A player of a video game is confronted with a series of 3 opponents and a(n) 77% probability of defeating each opponent. Assume that the results from opponents are independent (and that when the player is defeated by an opponent the game ends).

Round your answers to 4 decimal places.

a. What is the probability that a player defeats all 3 opponents in a game?

b. What is the probability that a player defeats at least two opponents in a game?

c. If the game is played 2 times, what is the probability that the player defeats all 3 opponents at least once?

Answer:

[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]

So then the value of 205 it's 0.76 deviations below the population mean on this case

Step-by-step explanation:

For this case we have the following data given:

177, 205, 210, 210, 232, 205, 185, 185, 178, 210, 206, 212, 184, 174, 185, 242, 188, 212, 215, 247, 241, 223, 220, 260, 245, 259, 278, 270, 280, 295, 275, 285, 290, 272, 273, 280, 285, 286, 200, 215, 185, 230, 250, 241, 190, 260, 250, 302, 265, 290, 276, 228, 265

And these values represent the weigths of all the team members of the Arizona Cardinals

Now if we organize the data values from the smallest to the largest we have:

174 177 178 184 185 185 185 185 188 190 200 205 205 206 210 210 210 212 212 215 215 220 223 228 230  232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290  290 295 302

For this case we can calculate the mean with the following formula:

[tex] \mu = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And if we replace we got:

[tex] \mu = 233.3396[/tex]

And for the deviation we can use the following formula:

[tex] \sigma =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

And if we replace we got:

[tex] \sigma = 37.498[/tex]

And in order to calculate How many standard deviations above or below the mean was he we can use the z score formula given by:

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

And we assume that x=205 and if we replace we have:

[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]

So then the value of 205 it's 0.76 deviations below the population mean on this case

Answer:

  P(t) = 100(3.8^t)

Step-by-step explanation:

You want a function P(t) that describes the exponential population growth of a bacteria culture from 100 cells to 380 in one hour, where t is in hours.

The function can be written in the form ...

  population = (initial population) × (growth factor)^(t/(growth period))

Here, the initial population is 100, and the growth factor in a period of 1 hour is 380/100 = 3.8. Since we want t in hours, this is ...

  population = 100 × 3.8^(t/1)

  P(t) = 100(3.8^t)

The mean of the number of different sweaters worn in a week is 2.76. This is calculated by multiplying each possible outcome (from 1 to 4 different sweaters) by its probability and then adding these up.

The random variable X here is the number of different sweaters you wear in a week. To find the mean (expected value) of a random variable, you need to take each possible outcome, multiply it by its probability, and then add up all these products.

Here are the possibilities:

So, the expected value or mean of X is:

1*(4/1024) + 2*(192/1024) + 3*(432/1024) + 4*(96/1024) + 4*(24/1024) = 2.76

So on an average week, you would expect to wear about 2 to 3 different sweaters.

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Answer: the cost of a DVD is $8 and the cost of a video game is $33

Step-by-step explanation:

Let x represent the cost of a video game.

Let y represent the cost of a DVD.

2 video games and 3 DVDs cost $90.00. This is expressed as

2x + 3y = 90 - - - - - - - - - - - 1

1 video game and 2 DVDs cost $49.00. This is expressed as

x + 2y = 49 - - - - - - - - - - - 2

Multiplying equation 1 by 1 and equation 2 by 2, it becomes

2x + 3y = 90 - - - - - - - - - - - -3

2x + 4y = 98 - - - - - - - - - - - - 4

Subtracting equation 4 from equation 3, it becomes

3y - 4y = 90 - 98

- y = - 8

y = 8

Substituting y = 8 into equation 2, it becomes

x + 2 × 8 = 49

x + 16 = 49

x = 49 - 16

x = 33

By solving a system of linear equations, it is determined that the cost of a video game is $33 and the cost of a DVD is $8.

The question involves solving a system of linear equations to determine the cost of a video game and a DVD.

Let's denote the cost of one video game as V and the cost of one DVD as D. Based on the given information, we can set up the following two equations:

To solve for D, we can multiply the second equation by 2 and subtract it from the first equation:

2V + 3D - (2V + 4D) = 90 - 98

This simplifies to -D = -8, which means D = $8.

Now that we know the cost of a DVD, we can substitute it back into the second equation:

V + 2(8) = 49

V + 16 = 49

V = 49 - 16

V = $33.

