvector corresponding to λ1, we solve the system of equations (C -                

                   λ1I)x  =  0, which gives us the equation  - x1  +  2x2  =  0. Choosing x2  =                    

                                           1, we get the eigenvector v1  =                                          

                            [2,1]. Similarly, for λ2 we get the eigenvector v2  =  [1, -                            

                              1]. We can then diagonalize C by forming the matrix P  =                              

                         [v1, v2] and the diagonal matrix D  =  [λ1 0; 0 λ2]. We have C  =                        

                      -                                    -                                                        

                   PDP 1. To compute A5, we first compute C 1 as [7  - 2;  - 3 1] / 4. Then, A  =                    

                         -         -   -                                  5       5       5                          

                      CDC 1  =  PDP 1DC 1P. We have D  =  [1 0; 0 7], so D   =  [1  0; 0 7 ]  =                      

                                           5       5 -                                                              

                    [1 0; 0 16807]. Thus, A   =  PD P 1  =  [2 1; 1  - 1][1 0; 0 16807][1 / 3  -                    

                              1 / 3; 1 / 3 2 / 3]  =  [11203 11202; 16804 16805] / 9.                                

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The mean of the data-set in this problem is given as follows:

19.56 pounds.

The mean of a data-set is given by the sum of all observations in the data-set divided by the number of observations, which is also called the cardinality of the data-set.

The dot plot shows the number of times that each observation appears, hence the observations are given as follows:

Hence the mean is given as follows:

Mean = (2 x 12 + 4 x 15 + 1 x 16 + 1 x 18 + 2 x 20 + 1 x 22 + 3 x 25 + 2 x 29)/(2 + 4 + 1 + 1 + 2 + 1 + 3 + 2)

Mean = 313/16

Mean = 19.56 pounds.

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The probability density function of Z is fZ(z) = (1/√(2π(4 + 2e^(-5)))) * e^(-z^2/(2(4 + 2e^(-5))))

Given that X(t) is a stationary Gaussian random process with mX(t) = 0 and RX(τ) = 2e^(-5|τ|).

We are interested in finding the probability density function (PDF) of Z = X(2) + X(3).

First, we need to find the mean and variance of Z:

E[Z] = E[X(2) + X(3)] = E[X(2)] + E[X(3)] = 0 + 0 = 0

Var(Z) = Var(X(2) + X(3)) = Var(X(2)) + Var(X(3)) + 2Cov(X(2), X(3))

Since X(t) is a stationary process, we have:

Var(X(2)) = Var(X(3)) = RX(0) = 2

Cov(X(2), X(3)) = RX(1) = 2e^(-5)

Therefore, Var(Z) = 2 + 2 + 2e^(-5) = 4 + 2e^(-5)

Now we can use the properties of Gaussian random variables to find the PDF of Z. Since Z is a linear combination of Gaussian random variables, it is also Gaussian with mean 0 and variance 4 + 2e^(-5).

Thus, fZ(z) = (1/√(2π(4 + 2e^(-5)))) * e^(-z^2/(2(4 + 2e^(-5)))).

Therefore, the probability density function of Z is fZ(z) = (1/√(2π(4 + 2e^(-5)))) * e^(-z^2/(2(4 + 2e^(-5))))

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Answer:

138°

235°

240°

Step-by-step explanation:

The first situation shows a straight angle, so the smaller angles that make it up should be supplementary. That means that their measures add up to 180°. My work for this first situation is shown below:

n + 42° = 180°
n + 42° - 42° = 180° - 42°
n = 138°

Now, let's look at the second situation. This one's pretty simple. It's just asking us to add up the measures of all the angles are given, so all we have to do is some addition, shown below:

23° + 40° + 92° + 80° = 235°

Finally, we're at the last situation. This situation shows a full angle, so the smaller angles that make it up should add up to 360°, since this is the amount of degrees in a full circle. Again, if we know the measure of one of these angles, we should be able to find the measure of the other one. See my work below:

n + 120° = 360°
n + 120° - 120° = 360° - 120°
n = 240°

And there are all your answers! Let me know if you need further clarification on all that. :)

Answer: i do not know but watch this: Find the Maclaurin series for f(x) = (1-x)^(-1) and associated radius of convergence by Ms. Shaws Math Class

Answer: 1: C=72.3822947387, A=416.560184761, 2: C=37.6991118431, A=113.097335529

Step-by-step explanation:

For the first circle, since the area of a circle is equal to pi multiplied by the radius squared, we can use this equation:

[tex]\pi11.52^{2}[/tex]

Now we can multiply the radius by 2 to get the diameter, which is 23.04, and multiply the diameter by pi to get the circumference, 72.3822947387.

