an irregular object with a mass of 1220g displaces 200 cubic cm of water when placed in a large overflow container. calculate the density of the object. what is the density in g/cm cubic

Answers

Answer 1

Answer:

[tex]6.1\frac{\text{ g}}{\text{ cm}^3}[/tex]

Step-by-step explanation:

We have been given that mass of an irregular object is 1220 g and it displaces 200 cubic cm of water when placed in a large overflow container. We are asked to find density of the object.

We will use density formula to solve our given problem.

[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]

Since the object displaces 200 cubic cm of water, so the volume of irregular object will be equal to 200 cubic cm.

Upon substituting our given values in density formula, we will get:

[tex]\text{Density}=\frac{1220\text{ g}}{200\text{ cm}^3}[/tex]

[tex]\text{Density}=\frac{61\times 20\text{ g}}{10\times 20\text{ cm}^3}[/tex]

[tex]\text{Density}=\frac{61\text{ g}}{10\text{ cm}^3}[/tex]

[tex]\text{Density}=6.1\frac{\text{ g}}{\text{ cm}^3}[/tex]

Therefore, the density of the irregular object will be 6.1 grams per cubic centimeters.


Related Questions

Samantha wants to sort her greetin cards into boxes that hold 24 cards each she has 312 greeting cards how many boxes will she need?

Answers

Answer:

Samantha will need 13 boxes.

Step-by-step explanation:

Samantha wants to sort her greeting cards in boxes.

Each box can hold 24 cards.

Samantha has 312 greeting cards.

To calculate the number of boxes she need we have to divide total number of greeting cards to 24.

Therefore, = [tex]\frac{312}{24}[/tex]

                 = 13 boxes

Samantha will need 13 boxes.

Sample methods that embody random sampling are often termed probability sampling methods. OTrue False

Answers

Answer:

The answer is true.

Step-by-step explanation:

Sample methods that embody random sampling are often termed probability sampling methods.

Yes this is a true statement.

Random sampling means picking up the samples randomly from a whole population with each sample having an equal chance of getting selected.

For example- selecting randomly 10 students from each class of a school, to survey for the food quality in school's cafeteria.

And this is a type of probability sampling methods. Other types are stratified sampling, cluster sampling etc.

Two fair, distinct dice (one red and one green) are rolled. Let A be the event the red die comes up even and B be the event the sum on the two dice is eight. Are A,B independent events?

Answers

Answer: No, A and B are not independent events.

Step-by-step explanation:

Since we have given that

Number of outcomes that a die comes up with = 6

A be the event that the red die comes up even.

A={2,4,6}

B be the event that the sum on the two dice is 8.

B={(2,4),(4,2),(5,3),(3,5),(44)}

P(A) = [tex]\dfrac{3}{6}=\dfrac{1}{2}[/tex]

P(B) = [tex]\dfrac{5}{36}[/tex]

P(A∩B) = [tex]\dfrac{3}{36}[/tex]

But,

P(A).P(B) ≠ P(A∩B)

[tex]\dfrac{1}{2}\times \dfrac{5}{36}\neq \dfrac{3}{36}\\\\\dfrac{5}{72}\neq\dfrac{1}{12}[/tex]

Hence, A and B are dependent events.

Joyce Meadow pays her three workers $160, $470, and $800, respectively, per week. Calculate what Joyce will pay at the end of the first quarter for (A) state unemployment and (B) federal unemployment. Assume a state rate of 5.6% and a federal rate of .6%. Base is $7,000. A. $950.64; $67.14 B. $655.64; $97.14 C. $755.64; $81.14 D. $850.64; $91.14

Answers

Answer:

D. $850.64; $91.14

Step-by-step explanation:

Joyce Meadow pays her three workers $160, $470, and $800, respectively, per week.

So for 1 quarter, he will pay :

[tex]13\times160[/tex] = $2080

[tex]13\times470[/tex] = $6110

[tex]13\times800[/tex] = $10400

Given is that the base is $7,000.

It implies that the unemployment rate is to be paid on the first $7000 only.

Given rates are :

State rate of 5.6% and a federal rate of 0.6%.

State unemployment = [tex](0.056\times2080)+(0.056\times6110)+(0.056\times7000)[/tex] = $850.64

Federal unemployment = [tex](0.006\times2080)+ (0.006\times6110)+ (0.006\times7000)[/tex]  = $91.14

So, the correct option is D.

If you roll two fair dice (one black die and one white die), in how many ways can you obtain a 1 on the white die? A 1 on the white die can be obtained in different ways. (u) More Enter your answer in the answer box and then click Check Answer. All parts showing Clear All

Answers

Answer:

6

Step-by-step explanation:

Sample space of the experiment

first number in the bracket is white die and second number in the bracket is black

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

As it can be seen that the first numbers in the bracket are (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

∴1 on the white die can be obtained in 6 ways

In the case of rolling two dice and trying to obtain a 1 on the white die, there are 6 ways to accomplish this because the black die outcome is irrelevant and it can show any number from 1 to 6 while pairing with a 1 on the white die.

