Find coordinates of the mid point AS if A is (-4,7) and 5,3

Answer

The figure shows three lines that intersect at point N.

3 lines intersect. Lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees.

Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL?

42°

48°

132°

138°

Step-by-step explanation:

42

The required measure of angle MNL is 42°. Option A is correct.

3 lines intersect lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees. Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL is to determine

The angle can be defined as the one line inclined over another line.
unit of measure of an angle is degree and radians.

Angle GNH is congruent to angle KNL.
∠KNL = 48°
Angle MNL is complementary to angle KNL
Since angle MNK = 90°
∠MNL + ∠KNL = 90°

∠MNL = 90-48
∠MNL = 42°

Thus, the required measure of angle MNL is 42°. Option A is correct.

Learn more about Angles here:
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Answer:

C

Step-by-step explanation:

2/3x - 5>3

Add 5 to each side

2/3x - 5+5>3+5

2/3x > 8

Multiply each side by 3/2

3/2 *2/3x > 8*3/2

x > 12

There is an open circle at 12 and the lines goes to the right

Answer:

[tex]=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]

Step-by-step explanation:

[tex]x^{\frac{1}{3}}\left(x^{\frac{1}{2}}+2x^2\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=x^{\frac{1}{3}},\:b=x^{\frac{1}{2}},\:c=2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+x^{\frac{1}{3}}\cdot \:2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\\mathrm{Simplify}\:x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}:\quad x^{\frac{5}{6}}+2x^{\frac{7}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=x^{\frac{5}{6}}[/tex]

[tex]x^{\frac{1}{3}}x^{\frac{1}{2}}\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=\:x^{\frac{1}{3}+\frac{1}{2}}\\=x^{\frac{1}{3}+\frac{1}{2}}\\\mathrm{Join}\:\frac{1}{3}+\frac{1}{2}:\quad \frac{5}{6}\\\frac{1}{3}+\frac{1}{2}\\\mathrm{Least\:Common\:Multiplier\:of\:}3,\:2:\quad 6\\Adjust\:Fractions\:based\:on\:the\:LCM\\=\frac{2}{6}+\frac{3}{6}[/tex]

[tex]\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\=\frac{2+3}{6}\\\mathrm{Add\:the\:numbers:}\:2+3=5\\=\frac{5}{6}\\=x^{\frac{5}{6}}\\2x^2x^{\frac{1}{3}}=2x^{\frac{7}{3}}\\=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]

Answer:

The correct option is (d).

Step-by-step explanation:

The complete question is:

The random variable x represents the number of computers that families have along with the corresponding probabilities. Use the probability distribution table below to find the mean and standard deviation for the random variable x.

    x :    0          1          2           3         4

p (x) : 0.49     0.05    0.32     0.07    0.07

(a) The mean is 1.39 The standard deviation is 0.80

(b) The mean is 1.39 The standard deviation is 0.64

(c)The mean is 1.18 The standard deviation is 0.64

(d) The mean is 1.18 The standard deviation is 1.30

Solution:

The formula to compute the mean is:

[tex]\text{Mean}=\sum x\cdot p(x)[/tex]

Compute the mean as follows:

[tex]\text{Mean}=\sum x\cdot p(x)[/tex]

         [tex]=(0\times 0.49)+(1\times 0.05)+(2\times 0.32)+(3\times 0.07)+(4\times 0.07)\\\\=0+0.05+0.64+0.21+0.28\\\\=1.18[/tex]

The mean of the random variable x is 1.18.

The formula to compute variance is:

[tex]\text{Variance}=E(X^{2})-[E(X)]^{2}[/tex]

Compute the value of E (X²) as follows:

[tex]E(X^{2})=\sum x^{2}\cdot p(x)[/tex]

          [tex]=(0^{2}\times 0.49)+(1^{2}\times 0.05)+(2^{2}\times 0.32)+(3^{2}\times 0.07)+(4^{2}\times 0.07)\\\\=0+0.05+1.28+0.63+1.12\\\\=3.08[/tex]

Compute the variance as follows:

[tex]\text{Variance}=E(X^{2})-[E(X)]^{2}[/tex]

             [tex]=3.08-(1.18)^{2}\\\\=1.6876[/tex]

Then the standard deviation is:

[tex]\text{Standard deviation}=\sqrt{\text{Variance}}[/tex]

                              [tex]=\sqrt{1.6876}\\\\=1.2990766\\\\\approx 1.30[/tex]

Thus, the mean and standard deviation for the random variable x are 1.18 and 1.30 respectively.