If you want an exact circumference, all you need to do is display the diameter multiplied by pi without simplification, which would look like this:

23.04π.

We can do the same thing for the second circle. Because we already have the diameter, I will start with the circumference.

12pi = 37.6991118431, which is just simplified to 12pi. The reason you want to have pi in your answer for an exact answer, is because otherwise your answer would have to be an irrational decimal constant, which cannot fit on one page.

Now for the area of the second circle. Divide the diameter by 2, and multiply pi by the radius squared. This is an equation that would look something like this:

[tex]\pi6^{2}[/tex], which can simplify to [tex]\pi36[/tex] as an exact answer.

(a) The probability that all 4 missiles will fire is 0.2098 and (b) The probability that at most 2 will not fire is 0.0099.

Here we need to use the concept of combinations to get our required answer.

Here we have been given that 4 out of 14 missiles are defective.

Hence 10 missiles are working.

A sample of 4 missiles was chosen

Hence the simple can be chosen in ¹⁴C₄ ways

a)

The probability that all 4 missiles will fire is

¹⁰C₄/¹⁴C₄

= 0.2098

b)

The probability that at most 2 will not fire is

1 - the probability of including all three defective missiles

= 10³C₃/¹⁴C₄

= 0.0099

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(a) Probability all 4 fire: 0.3297. (b) Probability at most 2 not fire: 0.9591.

(a) To track down the chance that all of the 4 rockets will fire, we actually need to do not forget the quantity of methods we can select 4 operating rockets out of the eleven that are not unsuitable, isolated by the all out range of approaches we will select four rockets out of the whole component.

The amount of ways of selecting four operating rockets out of the eleven that aren't improper is given by way of the combination recipe:

C(eleven,four) = 330

The absolute quantity of methods of selecting four rockets out of the entire part is given via:

C(14,4) = 1001

Subsequently, the chance that each one of the four rockets will hearth is:

P(all four hearth) = C(11,four)/C(14,4) = 330/1001 ≈ zero.3297

(adjusted to 4 decimal spots)

(b) To find the likelihood that at most 2 rockets might not fire, we need to remember every one of the potential conditions in which 0, 1, or 2 poor rockets are selected, and afterward song down the likelihood of every case and upload them up.

Case 1: No defective rockets are selected

The amount of ways of selecting four working rockets out of the eleven that are not faulty is given through the mixture recipe:

C(11,four) = 330

Subsequently, the chance of choosing no defective rockets is:

P(0 inadequate) = C(eleven,4)/C(14,four) ≈ 0.3297

Case 2: One defective rocket is chosen

The quantity of ways of selecting three operating rockets out of the eleven that are not poor is given with the aid of the combo equation:

C(eleven,three) = 165

The amount of ways of selecting 1 damaged rocket out of the three that are available is given through the mix equation:

C(three,1) = 3

Subsequently, the all out number of approaches of selecting 1 insufficient and three running rockets is:

C(eleven,three) × C(3,1) = 495

Thusly, the probability of choosing one deficient rocket and three running rockets is:

P(1 deficient) = C(eleven,three) × C(three,1)/C(14,4) ≈ 0.4655

Case three: Two poor rockets are selected

The amount of approaches of choosing 2 poor rockets out of the three which might be handy is given through the combination recipe:

C(3,2) = 3

The amount of ways of choosing 2 running rockets out of the 11 that aren't faulty is given via the mix recipe:

C(11,2) = 55

In this way, the all out number of methods of selecting 2 broken rockets and 2 working rockets is:

C(3,2) × C(eleven,2) = one hundred sixty five

Subsequently, the chance of choosing two incorrect rockets and two operating rockets is:

P(2 insufficient) = C(3,2) × C(11,2)/C(14,four) ≈ 0.1638

Subsequently, the likelihood that at maximum 2 rockets might not fireplace is:

P(at maximum 2 don't fire) = P(zero imperfect) + P(1 blemished) + P(2 deficient) ≈ 0.9591

(adjusted to 4 decimal spots).