The question asks about the probability of getting a specific result when rolling two fair dice, which is a problem in the realm of simple probability within mathematics.

Specifically, the question is focused on finding the number of ways to obtain a 1 on the white die.

When rolling two dice, there are a total of 6 different possible outcomes for the black die (since a standard die has 6 faces), and 1 specific outcome we're looking for on the white die, which is a 1.

Each outcome on the black die can be paired with a 1 on the white die, resulting in the combinations (1,1), (2,1), (3,1), (4,1), (5,1), and (6,1).

This gives us a total of 6 ways to achieve a 1 on the white die, regardless of what the black die shows.

7 = -x+3. Consider the line y = Find the equation of the line that is parallel to this line and passes through the point (-5, 6). Find the equation of the line that is perpendicular to this line and passes through the point (-5, 6)

Answers

Answer with explanation:

The equation of line is, y= -x +3

→x+y-3=0---------(1)

⇒Equation of line Parallel to Line , ax +by +c=0 is given by, ax + by +K=0.

Equation of Line Parallel to Line 1 is

  x+y+k=0

The Line passes through , (-5,6).

→ -5+6+k=0

→ k+1=0

→k= -1

So, equation of Line Parallel to line 1 is

x+y-1=0

⇒Equation of line Perpendicular  to Line , ax +by +c=0 is given by, bx - a y +K=0.

Equation of Line Perpendicular to Line 1 is

  x-y+k=0

The Line passes through , (-5,6).

→ -5-6+k=0

→ k-11=0

→k= 11

So, equation of Line Parallel to line 1 is

x-y+11=0

suppose the column of the square matrix is linearly independent. What are solutions of Ax=0?

Answers

Answer:

the solution is x = 0

Step-by-step explanation:

We know the fact that if the column of the square matrix is linearly independent then the determinant of the matrix is non zero.

Now, since the determinant of the matrix is not zero then the inverse of the matrix must exists.

Therefore, we have

[tex]A^{-1}A=I....(i)[/tex]

Now, for the equation Ax =0

multiplying both sides by [tex]A^{-1}[/tex]

[tex]A^{-1}Ax=A^{-1}\cdot0[/tex]

From equation (i)

[tex]Ix=0\\\\x=0[/tex]

Therefore, the solution is x = 0

A process produces a certain part with a mean diameter of 2 inches and a standard deviation of 0.05 inches. The lower and upper engineering specification limits are 1.6 inches and 3.5 inches. What is the Cp (measure of potential capability)?

Answers

Answer:

The value of Cp (measure of potential capability) is 6.33.

Step-by-step explanation:

Given information: Process average = 2 inches, process standard deviation = 0.05 inches, lower engineering specification limit = 1.6 inches and upper engineering specification limit =3.5 inches.

The formula for Cp (measure of potential capability) is

[tex]CP=\frac{USL-LSL}{6\sigma}[/tex]

Where, USL is upper specification limit, LSL is specification limit, σ is process standard deviation.

Substitute USL=3.5, LSL=1.6 and σ=0.05 in the above formula.

[tex]CP=\frac{3.5-1.6}{6(0.05)}[/tex]

[tex]CP=\frac{1.9}{0.3}[/tex]

[tex]CP=6.3333[/tex]

[tex]CP\approx 6.33[/tex]

Therefore the value of Cp (measure of potential capability) is 6.33.

Online jewelry sales have increased steadily. In 2003, sales were approximately 2 billion dollars, and in 2013 they were approximately 5.5 1 billion. Construct a model to predict online jewelery sales. Use your model, to find the predicted online jewelry sales for 2015? (Express your answer rounded correctly to the nearest tenth of a billion.)

Answers

Answer:

1) [tex]y=2(1+0.081)^x[/tex]

2) 6.4 billion.

Step-by-step explanation:

1) Let the model that is used to find online jewelry sales ( in billions ) after x years,

[tex]y=a(1+r)^x[/tex]

Let 2003 is the initial year,

That is, for 2003, x = 0,

The sales on 2003 is 2 billion, y = 2,

⇒ [tex]2=a(1+r)^0\implies a=2[/tex]

Now, in 2013 they were approximately 5.5 1 billion

i.e. if x = 13 then y = 5.51,

[tex]\implies 5.51 = a(1+r)^{13}[/tex]

[tex]5.51=2(1+r)^{13}[/tex]

With help of graphing calculator,

r = 0.081,

Hence, the model that represents the given scenario is,

[tex]y=2(1+0.081)^x[/tex]

2) For 2015, x = 15,

Hence, online jewelry sales for 2015 would be,

[tex]y=2(1+0.081)^{15}=6.43302745602\approx 6.4\text{ billion}[/tex]

Please help if you know the answer...if anyone can show the steps that'd be great​

