1 + 2 + 3 + 4 + 5 + . . . is an example of a finite series. True False

Final answer:

The probability that Gerald will first select a banana and then a pear from the bag without replacement is 2/45.

Explanation:

To determine the probability that Gerald selects a banana first and then a pear without replacement, we have to consider the total number of possible outcomes for each draw and the favorable outcomes for the event.

For the first draw, the total number of fruits is 10 (3 apples + 2 oranges + 1 banana + 4 pears). The favorable outcome of drawing a banana is 1 since there's only one banana.

The probability of drawing a banana on the first draw is therefore 1/10. After drawing the banana, there are 9 fruits left in the bag with 4 pears among them.

The probability of then drawing a pear is 4/9. To find the total probability of both events happening in sequence (a banana first and then a pear), multiply the two probabilities:

P(banana first and pear second) = P(banana first) × P(pear second)
= (1/10) × (4/9)
= 4/90
= 2/45.

The simplification process shows that the probability Gerald will first select a banana and then a pear is 2/45.

The indicated function y1(x) is a solution of the given differential equation.The general solution is [tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain the product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

Given differential equation is

y''-4y'+4y=0

and

[tex]y_1(x) = e^{2x}[/tex]

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx[/tex]

The general form of equation

y''+P(x)y'+Q(x)y=0

Comparing both the equation

So, P(x)= - 4

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx\\\\\\[/tex]

[tex]y_2(x) = e^{2x}\int e^{-4x} \, / e^{4x}dx[/tex]

[tex]y_2(x) = e^{2x}\int e^{-8x}dx\\\\y_2(x) = -e^{-6x}/8[/tex]

The general solution is

[tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

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To find a vector orthogonal to 5i + 12j, we can use the property that orthogonal vectors have a dot product of 0. By setting up equations and solving them accordingly, you can find a vector that is perpendicular to 5i + 12j.

Orthogonal vectors: To find a vector orthogonal to 5i + 12j, we need to find a vector with a dot product of 0 with 5i + 12j. Since the dot product of orthogonal vectors is zero, we can set up equations and solve them to find a vector that is perpendicular to 5i + 12j.

To find the equation of a line parallel to the given line, we can use the slope of the given line and the point-slope form of a line. The equation of the line parallel to 7−4x=7y and passing through the point (2,0) is y = (7/4)x - (7/2).

To find the equation of a line parallel to the given line, we need to find the slope of the given line first. The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept. Rearranging the given equation, we have y = (7/4)x - 1. Dividing the coefficient of x by the coefficient of y, we find that the slope of the given line is 7/4. Since the line we're looking for is parallel to this line, it will also have a slope of 7/4. Now, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point on the line. Substituting in the values (2, 0) and slope (7/4), we can solve for y to find the equation of the line.

Using the point-slope form, we have y - 0 = (7/4)(x - 2). Simplifying, we get y = (7/4)x - (7/2), which is the equation of the line parallel to the given line and passing through the point (2, 0).

Final answer:

The equation of the line parallel to 7 - 4x = 7y through the point (2, 0) is y = (-4/7)x + 8/7.

Explanation:

To find the equation of a line parallel to the line 7 - 4x = 7y, we need to find the slope of the given line. First, rearrange the equation in the form y = mx + b, where m is the slope. So, 7y = 7 - 4x becomes y = (-4/7)x + 1. The slope of this line is -4/7. Since the line we want is parallel, it will have the same slope.

Next, we have the point (2, 0) through which the line passes. To find the equation, we'll use the point-slope form: y - y1 = m(x - x1). Substituting the given values, we have y - 0 = (-4/7)(x - 2). Simplifying, we get y = (-4/7)x + 8/7.

Therefore, the equation of the line parallel to 7 - 4x = 7y through the point (2, 0) is y = (-4/7)x + 8/7.

The equivalent units of production for calculating conversion costs in Department W using FIFO are 16,900 units. This consists of 16,000 units transferred out and 900 equivalent units for the 1,800 units at half completion stage.

To calculate the equivalent units of production for unit conversion cost in Department W, using the FIFO method, we need to account only for the work done in the current period. Department W had 2,400 units at the beginning that were one-third completed, which means 800 units (2,400 units * 1/3) were already processed in the previous period. Therefore, these do not count for the current period. During the period, 16,000 units were transferred out. We also need to consider the 1,800 units at the end at one-half completion, which contributes 900 equivalent units (1,800 units * 1/2) for the current period.