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Answer:

130.63

Step-by-step explanation:

The formula for a circle is:

Area = πr²

Let's plug in our values.

Area = 3.14(6.45)²

= 3.14(41.6025)

= 130.63185

Now we round to the nearest hundredth to get 130.63.

Hope this helps and good luck on your homework!

Answer:

130.62

Step-by-step explanation:

The formula for the area of a circle is [tex]\pi r^{2}[/tex].

We use 3.14 for pi and 6.45 for r:

3.14 · 6.45²

3.14 · 41.6

130.624, rounded to the nearest hundredth 130.62

The theoretical probability of the spinner not landing on blue is 3/4

From the question, we have the following parameters that can be used in our computation:

Colors = 4 i.e. purple, yellow, blue and red

Blue = 1

Not blue = 3

So, we have

Theoretical probability = Not Blue /Colors

Substitute the known values in the above equation, so, we have the following representation

Theoretical probability = 3/4

Hence, theoretical probability is 3/4

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The variance and standard deviation can never be negative. However, they can be zero if there is no variability in the data. It is possible for the variance and standard deviation to be smaller or larger than the mean depending on the spread of the data.

The variance and standard deviation can never be negative.
1. Variance is a measure of how spread out the data points are from the mean. It is calculated by finding the average of the squared differences from the mean. Since squares are always positive or zero, the variance cannot be negative.

2. Standard deviation is the square root of the variance. Since the square root of a negative number is not a real number, the standard deviation cannot be negative either.

It is worth noting that both variance and standard deviation can be zero if all data points are the same, and they can be smaller or larger than the mean, depending on the data distribution.

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The equation of the line of best fit is y = 0.883x + 17.95.

We have,

To find the line of best fit, we want to find the equation of the line that comes closest to passing through all the points in the scatterplot.

One way to do this is to use linear regression analysis.

Using a calculator or statistical software,

We can find that the equation of the line of best fit for this data is:

y = 0.883x + 17.95

Thus,

The equation of the line of best fit is y = 0.883x + 17.95.

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The units/one's digit of 5^221 is 5.

To determine the units/one unit digit of a number raised to a power, we only need to consider the units/one's digit of the base. In this case, the unit digit of 5 is 5.

Now, we need to look for a pattern in the units digit of 5 raised to different powers.

5^1 = 5 (units digit is 5)
5^2 = 25 (units digit is 5)
5^3 = 125 (units digit is 5)
5^4 = 625 (units digit is 5)
5^5 = 3125 (units digit is 5)
. . .and so on.

We can see that the unit digit of 5 raised to any power is always 5. Therefore, the units/one's digit of 5^221 is 5, not 3.

So, the statement "the units/one's digit of 5^221 is 3" is disproved.

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Approximately 670 wrestlers weigh between 210 and 230 lbs.

a) To find the proportion/percentage of wrestlers that weigh less than 220 lbs, we need to standardize the weight value using the formula:

z = (x - μ) / σ

where x is the weight value, μ is the mean weight, and σ is the standard deviation.

So, for x = 220 lbs:

z = (220 - 225) / 20 = -0.25

Looking up the standard normal table or using a calculator, we find that the area/proportion to the left of z = -0.25 is 0.4013. Therefore, the proportion/percentage of wrestlers that weigh less than 220 lbs is:

0.4013 or 40.13%

b) To find the probability that a random wrestler weighs more than 250 lbs, we again need to standardize the weight value:

z = (250 - 225) / 20 = 1.25

Using the standard normal table or a calculator, we find that the area/proportion to the right of z = 1.25 is 0.1056. Therefore, the probability that a random wrestler weighs more than 250 lbs is:

0.1056 or 10.56%

c) To find the number of wrestlers that weigh between 210 and 230 lbs, we first need to standardize these weight values:

z1 = (210 - 225) / 20 = -0.75

z2 = (230 - 225) / 20 = 0.25

Next, we need to find the area/proportion between these two standardized values:

P(-0.75 < z < 0.25) = P(z < 0.25) - P(z < -0.75)

Using the standard normal table or a calculator, we find that P(z < 0.25) is 0.5987 and P(z < -0.75) is 0.2266. Therefore:

P(-0.75 < z < 0.25) = 0.5987 - 0.2266 = 0.3721

Finally, we can find the number of wrestlers by multiplying this proportion by the total population size:

0.3721 * 1800 = 669.78 or approximately 670 wrestlers

Therefore, approximately 670 wrestlers weigh between 210 and 230 lbs.

