Amount of water in a pool

The region of the concealed district will be 226.08 square centimeters.

It is the nearby bend of an equidistant point drawn from the middle. The sweep of a circle is the distance between the middle and the boundary.

Let d be the diameter of the circle. Then the area of the circle will be written as,

A = (π/4)d² square units

A ring-formed district's internal measurement is 14 cm and its external breadth is 22 cm. Then the region of the concealed district is given as,

A = (π / 4) (22² - 14²)

A = (3.14 / 4) (484 - 196)

A = 0.785 x 288

A = 226.08 square cm

The region of the concealed district will be 226.08 square centimeters.

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The trapezoid ABCD not isosceles because AB is not congruent to DC.

It is a polygon that has four sides. The sum of the internal angle is 360 degrees. In a trapezoid, one pair of opposite sides are parallel.

Quadrilateral ABCD has vertices A(-3,4), B(2,5), C(3,3), and D(-1,0).

The diagram is given below.

From the diagram, the line segment AD and BC are parallel to each other.

The length AB is given as,

AB² = (2 + 3)² + (4 - 5)²

AB = 5.1 units

The length CD is given as,

CD² = (3 + 1)² + (3 - 0)²

CD= 5 units

The trapezoid ABCD not isosceles because AB is not congruent to DC.

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Answer:6

Step-by-step explanation:

ƒ(4)=4+2=6

Therefore, the number that satisfies all the given conditions is 60,649.

In mathematics, an equation is a statement that asserts the equality of two expressions, typically separated by an equals sign ("="). The expressions on either side of the equals sign are called the left-hand side and the right-hand side of the equation, respectively. The purpose of an equation is to describe a relationship between two or more variables or quantities, such as x + 3 = 7 or y = 2x - 5. Equations can be used to solve problems and answer questions in various fields of study, such as algebra, geometry, physics, chemistry, and engineering. Solving an equation typically involves finding the value or values of the variable(s) that make the equation true. Some equations may have a unique solution, while others may have multiple solutions or no solutions at all. The study of equations and their properties is a fundamental topic in mathematics.

Here,

Let's break down the clues given in the problem and use them to find the unknown number:

6 ten thousands: The number must start with 6.

2 fewer thousands than ten thousands: The number of thousands is 2 less than the number of ten thousands. Since there are 6 ten thousands, there are 4 thousands.

Same number of hundreds as ten thousands: The number of hundreds is the same as the number of ten thousands, which is 6.

1 fewer ten than ten thousands: The number of tens is 1 less than the number of ten thousands, which is 6-1=5.

5 more ones than thousands: The number of ones is 5 more than the number of thousands, which is 4+5=9.

Putting all of these clues together, we get the number: 60,649

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The number of calories burned per minute be 23.33333.

Equations simultaneously, or a system of equations Several equations in algebra must be solved concurrently (i.e., the solution must satisfy all the equations in the system). There must be an equal number of equations and unknowns for a system to have a singular solution.

In order to locate the point where the lines intersect when the equations are graphed, systems of equations must be solved. The (x,y) ordered pair of this intersection point is regarded as the system's solution.

Let j be the number of calories burned by Shelly while running and c be the number of calories she burns while cycling.

The system of equations be

45j + 30c = 1350

30j + 45c = 1350

45j + 30(10)=1350

j = 23.33333

Therefore, the value j be 23.33333.

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When the mean is higher than the median, the distribution has a right-skewed pattern. Thus the statement 'a. the distribution is positively skewed' is the accurate one.

This indicates that the majority of the data are centered on the left and that the right is covered by a lengthy tail. The mean exceeds the median. In a positively skewed distribution, the mean overestimates the values that are most prevalent. The median is larger than the mean. In a distribution that is positively skewed, the mean—the average of all the values—is higher than the median since the data tends to be lower. The center value of the data is called the median, in comparison. As a result, if the data is more skewed towards the negative, the average will be higher than the middle number.