Therefore, the cost of a video game is $33, and the cost of a DVD is $8.

sin 40° = 0.6428

Step-by-step explanation:

The complementary ratio of sine is cos and vice-versa.

sin θ = cos (90 - θ)

cos θ = sin (90 - θ)

So cos 50° = sin (90 - 50)°

                   = sin 40°

sin 40° = 0.6428

This indicates that the complement of cos 50° is sin 40° which is equal to 0.6428.

Final answer:

Andy's expected profit per game is $1.73 when playing the described gambling game.

Explanation:

To find Andy's expected profit per game, we need to calculate the probability and corresponding profit for each possible outcome. Let's create a probability model:

- Andy draws a number card (2-10): This has a probability of 36/52 since there are 36 number cards in a standard deck of 52 cards. The profit for this outcome is $0.

- Andy draws a face card: This has a probability of 12/52 since there are 12 face cards in a standard deck. The profit for this outcome is $3 if the coin lands on heads, $2 if it lands on tails.

- Andy draws an ace: This has a probability of 4/52 since there are 4 aces in a standard deck. The profit for this outcome is $5, except if he draws the ace of clubs, in which case the profit is $25.

To find the expected profit, we multiply each probability by the corresponding profit and sum them up:

(36/52) * $0 + (12/52) * (($3 * 1/2) + ($2 * 1/2)) + (4/52) * (($5 * 3/4) + ($25 * 1/4)) = $0 + $0.69 + $1.04 = $1.73

Therefore, the expected profit per game for Andy is $1.73.

Answer:

your... syllabus is different from mine....

Step-by-step explanation:

which class u study

Answer:

A) P=2.28%

B) P=28.57%

C) I expect 10 students to be unable to complete the exam in the alloted time.

Step-by-step explanation:

In order to solve this problem, we will need to find the respective z-scores. The z-scores are found by using the following formula:

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

Where:

z= z-score

x= the value to normalize

[tex]\mu = mean[/tex]

[tex]\sigma[/tex]= standard deviation

The z-score will help us find the area below the normal distribution curve, so in order to solve this problem we need to shade the area we need to find. (See attached picture)

A) First, we find the z-score for 60 minutes, so we get:

[tex]z=\frac{60-80}{10}=-2[/tex]

So now we look for the z-score on our normal distribution table. Be careful with the table you are using since some tables will find areas other than the area between the mean and the desired data. The table I used finds the area between the mean and the value to normalize.

so:

A=0.4772 for a z-score of -2

since we want to find the number of students that take less than 60 minutes, we subtract that decimal number from 0.5, so we get:

0.5-0.4772=0.0228

therefore the probability that a student finishes the exam in less than 60 minutes is:

P=2.28%

B) For this part of the problem, we find the z-score again, but this time for a time of 75 minutes:

[tex]z=\frac{75-80}{10}=-0.5[/tex]

and again we look for this z-score on the table so we get:

A=0.1915 for a z-score of -0.5

Now that we got this area we subtract it from the area we found for the 60 minutes, so we get:

0.4772-0.1915=0.2857

so there is a probability of P=28.57% of chances that the students will finish the test between 60 and 75 minutes.

C) Finally we find the z-score for a time of 90 minutes, so we get:

[tex]z=\frac{90-80}{10}=1[/tex]

We look for this z-score on our table and we get that:

A=0.3413

since we need to find how many students will take longer than 90 minutes to finish the test, we subtract that number we just got from 0.5 so we get:

0.5-0.3413=0.1586

this means there is a 15.86% of probabilities a student will take longer than 90 minutes. Now, since we need to find how many of the 60 students will take longer than the 90 available minutes, then we need to multiply the total amount of students by the percentage we previously found, so we get:

60*0.1586=9.516

so approximately 10 Students will be unavailable to complete the exam in the allotted time.

Answer:

Part A:

Probability is P(z<-2)=1-0.9772=0.0228

Part B:

P(-2<z<-0.5)=0.2857

Part C:

Number of students unable to complete the exam=60-50=10 students

Step-by-step explanation:

Part A:

Mean=μ=80 min

Standard Deviation=σ=10 min

Formula:

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

where:

[tex]\bar x=60\ min[/tex]

[tex]z=\frac{60-80}{10}\\ z=-2[/tex]

Probability is P(z<-2)

From the probability distribution tables (Cumulative Standardized normal distribution table)

Probability is P(z<-2)=1-0.9772=0.0228

Part B:

For 75 min:

[tex]z=\frac{75-80}{10}\\ z=-0.5[/tex]

For [tex]\bar x=60\ min[/tex]

[tex]z=\frac{60-80}{10}\\ z=-2[/tex]

From the probability distribution tables (Cumulative Standardized normal distribution table)