Answers

A. y=2
B. x=5
C. x=-3
D. y=0
The work is attached below

Determine whether each statement is true or false in double-struck R3. (a) Two lines parallel to a third line are parallel. True False (b) Two lines perpendicular to a third line are parallel. True False (c) Two planes parallel to a third plane are parallel. True False (d) Two planes perpendicular to a third plane are parallel. True False (e) Two lines parallel to a plane are parallel. True False (f) Two lines perpendicular to a plane are parallel. True False (g) Two planes parallel to a line are parallel. True False (h) Two planes perpendicular to a line are parallel. True False (i) Two planes either intersect or are parallel. True False (j) Two lines either intersect or are parallel. True False (k) A plane and a line either intersect or are parallel. True False

Answers

Answer:

(a) True , (b) False , (c) True , (d) False , (e) False , (f) True , (g) False

(h) True , (i) True , (j) False , (k) True

Step-by-step explanation:

* Lets explain how to solve the problem

(a) Two lines parallel to a third line are parallel (True)

- Their direction vectors are scalar multiplies of the direction of the 3rd

 line, then they are scalar multiples of each other so they are parallel

(b) Two lines perpendicular to a third line are parallel (False)

- The x-axis and the y-axis are ⊥ to the z-axis but not parallel to

  each other

(c) Two planes parallel to a third plane are parallel (True)

- Their normal vectors parallel to the  normal vector of the 3rd plane,

  so these two normal vectors are parallel to each other and the

  planes are parallel

(d) Two planes perpendicular to a third plane are parallel (False)

- The xy plane and yz plane are not parallel to each other but both

 ⊥ to xz plane

(e) Two lines parallel to a plane are parallel (False)

- The x-axis and y-axis are not parallel to each other but both parallel

  to the plane z = 1

(f) Two lines perpendicular to a plane are parallel (True)

- The direction vectors of the lines parallel to the normal vector of

  the plane, then they parallel to each other , so the lines are parallel

(g) Two planes parallel to a line are parallel (False)

- The planes y = 1 and z = 1 are not parallel but both are parallel to

  the x-axis

(h) Two planes perpendicular to a line are parallel (True)

- The normal vectors of the 2 planes are parallel to the direction of

  line, then they are parallel to each other so the planes are parallel

(i) Two planes either intersect or are parallel (True)

(j) Two lines either intersect or are parallel (False)

- They can be skew

(k) A plane and a line either intersect or are parallel (True)

- They are parallel if the normal vector of the plane and the direction

  of the line are ⊥ to each other , otherwise the line intersect the plane

  at the angle 90° - Ф

This question is based on the properties of lines and planes. Therefore, (a) True , (b) False , (c) True , (d) False , (e) False , (f) True , (g) False

, (h) True , (i) True , (j) False , (k) True.

We have to choose correct statement and marked true or false.

Lets solve the problem.

(a) Two lines parallel to a third line are parallel. (True)

Reason - The direction vectors are scalar multiple of the direction of the third  line, then they are scalar multiple of each other. So, they are parallel.

(b) Two lines perpendicular to a third line are parallel. (False)

Reason- As we know that, x-axis and the y-axis are perpendicular to the z-axis but not parallel to  each other.

(c) Two planes parallel to a third plane are parallel (True)

Reason- The normal vectors of planes are parallel to the normal vector of the third  plane. So, these two normal vectors are parallel to each other and the  planes are parallel.

(d) Two planes perpendicular to a third plane are parallel. (False)

Reason- x-y plane and y-z plane are not parallel to each other. But they are perpendicular to x-z plane.

(e) Two lines parallel to a plane are parallel. (False)

Reason - Both  x-axis and y-axis are not parallel to each other. But, parallel  to the plane z = 1.

(f) Two lines perpendicular to a plane are parallel. (True)

Reason - The direction vectors of the lines parallel to the normal vector of  the plane, then they parallel to each other , so the lines are parallel.

(g) Two planes parallel to a line are parallel. (False)

Reason- The planes y = 1 and z = 1 are not parallel, but they are parallel to the x-axis.

(h) Two planes perpendicular to a line are parallel. (True)

Reason- The normal vectors of the two planes are parallel to the direction of  line. So, they are parallel to each other. Hence, they are parallel.

(i) Two planes either intersect or are parallel (True)

(j) Two lines either intersect or are parallel (False)

Reason- They can also be  skew.

(k) A plane and a line either intersect or are parallel (True)

Reason- They are parallel, if the normal vector of the plane and the direction  of the line are perpendicular to each other, otherwise the line intersect the plane  at the angle 90° [tex]\theta[/tex].

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Assume that the ages for first marriages are normally distributed with a mean of 26 years and a standard deviation of 4 years. What is the probability that a person getting married for the first time is in his or her twenties? (Round your answer to 4 decimal places.)

Answers

Answer:

0.7743

Step-by-step explanation:

Mean of age = u = 26 years

Standard Deviation = [tex]\sigma[/tex] = 4 years

We need to find the probability that the person getting married is in his or her twenties. This means the age of the person should be between 20 and 30. So, we are to find P( 20 < x < 30), where represents the distribution of age.