To determine the number of equivalent units for conversion costs, we perform the following calculation:

The inverse Laplace transform of [tex]\( \frac{1}{s^2 - 720s^7} \) is \( (1 - \frac{1}{720})t + e^{720t} \).[/tex]

To find the inverse Laplace transform of [tex]\( \frac{1}{s^2 - 720s^7} \),[/tex] we can use the method of partial fraction decomposition. First, factor the denominator:

[tex]\[ s^2 - 720s^7 = s^2(1 - 720s^5) \][/tex]

Now, we can write the partial fraction decomposition as:

[tex]\[ \frac{1}{s^2(1 - 720s^5)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs^5 + D}{1 - 720s^5} \][/tex]

Multiplying both sides by [tex]\( s^2(1 - 720s^5) \)[/tex], we get:

[tex]\[ 1 = As(1 - 720s^5) + Bs(1 - 720s^5) + (Cs^5 + D)s^2 \]\[ 1 = As - 720As^6 + Bs - 720Bs^6 + Cs^7 + Ds^2 \][/tex]

Equating coefficients:

For [tex]\( s^6 \):[/tex]

-720A - 720B = 0

A + B = 0

A = -B

For [tex]\( s^7 \):[/tex]

C = 0

For [tex]\( s^2 \):[/tex]

D = 1

Substituting back:

A = -B

D = 1

C = 0

So, the partial fraction decomposition is:

[tex]\[ \frac{1}{s^2(1 - 720s^5)} = \frac{-B}{s} + \frac{1}{s^2} + \frac{D}{1 - 720s^5} \][/tex]

Now, we can find the values of [tex]\( A \), \( B \), and \( D \):[/tex]

A = -B

D = 1

Now, we can use Theorem 7.2.1 to find the inverse Laplace transform:

[tex]\[ \mathcal{L}^{-1}\left( \frac{1}{s^2(1 - 720s^5)} \right) = -B \mathcal{L}^{-1}\left( \frac{1}{s} \right) + \mathcal{L}^{-1}\left( \frac{1}{s^2} \right) + D \mathcal{L}^{-1}\left( \frac{1}{1 - 720s^5} \right) \][/tex]

[tex]\[ = -B + t + D \mathcal{L}^{-1}\left( e^{720t} \right) \][/tex]

[tex]\[ = -B + t + De^{720t} \][/tex]

Since [tex]\( B = \frac{1}{720} \), \( D = 1 \)[/tex], the inverse Laplace transform is:

[tex]\[ \mathcal{L}^{-1}\left( \frac{1}{s^2(1 - 720s^5)} \right) = -\frac{1}{720}t + t + e^{720t} \][/tex]

[tex]\[ = \left( 1 - \frac{1}{720} \right)t + e^{720t} \][/tex]

Complete Question:

Use appropriate algebra and theorem 7.2.1 to find the given inverse laplace transform. (write your answer as a function of [tex]\( \frac{1}{s^2 - 720s^7} \).[/tex]

A.) The rocket reaches 720 feet in 5 seconds and 9 seconds.

B.) The rocket completes its trajectory and hits the ground in 14 seconds

A quadratic equation has the following general form:

[tex]ax^2 + bx + c = 0[/tex]

The formula to solve this equation is :

[tex]\large {\boxed {x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} } }[/tex]

Let's try to solve the problem now.

Given :

[tex]h = -16 t^2 + 224t[/tex]

The rocket reaches 720 feet → h = 720 feet

[tex]720 = -16 t^2 + 224t[/tex]

[tex]16 t^2 - 224t + 720 = 0[/tex]

[tex]16 (t^2 - 14t + 45 = 0)[/tex]

[tex]t^2 - 14t + 45 = 0[/tex]

[tex]t^2 - 9t - 5t + 45 = 0[/tex]

[tex]t(t - 9) - 5(t - 9) = 0[/tex]

[tex](t - 5)(t - 9) = 0[/tex]

[tex]t = 5 ~ or ~ t = 9[/tex]

The rocket reaches 720 feet in 5 seconds and 9 seconds.

The rocket hits the ground → h = 0 feet

[tex]0 = -16 t^2 + 224t[/tex]

[tex]16 (t^2 - 14t ) = 0[/tex]

[tex]t^2 - 14t = 0[/tex]

[tex]t( t - 14 ) = 0[/tex]

[tex]t = 0 ~ or ~ t = 14[/tex]

The rocket completes its trajectory and hits the ground in 14 seconds

Grade: College

Subject: Mathematics

Chapter: Quadratic Equation

Keywords: Quadratic , Equation , Formula , Rocket , Maximum , Minimum , Time , Trajectory , Ground

The  general form of the equation for the given circle centered at O(0, 0) is:

                                [tex]x^2+y^2-41=0[/tex]

We know that the standard form of circle is given by:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

where the circle is centered at (h,k) and the radius of circle is: r units

1)

[tex]x^2+y^2+41=0[/tex]

i.e. we have:

[tex]x^2+y^2=-41[/tex]

which is not possible.

( Since, the sum of the square of two numbers has to be greater than or equal to 0)

Hence, option: 1 is incorrect.

2)

[tex]x^2+y^2-41=0[/tex]

It could also be written as:

[tex]x^2+y^2=41[/tex]

which is also represented by:

[tex](x-0)^2+(y-0)^2=(\sqrt{41})^2[/tex]

This means that the circle is centered at (0,0).