P(-2<z<-0.5)=P(z<-0.5)-P(z<-2)

P(-2<z<-0.5)=(1-0.6915)-(1-0.9772)

P(-2<z<-0.5)=0.2857

Part C:

[tex]\bar x=90\ min[/tex]

[tex]z=\frac{90-80}{10}\\ z=1[/tex]

From the probability distribution tables (Cumulative Standardized normal distribution table)

Probability is P(z<1)=0.8413

Number of students=0.8413*60

Number of students to complete the exam=50.478≅50

Number of students unable to complete the exam=60-50=10 students

Answer:

1) m=[1,4]

2) m=[-3,0,1]

Step-by-step explanation:

for y= e^(m*x) , then

y′=m*e^(m*x)

y′′=m²*e^(m*x)

y′′′=m³*e^(m*x)

thus

1) y′′+3y′−4y=0

m²*e^(m*x) + 3*m*e^(m*x) - 4*e^(m*x) =0

e^(m*x) *(m²+3*m-4) = 0 → m²+3*m-4 =0

m= [-3±√(9-4*1*(-4)] /2 → m₁=-4 , m₂=1

thus m=[1,4]

2) y′′′+2y′′−3y′=0

m³*e^(m*x) + 2*m²*e^(m*x) - 3*m*e^(m*x) =0

e^(m*x) *(m³+2*m²-3m) = 0 → m³+2*m²-3m=0

m³+2*m²-3m= m*(m²+2*m-3)=0

m=0

or

m= [-2±√(4-4*1*(-3)] /2  → m₁=-3 , m₂=1

thus m=[-3,0,1]

Answer: he would need 77 scoops of fertilizer.

Step-by-step explanation:

The measure of Horalco's garden is

10 1/2 ft by 5 1/2 ft. Converting to improper fraction, it becomes 21/2 ft by 11/2 ft. The garden is rectangular in shape and the formula for determining the area of a rectangle is expressed as

Area = length × width.

Therefore, area of the garden is

21/2 × 11/2 = 231/4 ft²

He needs 1 1/3 scoops of fertilizer for each square foot of the garden. Converting to improper fraction, it becomes 4/3 scoops. Therefore, the number of scoops of fertilizer that Horaclo need for the entire garden is

4/3 × 231/4 = 77 scoops of fertilizer.

Answer:

Step-by-step explanation:

Hello!

You have a bag with 181 peanuts in it.

65 of the shells contain one peanut. ⇒ 16 are cracked and 49 are complete.

111 of the shells contain two peanuts. ⇒ 30 are cracked 81 are complete.

5 of the shells have three peanuts and none of them is cracked.

A. What is the probability the shell is cracked?

To calculate the probability of the shell bein cracked you have to add all possible cases in which it is cracked and divide it by the total of peanuts in the bag:

[tex]P(Cracked)= \frac{(16+30)}{181}= 0.25[/tex]

B. What is the probability the shell is not cracked (Cr) and it contains two peanuts(2p)?

This probability is an intersection P(Cr∩2p), to calculate it you have to divide the total of peanuts that fulfill these characteristics and divide it by the total number of peanuts in the bag.

[tex]P(Crn2p)= \frac{30}{181}= 0.17[/tex]

C. What is the probability the shell is not cracked (Cr') or it contains two peanuts(2p)?

This probability is the union between two events, the event "cracked" and the event "two peanuts", symbolically: P(Cr'∪2p), these two events are not mutually exclusive, this means that they can happen at the same time, you have to apply the following formula to calculate it:

[tex]P(Cr'u2p)= P(Cr') + P(2p) - P(Cr'n2p)[/tex]

Since these events aren't mutually exclusive, some of them fulfill both categories, this means they are counted when you calculate the probability of "not cracked" and again when you calculate the probability of "2 peanuts" therefore you need to subtract their intersection to obtain the correct probability.

[tex]P(Cr'u2p)= \frac{(49+81+5)}{181} + (\frac{111}{181} ) - (\frac{81}{181} )= 0.76+0.61-0.45=0.92[/tex]

D. What is the probability the shell is not cracked given that it contains two peanuts?

This is a conditional probability, this means that both events are dependant and the occurrence of "2p" modifies the probability of the peanut shell to be "Cr'", symbolically: P(Cr'/2p) and you calculate it using the following formula:

[tex]P(Cr'/2p)= \frac{P(Cr'n2p)}{P(2p)}= \frac{0.45}{0.61}= 0.74[/tex]

I hope it helps!

A) The probability the shell is cracked is 25.4%.

B) The probability the shell is not cracked and it contains two peanuts is 44.75%.

