-1/5+2/15 what is the answer the answer to that

Tim's fair share is 175 rupees and Tom's fair share is 125 rupees.

To find the fair share of both Tim and Tom depending on the work done, follow these steps:

Step 1: Find the total number of windows cleaned by both Tim and Tom.

Total number of windows cleaned = number of windows on ground floor + number of windows on first floor

Total number of windows cleaned = 7 + 8 = 15

Step 2: Find the ratio of the number of windows cleaned by Tim to the number of windows cleaned by Tom.

Ratio of Tim's work to Tom's work = number of windows cleaned by Tim / number of windows cleaned by Tom

Ratio of Tim's work to Tom's work = 7 / 8 = 0.875

Step 3: Divide the total money paid by the master (300 rupees) in the same ratio as the work done by Tim and Tom.

Tim's share = (0.875 / (0.875 + 1)) * 300 = 175 rupees

Tom's share = (1 / (0.875 + 1)) * 300 = 125 rupees

Step 4: Check the shares to ensure they add up to the total amount paid by the master.

Tim's share + Tom's share = 175 + 125 = 300 rupees

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Answer:L x b

              98ft x 56ft =5488

                                  =5488 / $3.75= $1463.47

Step-by-step explanation: Play ground is more like a rectangle so we use the formula for the rectrectangle to get total area . A=Lxb

Divide  the total with the cost since it say each per sqr feet

A=lxb

98x56=5488

5488/3.75= 1463.47

the probability that there are at most 2 defective components in the box is approximately 0.9044 and the probability of having at most 3 defective components out of 250 boxes is very close to zero.

a) Let X be the number of defective components in a box of 10 components. Then X follows a binomial distribution with n=10 and p=0.01, since the probability of a component being defective is 0.01. We want to find the probability that there are at most 2 defective components in the box, i.e., P(X ≤ 2).

Using the binomial probability formula, we get:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= (10 choose 0) × 0.01⁰ × 0.99¹⁰ + (10 choose 1) × 0.01¹ × 0.99⁹ + (10 choose 2) × 0.01² × 0.99⁸

= 0.90438222

Therefore, the probability that there are at most 2 defective components in the box is approximately 0.9044 (rounded to four decimal places).

b) We want to find the probability of having at most 3 defective components out of 250 boxes, each containing 10 components. Since np = 100.01 = 0.1 < 5 and n × (1-p)=10 × 0.99=9.9 > 5, we can use the normal approximation to the binomial distribution, with mean μ = np = 2.5 and standard deviation σ = √np(1-p) = 1.577.

Let X be the number of boxes with at most 3 defective components. Then X follows an approximate normal distribution with mean μ' = np=2.5250 = 625 and standard deviation σ' = √np(1-p)) = 12.5 × 1.577 = 19.712.

We want to find P(X ≤ 250), which can be written as P(X < 251) since X is a discrete variable. Using the continuity correction, we can approximate this probability as P(X < 251.5). Then we standardize the variable:

z = (251.5 - μ')/σ' = (251.5 - 625)/19.712 = -18.919

Using a standard normal table or calculator, we find that P(Z < -18.919) is a very small number, practically zero. Therefore, the probability of having at most 3 defective components out of 250 boxes is very close to zero.

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Complete Question

factory produces components of which 1% are defective. The components are packed in boxes of 10. A box is selected by random a) Find the probability that there are at most 2 defective components in the box b) Use a suitable approximation to find the probability of having at most 3 defective (inclusive 3 cases) components out of 250.

Answer:772

Step-by-step explanation:

SA=PH+2b

SA=(10+8+10+8)(17)+2(8x10)

SA=772

Answer:

Step-by-step explanatin

multiply all of them

Answer:

28/51

Step-by-step explanation:

First you add up all the values: 28 + 15 + 8 = 51. That is your denominator because the total is the denominator. Then you put 28 as your numerator because that's how many off all the people that only like pizza.

The intersection of two evens a and b is the event that both a and b occur that is option A is correct.

The intersection of two events A and B is defined as the event that occurs when both A and B occur simultaneously. It is denoted by A ∩ B, where the symbol ∩ represents intersection.

For example, if event A is "getting a head when flipping a coin" and event B is "rolling a 6 on a fair die", then the intersection of A and B would be the event "getting a head when flipping a coin and rolling a 6 on a fair die".