The correct option is (d).

80km / 48 min = 1 2/3 km per minute.

1 2/3 km per minute x 60 minutes(1 hour) = 100 km per hour

Answer:

90° and 180°

Step-by-step explanation:

An obtuse angle is any angle larger than 90° and smaller than 180°

Answer:

The Pythagorean Theorum

Step-by-step explanation:

you can literally search it and see that it is right!

Answer:

A

Step-by-step explanation:

They are congruent because of the SSS theorem.  The chords are congruent because the angles are congruent (angles are congruent because they are vertical angles).  Congruent central angles have congruent chords.  The other two sides are congruent because they are all radii of the circle and radius are always congruent.

Answer:

A

Step-by-step explanation:

The triangles are isosceles and congruent  too. ( both have 2  sides congruent  as  radius and the angles between them are congruent- SAS)

Answer:

6e+52

Step-by-step explanation:

cAlCuLaToR

Answer:

6e+52

Step-by-step explanation:

multiply

Answer:

x = 6

Step-by-step explanation:

x + 6 = x + x

Combine like terms

x+6 =2x

Subtract x from each side

x+6-x = 2x-x

6 = x

Answer:

6163.2 years

Step-by-step explanation:

A_t=A_0e^{-kt}

Where

A_t=Amount of C 14 after “t” year

A_0= Initial Amount

t= No. of years

k=constant

In our problem we are given that A_t is 54% that is if A_0=1 , A_t=0.54

Also , k=0.0001

We have to find t=?

Let us substitute these values in the formula

0.54=1* e^{-0.0001t}

Taking log on both sides to the base 10 we get

log 0.54=log e^{-0.0001t}

-0.267606 = -0.0001t*log e

-0.267606 = -0.0001t*0.4342

t=\frac{-0.267606}{-0.0001*0.4342}

t=6163.20

t=6163.20 years

PLEASE MARK BRAINLY

Answer:

(7, 5) is the final reflection of the point.

Step-by-step explanation:

We are given point A(-7, 5) which is first reflected over the line [tex]x= -5[/tex].

The minimum distance of the point A(-7, 5) from the line [tex]x= -5[/tex] is 2 units across the horizontal path (No change in y coordinate).

Point A lies 2 units on the left side of the line [tex]x= -5[/tex].

So, its reflection will be 2 units on the right side of [tex]x= -5[/tex].

Let its reflection be A' which has coordinates as (-5+2,5) i.e. (-3, 5).

Now A'(-3, 5) is reflected on the line [tex]x=2[/tex].

The minimum distance of the point A'(-3, 5)  from the line [tex]x=2[/tex] is 5 units across the horizontal path (No change in y coordinate).

Point A' lies 5 units on the left side of the line [tex]x=2[/tex].

So, its reflection will be 5 units on the right side of [tex]x=2[/tex].

Let its reflection be A'' which has coordinates as (2+5, 5) i.e (7, 5) is the final reflection of the point..

Please find attached image.

(7, 5) is the final reflection of the point.

Answer:C

Step-by-step explanation:

the x value  5  corresponds to two difference y-values.

Answer:

[tex]x+6y[/tex] where x is the cost of one adult ticket and y is the cost of one child ticket.

Step-by-step explanation:

This is an incomplete question since we would need to know the cost of the adult ticket and the cost of the children ticket.

However, let's say that the price is x dollars per adult and y dollars per child.

Now, we need to find out how much one adult and 6 children paid.

Thus, we would have to multiply the cost per adult by the number of adults and the cost per child per number of children and then sum up these two results.

Writing this in an algebraic way we would have:

[tex]1(x)+6y\\x+6y[/tex]

Thus, 1 adult and 6 children would have paid x + 6y dollars where x is the cost of the adult ticket and y is the cost of the children ticket.

(For example, if an adult ticket is 6 dollars and a child ticket is 4 dollars we would have that they paid 6 + 6(4) = 6 + 24 = 30 dollars)

Answer:

(a) A cross sectional study (b) The parameter can be computed as follows: Non-drinkers who agree exposed to obesity, Drinkers who are exposed or vulnerable to obesity (c) A postie relationship is established from the experiment between drinkers who are exposed to obesity and non drinkers who are exposed to obesity

Step-by-step explanation:

(a) The type of design is refereed to as a cross sectional study

(b) Now, because 50 men are beer drinkers out of 891 men.