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This is because as the level of confidence decreases, the corresponding z-score becomes smaller, which in turn results in a smaller margin of error.

(a) To find a 90% confidence interval for the proportion p, we can use the formula:

CI = p ± z*(sqrt(p*(1-p)/n))

where p is the sample proportion, n is the sample size, and z is the z-score corresponding to the desired level of confidence (in this case, 90%).

We are given that p = 330/550 = 0.6 and n = 550. Using a standard normal distribution table, the z-score for a 90% confidence interval is approximately 1.645.

Substituting these values into the formula, we get:

CI = 0.6 ± 1.645*(sqrt(0.6*(1-0.6)/550))

= 0.6 ± 0.042

= (0.558, 0.642)

Therefore, a 90% confidence interval for the proportion of voters who indicated their preference for the candidate is 0.558 to 0.642.

(b) The margin of error for this 90% confidence interval is given by:

ME = z*(sqrt(p*(1-p)/n))

where z is the z-score corresponding to the desired level of confidence and p and n are as before.

Substituting the values we obtained earlier, we get:

ME = 1.645*(sqrt(0.6*(1-0.6)/550))

= 0.042

Therefore, the margin of error for this 90% confidence interval is 0.042.

(c) Without doing any calculations, we can see that the margin of error for an 80% confidence interval will be smaller than that for a 90% confidence interval. This is because as the level of confidence decreases, the corresponding z-score becomes smaller, which in turn results in a smaller margin of error.

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The radius of the circle represented by the equation x² + y² - 4x - 8y - 16 = 0 is given as follows:

6 units.

The equation of a circle of center [tex](x_0, y_0)[/tex] and radius r is given by:

[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]

The equation for this problem is given as follows:

x² + y² - 4x - 8y - 16 = 0.

To obtain the radius of the circle, we must complete the squares from the equation, as follows:

x² - 4x + y² - 8y = 16

(x - 2)² + (y - 4)² = 16 + 2² + 4²

(x - 2)² + (y - 4)² = 36.

Hence the center and the radius of the circle are given as follows:

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(a) The dependent variable is the number of accidents, y. The independent variable is the difference between the width of the bridge and roadway, x.

(b) To estimate the number of accidents if the difference of the width is 8 feet, we substitute x = 8 into the regression equation:

y = 74.7 - 6.44(8) = 24.58

Therefore, we estimate that there would be 24.58 accidents if the difference of the width is 8 feet.

As for the earlier question, the answer would be:

We need to calculate the test statistic to determine if we can reject the null hypothesis. The test statistic is calculated as:

test statistic = (sample mean - null hypothesis value) / (standard error of the sample mean)

Since the question does not provide any sample mean or standard error, we cannot calculate the test statistic. Therefore, we cannot determine if we can reject the null hypothesis based on the critical value alone.

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To represent the hourly wage of an employee at the antique store, we can use the following equation:
y = 15 + 0.75x
where y represents the worker's hourly wage, and x represents the number of years the employee has worked at the store. In this equation, 15 is the initial hourly wage, and 0.75 is the annual raise.

The equation that shows how a worker's hourly wage, y, depends on the number of years he or she has worked at the store can be written as:

y = 15 + 0.75x

where x represents the number of years the employee has worked at the antique store.

This equation takes into account the starting wage of $15 per hour and the $0.75 raise that the employee receives each year they work at the store.

So, for example, if an employee has worked at the store for 5 years, their hourly wage would be:

y = 15 + 0.75(5) = 18.75

where y represents the worker's hourly wage, and x represents the number of years the employee has worked at the store.

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The best data set that represents the histogram is (4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42, option C is correct.