Since the data is normally distributed we can use the z distribution to solve this problem. The formula to calculate the z score is:

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

20 converted to z score will be:

[tex]z=\frac{20-26}{4}=-1.5[/tex]

30 converted to z score will be:

[tex]z=\frac{30-26}{4}=1[/tex]

So, now we have to find the probability that the z value lies between -1.5 and 1.

P( 20 < x < 30) = P( -1.5 < z < 1)

P( -1.5 < z < 1 ) = P(z < 1) - P(z<-1.5)

From the z-table:

P(z < 1) = 0.8413

P(z < -1.5) =0.067

So,

P( -1.5 < z < 1 ) = 0.8413 - 0.067 = 0.7743

Thus,

P( 20 < x < 30) = 0.7743

So, we can conclude that the probability that a person getting married for the first time is in his or her twenties is 0.7743

A recent report in a women magazine stated that the average age for women to marry in the United States is now 25 years of age, and that the standard deviation is assumed to be 3.2 years. A sample of 50 U.S. women is randomly selected. Find the probability that the sample mean age for 50 randomly selected women to marry is at most 24 years

Answers

Answer:  0.0136

Step-by-step explanation:

Given : Mean : [tex]\mu=\ 25[/tex]

Standard deviation : [tex]\sigma= 3.2[/tex]

Sample size : [tex]n=50[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x = 24

[tex]z=\dfrac{24-25}{\dfrac{3.2}{\sqrt{50}}}=-2.20970869121\aprox-2.21[/tex]

The P-value =[tex]P(z\leq24)=0.0135526\approx0.0136[/tex]

Hence, the probability that the sample mean age for 50 randomly selected women to marry is at most 24 years = 0.0136

Solve the following simultaneous linear congruences.

a) x ? 1 (mod 3), x ? 2 (mod 4), x ? 3 (mod 5).
b) x ? 4 (mod 10), x ? 8 (mod 12), x ? 6 (mod 18).

Answers

a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let

[tex]x=4\cdot5+3\cdot5+3\cdot4[/tex]

Taken mod 3, the last two terms vanish, and [tex]20\equiv2\pmod3[/tex] so we need to multiply by the inverse of 2 modulo 3 to end up with a remainder of 1. Since [tex]2\cdot2\equiv4\equiv1\pmod3[/tex], we multiply the first term by 2.

[tex]x=4\cdot5\cdot2+3\cdot5+3\cdot4[/tex]

Taken mod 4, the first and last terms vanish, and [tex]15\equiv3\pmod4[/tex]. Multiply by the inverse of 3 modulo 4 (which is 3 because [tex]3\cdot3\equiv9\equiv1\pmod4[/tex]), then by 2 to ensure the proper remainder is left.

[tex]x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4[/tex]

Taken mod 5, the first two terms vanish, and [tex]12\equiv2\pmod5[/tex]. Multiply by the inverse of 2 modulo 5 (3, since [tex]3\cdot2\equiv6\equiv1\pmod5[/tex]) and again by 3.

[tex]x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4\cdot3\cdot3[/tex]

[tex]\implies x=238[/tex]

By the CRT, we have

[tex]x\equiv238\pmod{3\cdot4\cdot5}\implies x\equiv-2\pmod{60}\implies\boxed{x\equiv58\pmod{60}}[/tex]

i.e. any number [tex]58+60n[/tex] (where [tex]n[/tex] is an integer) satisifes the system.

b. The moduli are not coprime, so we need to check for possible contradictions. If [tex]x\equiv a\pmod m[/tex] and [tex]x\equiv b\pmod n[/tex], then we need to have [tex]a\equiv b\pmod{\mathrm{gcd}(m,n)}[/tex]. This basically amounts to checking that if [tex]x\equiv a\pmod m[/tex], then we should also have [tex]x\equiv a\pmod{\text{any divisor of }m}[/tex].

[tex]x\equiv4\pmod{10}\implies\begin{cases}x\equiv4\equiv0\pmod2\\x\equiv4\pmod5\end{cases}[/tex]

[tex]x\equiv8\pmod{12}\implies\begin{cases}x\equiv0\pmod2\\x\equiv2\pmod3\end{cases}[/tex]

[tex]x\equiv6\pmod{18}\implies\begin{cases}x\equiv0\pmod2\\x\equiv0\pmod3\end{cases}[/tex]

The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.

(1 point) The random variables X and Y have the joint density: fX,Y(x,y)={2−x−y00

Answers

Answer:

. Let fX,Y(x,y) = 10xy^2 for 0 < x < y < 1 be the joint density function of the random pair (X, Y). (a) Obtain the marginal density f(y) of Y. (b) Obtain the conditional density fx|y(x|y) of X given Y = y. (c) Evaluate the conditional expectation of X, given Y=y

Y = y.