Hence we can deduce form this that  500/891 gives us 0.56%.

This suggest that 0.56% men are beer drinkers out of which 325 have obesity, lets take for example 235/500 = 0.65% are exposed to obesity in which 80/ (89-500) = 80/491 = 0.16%

The non drinkers are 0.16% and are not exposed to obesity

Thus,

The parameters to be calculated is stated below:

(c) The next step is to determine and explain the results.

In this case we can say there is a positive relationship between drinkers and non drinkers, since from the experiment 0.65% are exposed to obesity and 0.16$ non drinkers are exposed to obesity.

Answer:

s(t)=8t

Step-by-step explanation:

The length of the sides of a square are initially 0 cm and increase at a constant rate of 8 cm per second.

Let the length of the Square = s

[tex]\dfrac{ds}{dt}=8 $cm/seconds, s_0=0 cm[/tex]

We solve the differential equation above subject to the given initial condition.

[tex]\dfrac{ds}{dt}=8\\ds=8$ dt\\Take the integral of both sides\\\int ds=\int 8$ dt\\s(t)=8t+C, where C is the constant of integration\\When t=0, s=0cm\\s(0)=0=8(0)+C\\C=0\\Therefore, s(t)=8t[/tex]

The formula that expresses the side length of the square, s (in cm), in terms of the number of seconds, t , since the square's side lengths began growing is:

s(t)=8t (in cm)

Answer:

B:

Step-by-step explanation:

According to theorem, "the angle in a semi-circle is a right angle" So,

<O = 90°

<M = 54

<K = 180-90-54

<OKM = 36°

Answer:

2.964

Step-by-step explanation:

Answer:

The estimated temperature at the point (2.03, 0.96) is 131

Step-by-step explanation:

In this question, we are to estimate the temperature at the given point using the temperature of the unevenly heated plate.

We proceed as follows;

In the question, we identify that the temperature at the point T(2,1) = 130 degrees celcius

Now, let’s look at how the temperature changes. There is a positive change of 16 units when we move across the x-axis and a negative decrease when we move up the y-axis to a tune of 16 units(negative)

Now, how does the problem wants us to move using the notation of change?

Look at the point (2.03, 0.96), since movement across the x-axis is positive, the motion here in terms of x i.e Δx is 0.03 while the corresponding motion in terms of y(albeit negative) is Δy = -0.04

Mathematically, the change in temperature is proportional to the distance traveled. What this means is that we need to multiply the changes in direction by the corresponding temperature. This is shown below;

ΔTx =Δx*Tx(2,1) => ΔTx = (0.03)*(16) = 0.48

ΔTy = Δy*Ty(2,1) => ΔTy = (-0.04)*(-13) = 0.52

We can now combine the equations above to form a single one as follows; which is an approximation;

ΔT = Δx*Tx(2,1) + Δy*Ty(2,1) => ΔT = (0.03)*(16) + (-0.04)*(-13) = 1

To arrive at the final answer, we add the change in temperature to the staring temperature which is ;

T(2.03,0.96) = T(2,1) + ΔT = 130 + 1= 131

Answer:

£135 is the correct answer.

Step-by-step explanation:

Let C be the cost of table.

And let R be the radius of table.

Cost of table is directly proportional to square of radius.

As per question statement:

[tex]C\propto R^{2}[/tex] or

[tex]C=a\times R^2 ....... (1)[/tex]

where [tex]a[/tex] is the constant to remove the [tex]\propto sign[/tex].

It is given that

[tex]C_1 =[/tex] £60 and [tex]R_1 = 50\ cm[/tex]

[tex]C_2 = ?[/tex] when [tex]R_2= 75\ cm[/tex]

Putting the values of [tex]C_1[/tex] and [tex]R_1[/tex] in equation (1):

[tex]60=a \times 50^2 ....... (2)[/tex]

Putting the values of [tex]C_2[/tex] and [tex]R_2[/tex] in equation (1):

[tex]C_2=a \times 75^2 ....... (3)[/tex]

Dividing equation (2) by (3):

[tex]\dfrac{60}{C_2}= \dfrac{a \times 50^2}{a \times 75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{50^2}{75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{2^2}{3^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{4}{9}\\\Rightarrow C_2 = 15 \times 9 \\\Rightarrow C_2 = 135[/tex]

So, £135 is the correct answer.