From the histogram, we can see that there were 3 data points in the interval 1-10, 3 data points in the interval 11-20, 7 data points in the interval 21-30, 4 data points in the interval 31-40, and 3 data points in the interval 41-50.

Therefore, the best data set that represents the histogram is the one that has 3 data points in the range 1-10, 3 data points in the range 11-20, 7 data points in the range 21-30, 4 data points in the range 31-40, and 3 data points in the range 41-50.

4, 5, 7, 8, 10, 12, 13, 15, 18, 21, 23, 28, 32, 34, 36, 40, 41, 41, 42, 42 does not fit this pattern since it has more than 3 data points in some of the intervals.

4, 7, 10, 13, 14, 19, 22, 24, 26, 27, 29, 31, 33, 35, 36, 38, 40, 42, 42, 42 also does not fit the pattern since it has more than 3 data points in some of the intervals.

4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42 fits the pattern and has 3 data points in each interval. This is the correct answer.

4, 6, 9, 12, 16, 18, 21, 24, 25, 26, 28, 29, 30, 32, 35, 36, 38, 41, 41, 42 does not fit the pattern since it has more than 3 data points in some of the intervals.

Therefore, the best data set that represents the histogram is  (4, 5, 7, 8, 12, 13, 15, 18, 19, 21, 24, 25, 26, 28, 29, 30, 32, 33, 35, 42

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The slope of the line passing through (7,1) and (-2,3) is -2/9.

We use the following formula to get the slope of a line through two specified points:

slope = (y2 - y1) / (x2 - x1)

where (x1, y1) and (x2, y2) are the coordinates of the two points.

We can calculate the slope of the line passing through the points (7, 1) and (-2, 3) using this formula:

slope = (3 - 1) / (-2 - 7) = 2 / (-9) = -2/9

Therefore, the slope of the line passing through the points (7, 1) and (-2, 3) is -2/9.

The slope of a line, in geometric terms, is the ratio of the vertical change (rise) to the horizontal change (run). If the slope is negative, the line is decreasing as we move from left to right. With a slope of 2 units downward for every 9 units to the right, the line is sloping downward from left to right.

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The solution is, the probability that exactly two of the dice show an even number is 0.3125.

We will use the Binomial Probability formula to find the answer

The formula is given by ⁿCₓ (p)ˣ (1-p)ⁿ⁻ˣ

Where:

n, the number of trials = 5

x, the number of success that we aim = 2

p, the probability of success = 0.5 ⇒ there are six out of twelve numbers on the dice that are even

Substitute these values into the formula, we have

P(2) = ⁵C₂ (0.5)² (1-0.5)⁵⁻²

P(2) = ⁵C₂ (0.5)² (0.5)³

P(2) = 0.3125

The solution is, the probability that exactly two of the dice show an even number is 0.3125.

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complete question:

Ben rolls 5 fair 12-sided dice. the 12 faces of each die are numbered from 1 to 12. what is the probability that exactly two of the dice show an even number?

X is in Bs(b+1/2c), which implies Br(α+1/2c)⊂Bs(b+1/2c).

a. We want to show that Br(α+2c)⊂Bs(b+2c) given that Br(α)⊂Bs(b).

Let x be any element in Br(α+2c), then we have ||x-(α+2c)|| < r.

Using the triangle inequality, we get:

||x-(α+2c)|| = ||(x-α)-2c|| ≤ ||x-α||+2||c|| < r+2||c|| = 8 (since r > 0 and ||c|| < 4).

So, ||x-α|| < 8 - 2||c|| < 2.

Thus, x is also in Br(α) ⊂ Bs(b), which implies ||x-b|| < r.

Using the triangle inequality again, we have:

||x-(b+2c)|| = ||(x-b)-2c|| ≤ ||x-b||+2||c|| < r+2||c|| = 8.

Therefore, x is in Bs(b+2c), which implies Br(α+2c)⊂Bs(b+2c).

b. We want to show that Br(α+1/2c)⊂Bs(b+1/2c) given that Br(α)⊂Bs(b).

Let x be any element in Br(α+1/2c), then we have ||x-(α+1/2c)|| < r.

Using the triangle inequality, we get:

||x-(α+1/2c)|| = ||(x-α)-1/2c|| ≤ ||x-α||+1/2||c|| < r+1/2||c|| = 4 (since r > 0 and ||c|| < 8).