Step-by-step explanation:

Find the lengths of the sides of the triangle PQR. P(0, 1, 5), Q(2, 3, 4), R(2, −3, 1) |PQ| = Correct: Your answer is correct. |QR| = Correct: Your answer is correct. |RP| = Correct: Your answer is correct. Is it a right triangle? Yes No Is it an isosceles triangle? Yes No

Answers

Answer:

1. The values of |PQ|, |QR| and |RP| are 3, 3√5 and 6 respectively.

2. No.

3. No.

Step-by-step explanation:

The vertices of given triangle are P(0, 1, 5), Q(2, 3, 4), R(2, −3, 1).

Distance formula:

[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}[/tex]

Using distance formula we get

[tex]|PQ|=\sqrt{(2-0)^2+(3-1)^2+(4-5)^2}=\sqrt{9}=3[/tex]

[tex]|QR|=\sqrt{(2-2)^2+(-3-3)^2+(1-4)^2}=\sqrt{45}=3\sqrt{5}[/tex]

[tex]|RP|=\sqrt{(0-2)^2+(1-(-3))^2+(5-1)^2}=\sqrt{36}=6[/tex]

The values of |PQ|, |QR| and |RP| are 3, 3√5 and 6 respectively.

In a right angled triangle the sum of squares of two small sides is equal to the square of third side.

[tex](3)^2+(3\sqrt{5})^2=54\neq 6^2[/tex]

Therefore PQR is not a right angled triangle.

In an isosceles triangle, the length of two sides are equal.

The measure of all sides are different, therefore PQR is not an isosceles triangle.

A flagpole 94.9 ft tall is on the top of a building. From a point on level​ ground, the angle of elevation of the top of the flagpole is 35.3degrees​, while the angle of elevation of the bottom of the flagpole is 26.2degrees. Find the height of the building.

Answers

Answer:

214 ft

Step-by-step explanation:

Height of building = 94.9 ft

The angle of elevation of the top of the flagpole = θ₁ = 35.3°

The angle of elevation of the bottom of the flagpole = θ₂ = 26.2°

Let,

Height of building = x

Distance from observation point to base of building = y

[tex]tan 26.2 =\frac{x}{y}\\\Rightarrow y=\frac{x}{tan26.2}[/tex]

[tex]tan 35.3 =\frac{94.9+x}{y}\\\Rightarrow tan 35.3 =\frac{94.9+x}{\frac{x}{tan26.2}}\\\Rightarrow \frac{x}{tan26.2}tan35.3=94.9+x\\\Rightarrow \frac{tan35.3}{tan26.2}x-x=94.9\\\Rightarrow x=\frac{94.9}{\frac{tan35.3}{tan26.2}-1}\\\Rightarrow x=214.84/ ft[/tex]

I have used the exact values from the calculator.

∴ Height of the building is 214.84 ft

Answer:

The height of the building is 214.84 ft.

Step-by-step explanation:

Given information:

The height of the flagpole = 94.9 ft.

The angle of elevation of top = θ[tex]_1[/tex] = [tex]35.3^o[/tex]

The angle of elevation of bottom = θ[tex]_2=26.2^o[/tex]

If the height of building is [tex]x[/tex]

Then,

[tex]tan 26.2=x/y\\y=x/(tan26.2)\\[/tex]

And:

[tex]tan 35.3=(94.9+x)/y\\[/tex]

[tex]94.9+x=y \times tan35.3[/tex]

On putting the value in above equation:

[tex]x=\frac{94.9}{\frac{tan35.3}{tan26.2}-1 }[/tex]

solving the above equation:

[tex]x=214.84 ft.[/tex]

Hence, the height of the building is 214.84 ft.

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Find parametric equations for the line. (Use the parameter t.) The line through the origin and the point (2, 6, −1) (x(t), y(t), z(t)) Find the symmetric equations.

Answers

The symmetric equations of the line are: x/(-2) = y/(-6) = z/1 and the x(t) = 2t, y(t) = 6t and z(t) = -t are the parametric equations.

What is Parametric Equation?

a parametric equation defines a group of quantities as functions of one or more independent variables called parameters.

To find the parametric equations for the line through the origin and the point (2, 6, -1), we can use the vector equation of a line:

r(t) = r₀ + tv

where r(t) = (x(t), y(t), z(t)) is the position vector of a point on the line,

r₀ = (0, 0, 0) is the position vector of the origin,

t is a parameter, and v is the direction vector

v = (2, 6, -1) - (0, 0, 0) = (2, 6, -1)

Now r(t) = (0, 0, 0) + t(2, 6, -1) = (2t, 6t, -t)

Therefore, the parametric equations of the line are:

x(t) = 2t

y(t) = 6t

z(t) = -t

To find the symmetric equations of the line, we can eliminate the parameter t from the parametric equations.

t = -z

Substituting this into the x(t) and y(t) equations, we get:

x = 2t = -2z

y = 6t = -6z

Therefore, the symmetric equations of the line are:

x/(-2) = y/(-6) = z/1

Hence, the symmetric equations of the line are: x/(-2) = y/(-6) = z/1

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Final answer:

Parametric equations for a line passing through the origin and a point are found by using the components of the point as coefficients for the parameter t. Symmetric equations are obtained by equating the ratios of each component to their corresponding coefficients in the direction vector.