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The length which lies from K Street between C Street and D Street is 337.5 ft.

Given that,

E Street, W Street, C Street, and D Street are parallel streets that cross K Street and M Street in a particular city. It is unknown how far K Street is from C Street and D Street.

Let the distance of street K between C and D be x,

Now,

Taking the equality of the proportionality expression of triangles,

600 / 400 = 600 + x / 400 + 250

6 / 4 = 600 + x / 625

3750/4 = 600 + x

937.5 = 600 + x

x = 337.5 ft

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The solution is, 7x - 31 + 4y + 27 + 5x - 8 + 63​ = 12x+4y+51.

Simplify simply means to make it simple. In mathematics, simply or simplification is reducing the expression/fraction/problem in a simpler form. It makes the problem easy with calculations and solving.

here, we have,

now, we have to solve the given expression, using addition,

given that,

7x - 31 + 4y + 27 + 5x - 8 + 63​

=12x+4y+51

here, we add the terms with x with each other, then add the terms with y, and, add the numbers & subtract them from given terms with minus sign.

Hence, The solution is, 7x - 31 + 4y + 27 + 5x - 8 + 63​ = 12x+4y+51.

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Since the p-value is less than .05, we reject the null hypothesis that the means of the three assembly methods are equal.

Therefore, we can conclude that there is a significant difference in the means of the three assembly methods.

Source Variation | Squares' Sum | Degrees of Freedom | Mean Square | F

Treatments 4560     2        2280     9.87

Error         6240     27      231.11

Total         10800   29

Using Alpha = .05 to test for any significant difference in the means for the three assembly methods.

The value of the test statistic is 9.87

The p-value is: less than .01

Conclusion not all means of the three assembly methods are equal.

Complete Question:

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,800; SSTR = 4560.

Set up the ANOVA table for this problem (to 2 decimals, if necessary).

The probability that a newly arrived email which does mention "free gift card", is not spam is 0.0165.

Let S be the event that an email is spam, and let F be the event that an email mentions "free gift card". We want to find P(S' | F), the probability that the email is not spam given that it mentions "free gift card".

We can use Bayes' theorem to find P(S' | F):

P(S' | F) = P(F | S') P(S') / P(F)

We can calculate each of these probabilities as follows:

P(F | S') = the probability that a non-spam email mentions "free gift card". From the problem, we know that this probability is 0.005 (i.e., 99.5% of non-spam emails do not mention "free gift card").

P(S') = the probability that an email is not spam. From the problem, we know that this probability is 0.4 (i.e., 60% of emails are spam, so 40% are not spam).

P(F) = the probability that an email mentions "free gift card". This can be calculated using the law of total probability:

P(F) = P(F | S) P(S) + P(F | S') P(S')

= 0.20 × 0.60 + 0.005 × 0.40

= 0.121

In the first term, we use the given probability that 20% of spam emails mention "free gift card". In the second term, we use the probability that 0.5% of non-spam emails mention "free gift card".

Plugging these values into Bayes' theorem, we get:

P(S' | F) = 0.005 × 0.4 / 0.121 ≈ 0.0165

Therefore, the probability that a newly arrived email which does mention "free gift card" is not spam is approximately 0.0165 or 1.65%.

Correct Question :

A spam filer is designed by screening commonly occurring phrases in spam. Suppose that 60% of email is spam. In 20% of the spam emails, the phrase “free gift card” is used. In non-spam emails, 99.5% of them do not mention “free gift card”. What is the probability that a newly arrived email which does mention “free gift card”, is not spam?

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A compound event can be described for flipping a coin two times and getting at least 1 head.

Compound event is defined as the events in which there are more than one event happens together.

All the outcomes of the events sum up the probability to 1.

We have to describe a compound event where the probability is in between 50% and 80%.

That is, probability is in between 0.5 and 0.8.