So, ||x-α|| < 4 - 1/2||c|| < 3.

Thus, x is also in Br(α) ⊂ Bs(b), which implies ||x-b|| < r.

Using the triangle inequality again, we have:

||x-(b+1/2c)|| = ||(x-b)-1/2c|| ≤ ||x-b||+1/2||c|| < r+1/2||c|| = 4.

Therefore, x is in Bs(b+1/2c), which implies Br(α+1/2c)⊂Bs(b+1/2c).

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The value of the angle m<QPS cannot be determined. Option D

To determine the value of the angle, we need to take note of the properties of a parallelogram

From the information given, we have that;

m<R = 6x -14

m<Q = 3x + 8

m< S= 5x + 4

The value of the angle , QPS cannot be determined because the angle is not represented with any variable for calculation

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No, the value would not be included if we want at most 4. In statistics, a variable is a characteristic or attribute that can be measured or observed.

A random variable is a variable whose value is determined by chance or probability. The possible values of a random variable are called its values. In this case, the random variable X has possible values of 1-6. If we want at most 4, this means we want all the values of X that are less than or equal to 4. Therefore, the value in question (which we don't know) would only be included if it is less than or equal to 4. If it is greater than 4, then it would not be included. To summarize, whether the value of X is included or not depends on whether it is less than or equal to 4, which is the condition we have set.

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Answer:

The density graph for all possible temperatures from 60 degrees to 160 degrees can be used to find the probability of a temperature falling within a certain range.

Option (A) is incorrect because it includes temperatures that are outside the range of the graph.

Option (B) is incorrect because it includes temperatures that are outside the range of the graph.

Option (C) is incorrect because it includes temperatures that are outside the range of the graph.

Option (D) is the only option that falls within the range of the graph. Therefore, the density graph can be used to find the probability of a temperature from 90 degrees to 120 degrees, which is option (D).

Step-by-step explanation:

The equation of the line that is parallel to 2y = 3x - 1 and passes through the point (4, 2) is: y = (3/2)x - 4

To find the equation of a straight line that is parallel to the line 2y = 3x - 1, we need to remember that parallel lines have the same slope.

First, let's rearrange the given equation into slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept:

2y = 3x - 1

y = (3/2)x - 1/2

So the slope of this line is 3/2.

Now, if we want to find the equation of a line that is parallel to this line, we just need to use the same slope. Let's call the new line y = mx + b, where m is the slope we just found and b is the y-intercept we need to find.

So the equation of the parallel line is:

y = (3/2)x + b

To find the value of b, we need to use a point on the line. Let's say we want the line to go through the point (4, 2):

2 = (3/2)(4) + b

2 = 6 + b

b = -4

So the equation of the line that is parallel to 2y = 3x - 1 and passes through the point (4, 2) is: y = (3/2)x - 4

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So, 0.25 or 25% of the radium sample will remain after 3240 years.

The decay chain for radium-226 is as follows: radium-226 has a half-life of 1600 years and produces an alpha particle and radon-222; radon-222 has a half-life of 3.82 days and produces an alpha particle and polonium-218; polonium-218 has a half-life of 3.05 minutes and produces an alpha particle and lead-214; lead-214 has a half-life of 26.8 minutes and produces.

The half-life of radium is 1620 years, which means that after 1620 years, half of the radium sample will decay, and the remaining half will remain. After another 1620 years (3240 years total), the remaining half will decay, and half of that half, or one-fourth of the original sample, will remain.

Therefore, after 3240 years, the fraction of the radium sample that will remain is:

Formula used :[tex]N(t)=2^{-t/1620}[/tex]

[tex]N(t)=2^{-3240/1620}[/tex]

= 1/4

= 0.25

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False. A set is considered closed under an operation if the result of that operation on any two elements in the set also belongs to the set.

A set is considered closed if it contains all of its limit points. In other words, if a sequence of points in the set converges to a point that is also in the set, then the set is closed. Another equivalent definition is that the complement of the set.

In mathematics, sets are collections of distinct objects. These objects can be anything, including numbers, letters, or even other sets. The concept of sets is fundamental in mathematics and is used to define many other mathematical structures.