For example, suppose A represents the event "rolling an even number on a dice" and B represents the event "rolling a number greater than 3 on a dice". The intersection of A and B is the event that "rolling a number that is both even and greater than 3 on a dice", which consists of the outcomes {4, 6}.

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Complete Question:

the intersection of two events a and b is the event that:

a) both a and b occur.

b) the union of ac and bc occurs.

c) the union of a and b does not occur.

d) either a or b or both occur.

e) either a or b, but not both.

f) none of the above.

let's firstly convert the mixed fractions to improper fractions and then add them up.

[tex]\stackrel{mixed}{1\frac{12}{15}}\implies \cfrac{1\cdot 15+12}{15}\implies \stackrel{improper}{\cfrac{27}{15}}~\hfill \stackrel{mixed}{1\frac{5}{15}} \implies \cfrac{1\cdot 15+5}{15} \implies \stackrel{improper}{\cfrac{20}{15}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{27}{15}~~ + ~~\cfrac{20}{15}\implies \cfrac{27~~ + ~~20}{\underset{\textit{denominator is the same}}{15}}\implies \cfrac{47}{15}\implies 3\frac{2}{15}[/tex]

The upper and lower 3-sigma control chart limits for this DNA sample data are UCL = 18.53 and LCL = 0

When we are given data on the DNA samples taken over the past 10 days, we can calculate the upper and lower 3-sigma control chart limits. The sample size is 100, and the defective number of DNA samples is given for each of the ten days.

The 3-sigma control limits for a process control chart can be calculated using the following formula: Upper control limit (UCL) = Mean + (3 × Standard Deviation) and Lower control limit (LCL) = Mean - (3 × Standard Deviation),where the mean is the average value of the data, and the standard deviation is the spread of the data around the mean.

Here, we need to calculate the mean and standard deviation of the defective samples for the past 10 days. The average number of defectives per day (mean) is (7+9+9+11+7+8+0+11+13+2) / 10 = 77 / 10 = 7.7

To calculate the standard deviation, firstly, calculate the variance: variance = sum of the squared differences from the mean divided by the number of samples = [tex][(7-7.7)^2 + (9-7.7)^2 + ... + (2-7.7)^2] / 10[/tex]  ≈ 13.01 2.. Now, take the square root of the variance which gives standard deviation. So, standard deviation = [tex]\sqrt{13.01\\}[/tex] ≈ 3.61

Using the 3-sigma rule UCL = Mean + (3 * Standard Deviation) = 7.7 + (3 * 3.61) ≈ 18.53    and   LCL = Mean - (3 * Standard Deviation) = 7.7 - (3 * 3.61) ≈ -3.13. However, since the LCL cannot be negative, we set it to 0 in this context. So, the upper and lower 3-sigma control chart limits are: UCL = 18.53 LCL = 0

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The student should record the area of the paper as option (c) 602 cm^2

The student measured the length and width of a single sheet of paper and was asked to calculate its surface area. The surface area of the paper is the product of its length and width, which can be calculated by multiplying the two measurements together.

The surface area of the paper can be calculated as the product of the length and the width

Surface area = length × width

Substituting the given measurements, we get

Surface area = 43 cm × 14 cm

Surface area = 602 cm^2

Therefore, the correct option is (c) 602 cm^2

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The given question is incomplete, the complete question is;

A student is calculating the surface area of a single sheet of paper. he measures the length to be 43 cm he measures the width to be 14cm the student should record the area of the paper as (a) 602.64 cm^2 . (b) 602.6 cm^2 . (c) 602 cm^2 . (d) 603 cm^2 .

If lisa is on her way home in her car, she has driven 24 miles so far, which is three-fourths of the way home, Lisa's total drive is 32 miles.

Let's represent the total length of Lisa's drive as x. We know that she has driven 24 miles so far, which is three-fourths of the total length. We can write this information as:

24 = (3/4) x

To find x, we need to isolate it on one side of the equation. We can start by multiplying both sides by 4/3 to get rid of the fraction:

24 * (4/3) = x

Simplifying, we get:

32 = x

We can say that Lisa has already driven 24 miles, which is three-fourths of the total distance. To find the total distance, we use the equation 24 = (3/4) x, where x represents the total distance.

To solve for x, we multiply both sides of the equation by 4/3 to cancel out the fraction, giving us 32 = x. Therefore, Lisa's total drive is 32 miles.

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Therefore, the average velocity of the person over the interval 0 ≤ t ≤ 12 can be calculated as follows:Average Velocity = Total distance travelled / Total time taken= 6.6 / 12= 0.55 m/s.