Answer:

The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

For the third class 78 - 83 ; the upper limit = 83

The class width for this histogram 5

The number of students that took the exam simply refers to the frequency is  84

The percentage of students that scored higher than 95.5 is 53%

Step-by-step explanation:

The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:

NOW;

The table given in the question can be illustrated as follows:

S/N           Class                Score          Frequency

1                 68  - 73            70.5           0

2                73 - 78             75.5           4

3                78 - 83             80.5           8

4                83 - 88             85.5           12

5                88 - 93            90.5           16

6                93 - 98            95.5           20

7                98 - 103           100.5          24                            

TOTAL:                                                 84

a) The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

b) For the third class 78 - 83 ; the upper limit = 83  ( since the upper limit is derived by addition of  5 to the last number showing  in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))

c)   The class width for this histogram 5 ; since it  is the difference  between the upper and lower boundaries  limit of the given class.

So , from above the difference in any of the class will definitely result into 5

d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84

e) Lastly; the percentage of students that scored higher than 95.5 is ;

⇒[tex]\dfrac{20+24}{84} *100[/tex]

= 0.5238095 × 100

= 52.83

To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%

Answer:

1. 98-103 (6th class)

2. 88

3. 5

4. 84

5. 52%

Step-by-step explanation:

Find attached the frequency table.

The class of exam scores falls between (1, 2, 3, 4, 5, or 6).

The exam score ranged from 68-103

1) The largest number of exam scores = 24

The largest number of exam scores is in the 6th class = 98 -103

Step 2 of 5:

The upper class limit is the higher number in an interval. Third class interval is 83-88

The upper class limit of the third class 88.

Step 3 of 5:

Class width = upper class limit - lower class limit

We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78

Class width = 78 - 73

Class width for the histogram = 5

Step 4 of 5:

The total of students that took the test = sum of all the frequency

= 0+4+8+12+16+20+24 = 84

The total of students that took the test = 84

Step 5 of 5:Find the percentage of students that scored higher than 95.5

Number of student that scored higher than 95.5 = 20 + 24 = 44

Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100

= (44/84) × 100 = 0.5238 × 100 = 52.38%

Percentage of students that scored higher than 95.5 = 52% (nearest percent)

Answer:

73.24% probability that 6 or more people from this sample are unemployed

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 100, p = 0.071[/tex]

So

[tex]\mu = E(X) = np = 10*0.071 = 7.1[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.071*0.929} = 2.5682[/tex]

What is the probability that 6 or more people from this sample are unemployed

Using continuity correction, this is [tex]P(X \geq 6 - 0.5) = P(X \geq 5.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 5.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{5.5 - 7.1}{2.5682}[/tex]

[tex]Z = -0.62[/tex]

[tex]Z = -0.62[/tex] has a pvalue of 0.2676

1 - 0.2676 = 0.7324

73.24% probability that 6 or more people from this sample are unemployed

Answer:

f\left(x\right)=x^3-x

Step-by-step explanation:

Answer:

A. (I) v = 46.42 m/s; (ii) v = 47.35 m/s; (III) v = 48.09 m/s; (iv) v = 48.26 m/s; (v) v = 58.28 m/s

B. v = 48.28 m/s

Note: the question is missing some values. The full Question is provided below:

If an arrow is shot upward on Mars with a speed of 52 m/s, its height in meters t seconds later is given by y = 52t − 1.86t2. (Round your answers to two decimal places.)

