A game require rolling a six sided die numbered fro 1 to 6. What is the probability of rolling a 1 or a 2?

Answer:

I would say the second one

Step-by-step explanation:

f(x) has a range of y<0, because it is reflected over the x axis

g(x) = -5/7(3/5)^-x is also reflected over the x axis, except also in the y axis. Regardless of the reflection in the y-axis, y still cannot be equal to or greater than 0. Therefore, I believe it is the second choice.

(The third and forth choice are the same, which rules them both out. The first on reflects it over the y-axis, meaning that x can be greater than 0.)

Answer:x>-1

Step-by-step explanation:

Step 1: Subtract 3 from both sides.

-3x+3-3<6-3

-3x<3

Step 2: Divide both sides by -3.

-3x/-3<3/3

X>-1

Answer:

X = 5

Step-by-step explanation:

If 4x = 20

And we are asked to find the solution.

It simply means looking for the value of x

So

4x = 20

X = 20/4

X = 5

X is simply the solution

X = 5

Answer:

D 2.16

Step-by-step explanation:

a p e x just use log

Answer:

0.495 probability that Jim and Molly take counters of different colours

Step-by-step explanation:

For each trial, there are only two possible outcomes. Either a blue counter is picked, or a red counter is picked. The counter is put back in the bag after it is taken, which means that we can 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.

The probability that the counter is red is 0.45

This means that [tex]p = 0.45[/tex]

Jim taken a counter, then Molly:

Two trials, so [tex]n = 2[/tex]

What is the probability that Jim and Molly take counters of different colours?

One red and one blue. So this is P(X = 1).

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

[tex]P(X = 1) = C_{2,1}.(0.45)^{1}.(0.55)^{1} = 0.495[/tex]

0.495 probability that Jim and Molly take counters of different colours

Answer:

[tex]N(T(t)) = 1408t^2 - 385.6t - 105.52[/tex]

Time for bacteria count reaching 8019: t = 2.543 hours

Step-by-step explanation:

To find the composite function N(T(t)), we just need to use the value of T(t) for each T in the function N(T). So we have that:

[tex]N(T(t)) = 22 * (8t + 1.7)^2 - 123 * (8t + 1.7) + 40[/tex]

[tex]N(T(t)) = 22 * (64t^2 + 27.2t + 2.89) - 984t - 209.1 + 40[/tex]

[tex]N(T(t)) = 1408t^2 + 598.4t + 63.58 - 984t - 169.1[/tex]

[tex]N(T(t)) = 1408t^2 - 385.6t - 105.52[/tex]

Now, to find the time when the bacteria count reaches 8019, we just need to use N(T(t)) = 8019 and then find the value of t:

[tex]8019 = 1408t^2 - 385.6t - 105.52[/tex]

[tex]1408t^2 - 385.6t - 8124.52 = 0[/tex]

Solving this quadratic equation, we have that t = 2.543 hours, so that is the time needed to the bacteria count reaching 8019.

Answer:

a) The volume of the wooden block is 240 cm^3.

b) The density of the wooden block is 0.7 g/cm^3.

Step-by-step explanation:

The volume of the rectangular wooden block can be calculated as the multiplication of the length in each dimension: length, wide and height.

With dimensions 10 cm x 3 cm x 8 cm, the volume is:

[tex]V=L\cdot W\cdot H = 10\cdot 3\cdot 8=240[/tex]

The volume of the wooden block is 240 cm^3.

If we know that the mass of the wooden block is 168 g, we can calculate the density as:

[tex]\rho = \dfrac{M}{V}=\dfrac{168}{240}=0.7[/tex]

The density of the wooden block is 0.7 g/cm^3.

Answer:

36=2×3×3×3

Answer

[tex]36 = 2 \times 2 \times 3 \times 3 \\ \: \: \: \: \: \: \: \: = {2}^{2} \times {3}^{2} [/tex]

Step-by-step explanation:

First write the prime factors of 36 that you can see here

[tex]2 \: \: \: 2 \: \: \: 3 \: \: \: 3[/tex]

Now write 36 as a product of its prime factors.