In the given question, we need to find the average velocity of a person running in a straight line across a field, given differentiable function v(t) on the interval [0,12]. Therefore, to calculate the average velocity of a person, we use the following formula:Average Velocity = Total distance travelled / Total time takenWe have a graph with the velocity of the person, which is a differentiable function v(t) given above on the interval 0 ≤ t ≤ 12.

We need to find the distance travelled by the person. Therefore, we use the following formula:Distance travelled = ∫v(t)dt From the given graph, the velocity of the person is zero when t = 0 and when t = 5. Similarly, the velocity of the person is 0 when t = 10 and when t = 12.So, we have to calculate the distance travelled from 0 to 5, from 5 to 10, and from 10 to 12 to determine the total distance travelled by the person over the given interval .Distance travelled from 0 to 5 can be calculated as follows :

Distance travelled from 0 to 5 = ∫v(t)dt from [tex]0 to 5= 5 x 0.6 = 3[/tex]Distance travelled from 5 to 10 can be calculated as follows :Distance travelled from 5 to 10 = [tex]∫v(t)dt[/tex] from [tex]5 to 10= 5 x 0.4 = 2[/tex]

Distance travelled from 10 to 12 can be calculated as follows: Distance travelled from 10 to 12 = ∫v(t)dt from 10 to 12= 2 x 0.8 = 1.6Total distance travelled = Distance travelled from 0 to 5 + Distance travelled from 5 to 10 + Distance travelled from 10 to [tex]12= 3 + 2 + 1.6= 6.6[/tex]

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The probability that the coin will land on the same side in all three tosses is 1/8.

There are two possible outcomes for each coin flip: heads or tails. Therefore, there are 2 × 2 × 2 = 8 possible outcomes for flipping a coin three times in a row.To find the probability that the coin will land on the same side in all three tosses, we need to count the number of outcomes that satisfy this condition.

There are only two such outcomes: either all three tosses are heads or all three tosses are tails. Therefore, the probability of this happening is 2/8 or 1/4.But we are asked for the probability that the coin will land on the same side in all three tosses, not just one specific side.

Therefore, we need to divide our previous result by 2 (the number of sides of the coin) to get the final answer: 1/4 ÷ 2 = 1/8. The probability that the coin will land on the same side in all three tosses is 1/8.

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Answer:

a To$4198i0 multiple times a week and prosperity in 44 approximate length of 4X5

The probability of getting exactly 2 tails in 7 flips of a coin with a probability of tails being 3/4 is 189/16384 or approximately 0.0115.

We can use the binomial distribution formula to solve this problem. Let X be the number of tails that come up in 7 flips of the coin. Then X follows a binomial distribution with n = 7 and p = 3/4 (since the probability of tails is 3/4).

The probability of getting exactly 2 tails is given by:

P(X = 2) = (7 choose 2) * (3/4)^2 * (1/4)^5

= (21 * 9/16 * 1/1024)

= 189/16384

So the probability that George flips exactly 2 tails is 189/16384, or approximately 0.0115.

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Based on the boxplot, approximately 25% of these Ayrshire cattle had a butterfat percentage higher than 8.3%.

To determine this, we can look at the boxplot and see that the third quartile (Q3) is at 8.3%. Since 25% of the data is greater than 8.3%, we can infer that 25% of these Ayrshire cattle had a butterfat percentage higher than 8.3%. To further explain, a boxplot is a graphical representation of the five-number summary of a dataset, which consists of the minimum, first quartile (Q1), median (Q2), third quartile (Q3), and maximum.

The boxplot is divided into two parts, the box and the whiskers. The box indicates the interquartile range (IQR) which is the difference between the third and first quartile. The whiskers represent the minimum and maximum of the dataset. The median is shown by a line within the box. In this boxplot, the first quartile (Q1) is 6.7%, the median (Q2) is 7.4%, and the third quartile (Q3) is 8.3%. Therefore, based on the boxplot, about 25% of these Ayrshire cattle had a butterfat percentage that was higher than 8.3%.

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Complete Question : Based on the boxplot, about 25% of these ayrshire cattle had a butterfat percentage that was higher than?

Answer:

15 Full orbits per day

Step-by-step explanation:

To work out the number of full orbits the ISS does in 1 day, we need to know how long it takes for the ISS to complete one orbit around the Earth.