That is, there is more than half of the chance to occur the event.

Suppose that we flip a coin two times.

Sample space = {HH, HT, TH, TT}

Find the probability of getting at least 1 head.

There are 3 outcomes out of 4 of getting at least 1 head.

Probability = 3/4 = 75%

Hence there is a 75% probability for getting at least 1 head when flipped a coin two times.

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Answer:

The final height of the tower is 114 feet 8 inches.

To solve this problem, we need to convert the additional 8 inches into feet. 8 inches is equal to 0.67 feet, so the new height of the tower is equal to 114 feet + 0.67 feet which is equal to 114 feet 8 inches.

a) To compare the overall percentage of late deliveries for each delivery service, we need to calculate the total number of late packages and total number of deliveries for each service, and then divide the number of late packages by the total number of deliveries, and multiply by 100 to get the percentage.

For Pack Rats, the total number of deliveries is 400 + 100 = 500, and the total number of late packages is 12 + 16 = 28. Therefore, the percentage of late deliveries for Pack Rats is:

percentage of late deliveries for Pack Rats = (28/500) x 100 = 5.6%

percentage of late deliveries for Boxes R Us = (30/500) x 100 = 6%

Therefore, based on these calculations, Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us.

b) While Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us, it's important to note that this decision should not be based solely on this one comparison. It's possible that there are other factors that the company needs to consider, such as the cost of each delivery service or the quality of customer service provided by each company.

Furthermore, the sample size in this trial period may not be large enough to draw a definitive conclusion about the reliability of each service. Therefore, while Pack Rats may be a good choice based on the available data, the company should consider other factors and conduct further research before making a final decision.

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Using relational algebra:

a. flight ⋈ airport ⋈ ρ destination  airport

b. flight ⋈ airport

c. passenger ⋈ airport

d. airline ⋈ airport ⋈ flight

e. airline ⋈ airport ⋈ flight

f. airline ⋈ airport ⋈ flight ⋈ ρ destination airport

g. flight ⋈ passenger ⋈ booking

QUESTION (a):

- using EQUI join (⋈) join relations "Flight, Airport and Airport as destination"

- and using select Operation (σ), relational operators (=, >) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "Flight.fID, Airport.code, destination.code, Flight.fare" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π flight . fid, airport . code, destination . code , flight.fare   σ flight . airport = airport . code

    AND flight . destination = destination . code  (flight ⋈ airport ⋈ ρ destination  airport)

QUESTION (b):

- using EQUI join (⋈) join relations "Flight, Airport"

- and using select Operation (σ), relational operators (=, >=, <=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airport . city, airport . code, flight . departure" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airport . city, airport . code, flight . departure    σ flight . airport = airport . code

     AND (4 <= flight . departure AND flight . departure <= 10)  (flight ⋈ airport)

QUESTION (c):

- using EQUI join (⋈) join relations "passenger, Airport"

- and using select Operation (σ), relational operators (<>) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "passenger . name, passenger . hometown" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π passenger . name, passenger . hometown    σ passenger . hometown <> airport . city  (passenger ⋈ airport)

QUESTION (d):

- using EQUI join (⋈) join relations "airline, airport, flight"

- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

    AND flight . airport = airport . code

    AND airport . city = "Toronto"  (airline ⋈ airport ⋈ flight)

QUESTION (e):

- using EQUI join (⋈) join relations "airline, airport, flight"

- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

     AND flight . airport = airport . code

    AND airport . city <> "Toronto"  (airline ⋈ airport ⋈ flight)

QUESTION (f):

- using EQUI join (⋈) join relations "airline, airport, flight and airport as destination"

- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "airline . name" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π airline . name    σ airline . aid = flight . airline

    AND flight . airport = airport . code

    AND flight . destination = destination . code

    AND airport . city = "Toronto"

    AND destination . city = "vancouver"  (airline ⋈ airport ⋈ flight ⋈ ρ destination airport)

QUESTION (g):

- using EQUI join (⋈) join relations "flight, passenger, Booking"

- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations

- Then using Project Operation (∏) projects columns "passenger . name, airport, destination" from the relation

- Below shows the SQL query, relation algebra query and relation algebra tree

π passenger . name, airport, destination    σ passenger . pid = booking . pid

    AND booking . fid = flight . fid

    AND destination . city = "vancouver"  (flight ⋈ passenger ⋈ booking)

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Question correction:

See on the attached image.