[tex]36 = 2 \times 2 \times 3 \times 3 \\ \: \: \: \: \: \: \: \: = {2}^{2} \times {3}^{2} [/tex]

Answer:

  56.25°

Step-by-step explanation:

The definition of the cosine function tells you that

  cos(B) = BC/BA

  B = arccos(BC/BA) = arccos(5/9)

  B ≈ 56.25°

Answer:

The probability that at exactly one of them does exactly two language classes is 0.32.

Step-by-step explanation:

We can model this variable as a binomial random variable with sample size n=2.

The probability of success, meaning the probability that a student is in exactly two language classes can be calculated as the division between the number of students that are taking exactly two classes and the total number of students.

The number of students that are taking exactly two classes is equal to the sum of the number of students that are taking two classes, minus the number of students that are taking the three classes:

[tex]N_2=F\&S+S\&G+F\&G-F\&S\&G=12+4+6-2=20[/tex]

Then, the probabilty of success p is:

[tex]p=20/100=0.2[/tex]

The probability that k students are in exactly two classes can be calcualted as:

[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{2}{k} 0.2^{k} 0.8^{2-k}\\\\\\[/tex]

Then, the probability that at exactly one of them does exactly two language classes is:

[tex]P(x=1) = \dbinom{2}{1} p^{1}(1-p)^{1}=2*0.2*0.8=0.32\\\\\\[/tex]

Answer:

55.82 cm

Step-by-step explanation:

d1= 8.72 cm

a= 15.6 cm

A rhombus= 1/2*d1*d2 = A square

A square= 15.6²= 243.36 cm²

d2= 2A/d1= 2*243.36/8.72 ≈55.82 cm

Answer:

15.74% of the player's serves were between 115 mph and 145 mph

Step-by-step explanation:

When the distribution is normal, we use 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.

In this question, we have that:

[tex]\mu = 100, \sigma = 15[/tex]

What percentage of the player's serves were between 115 mph and 145 mph

This is the pvalue of Z when X = 145 subtracted by the pvalue of Z when X = 115.

X = 145

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

[tex]Z = \frac{145 - 100}{15}[/tex]

[tex]Z = 3[/tex]

[tex]Z = 3[/tex] has a pvalue of 0.9987

X = 115

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

[tex]Z = \frac{115 - 100}{15}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8413

0.9987 - 0.8413 = 0.1574

15.74% of the player's serves were between 115 mph and 145 mph

Answer:

  see attached

Step-by-step explanation:

The attachments show the initial (brown) and final (blue) positions of the phone. The spreadsheet shows all the intermediate locations and the formulas used to determine them.

The two reflections cancel the total of 180° of CW rotation, so the net result is simply a translation. That translation is up by 9 units.

__

Translation up adds to the y-coefficient; translation right adds to the x-coefficient. Down or left use negative values.

90° CW does this: (x, y) ⇒ (y, -x)

Reflection across y does this: (x, y) ⇒ (-x, y)

Reflection across x does this: (x, y) ⇒ (x, -y)

Answer:

[tex]n=(\frac{2.17(0.143)}{0.023})^2 =182.03 \approx 183[/tex]

So the answer for this case would be n=183 rounded up to the nearest integer

Step-by-step explanation:

Information provided

[tex]\bar X[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma = 0.143[/tex] represent the population standard deviation

n represent the sample size  

[tex] ME = 0.023[/tex] the margin of error desired

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =0.023 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

The confidence level is 97% or 0.97 and the significance would be [tex]\alpha=1-0.97=0.03[/tex] and [tex]\alpha/2 = 0.015[/tex] then the critical value would be: [tex]z_{\alpha/2}=2.17[/tex], replacing into formula (5) we got:

[tex]n=(\frac{2.17(0.143)}{0.023})^2 =182.03 \approx 183[/tex]

So the answer for this case would be n=183 rounded up to the nearest integer

Hey there! :)

Answer:

a. 3

b. -22

c. -2

d. -2

e. 5a + 8

f. a² + 6a + 3

Step-by-step explanation:

Calculate the answers by substituting the values inside of the parenthesis for 'x':

a. f(1) = 5(1) - 2 = 3

b. f(-4) = 5(-4) - 2  = -22

c. g(-3) = (-3)² + 2(-3) - 5 = 9 - 6 - 5 = -2

d. g(1) = 1² + 2(1) - 5 = 1 + 2 -5 = -2

e. f(a+ 2) = 5(a+2) - 2 = 5a + 10 - 2 = 5a + 8

f. g(a + 2) = (a + 2)² + 2(a + 2) - 5 = a² + 4a + 4 + 2a + 4 - 5 =

a² + 6a + 3

Answer:

Minimum = 25

First quartile = 58

Second quartile = 72

Third quartile = 80

Maximum = 98

Step-by-step explanation:

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

Answer:

You're pretty sure that your candidate for class president has about 55% of the votes in the entire school. but you're worried that only 100 students will show up to vote. how often will the underdog (the one with 45% support) win? to find out, you set up a simulation.

a. describe-how-you-will-simulate a component.

b. describe-how-you-will-simulate a trial.

c. describe-the-response-variable

Step-by-step explanation:

Part A:

A component is one voter's voting. An outcome is a vote in favor of our candidate.

Since there are 100 voters, we can stimulate the component by using two random digits from 00 - 99, where the digits 00 - 64 represents a vote for our candidate and the digits 65 - 99 represents a vote for the under dog.

Part B:

A trial is 100 votes. We can stimulate the trial by randomly picking 100 two-digits numbers from 00 - 99.

And counted how many people voted for each candidate.  Whoever gets the majority of the votes wins the trial.

Part C:

The response variable is whether the underdog  wins or not.

To calculate the experimental probability, divide the number of trials in which the simulated underdog wins by the total number of trials.

Answer:

x = -9

Step-by-step explanation:

Step 1: Write out expression

5(x + 13) = 20

Step 2: Distribute

5x + 65 = 20

Step 3: Isolate x

5x = -45

x = -9

And we have our answer!

Answer:

-9

Step-by-step explanation:

Let the number be x.

5(x+13) = 20

Expand.

5x+65 = 20

Subtract 26 on both sides.

5x = 20 - 65

5x = -45

Divide 5 into both sides.

x = -45/5

x = -9

The number is -9.

Answer:

a. p=0.562

b. E = 0.0253

c. The 90% confidence interval for the population proportion is (0.537, 0.587).

d. We have 90% confidence that the interval (0.537, 0.587) contains the true value of the population proportion.

Step-by-step explanation:

We have to calculate a 90% confidence interval for the proportion.

The sample proportion is p=0.562.

[tex]p=X/n=583/1038=0.562[/tex]

The standard error of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.562*0.438}{1038}}\\\\\\ \sigma_p=\sqrt{0.000237}=0.0154[/tex]

The critical z-value for a 90% confidence interval is z=1.645.

The margin of error (MOE) can be calculated as:

[tex]MOE=z\cdot \sigma_p=1.645 \cdot 0.0154=0.0253[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=p-z \cdot \sigma_p = 0.562-0.0253=0.537\\\\UL=p+z \cdot \sigma_p = 0.562+0.0253=0.587[/tex]

The 90% confidence interval for the population proportion is (0.537, 0.587).

We have 90% confidence that the interval contains the true value of the population proportion.

Answer:

1.2kg

Step-by-step explanation:

6 identical toys weigh 1.8kg.

1 toy would weigh:

1.8/6 = 0.3

0.3 kg.

Multiply 0.3 with 4 to find how much 4 identical toys would weigh.

0.3 × 4 = 1.2

4 identical toys would weigh 1.2kg

Answer:

[tex]1.2kg[/tex]

Step-by-step explanation:

6 identical toys weigh = 1.8kg

Let's find the weight of 1 toy ,

[tex]1.8 \div 6 = 0.3[/tex]

Now, lets find the weigh of 6 toys,

[tex]0.3 \times 4 = 1.2kg[/tex]

Answer:

The number of employees classified into groups as shown below:

1 - 10: 3 6 (2companies)

11-20: 16 (1 company)

21-30: 25, 26, 27 (3 companies)

31-40: 34, 35, 35, 35, 36 (5 companies)

41-50: 41, 43, 48, 48 (4 companies)

Hope this helps!