We can use the information given to us to calculate the time it takes for the ISS to complete one orbit:

Distance traveled in one orbit = 42,600 kilometers

Speed of the ISS = 27,576 kilometers per hour

To calculate the time taken for one orbit:

Time taken = Distance traveled / Speed

Time taken = 42,600 kilometers / 27,576 kilometers per hour

Time taken = 1.54 hours (rounded to 2 decimal places)

So, the ISS takes approximately 1.54 hours to complete one orbit around the Earth.

Now, we can calculate the number of orbits the ISS does in one day:

Number of orbits per day = 24 hours / Time taken for one orbit

Number of orbits per day = 24 hours / 1.54 hours

Number of orbits per day = 15.58 (rounded to 2 decimal places)

Therefore, the ISS completes approximately 15 full orbits around the Earth in one day.

By the central limit theorem, with a large random sample, the sample histogram will not closely resemble the normal curve but with a large random sample, the probability density function of the sample mean closely resembles the normal curve.

The central limit theorem for samples says that if we keep drawing larger and larger samples and calculating their means, the sample forms their own normal distribution (the sampling distribution). The normal distribution will have the same mean as the original distribution and a variance that equals the original variance divided by the sample size. The variable n is the number of values that are averaged together,and not the number of times the experiment is done.

Hence,with a large random sample, the sample histogram will not resemble the normal curve but with a large random sample, the probability density function of the sample mean will closely resemble the normal curve.

The complete question is-

true or false: (justify/explain your answer) state whether a or b is the true statement below and then explain why the other statement is false. a. with a large random sample, the sample histogram will closely resemble the normal curve. b. with a large random sample, the probability density function of the sample mean will close resemble the normal curve.

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To find:-

Answer:-

We are here given that 1 gallon of paint can be used to paint 400 sq ft. of an area .

There are 4 walls in a room and the area of opposite walls are equal.

So let's find out the area of wall ABCD , its area would be equal to the area of the wall EFGH . So ,

Area of wall ABCD and EFGH :-

Area = lb

Area = 18ft * 8ft

Area = 144ft.²

Therefore area of wall EFGH will also be 144ft.² .

===============================================

Again,

Area of wall DAHE and CBGF :-

Here both the areas of the wall would again be equal, as they have same dimensions of 14ft × 8ft .

Area = lb

Area = 14ft * 8ft

Area = 112ft.²

Therefore area of walls DAHE and CBGF is 112ft.²

==============================================

Calculating total area of the four walls :-

Total area = 2( 112ft.² + 144ft.² )= 512ft.²

Hence total area of four walls is 512ft.²

Now since there will be two coats of paints , total area to be painted would become double that is 1024ft.²

Now we may use Unitary Method :-

To paint 400ft.² 1 gallon of paint is needed.

To paint 1ft.² 1/400 gallon of paint is needed.

To paint 1024ft.² , 1/400*1024 = 2.56 gallons of paint would be needed.

Hence the total paint required is 2.56 gallons.

and we are done!

Answer:

3 one-gallon containers of paint.

Step-by-step explanation:

First, find the total surface area that needs to be painted.

To find the surface area of the walls of a rectangular room, calculate the area of each of the four walls and then add them together.

The area of a rectangle is the product of its width and length.

For the walls of the given room, the "length" is the height of 8 ft, since we want to paint from the floor to the ceiling.

The room has four walls:

Therefore, the total surface area is:

[tex]\begin{aligned}\textsf{Total surface area}&=2(14 \times 8)+2(18 \times 8)\\&=2(112)+2(144)\\&=224+288\\&=512\;\sf ft^2\end{aligned}[/tex]

Since we want to apply two coats of paint, we need to double the surface area:

[tex]\implies 2 \times 512=1024 \; \sf ft^2[/tex]

Now, we can find how many gallons of paint are needed by dividing the total surface area by the coverage per gallon:

[tex]\implies \dfrac{1024}{400}=2.56\; \sf gallons[/tex]

Therefore, we need 2.56 gallons of paint to paint two coats of each wall of the rectangular room.

Since we cannot buy a fraction of a gallon, we need to round up to the nearest gallon. Therefore, we need to buy 3 one-gallon containers of paint.

End of unit 4 assessment on right triangle trigonometry is an evaluation of a student's understanding of the basic concepts and applications of trigonometry involving right triangles.

This assessment may cover topics such as the trigonometric functions, Pythagorean theorem, special right triangles, and solving right triangles.

Trigonometry is the study of the relationships between the angles and sides of triangles, particularly right triangles. It is a branch of mathematics that has numerous applications in fields such as physics, engineering, and astronomy.