The missing value is given as follows:

P(X = 28) = 0.31.

The mean and the standard deviation are given as follows:

The sum of the probabilities of all the outcomes is of one, hence the missing value is obtained as follows:

0.14 + 0.18 + 0.21 + P(X = 28) + 0.16 = 1

0.69 + P(X = 28) = 1

P(X = 28) = 0.31.

The mean is given by the sum of all outcomes multiplied by their respective probabilities, hence:

E(X) = 25 x 0.14 + 26 x 0.18 + 27 x 0.21 + 28 x 0.31 + 29 x 0.16

E(X) = 27.17.

The standard deviation is given by the square root of the sum of the difference squared between each observation and the mean, multiplied by their respective probabilities, hence:

S(X) = sqrt((25-27.17)² x 0.14 + (26-27.17)² x 0.18 + (27-27.17)² x 0.21 + (28-27.17)² x 0.31 + (29-27.17)² x 0.16)

S(X) = 1.289.

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The probability is P(k) = (q^(k-1)) * p for k>=1, and P(0) = q. The expected number of times the loop will be executed is 1/p.The expected number of times that the repeat loop will be executed is 1/p.

To find the probability that the loop will be executed k times, we can consider the probability of the event that B is false k-1 times followed by B being true. This probability is q^(k-1) * p.

The event of the loop not being executed at all corresponds to B being true in the first trial, which has a probability of q. Therefore, the probability that the loop will be executed k times is P(k) = (q^(k-1)) * p for k>=1, and P(0) = q.

The expected number of times the loop will be executed is the sum of the probabilities of executing the loop k times, weighted by k, i.e., E = Sum(kP(k)) for k>=1, and E = 0 if P(0) = q.

By using the expression for P(k), we can simplify this to E = Sum(kq^(k-1)*p) for k>=1, and E = 0 if P(0) = q. By applying the formula for the sum of a geometric series, we get E = 1/p.

For the "repeat S until B" loop, the expected number of times that the loop will be executed is the expected number of trials in a Bernoulli process until the first success, where the success probability is p. By using the formula for the expected value of a geometric distribution, we get E = 1/p.

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The integral for the volume of the solid is obtained by rotating the region bounded by the given curves about the specified axis is  V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]

When we rotate a thin vertical strip, about the y-axis.

We get a cylindrical shell  with an inner of radius an x and an outer of radius x + dx

The height of the cylinder shell is tan (x) - x

The volume of the cylindrical shell is

     dV = π [tex](Outer Radius)^{2} (Height)[/tex] - π[tex](Inner Radius)^{2} (Height)[/tex]

     dV = π [tex](x + dx )^{2} (tan(x) - x)[/tex] - π [tex](x )^{2} (tan(x) - x)[/tex]

dV = π[tex](x^{2} + 2 xdx + (dx)^{2} ) (tan(x) - x)[/tex] - π [tex](x)^{2} (tan(x) - x)[/tex]

assume [tex]dx^{2}[/tex]≈ 0

            dV = π[tex](x^{2} + 2xdx + 0 - x^{2} ) (tan(x) - x)\\[/tex]

                    dV = 2πx (tan(x) - x) dx

                   V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]

Therefore, the integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis is  

V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex] .

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The correct question is:

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=tanx,y=x,x=π/3; about the y-axis

The relative frequency of landing on heads is 0.4, then the correct option is D.