C) The probability the shell is not cracked or it contains two peanuts is 91.16%.

D) The probability the shell is not cracked given that it contains two peanuts is 72.97%.

Since a bag of peanuts in their shells contains 181 peanuts. 65 of the shells contain one peanut, 111 of the shells contain two peanuts, and the rest contain three peanuts, and, of the shells with one peanut, 16 of them are cracked, and of the shells with two peanuts, 30 of them are cracked, while none of the shells with three peanuts are cracked, and one peanut shell is randomly selected from the bag, to determine A) what is the probability the shell is cracked; B) what is the probability the shell is not cracked and it contains two peanuts; C) what is the probability the shell is not cracked or it contains two peanuts; and D) what is the probability the shell is not cracked given that it contains two peanuts; the following calculations must be performed:

A)

B)

C)

D)

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Ansoogawer:

Step-by-step explanation:

Answer:

y = -3000x + 25,635

Step-by-step explanation:

well if the inital value of the car is $25,635 this means that when

x = 0   y = 25,635

(0 , 25635)

this will be our first point

Now if you tell us that in 1 year depreciate in value $3,000

this means thaw when

x = 1  y = 25,635 - 3,000

x = 1 y = 22,635

(1, 22635)

Now that we have 2 points we can have the equation

First we take the slope as follows

m = (y2 - y1) / (x2 - x1)

m = (22,635 - 25,635) / (1 - 0)

m = -3000 / 1

m = -3000

after calculating the slope we have to replace it in the following formula

(y - y1) = m (x - x1)

y - 25,635 = -3000 ( x - 0)

y - 25,635 = -3000x

y = -3000x + 25,635

Finally we replace the value of y by 10000

10,000 = -3000x + 25,635

10,000 - 25,635 = -3000x

-15,635 = -3000x

-15,635/-3000 = x

5.21167 = x       years

These are the years it would take for the value to be 10,000

to know the days we simply multiply by 365

5.21167 * 365 = 1902.26      days

Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x [tex]10^{-5}[/tex] .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, [tex]\sigma[/tex] = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm   and Sample size, n = 15

(a) Let Null Hypothesis, [tex]H_0[/tex] : Mean Diameter, [tex]\mu[/tex] = 8.2500 cm

 Alternate Hypothesis, [tex]H_1[/tex] : Mean Diameter,[tex]\mu[/tex] [tex]\neq[/tex] 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

                           [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows Standard Normal distribution

After putting each value, Test Statistics = [tex]\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }[/tex] = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172  and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x [tex]10^{-5}[/tex] .

(c) For constructing Two-sided confidence interval we know that:

    Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 <  [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P([tex]-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < [tex]1.96\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

P([tex]-Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]-\mu[/tex] < [tex]1.96\frac{\sigma}{\sqrt{n} }-Xbar[/tex] ) = 0.95

P([tex]Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]Xbar+1.96\frac{\sigma}{\sqrt{n} }[/tex]) = 0.95

So, 95% confidence interval for [tex]\mu[/tex] = [[tex]Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] , [tex]Xbar+1.96\frac{\sigma}{\sqrt{n} }[/tex]]

                                                        = [[tex]8.2535-1.96\frac{0.0020}{\sqrt{15} }[/tex] , [tex]8.2535+1.96\frac{0.0020}{\sqrt{15} }[/tex]]

                                                        = [8.2525 , 8.2545]

Here [tex]\mu[/tex] = mean diameter.

Therefore, 95% two-sided confidence interval on the mean diameter

           =  [8.2525 , 8.2545] .

Answer:

a

b

c

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Since the figure was not supplied, let's focus on that principle.

To calculate it simply subtract the area of the square minus the area of the circle. Given the side of the square 20 ft

1. Square Area

[tex]S= s^{2}\\S=20^{2} \Rightarrow S=400 ft^{2}[/tex]

2.Circle Area

Notice the radius is half the square side, i.e. 10 ft

[tex]S=\pi*R^{2}\\S=3.14*(10)^{2}\\S=314 \:ft^{2}\\[/tex]

Subtracting the area of the square and the area of the circle:

[tex]400-314=86 ft^{2}[/tex]

The area of the shaded is 86 feet².

Thus, option (B) is correct.

Let's assume the side length of the square is "s".

The area of the square is then given by s².

Substitute the side s = 20 into above formula as

Area of square = 20 x 20

                         = 400 square feet

Now, Area of circle = πr²

                                = 3.14 x (10)²

                                = 3.14 x 100

                                = 314 square feet.