Answer:

11-20 is 1

31-40 is 5

Step-by-step explanation:

Just count the amount

Hope that helps :D

Step-by-step explanation:

can u give image PlZzzzz ....

Answer:

Hey!

Your answer should be Y=2x+4

Step-by-step explanation:

Hope this helps!

Answer:

D.

Step-by-step explanation:

One slope is positive and one negative, so one line should go up and one down. B or D.

y = 1/2 x - 1 line goes up and y-int. = - 1.  Answer D.

y = - 1/2 x + 3 line goes up and y-int. = 3.  Answer D.

The missing part in the question;

and the payoff is $2.20 won for every dollar bet. (As the player’s probability of winning in this case is [tex]\dfrac{1}{4}[/tex]........

Also:

For such a bet, the casino pays off as shown in the following table.

The table can be shown as:

Keno Payoffs in 10 Number bets

Number of matches        Dollars won for each $1 bet

0  -   4                                        -1

5                                                  1

6                                                  17

7                                                  179

8                                                 1299

9                                                 2599

10                                               24999

Answer:

Step-by-step explanation:

Given that:

Twenty numbers are selected at random by the casino from the set of numbers 1 through 80

A player can select from 1 to 15 numbers; a win occurs if some fraction of the player’s chosen subset matches any of the 20 numbers drawn by the house

Let assume X to represent the numbers of player chooses which are in the Casino-selected-set of 20.

Let assume the random variable X has a hypergeometric distribution with parameters  N= 80 and m =20.

Then, the probability mass function of a hypergeometric distribution can be defined as:

[tex]P(X=k)=\dfrac{(^m_k)(^{N-m}_{n-k})}{(^N_n)}, k =1,2,3 ... n[/tex]

Now; the probability that i out of  n numbers chosen by the player among 20  can be expressed as:

[tex]P(X=k)=\dfrac{(^{20}_k)(^{60}_{n-k})}{(^{80}_n)}, k =1,2,3 ... n[/tex]

Also; given that ; When the player selects 2 numbers, a payoff (of odds) of $12 won for every $1 bet is made when both numbers are among the 20

So; n= 2; k= 2

Then :

Probability P ( Both number in the set 20)  [tex]=\dfrac{(^{20}_2)(^{60}_{2-2})}{(^{80}_2)}[/tex]

Probability P ( Both number in the set 20) [tex]= \dfrac{20*19}{80*79}[/tex]

Probability P ( Both number in the set 20) [tex]=\dfrac{19}{316}[/tex]

Probability P ( Both number in the set 20) [tex]=\dfrac{1}{16.63}[/tex]

Thus; the payoff odd for [tex]=\dfrac{1}{16.63}[/tex] is 16.63:1 ,as such fair payoff in this case is $16.63

Again;

Let assume X to represent the numbers of player chooses which are in the Casino-selected-set of 20.

Let assume the random variable X has a hypergeometric distribution with parameters  N= 80 and m =20.

The probability mass function of the hypergeometric distribution can be defined as :

[tex]P(X=k)=\dfrac{(^m_k)(^{N-m}_{n-k})}{(^N_n)}, k =1,2,3 ... n[/tex]

Now; the probability that i out of  n numbers chosen by the player among 20  can be expressed as:

[tex]P(n,k)=\dfrac{(^{20}_k)(^{60}_{n-k})}{(^{80}_n)}, k =1,2,3 ... n[/tex]

From the table able ; the expected payoff can be computed as shown in the attached diagram below. Thanks.

Answer:

The sample size for the data set = 56

Step-by-step explanation:

The sample size or number of individuals (n) is gotten from a histogram by summing up the total frequencies of occurrences.