When we have an experiment with some outcomes, and we perform the experiment N times, and in K of these N times we get a particular outcome, then the relative frequency for that outcome is K/N

In this case the coin is flipped 35 times and it lands on heasd 14 times, then the relative frequency of landing on heads is:

R = 14/35 = 0.4

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Answer:

29

Step-by-step explanation:

trust

Answer: okay good!!!!!!

Answer: Given the formula:

V = 6pirt + 4pir^2

To solve for t, we'll isolate t by rearranging the equation.

First, subtract 4pir^2 from both sides:

V - 4pir^2 = 6pirt

Next, divide both sides by 6pi:

(V - 4pir^2)/6pi = t

So, t = (V - 4pir^2)/6pi.

This gives us the value of t in terms of V and the radius of the cylinder, r.

Step-by-step explanation:

Answer:

$1,300

Step-by-step explanation:

Let original price be $x

Computer has been marked down by 45%

45% = 45/100 = 0.45

So discount = 45% of x = 0.45 x

Original price - Discount = Sale price
x - 0.45x = $715

0.55x = 715

x = 715/0.55

x = $1,300

which is the original price

The probability of getting 3 heads in 4 coin tosses, given that we get at least 2 heads, is 4/11 or 0.364.

To solve this problem, we can use the conditional probability formula. Let A be the event of getting 3 heads in 4 coin tosses, and let B be the event of getting at least 2 heads in 4 coin tosses. Then we want to find P(A|B), the probability of getting 3 heads in 4 coin tosses given that we get at least 2 heads.

By the definition of conditional probability, we have:

P(A|B) = P(A and B) / P(B)

To find P(B), the probability of getting at least 2 heads in 4 coin tosses, we can use the complement rule and find the probability of getting 0 or 1 heads:

P(B) = 1 - P(0 heads) - P(1 head)

To find P(0 heads), the probability of getting 0 heads in 4 coin tosses, we use the binomial probability formula:

P(0 heads) = (4 choose 0) * (0.5)^0 * (1-0.5)^(4-0) = 1/16

Similarly, we can find P(1 head):

P(1 head) = (4 choose 1) * (0.5)^1 * (1-0.5)^(4-1) = 4/16

So,

P(B) = 1 - P(0 heads) - P(1 head) = 11/16

To find P(A and B), the probability of getting 3 heads in 4 coin tosses and getting at least 2 heads, we can use the binomial probability formula again:

P(A and B) = (4 choose 3) * (0.5)^3 * (1-0.5)^(4-3) = 4/16

Therefore,

P(A|B) = P(A and B) / P(B) = (4/16) / (11/16) = 4/11

So the probability of getting 3 heads in 4 coin tosses, given that we get at least 2 heads, is 4/11 or approximately 0.364.

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Answer: If the rate at which the flu spreads is proportional to the number of interactions between students who have the flu and students who have not been exposed to it, we can write:

dx/dt = kx(1000 - x)

where x is the number of students who have contracted the flu, k is the constant of proportionality, and t is time. The term x(1000 - x) represents the number of interactions between the students who have the flu and the students who have not been exposed to it. The factor of 1000 - x represents the number of students who have not yet been exposed to the flu.

Thus, the differential equation models the spread of the flu on the isolated college campus as the number of people who have contracted the flu changes over time due to the number of interactions between the students who have the flu and the students who have not been exposed to it.

Step-by-step explanation:

Answer:

Step-by-step explanation:

c

By rewriting the equation in slope-intercept form and setting Celsius to 0, we can find the y-intercept of the corresponding line, which represents the temperature in Kelvin when Celsius is 0°C.

When studying linear relationships between two variables, it is often useful to identify the y-intercept of the line that represents such relationship.

The equation K = °C + 273.15 expresses the relationship between the temperature in Celsius (°C) and the temperature in Kelvin (K). It tells us that to convert a temperature from Celsius to Kelvin, we need to add 273.15 to the original value.