The trigonometric functions are sine, cosine, and tangent, which are ratios of the sides of a right triangle. These functions can be used to solve problems involving angles and sides of right triangles, such as finding the missing side or angle.

The Pythagorean theorem is another fundamental concept in right triangle trigonometry. It states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

Special right triangles, such as the 30-60-90 triangle and the 45-45-90 triangle, have specific ratios of their side lengths that can be used to solve problems more easily.

Solving right triangles involves finding the measures of all the angles and sides of a right triangle given certain information, such as the length of one side and the measure of one angle.

In conclusion, the end of unit 4 assessment on right triangle trigonometry evaluates a student's understanding of the basic concepts and applications of trigonometry involving right triangles. This assessment is important for students to demonstrate their mastery of the subject and to prepare them for further studies in mathematics and related fields.

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End of unit 4 assessment right triangle trigonometry describe the importance of Side ratios in right triangles as a function of the angles ?

Therefore , the solution of the given problem of unitary method comes out to be  the composite shape's overall size is 30  square units.

Utilizing previously well-known variables, this uniform convenience, or all crucial elements from a prior flexible study that followed a particular methodology event can all be used to achieve the goal. It will be possible to contact the entity again if the anticipated assertion outcome actually happens; if it doesn't, both important systems will surely miss the statement.

Here,

We must divide the composite shape into smaller shapes and sum up their areas in order to determine the area of the composite shape.

We can see that the composite form is made up of a triangle with a base of four and a height of three, and a rectangle with dimensions of six by four.

=> length times breadth equals six by four, or 24 square units, for a rectangle.

Triangle's area is equal to

=>  (1/2) x base times height, or (1/2) x 4 times 3, or 6 square units.

As a result, the composite shape's overall size is:

=> 24 + 6 = 30  square units.

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a.) The mode for the chart is 24.

b.) The probability that the winning score will be 25 = 7/50

C.)The probability that the winning score will be 23 or more = 37/50.

The number of times the game is played = 50 times

The number of games that showed the score of 25= 7

The probability of winning a score of 25 = 7/50

The scores that are 23 and above; 10+14+7+4+2= 37

The probability of winning a score of 23 and above = 37/50

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To prove that the functions e^(r1*t), e^(r2*t), and e^(r3*t) are linearly independent on (-∞,∞) for distinct real numbers r1, r2, and r3, we will follow the hint provided:

1. Assume to the contrary that e^(r1*t) = C1*e^(r2*t) + C2*e^(r3*t) for all t, where C1 and C2 are constants.
2. Divide both sides of the equation by e^(r1*t) to obtain: 1 = C1*e^((r2-r1)*t) + C2*e^((r3-r1)*t).
3. Differentiate both sides of the equation with respect to t:
  0 = C1*(r2-r1)*e^((r2-r1)*t) + C2*(r3-r1)*e^((r3-r1)*t).
4. Now, observe that e^((r2-r1)*t) and e^((r3-r1)*t) are linearly dependent, which is a contradiction, since we know that r1, r2, and r3 are distinct real numbers, and the exponential functions with distinct exponents are linearly independent.Thus, the functions e^(r1*t), e^(r2*t), and e^(r3*t) are linearly independent on (-∞,∞) for distinct real numbers r1, r2, and r3.

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We must prioritize biblical teachings and values over the cultural trends of the day. We should be salt and light to the world and live our lives in a way that honours God.

As Christians, it is our responsibility to be salt and light to the world. The dominant culture may be a reflection of the majority of the population's thoughts, but it may not necessarily align with the Bible's teachings. We should not be swayed by the current trends and instead prioritize God's standards.

The counterculture of the 1960s, according to some, promoted ideas that are contrary to Christian teachings. Many people who were part of this movement advocated for sexual freedom, drug use, and rebellion against authority. Christians should not have supported these behaviours or beliefs since they are inconsistent with biblical principles.

Likewise, in today's world, there are areas in which Christians must not conform to the dominant culture. One of the primary areas where Christians may be tempted to conform is in terms of sexual morality. Many people believe that premarital sex and homosexuality are acceptable practices, but these beliefs are not supported by the Bible.

As a result, Christians should not follow these cultural trends, but rather stick to biblical teachings and live their lives according to God's standards.Another example is in regards to materialism. The desire to accumulate wealth and material possessions is prevalent in modern culture.