Now, the area of the shaded region can be calculated as:

Area of shaded region = Area of square - Area of circle

                                      = 400 - 314

                                      = 86 feet²

Thus, option (B) is correct.

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Answer:

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.

Answer:

A) True

B) True

C) False

Step-by-step explanation:

Knowing that the collinear points are all those that pass through a line, we have:

A) given two points they form a line, by themselves they are collinear  (graph 1)

B) Can be or not can be (graph 2)

C) Can be not must be (graph 3)

We want to see if the given statements are true or false.

We will see that:

Two or more points are collinear if we can draw a line that connects them.

A) Whit that in mind, the first statement is true, 2 points is all we need to draw a line, thus two different points are always collinear, so the first statement is true.

B) For the second statement suppose you have a line already drawn, then you can draw 4 points along the line, if you do that, you will have 4 collinear points, so yes, 4 points can be collinear.

C) For the final statement, again assume you have a line, you used 2 points to draw that line (because two points are always collinear). Now you could have more points outside the line, thus, the set of all the points is not collinear (not all the points are on the same line).

So sets of 3 or more points can be collinear, but not "must" be collinear, so the last statement is false.

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Answer:

8

Step-by-step explanation:

Sin 30 = k/16

k = 16 x sin 30

k = 16 x (0.5) = 8

The largest intervals on which the existence and uniqueness theorem guarantees a unique solution for the given initial conditions.

The interval a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution depends on the initial conditions. For the given initial conditions:

Answer:

1/2

Step-by-step explanation:

Well if we think logically it is quite simple.

Since there is a 1-in-1 ratio of men and women, this means that there are the same number of men and women.

Now taking a random value such as 6 employees, this means that you will see 3 men and 3 women, now in relation to the total it is 3/6

3/6 = 1/2

Answer:

P ( disease | positive ) = 0.68

Step-by-step explanation:

This is a classic question of Bayes' Theorem, which calculates probability in a scenario where one event has already occured.

In this case, we have the following data:

P ( disease ) = 0.10

P (no disease) = 0.90

P ( positive | disease ) = 0.95

P ( positive | no disease) = 0.05

Formula to use:

[tex]P ( disease\ |\ positive ) = \frac{P(disease)\ P(positive\ |\ disease)}{P(disease)\ P(positive\ |\ disease) + P(no\ disease)\ P(positive\ |\ no\ disease)}[/tex]We substitute the values from our data in this formula:

[tex]\frac{(0.10) \times (0.95)}{(0.10)\times(0.95) + (0.90)\times(0.05)} \\\\\frac{0.095}{0.095+0.045}\\\\\frac{0.095}{0.14}\\ \\\\\=0.68[/tex] (Rounded off to two decimal places)

Hence, the probability of a person testing positive once they have the disease is 68%.

Answer:

Pencils cost $2

Markers cost $1

Erasers cost $0.25

Step-by-step explanation:

Let pencils be denoted by P, markers by M and erasers by E. The following linear system can be modeled from the data provided:

[tex]5P+7M+8E = 19\\7P+5M+8E=21\\4P+8M+7E =17.75[/tex]

Solving the linear system:

[tex]2P-2M= 2\\P=1+M\\4+4M+8M+7E =17.75\\\\12M+7E = 13.75\\12M+8E=14\\E=0.25\\M=\frac{14-0.25*8}{12}\\ M=1\\P=1+1=2[/tex]

Pencils cost $2

Markers cost $1

Erasers cost $0.25

Answer:

Step-by-step explanation:

Answer:

d. 4.6 years

Step-by-step explanation:

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.

In this problem, we have that:

[tex]\mu = 5.8, \sigma = 0.9[/tex]

What warranty should be provided so that the company is replacing at most 10% of their toasters sold?

Only those on the 10th percentile or lower will be replaced.

So the warranty is the value of X when Z has a pvalue of 0.10.

So it is X when [tex]Z = -1.28[/tex]

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

[tex]-1.28 = \frac{X - 5.8}{0.9}[/tex]

[tex]X - 5.8 = -1.28*0.9[/tex]

[tex]X = 4.6[/tex]

So the correct answer is:

d. 4.6 years

Final answer:

To find the warranty length, we need to determine the value that corresponds to the 10th percentile of the normally distributed life span. By solving the equation using the z-score formula, we find that the warranty should be provided for at least 4.6 years.

Explanation:

In order to determine the warranty that should be provided to the toasters, we need to find the value that corresponds to the 10th percentile of the normally distributed life span. To do this, we can use a standard normal distribution table or a calculator. The z-score for the 10th percentile is approximately -1.28. Using the formula z = (x - mean) / standard deviation, we can solve for x to find the warranty length.