We can plot the Celsius and Kelvin values as two variables on a graph, with Celsius on the x-axis and Kelvin on the y-axis. Since the equation K = °C + 273.15 is a linear equation (meaning that the graph is a straight line), we can use the slope-intercept form of the equation to find the y-intercept, which is the point where the line intersects the y-axis.

The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept. In the case of K = °C + 273.15, we can rewrite it as K = 1°C + 273.15, which tells us that the slope of the line is 1. This means that for every 1°C increase in temperature, there is a corresponding 1 K increase in temperature.

To find the y-intercept, we can simply set the value of Celsius to 0, since this is the point where the line intersects the y-axis. Plugging in 0 for °C in the equation K = °C + 273.15, we get K = 273.15. This means that the y-intercept of the line is (0, 273.15), which corresponds to the temperature in Kelvin when the temperature in Celsius is 0°C.

In summary, the equation K = °C + 273.15 provides a linear relationship between the temperature in Celsius and the temperature in Kelvin.

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Answer:

90c^2 -30c - 420

Step-by-step explanation:

-3(-3c+7)5(4+2c)

(9c - 21) (20 + 10c)

180c + 90c^2 - 420 - 210c

90c^2 -30c - 420

The moment generating function of a negative binomial distribution with parameters r and p is[tex]M_X(t) = [(1-p)/(1-pe^t)]^r.[/tex]

In probability theory, the negative binomial distribution is a discrete probability distribution that describes the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs.

The moment generating function (MGF) of a random variable X is defined as

[tex]M_X(t) = E(e^{tX})[/tex]

To derive the MGF of a negative binomial distribution, we can use the fact that a negative binomial random variable can be expressed as a sum of independent geometric random variables.

Let X ~ NB(r, p) be a negative binomial random variable, where r is the number of failures and p is the probability of success in each trial. Let Y ~ Geo(p) be a geometric random variable representing the number of failures before the first success.

We know that the MGF of a geometric distribution is

[tex]M_Y(t) = E(e^{tY}) = (1-p)/(1-pe^t).[/tex]

Using the fact that a negative binomial random variable is a sum of r independent geometric random variables, we can derive the MGF of X as follows:

[tex]M_X(t) = E(e^{tX}) = E(e^{tY_1 + tY_2 + ... + tY_r})\\\\= E(e^{tY_1} * e^{tY_2} * ... * e^{tY_r})[/tex]

(since the Y's are independent)

[tex]= E(e^{tY_1} * e^{tY_2} * ... * e^{tY_r})[/tex]

(since the Y's have the same distribution)

[tex]= [M_Y(t)]^r\\\\= [(1-p)/(1-pe^t)]^r[/tex]

Therefore, the moment generating function of a negative binomial distribution with parameters r and p is[tex]M_X(t) = [(1-p)/(1-pe^t)]^r.[/tex] This function can be used to derive moments and other statistical properties of the negative binomial distribution.

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The proof of LMNO is a rhombus is shown below.

A parallelogram is a particular instance of a rhombus. The opposing sides and angles in a rhombus are parallel and equal. A rhombus also has equal-length sides on each side, and its diagonals meet at right angles to form its shape. The rhombus is also referred to as a diamond or rhombus.

Given:

|LO|=|MN| and |LM|=|ON|

Since Opposite sides of a parallelogram are equal.

Now, LN⊥OM

So, ∠LPO = ∠NPO = 90° ( by definition of perpendicular lines)

LPO ≅ ∠NPO (by definition of congruent angles)

|LP|=|PN| (diagonals of a parallelogram bisect each other)

Thus, LMNO is a rhombus

Learn more about Rhombus here:

https://brainly.com/question/17044202

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Answer:

♣: ✔ All right angles are congruent.

♦: ✔ reflexive property

♠: ✔ Opposite sides of a parallelogram are congruent.