3)

[tex]x^2+y^2+x+y-41=0[/tex]

It could be written in standard form by:

[tex](x+\dfrac{1}{2})^2+(y+\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2}})^2[/tex]

Hence, the circle is centered at [tex](-\dfrac{1}{2},-\dfrac{1}{2})[/tex]

Hence, option: 3 is incorrect.

4)

[tex]x^2+y^2+x-y=41[/tex]

In standard form it could be written by:

[tex](x+\dfrac{1}{2})^2+(y-\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2})^2[/tex]

Hence, the circle is centered at:

[tex](\dfrac{-1}{2},\dfrac{1}{2})[/tex]

The smallest numeral that can be created from the digits 9, 3, and 6 is 369. This is achieved by arranging the digits in ascending order.

The smallest numeral that can be formed using the digits 9, 3, and 6 is 369. In mathematics, when we are to create the smallest possible numeral from a given set of digits, we arrange the digits in increasing order from left to right, that means the smallest digit will be on the left-most side and the largest digit will be on the right-most side.

So, with the digits 9, 3, and 6, we place 3 first as it's the smallest, then 6 as it's the next smallest, and finally 9, resulting in the smallest numeral 369.

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The smallest numeral possible using the digits 9, 3, and 6 is 369, arranged in ascending order.

To write the smallest numeral possible using the digits 9, 3, and 6, we arrange the digits in ascending order. The smallest digit is placed at the beginning, followed by the larger ones. Therefore, the smallest numeral we can create is 369.

To solve this problem, we use the z statistic. The formula for z score is given as:

z = (x – u) / s

Where,

x = sample score

u = the average score = 500

s = standard deviation = 50

First, we calculate for z when x = 500

z = (500 – 500) / 50

z = 0 / 50

z = 0

Using the standard z table, at z = 0, the value of P is: (P = proportion)

P (z = 0)= 0.5

Secondly, we calculate for z when x = 600

z = (600 – 500) / 50

z = 100 / 50

z = 2

Using the standard z table, at z = 2, the value of P is: (P = proportion)

P (z = 2) = 0.9772

Since we want to find the proportion between 500 and 600, therefore we subtract the two:

P (500 ≥ x ≥ 600) = 0.9772 – 0.5

P (500 ≥ x ≥ 600) = 0.4772

Answer:

Around 47.72% of students have score from 500 to 600.

Answer:

To solve this problem, we use the z statistic. The formula for z score is given as:

z = (x – u) / s

Where,

x = sample score

u = the average score = 500

s = standard deviation = 50

First, we calculate for z when x = 500

z = (500 – 500) / 50

z = 0 / 50

z = 0

Using the standard z table, at z = 0, the value of P is: (P = proportion)

P (z = 0)= 0.5

Secondly, we calculate for z when x = 600

z = (600 – 500) / 50

z = 100 / 50

z = 2

Using the standard z table, at z = 2, the value of P is: (P = proportion)

P (z = 2) = 0.9772

Since we want to find the proportion between 500 and 600, therefore we subtract the two:

P (500 ≥ x ≥ 600) = 0.9772 – 0.5

P (500 ≥ x ≥ 600) = 0.4772

Answer:

Around 47.72% of students have score from 500 to 600.

Step-by-step explanation:

Final answer:

The length of the fourth side on quadrilateral ABCD is 120 feet.

Explanation:

Given that quadrilateral ABCD is similar to quadrilateral EFGH, we can use the property of similar figures to find the length of the fourth side on quadrilateral ABCD.

If the two shortest sides of quadrilateral EFGH are 6 feet and 12 feet long, we can set up a proportion using the corresponding sides of the two quadrilaterals.

Let x be the length of the fourth side on quadrilateral ABCD.

Using the property of similar figures, we have:

(60/6) = (x/12)

Cross multiplying, we get:

6x = 720

Dividing both sides by 6, we find:

x = 120

Therefore, the length of the fourth side on quadrilateral ABCD is 120 feet.

Answer:

b

Step-by-step explanation:

The probability with the condition of the spinner landing on a number less than 3 is 0.2

A conditional probability is a probability of an event occuring with a condition that another event had previously occurred. The event in the question is spinning the spinner once while the condition is that the number landed on is less than 3.

The spinner has 10 equal sections numbered 1 through 10.

The conditional probability of landing on a number less than 3 is the same as the probability of landing on either 1 or 2.

There are two sections out of ten that corresponds to numbers less than 3. The probability of landing on a number less than 3 is therefore;

P(Landing on a number less than 3) = P(Landing on 1) + P(Landing on 2)

P((Landing on 1) = 1/10

P(Landing on 2) = 1/10

P(Landing on 1) + P(Landing on 2) = (1/10) + (1/10) = 2/10

(1/10) + (1/10) = 2/10 = 0.2

The probability of landing on a number less than 3 is 0.2

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