-1.28 = (x - 5.8) / 0.9

x - 5.8 = -1.28 × 0.9

x - 5.8 = -1.152

x = 5.8 - 1.152

x = 4.648

Therefore, the warranty should be provided for at least 4.648 years, which is approximately 4.6 years. The correct answer is option d. 4.6 years.

The Possible outcomes and associated values in the sample space along with X are SSS (X = 3), SSF, SFS, FSS, SFF (X = 1), FSF, FFS,

FFF (X = 0).

We have,

Let's list all the possible outcomes in the sample space when three concrete beams are selected and their failure types (shear or flexure) are determined.

For each outcome, we'll also determine the value of X, which represents the number of beams that failed by shear.

Let S represent shear failure and F represent flexure failure.

Possible outcomes and associated values of X:

SSS (All three beams failed by shear)

X = 3

SSF (Two beams failed by shear, one by flexure)

X = 2

SFS (Two beams failed by shear, one by flexure)

X = 2

FSS (Two beams failed by shear, one by flexure)

X = 2

SFF (One beam failed by shear, two by flexure)

X = 1

FSF (One beam failed by shear, two by flexure)

X = 1

FFS (One beam failed by shear, two by flexure)

X = 1

FFF (All three beams failed by flexure)

X = 0

These are the eight possible outcomes in the sample space along with the associated values of X, representing the number of beams that failed by shear in each outcome.

Thus,

Possible outcomes and associated values of X: SSS (X = 3), SSF, SFS, FSS, SFF (X = 1), FSF, FFS, FFF (X = 0).

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The outcomes in the sample space for three concrete beams failing are: (S,S,S) with X = 3, (S,S,F),(S,F,S),(F,S,S) with X = 2, (S,F,F),(F,F,S),(F,S,F) with X = 1, and (F,F,F) with X = 0. S represents a shear failure, F a flexure failure, and X the number of shear failures.

The sample space for this problem includes all possible outcomes for the three concrete beams that can fail. The possible outcomes are:

Here, S represents a shear failure and F represents a flexure failure. The number specified by X in each scenario represents how many beams among the three randomly selected ones failed by shear.

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Answer:395.75lb

Step-by-step explanation:see attachment

Answer:

a) [tex] E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_1[/tex] is an unbiased estimator of [tex]\mu[/tex]

[tex] E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_2[/tex] is an unbiased estimator of [tex]\mu[/tex]

b) [tex] Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2} [/tex]

[tex] Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8} [/tex]

Step-by-step explanation:

For this case we know that we have two random variables:

[tex] X_1 , X_2[/tex] both with mean [tex]\mu = \mu[/tex] and variance [tex] \sigma^2[/tex]

And we define the following estimators:

[tex] \hat \theta_1 = \frac{X_1 + X_2}{2}[/tex]

[tex] \hat \theta_2 = \frac{X_1 + 3X_2}{4}[/tex]

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

[tex] E(\hat \theta_i) = \mu , i = 1,2 [/tex]

So let's find the expected values for each estimator:

[tex] E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})[/tex]

Using properties of expected value we have this:

[tex] E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_1[/tex] is an unbiased estimator of [tex]\mu[/tex]

For the second estimator we have:

[tex]E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})[/tex]

Using properties of expected value we have this:

[tex] E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu[/tex]

So then we conclude that [tex] \hat \theta_2[/tex] is an unbiased estimator of [tex]\mu[/tex]

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

[tex] Var(aX) = a^2 Var(X)[/tex]

For the first estimator we have:

[tex] Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})[/tex]

[tex] Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)] [/tex]

Since both random variables are independent we know that [tex] Cov(X_1, X_2 ) = 0[/tex] so then we have:

[tex] Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2} [/tex]

For the second estimator we have:

[tex] Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})[/tex]

[tex] Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)] [/tex]

Since both random variables are independent we know that [tex] Cov(X_1, X_2 ) = 0[/tex] so then we have:

[tex] Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8} [/tex]

Both θ₁^ and θ₂^ are unbiased estimators of μ. The variance of θ₁^ is σ² / 2, while the variance of θ₂^ is 5σ² / 8.

Let's analyze the provided estimators of the mean (">").

a) Unbiased Estimators

An estimator  heta is unbiased if E[θ] = μ. We have:

θ₁^ = (X₁ + X₂) / 2

The expected value of θ₁^ is:

E[θ₁^] = E[(X₁ + X₂) / 2] = (E[X₁] + E[X₂]) / 2 = (μ + μ) / 2 = μ

Thus, θ₁^ is an unbiased estimator of μ.