In this example, the frequencies are: 2 4 6 8 10 12 14

Therefore, the sample size (n) is calculated as follows:

n = 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56

Therefore the sample size for the data set = 56

The sample size for the data set = 56

Given that,

The calculation is as follows:

= 2 + 4 + 6 + 8 + 10 + 12 + 14

= 56

Learn more: https://brainly.com/question/15622851?referrer=searchResults

Answer:

[tex] P(X>14)= 1-P(X<14) =1- F(14)[/tex]

And replacing we got:

[tex] P(X>14)= 1- \frac{14-8}{22-8}= 0.5714[/tex]

The probability that a randomly selected barber needs at least 14 minutes to complete the haircut is 0.5714

Step-by-step explanation:

We define the random variable of interest as x " time it takes a barber to complete a haircuts" and we know that the distribution for X is given by:

[tex] X \sim Unif (a= 8, b=22)[/tex]

And for this case we want to find the following probability:

[tex] P(X>14)[/tex]

We can find this probability using the complement rule and the cumulative distribution function given by:

[tex] P(X<x) = \frac{x-a}{b-a} ,a \leq x \leq b[/tex]

Using this formula we got:

[tex] P(X>14)= 1-P(X<14) =1- F(14)[/tex]

And replacing we got:

[tex] P(X>14)= 1- \frac{14-8}{22-8}= 0.5714[/tex]

The probability that a randomly selected barber needs at least 14 minutes to complete the haircut is 0.5714

Answer:

3m + 4c

Step-by-step explanation:

Whenever a word problem says the word earn that means the slope, also known as the rate of change, will be positive. Knowing this you can determine that both the caramel and milk chocolate slopes will be positive. After figuring all that out the only thing left to do is to make the equation. You know you have two slopes, and each slope needs a variable, so you will have to look back at the question. It is given that m represents the milk chocolate and c represents the caramel. Now all you have to do is make the slope the coefficient to the corresponding variable. The milk chocolates are 3 dollars, so the 3 goes in front of the m and the caramel chocolates are 4 dollars, so teh 4 goes in front of the 4. Since both slopes are positive no negatives or minus signs will be used in the equation. Knowing all this information you can now create the expression 3m + 4c.

Answer:

D

Step-by-step explanation:

3m + 4c

Answer:

The x-coordinate is the solution to x - 4 = 0, which is x = 4 and the y-coordinate is 3 so the answer is (4, 3).

Answer:

98

Step-by-step explanation:

Z as Zach; W as Wendy; L as Lee; C as Chen

We know that average score of Z,W, and L is 91, so:

(z + w + l)/3 = 91

z + w + l = 273

Average score W, L, C = 89, so:

(w + l + c)/3 = 89

w + l + c = 267

We take both:

(z + w + l) – (w + l + c) = 273 – 267

z – c = 6

Average score Z and C = 95

(z + c)/2 = 95

z + c = 190

(z + c) – (z – c) = 184

2c = 184

c = 92

z + c = 190

z + 92 = 190

z = 98

So, Zachs score is 98

Answer:

The probability that none of the households are tuned to 50 Minutes is 0.04398.

Step-by-step explanation:

We are given that the television show 50 Minutes has been successful for many years. That show recently had a share of 20, meaning that among the TV sets in use, 20% were tuned to 50 Minutes.

A pilot survey begins with 14 households have TV sets in use at the time of a 50 Minutes broadcast.

The above situation can be represented through binomial distribution;

[tex]P(X = r)= \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ;x = 0,1,2,3,.........[/tex]

where, n = number of samples (trials) taken = 14 households

r = number of success = none of the households are tuned to 50 min

p = probability of success which in our question is probability that households were tuned to 50 Minutes, i.e. p = 20%

Let X = Number of households that are tuned to 50 Minutes

So, X ~ Binom(n = 14, p = 0.20)

Now, the probability that none of the households are tuned to 50 Minutes is given by = P(X = 0)

               P(X = 0)  =  [tex]\binom{14}{0} \times 0.20^{0} \times (1-0.20)^{14-0}[/tex]

                              =  [tex]1 \times 1 \times 0.80^{14}[/tex]

                              =  0.04398