Still, Christians must resist the temptation to make money and possessions their god, as this is against the Bible's teachings. Instead, we should focus on using our resources to glorify God and help others.In conclusion, Christians must be cautious about conforming to the dominant culture.

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Hope this helps! You just submitted the picture and never really showed which side was a b c or anything!

By the angle bisector theorem,

[tex]\frac{5}{9}=\frac{2}{x-2}[/tex]

After cross multiplying,

5(x-2) = 2(9)

5x-10 = 18

5x = 28

x = 5.6

Increasing the sample size leads to a more accurate estimation of the population mean.

What is Probability ?

Probability can be defined as ratio of number of favourable outcomes and total number outcomes.

As the sample size, n, increases, the center of the sampling distribution of sample means becomes more precise and closer to the true population mean. This is known as the central limit theorem, which states that as the sample size increases, the distribution of sample means becomes approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

In other words, as we take larger and larger samples, we are more likely to obtain sample means that are closer to the true population mean. This is because larger samples are less affected by random fluctuations and more likely to provide a representative picture of the population as a whole.

Therefore, increasing the sample size leads to a more accurate estimation of the population mean.

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The device that indicates whether two AC sources to be connected in parallel are in the correct phase relationship is called a synchronizing device.

A synchronizing device is a mechanism that ensures that two AC sources are in sync when they are connected in parallel. It's used to match the voltage, frequency, and phase angle of two alternating current (AC) sources.

It guarantees that the power supplied by both generators is synchronized, allowing them to be combined into a single electrical system without disrupting the balance of the current or causing a short circuit.

As a result, it is critical to the safe and efficient operation of power systems. A phase sequence indicator (PSI) or a synchroscope is often used as a synchronizing device. It works by providing an indication of the voltage difference, the phase angle difference, and the frequency difference between two AC sources that are to be synchronized.

Therefore, a synchronizing device is an instrument that determines whether two alternating current (AC) sources to be connected in parallel are in the appropriate phase relationship.

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Answer:

[tex]\huge\boxed{\sf y = \frac{9-4x}{6}}[/tex]

Step-by-step explanation:

9 = 4x + 6y

9 - 4x = 6y

[tex]\displaystyle \frac{9-4x}{6} = y\\\\OR\\\\y = \frac{9-4x}{6} \\\\\rule[225]{225}{2}[/tex]

Answer:

Step-by-step explanation:

Now we have to,

→ Find the required value of y.

The equation is,

→ 9 = 4x + 6y

Then the value of y will be,

→ 9 = 4x + 6y

→ 4x + 6y = 9

→ 6y = 9 - 4x

→ 6y = -4x + 9

→ y = (-4x + 9)/6

→ y = (-4x/6) + (9/6)

→ y = (-2x/3) + (3/2)

Hence, this is the answer.

The prοbability that mοre than 2 animals need tο stay οvernight is apprοximately 0.9124, οr 91.24%.

The binοmial distributiοn is a prοbability distributiοn that describes the number οf successes in a fixed number οf independent trials, each with the same prοbability οf success. The distributiοn is characterized by twο parameters: the number οf trials (n) and the prοbability οf success (p) fοr each trial.

Tο sοlve this prοblem, we need tο use the binοmial distributiοn since we have a fixed number οf trials (25 animals brοught in) and each trial (animal) can either be a success (needs tο stay οvernight) οr a failure (dοesn't need tο stay οvernight).

Let X be the number οf animals that need tο stay οvernight. Then X fοllοws a binοmial distributiοn with n = 25 trials and p = 0.2 prοbability οf success (animal needing tο stay οvernight).

We want tο find the prοbability that mοre than 2 animals need tο stay οvernight, which can be expressed as:

nοw, P(X > 2) = 1 - P(X ≤ 2)

Tο calculate P(X ≤ 2), we can use the binοmial cumulative distributiοn functiοn (CDF) οr simply add up the prοbabilities οf X = 0, X = 1, and X = 2:

similarly, P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

[tex]= (0.8)^{25} + 25(0.2)(0.8)^{24} + (25\ \text{choose}\ 2)(0.2)^{2}(0.8)^{23}[/tex]

≈ 0.0876

Therefοre, P(X > 2) = 1 - P(X ≤ 2) ≈ 1 - 0.0876 = 0.9124

Sο the prοbability that mοre than 2 animals need tο stay οvernight is apprοximately 0.9124, οr 91.24%.

To learn more about binomial distribution, visit the link:

https://brainly.com/question/29163389

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