Sets can be denoted in various ways, including listing the elements inside curly braces { }, using set-builder notation, or using set operations to define new sets from existing ones. Some common set operations include union, intersection, difference, and complement.

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Answer:

B

Step-by-step explanation:

Using the same yearly contribution of $5,000 and an average annual rate of return of 6.5%, starting at age 40 instead of 35, the total balance at retirement age of 65 would be $359,978.25.

To calculate this, we can use the future value formula:

FV = PV x (1 + r)^n

where FV is the future value, PV is the present value (initial contribution), r is the interest rate per period, and n is the number of periods.

If we start at age 40 and contribute $5,000 per year for 25 years (until age 65), the present value would be $0 (since we haven't made any contributions yet) and the number of periods would be 25. The interest rate per period would be 6.5% / 1 = 0.065.

Using these values in the future value formula, we get:

FV = $5,000 x ((1 + 0.065)^25 - 1) / 0.065 = $359,978.25

Therefore, the difference in the account balances between starting at age 35 and starting at age 40 would be:

$431,874.32 - $359,978.25 = $71,896.07

So the correct answer is option B: $359,978.25.

The different ways to choose the ones digit A are:

- A can be any digit if the sum of the first 6 digits is divisible by 3.
- A can be 2, 5, or 8 if the sum of the first 6 digits leaves a remainder of 1 when divided by 3.
- A can be 1, 4, 7, or 0 if the sum of the first 6 digits leaves a remainder of 2 when divided by 3.

To determine if a number is divisible by 3, we can add up the digits and check if the sum is divisible by 3. For the number 572435A, the sum of the first 6 digits is 5 + 7 + 2 + 4 + 3 + 5 = 26. We need to add a number A to this sum to make it divisible by 3.

There are three possible scenarios to make the sum divisible by 3:

1. If the sum of the first 6 digits is already divisible by 3, then any digit A can be added to the end to make the number divisible by 3. In this case, the sum of the first 7 digits will also be divisible by 3.

2. If the sum of the first 6 digits leaves a remainder of 1 when divided by 3, then A must be 2, 5, or 8. Adding any other digit to the end will result in a sum that is not divisible by 3. In this case, adding 2, 5, or 8 to the end of the number will result in a sum of digits that is divisible by 3.

3. If the sum of the first 6 digits leaves a remainder of 2 when divided by 3, then A must be 1, 4, 7, or 0. Adding any other digit to the end will result in a sum that is not divisible by 3. In this case, adding 1, 4, 7, or 0 to the end of the number will result in a sum of digits that is divisible by 3.

Therefore, the different ways to choose the ones digit A in the number 572435A so that the number will be divisible by 3 are:

- A can be any digit if the sum of the first 6 digits is divisible by 3.
- A can be 2, 5, or 8 if the sum of the first 6 digits leaves a remainder of 1 when divided by 3.
- A can be 1, 4, 7, or 0 if the sum of the first 6 digits leaves a remainder of 2 when divided by 3.

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Answer:

x = 3/2 or 1.5

Step-by-step explanation:

[tex]ax^2+bx+c[/tex]

[tex]x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex], where x is the root(s)

[tex]b^2-4ac[/tex]

(a.) For 4x^2 + 16x - 9b, 4 is our a value, 16 is our b value and -9 is our c value:

[tex]16^2-4(4)(-9)\\256+144\\400 > 0[/tex]

(b.) For 2x^2 + 80x + 400, 2 is our a value, 80 is our b value, and 400 is our c value:

[tex]80^2-4(2)(400)\\6400-3200\\3200 > 0[/tex]

(c.) For x^2 - 6x - 9, 1 is our a value, -6 is our b value and -9 is our c value

[tex](-6)^2-4(1)(-9)\\36+36\\72 > 0[/tex]

(d.) For 4x^2 - 12x + 9, 4 is our a value, -12 is our b value, and 9 is our c value:

[tex](-12)^2-4(4)(9)\\144-144\\0=0[/tex]

[tex]x=\frac{-(-12)+/-\sqrt{(-12)^2-4(4)(9)} }{2(4)}\\ \\x=\frac{12+/-\sqrt{0} }{8} \\\\x=12/8=3/2\\or\\x=1.5[/tex]