Explanation:

To find the parametric equations for the line passing through the origin (0,0,0) and the point (2, 6, −1), we can use the position vector of the point (2,6,−1) and multiply it by the parameter t. This gives us the parametric equations:

x(t) = 2ty(t) = 6tz(t) = -t

The symmetric equations of the line can be obtained by eliminating the parameter t. Since t = x/2 = y/6 = z/(−1), the symmetric equations are:

x/2 = y/6 = z/(−1).

The half-life of Radium-226 is 1590 years. If a sample contains 500 mg, how many mg will remain after 2000 years? Preview mg Give your answer accurate to at least 2 decimal places. Get help: Video Video

Answers

Answer:

[tex]a_{n}[/tex]=209.09 mg

Step-by-step explanation:

given: material= radium

half life= 1590 years

initial mass [tex]a_{0}[/tex] =500mg

we know that to calculate the amount left we use

[tex]a_{n}[/tex] = [tex]a_{0}[/tex][tex]\left ( 0.5\right )^{n}[/tex]

[tex]n=\frac{2000}{1590} = 1.2578[/tex]

therefore

[tex]a_{n}[/tex] = [tex]500\times0.5^{1.2578}[/tex]

[tex]a_{n}[/tex]=209.09058407921 mg

[tex]a_{n}[/tex]=209.09 mg amount left after 2000 years

please help, its formula manipulation, only respond if u know how to get the answer, thank you​

Answers

Answer:

Problem 1:

[tex]r=\frac{3V}{2 \pi h^2}[/tex]

Problem 2:

[tex]h=\frac{3V}{b^2}[/tex]

Problem 3:

The radius is [tex]\frac{25}{\pi}[/tex] cm.

Problem 4:

The width is 15 cm.

Step-by-step explanation:

Problem 1:

We want to solve [tex]V=\frac{2\pi rh^2}{3}[/tex] for [tex]r[/tex].

[tex]V=\frac{2\pi rh^2}{3}[/tex]

Multiply both sides by 3:

[tex]3V=2\pi r h^2[/tex]

Rearrange the multiplication using commutative property:

[tex]3V=2\pi h^2 \cdot r[/tex]

We want to get [tex]r[/tex] by itself so divide both sides by what [tex]r[/tex] is being multiplied by which is [tex]2\pi h^2[/tex].

[tex]\frac{3V}{2 \pi h^2}=r[/tex]

[tex]r=\frac{3V}{2 \pi h^2}[/tex]

Problem 2:

We want to solve for [tex]h[/tex] in [tex]V=\frac{b^2h}{3}[/tex].

Multiply both sides by 3:

[tex]3V=b^2h[/tex]

We want [tex]h[/tex] by itself so divide both sides by what [tex]h[/tex] is being multiply by; that is divide both sides by [tex]b^2[/tex].

[tex]\frac{3V}{b^2}=h[/tex]

[tex]h=\frac{3V}{b^2}[/tex]

Problem 3:

The circumference formula for a circle is [tex]2\pi r[/tex]. We are asked to solve for the radius when the circumference is [tex]50[/tex] cm.

[tex]2\pi r=50[/tex]

Divide both sides by what r is being multiply by; that is divide both sides by [tex]2\pi[/tex]:

[tex]r=\frac{50}{2\pi}[/tex]

Reduce fraction:

[tex]r=\frac{25}{\pi}[/tex]

The radius is [tex]\frac{25}{\pi}[/tex] cm.

Problem 4:  

The perimeter of a rectangle is [tex]2w+2L[/tex] where [tex]w[/tex] is the width and [tex]L[/tex] is the length.

We are asked to find w, the width, for when L, the length, is 5, and the perimeter is 40.

So we have this equation to solve for w:

[tex]40=2w+2(5)[/tex]

Simplify the 2(5) part:

[tex]40=2w+10[/tex]

Subtract both sides by 10:

[tex]30=2w[/tex]

Divide both sides by 2:

[tex]\frac{30}{2}=w[/tex]

Simplify the fraction:

[tex]15=w[/tex]

The width is 15 cm.

2. Using the Euclidean Algorithm, Calculate the Greatest Common Divisor of the num- bers 11655 and 11340.

Answers

Only two steps needed:

11,655 = 1*11,340 + 315

11,340 = 36*315 + 0

This shows that [tex]\mathrm{gcd}(11,655,\,11,340)=315[/tex].

Final answer:

The Greatest Common Divisor (GCD) of the numbers 11655 and 11340 is found using the Euclidean Algorithm by first dividing 11655 by 11340 to get a remainder of 315. This remainder becomes the divisor in the next step and the process repeats until the remainder is zero. Hence, the GCD of 11655 and 11340 is 315.

Explanation:

To calculate the Greatest Common Divisor (GCD) of the numbers 11655 and 11340 using the Euclidean Algorithm, follow this process:

Divide 11655 by 11340. The quotient is 1 and the remainder is 315. Now, take the divisor 11340 and divide it by the remainder from the previous step (315). The quotient is 36 with a remainder of 0. The remainder is now zero, which means our last non-zero remainder (315) is the Greatest Common Divisor (GCD).