(a) Find the average speed over the given time intervals. (i) [1, 2] m/s (ii) [1, 1.5] m/s (iii) [1, 1.1] m/s (iv) [1, 1.01] m/s (v) [1, 1.001] m/s

(b) Estimate the speed when t = 1. m/s

Step-by-step explanation:

Height, y = 52t - 1.86t²

Velocity = ∆y/∆t = 52 - 1.86 * 2t = 52- 3.72t

A. Average velocity = (v1 + v2)/2

(i) At t = 1, 2

Average velocity = (52 - 3.72*1 + 52 -3.72*2)/2 = 46.42 m/s

(ii) At t = 1,1.5

Average velocity = (52 - 3.72*1 + 52 - 3.72*1.5)/2 = 47.35 m/s

(iii) At t = 1,1.1

Average velocity = (52 - 3.72*1 + 52 -3.72*1.1)/2 = 48.09m/s

(iv) At to = 1, 1.01

Average velocity = (52 - 3.72*1 + 52 - 3.72*1.01)/2 = 48.26 m/s

(iv) At t = (1, 1.001)s

Average velocity = (52 - 3.72*1 + 52 - 3.72*1.001)/2 = 48.28 m/s

B. Speed at t = 1s

Velocity = 52 - 3.72 * 1 = 48.28 m/s

Answer:

0.6042 or 29/48

Step-by-step explanation:

-5/16 = -0.3125

7/24 = 0.2917

0.2917 - -0.3125 = 0.6042

0.6042 ≅ 29/48

Answer:

29/48

Step-by-step explanation:

-5/16 + x= 7/24

x= 7/24-(-5/16)

x=7/24+5/16

x= 2*7/2*24+ 3*5/3*16

x=29/48

Answer:

63 metres

Step-by-step explanation:

A rectangle has 4 sides

2 of these sides are the lengths

The other 2 sides are the width

If the length of one side is 97 metres, the other side length must also be 97 metres

The two lengths then add together (97 + 97) to become 194 metres

Now we can use this information to calculate the width

320 (the total perimeter) subtract 194 (The total length) = 126 metres

This means that 126 metres is the total width

Because there are two sides which add up to the total width we divide 126 by 2

This allows us to get the measurement of the width

126 divided by 2 = 63 metres

Answer:

There were 210 downloads of the standard version.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

x is the number of downloads of the standard version.

y is the number of downloads of the high-quality version.

The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.2 MB. The total size downloaded for the two versions was 4074 MB.

This means that:

[tex]2.6x + 4.2y = 4074[/tex]

Yesterday, the high-quality version was downloaded four times as often as the standard version.

This means that [tex]y = 4x[/tex]

How many downloads of the standard version were there?

This is x.

[tex]2.6x + 4.2y = 4074[/tex]

Since [tex]y = 4x[/tex]

[tex]2.6x + 4.2*4x = 4074[/tex]

[tex]19.4x = 4074[/tex]

[tex]x = \frac{4074}{19.4}[/tex]

[tex]x = 210[/tex]

There were 210 downloads of the standard version.

Answer: There is a 0.88% chance of pulling three red marbles in a row.

Step-by-step explanation:

First pull = 4/15 (26.67%)

second pull = 3/14 (21.43%)

Third pull = 2/13 (15.38%)

You need to multiply these three fractions to get the probability of pulling three reds in a row, doing that will get you 4/455 or 0.88%

Tell me if it is right

Answer:

213, 123

Step-by-step explanation:

Answer:

a) P(6) = 0.0097

b) P(More than 3) = 0.1611

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is guessed correctly, or it is not. Questions are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

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

And p is the probability of X happening.

A student takes a multiple-choice test that has 11 questions.

This means that [tex]n = 11[/tex]

Each question has five choices.

This means that [tex]p = \frac{1}{5} = 0.2[/tex]

(a) Find P (6)

This is P(X = 6).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 6) = C_{11,6}.(0.2)^{6}.(0.8)^{5} = 0.0097[/tex]

P(6) = 0.0097

(b) Find P (More than 3).

Either P is 3 or less, or it is more than three. The sum of the probabilities of these outcomes is 1. So

[tex]P(X \leq 3) + P(X > 3) = 1[/tex]

We want P(X > 3). So

[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]

In which

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{11,0}.(0.2)^{0}.(0.8)^{11} = 0.0859[/tex]

[tex]P(X = 1) = C_{11,1}.(0.2)^{1}.(0.8)^{10} = 0.2362[/tex]

[tex]P(X = 2) = C_{11,2}.(0.2)^{2}.(0.8)^{9} = 0.2953[/tex]

[tex]P(X = 3) = C_{11,3}.(0.2)^{3}.(0.8)^{8} = 0.2215[/tex]

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0859 + 0.2362 + 0.2953 + 0.2215 = 0.8389[/tex]

Then

[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8389 = 0.1611[/tex]

P(More than 3) = 0.1611