θ₂^ = (X₁ + 3X₂) / 4

The expected value of θ₂^ is:

E[θ₂^] = E[(X₁ + 3X₂) / 4] = (E[X₁] + 3E[X₂]) / 4 = (μ + 3μ) / 4 = 4μ / 4 = μ

Thus, θ₂^ is also an unbiased estimator of μ.

b) Variance of Each Estimator

The variance of an estimator θ is given by Var(θ). Considering the given variances of X₁ and X₂ (both σ²):

Var(θ₁^) = Var[(X₁ + X₂) / 2] = (1/2)²Var(X₁) + (1/2)²Var(X₂) = (1/4)σ² + (1/4)σ² = σ² / 2

Var(θ₂^) = Var[(X₁ + 3X₂) / 4] = (1/4)²Var(X₁) + (3/4)²Var(X₂) = (1/16)σ² + (9/16)σ² = (1/16 + 9/16)σ² = (10/16)σ² = 5σ² / 8

Thus, the variance of θ₁^ is σ² / 2 and the variance of θ₂^ is 5σ² / 8.

Answer:

15.9% of babies are born with birth weight under 6.3 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.8 pounds

Standard Deviation, σ = 0.5

We are given that the distribution of  birth weights is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(birth weight under 6.3 pounds)

P(x < 6.3)

[tex]P( x < 6.3) = P( z < \displaystyle\frac{6.3 - 6.8}{0.5}) = P(z < -1)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < -1) = 0.159 = 15.9\%[/tex]

15.9% of babies are born with birth weight under 6.3 pounds.

To find the proportion of babies born with a weight less than 6.3 pounds, we calculate the Z-score which results in -1. This Z-score corresponds to about 16% of the population in a standard normal distribution. Therefore, roughly 16% of babies are born weighing less than 6.3 pounds.

To solve this problem, you can use the properties of a normal distribution. In a normal distribution, scores fall within one standard deviation of the mean approximately 68% of the time, within two standard deviations about 95% of the time, and within three standard deviations about 99.7% of the time.

In this scenario, we would find the Z-score, a measure that describes a value's relationship to the mean of a group of values. The formula for the Z-score is (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

With a mean (μ) = 6.8 pounds, standard deviation (σ) = 0.5 pounds, and X = 6.3 pounds, the calculation for the Z-score would be (6.3-6.8)/0.5 = -1. This means that 6.3 pounds is one standard deviation below the mean. Referring to the standard normal distribution table, a Z-score of -1 corresponds to approximately 16% or 0.16 of the population. Therefore, approximately 16% of babies are born with a weight of less than 6.3 pounds.

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To find out how many inches must fall in December so that the average monthly precipitation for October, November, and December exceeds 3.11 inches, subtract the precipitation from October and November from the total precipitation. Therefore, 3.3 inches must fall in December.

To find out how many inches must fall in December so that the average monthly precipitation for October, November, and December exceeds 3.11 inches, we can use the formula for calculating average:

Average = (Total precipitation) / (Number of months)

Let's solve for the total precipitation:

Total precipitation = Average precipitation * Number of months

Total precipitation = 3.11 inches * 3 months = 9.33 inches

To find out how many inches must fall in December, we subtract the precipitation from October and November from the total precipitation:

December precipitation = Total precipitation - (October precipitation + November precipitation)

December precipitation = 9.33 inches - (2.98 inches + 3.05 inches) = 3.3 inches

Therefore, in order for the average monthly precipitation for these months to exceed 3.11 inches,

3.3 inches

must fall in December.

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Answer: 2

Step-by-step explanation:

16/9=1.7

=Approximately=2

Answer:

2 is the largest possible value.

Step-by-step explanation:

Given Data:

Length of one side       = 9

Length of second side = 16

Let AB and BC sides of triangle be of length 9 and 16 respectively as shown in figure attached.

Now, let sides AD and CF be the respective altitudes.

Also, ∠ABC = ∅ (as shown in figure)                                    

If AD and CF are the respective altitudes then,

we have

       AD = 9Sin∅ ;

       CF = 16Sin∅;

By dividing both sides, we get

       AD/CF = 9/16

This equation shows that is independant of angle ∅.

Now, let ∠BAC = α

Now we have, BE = 9 sinα

and                   FC = AC sinα

By dividing both sides, we get

       BE/FC=9/AC

Similarly we also have,

       BE/AD = 16/AC

ABC is a triangle as long as length of AC is ithin range of 8 to 24 i.e, 8≤AC≤24 (because sum of any two sides of triangle should be greater than length of third side)

Using these values we get ranges of:

9/24 ≤ BF/FC ≤ 9/8  ;  2/3 ≤ BE/AD ≤ 2

So,

2 is the largest possible value of the ratio of any two of these altitudes.