Therefore, the Greatest Common Divisor of the numbers 11655 and 11340 using the Euclidean Algorithm is 315.

Learn more about Greatest Common Divisor here:

https://brainly.com/question/23270841

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The claim is that the IQ scores of statistics professors are normally​ distributed, with a mean greater than 116. A sample of 20 professors had a mean IQ score of 121 with a standard deviation of 11. Find the value of the test statistic.

Answers

Answer:   t= 2.032

Step-by-step explanation:

Given : Sample size : [tex]n=20[/tex]

Sample mean : [tex]\overline{x}=121[/tex]

Standard deviation : [tex]\sigma= 11[/tex]

Claim : The IQ scores of statistics professors are normally​ distributed, with a mean greater than 116.

Let [tex]\mu [/tex] be the mean scores of statistics professors.

Then the set of hypothesis for the given situation will be :-

[tex]H_0:\mu\leq116\\\\H_1:\mu>116[/tex]

As the alternative hypothesis is right tailed , thus the test would be right tail test.

Since the sample size is less than 30, therefore the test would be t-test .

The test statistics for the given situation will be :-

[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]\Rightarrow\ t=\dfrac{121-116}{\dfrac{11}{\sqrt{20}}}=2.03278907045\approx2.032[/tex]

Hence, the value of the test statistic : t= 2.032

Answer:

It's 2.05.

Step-by-step explanation:

To prove a hypothesis, we have to use test statisticians like the z-value which is used in normally distributed data, and this is the case.

To calculate the z-value we use: [tex]z=\frac{x-u}{\frac{o}{\sqrt{n} } }[/tex]; where x is the sample mean, u is the population mean, o is the standard deviation and n the sample size.

Replacing all values:

[tex]z=\frac{121-116}{\frac{11}{\sqrt{20} } } =\frac{5}{2.44}=2.05[/tex]

Therefore the value of the test statistic is 2.05.

(It's important to clarify that the problem isn't asking about the hypothesis, or the probability value, it's just asking for the test parameter, which in this case is just a z-value).

Which of the following is NOT a requirement of the Combinations​ Rule, Subscript n Baseline Upper C Subscript requalsStartFraction n exclamation mark Over r exclamation mark left parenthesis n minus r right parenthesis exclamation mark EndFraction ​, for items that are all​ different? Choose the correct answer below. A. That r of the n items are selected​ (without replacement). B. That there be n different items available. C. That order is not taken into account​ (consider rearrangements of the same items to be the​ same). D. That order is taken into account​ (consider rearrangements of the same items to be different​ sequences).

Answers

Answer:

C. That order is taken into account

Answer:

D. That order is taken into account​ (consider rearrangements of the same items to be different​ sequences).

Step-by-step explanation:

[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

For example:

Combinations of 2 from a set a,b,c: ab=ba, ac = ca, bc = cb, meaning that the order is not important(ab and ba is the same sequence).

aa, bb, cc are not valid, since there is no replacement.

So the answer for this qustion is:

D. That order is taken into account​ (consider rearrangements of the same items to be different​ sequences).

Solve the separable initial value problem. 1. y' = ln(x)(1 + y2), y(1) = 3 = y= tan(xlnx-x+1+arctan(3) 2. y' = 9x? V1 + x? (1 + y2), y(0) = 3 = y=

Answers

Final answer:

The solution to the separable initial value problem is [tex]\( y = \tan(x \ln(x) - x + 1 + \arctan(3)) \).[/tex]

Explanation:

A differential equation is a mathematical equation that relates a function to its derivatives. It describes how a function's rate of change is related to its current value and possibly other variables. Differential equations are used to model various physical, biological, and social phenomena in fields such as physics, engineering, biology, and economics.

To solve the differential equation [tex]\( y' = \ln(x)(1 + y^2) \)[/tex] with the initial condition [tex]\( y(1) = 3 \)[/tex], we separate variables and integrate both sides.

After integration, we get [tex]\( \tan(y) = x\ln(x) - x + C \)[/tex], where [tex]\( C \)[/tex] is the constant of integration. Using the initial condition, we find [tex]\( C = 1 + \arctan(3) \)[/tex].

Substituting this value back into the equation, we obtain the solution[tex]\( y = \tan(x \ln(x) - x + 1 + \arctan(3)) \).[/tex]This function satisfies the given differential equation and initial condition.

A bag of ice pops contains 2 flavored raspberry, 5 flavored lemon, and 3 flavored lime. Find the probability of each event.

a) Picking 2 lemon ice pops at random.

b) picking 2 lime and 1 raspberry at random.

Answers

Answer:

A) 2/10 or 20%

B) 3/10 or 30%

Step-by-step explanation:

There are 10 ice pops.

10 is going to be your denominator

Your numerator is going to be whatever amount of ice pops is.

After finding your fraction, divide the numerator by the denominator.

Lastly turn your decimal into a fraction.