Answer:

a) -1.013

b) 2.32

c) -2

d) 0.9867

e) 2.9867                                              

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $775

Standard Deviation, σ = $75

We are given that the distribution of  average price of a laptop is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find z-score for corresponding

a. $699

[tex]z = \displaystyle\frac{699 - 775}{75} = -1.013[/tex]

b. $949

[tex]z = \displaystyle\frac{949 - 775}{75} = 2.32[/tex]

c. $625

[tex]z = \displaystyle\frac{625 - 775}{75} = -2[/tex]

d. $849

[tex]z = \displaystyle\frac{849 - 775}{75} = 0.9867[/tex]

e. $999

[tex]z = \displaystyle\frac{999 - 775}{75} = 2.9867[/tex]

a. The z-score for $699 is -1.20.

b. The z-score for $949 is 2.13.

c. The z-score for $625 is -2.33.

d. The z-score for $849 is 0.67.

e. The z-score for $999 is 3.13.

To calculate the z-score for each price, you can use the formula:

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

Where:

- [tex]\(Z\)[/tex] is the z-score.

- [tex]\(X\)[/tex] is the value you want to calculate the z-score for (the given price).

- [tex]\(\mu\)[/tex] is the mean (average) price of laptops, which is $775 in this case.

- [tex]\(\sigma\)[/tex]is the standard deviation of laptop prices, which is $75.

Now, let's calculate the z-scores for the given prices:

a. For $699: [tex]\(Z = \frac{699 - 775}{75} = -1.20\)[/tex]

b. For $949:[tex]\(Z = \frac{949 - 775}{75} = 2.13\)[/tex]

c. For $625: [tex]\(Z = \frac{625 - 775}{75} = -2.33\)[/tex]

d. For $849: [tex]\(Z = \frac{849 - 775}{75} = 0.67\)[/tex]

e. For $999: [tex]\(Z = \frac{999 - 775}{75} = 3.13\)[/tex]

The z-score measures how many standard deviations a particular value is away from the mean.

A positive z-score indicates a value above the mean, while a negative z-score indicates a value below the mean.

The absolute value of the z-score tells you how many standard deviations the value is from the mean.

In this context, it helps you understand how each laptop price compares to the average price in terms of standard deviations.

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Answer:

a) y= y₀+vy*t , x= x₀+vx*t

b) they are closest at t= 29.23 s

c) r min = 63.79 ft

Step-by-step explanation:

a) denoting v as velocities and "₀" as initial conditions , then the position of Sven is given by the coordinate (0,y) where

y= y₀+vy*t

and the position of Rudyard is given by the coordinate (x,0) where

x= x₀+vx*t

b) the distance r between Sven and Rudyard  is given by

r²=x²+y²

the distance will be minimum when the derivative of r with respect to the time is 0 . Then taking the derivative of the equation above

2*r*dr/dt = 2*x*dx/dt + 2*y*dy/dt

since dx/dt= vx and dy/dt= vy , then

r*dr/dt = x*vx+ y*vy

dr/dt = (x*vx+ y*vy)/r

assuming that r cannot be 0 , then

dr/dt =0 → x*vx+ y*vy = 0

(x₀+vx*t)*vx + (y₀+vy*t)*vy = 0

-(x₀*vx + y₀*vy) = (vx²+vy²)*t

t= -(x₀*vx + y₀*vy)/(vx²+vy²)

replacing values

t= -(x₀*vx + y₀*vy)/(vx²+vy²) = -[ 140 ft*(-6ft/s) + (-170 ft)*4 ft/s]/[ (-6ft/s)²+ (4 ft/s)²] = 29.23 s

then they are closest at t= 29.23 s

and the minimum distance will be

x = x₀ + vx*t = 140 ft+(-6ft/s)*29.23 s = -35.38 ft

y= y₀+vy*t = (-170 ft)+ 4 ft/s*29.23 s = -53.08 ft

r min = √(x²+y²)= 63.79 ft

r min = 63.79 ft

Note

to prove our assumption that r is not 0 , then x and y should be 0 at the same time. thus

0= y₀+vy*t → t = (-y₀)/vy = -140 ft/(-6ft/s) = 26.33 s

0= x₀+vx*t → t= (-x₀)/vx = -(-170 ft)/4 ft/s = 42.5 s

then r is never 0