2/10=0.2=20%

Hope I helped you.

We have three coins: one with heads on both faces, the second with tails on both faces, and the third a regular one. We choose one at random, toss it, and the outcome is heads. What is the probability that the opposite face of the tossed coin is tails?

Answers

Answer:

Probability: [tex]\frac{1}{2}[/tex] = 0.5 = 50%

Step-by-step explanation:

Based on the question one coin is chosen at random and tossed. That coin then lands and is heads. Since the coin landed on heads we can eliminate the possibility of the coin that was chosen being the coin with double tails.

The following possibilities are that the coin has double heads or is a regular coin with both tails and heads. Seeing as the coin landed on heads, there are only two possible out comes for the other side of the coin

The other side is either Heads or Tails. That gives us a 50% chance of the other side being tails.

[tex]\frac{1}{2}[/tex] = 0.5 = 50%

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (1, 0, 9) and perpendicular to the plane x + 2y + z = 7

Answers

Answer:

r=<1,0,9>+t<1,2,1>

and

x=1+t

y=2t

z=9+t

Step-by-step explanation:

A vector perpendicular to the plane :

[tex]ax+by+cz+d=0[/tex]

is given by (a,b,c)

So a vector perpendicular to given plane will have :

(1,2,1)

[tex]The\ parametric\ equation\ of\ a\ line\ through\ (1,0,9)\ and\ parallel\ to\ vector\ (a,b,c) is\ given\ by:\\x=x_0+ta\\y=y_0+tb\\z=z_0+tb\\x=1+t(1)\\x=1+t\\y=0+t(2)\\y=2t\\z=9+t(1)\\z=9+t\\The\ vector\ form\ is:\\r=<1,0,9>+t<1,2,1>[/tex] ..

The vector equation r(t) is (1, 0, 9) + t(1, 2, 1) and the parametric equations are x = 1 + t, y = 2t, z = 9 + t.

To find the vector equation and parametric equations for the line that passes through the point (1, 0, 9) and is perpendicular to the plane given by x + 2y + z = 7, follow these steps :

1. Find the Normal Vector to the Plane :

The normal vector of the plane x + 2y + z = 7 is given by the coefficients of x, y, and z in the plane equation. Therefore, the normal vector n is (1, 2, 1).

2. Vector Equation of the Line :

A line passing through point (1, 0, 9) in the direction of normal vector (1, 2, 1) can be written in vector form as :r(t) = (1, 0, 9) + t(1, 2, 1) where t is the parameter.

3. Parametric Equations :

Extract the parametric equations from the vector equation :

x(t) = 1 + ty(t) = 0 + 2tz(t) = 9 + t

Thus, the parametric equations for the line are :

x = 1 + ty = 2tz = 9 + t

Find X if AX+B=AC+D

Would it be (D-B/A)+C?

Answers

Answer:

it will be x=(AC-B+D)/A

Step-by-step explanation:

AX+B=AC+D

AX+B-B=AC+D-B

AX/A=(AC+D-B)/A

X=(AC+D-B)/A

Suppose you first walk 28.6 m in a direction 20 degrees west of north and then 22 m in a direction 40 degrees south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position?

Answers

Answer:

  29.5 m 64° west of north

Step-by-step explanation:

A suitable vector calculator can add the vectors for you. (See attached.) Here, we have used North as the 0° reference and positive angles in the clockwise direction (as bearings are measured).

___

You can also use a triangle solver (provided by many graphing calculators and stand-alone apps). For this, and for manual calculation (below) it is useful to realize the angle difference between the travel directions is 70°.

___

Using the Law of Cosines to find the distance from start (d), we have (in meters) ...

  d² = 28.6² + 22² -2·28.6·22·cos(70°) ≈ 871.561

  d ≈ √871.561 ≈ 29.52 . . . . meters

The internal angle between the initial travel direction and the direction to the end point is found using the Law of Sines:

  sin(angle)/22 = sin(70°)/29.52

  angle = arcsin(22/29.52×sin(70°)) ≈ 44.44°

This angle is the additional angle the destination is west of the initial travel direction, so is ...

  20° west of north + 44.44° farther west of north = 64.44° west of north

__

In the second attachment, North is to the right, and West is down. This is essentially a reflection across the line y=x of the usual map directions and angles. Reflection doesn't change lengths or angles, so the computations are valid regardless of how you assign map directions to x-y coordinates.

6 × 0.9 + 0.1 ÷ 4-0.23

Answers

Answer:

5.195

Step-by-step explanation:

To solve the different operations of a mathematical expression there are rules that must always be followed, that is, there is a hierarchy of operations that must be respected.

The hierarchy of operations works as follows:

First, operations in brackets, brackets or braces are resolved independently.

Next, the powers and roots are realized.

It continues to solve multiplications and divisions.

Finally, the addition and subtraction are carried out, regardless of the order.

6 x 0.9 + 0.1 / 4 -0.23 =

First, multiplications and divisions:

5.4 + 0.025 - 0.23.

Then, addition and subraction:

5.195

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