A sales representative makes presentations about a product in three homes per day. In each home, there may be a sale (denote by S) or there may be no sale (denote by F).
Determine the sample space for the experiment.

Answer:

6.56% probability that a real emergency situation exists.

Step-by-step explanation:

We have these following probabilities:

A 0.4% probability that a real emergency situation exists.

A 99.6% probability that a real emergency situation does not exist.

If an emergency situation exists, a 95% probability that the alarm sounds.

If an emergency situation does not exist, a 2% probability that the alarm sounds.

The problem can be formulated as the following question:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem:

What is the probability of a real emergency situation existing, given that the alarm has sounded.

P(B) is the probability of there being a real emergency situation. So [tex]P(B) = 0.004[/tex].

P(A/B) is the probability of the alarm sounding when there is a real emergency situation. So P(A/B) = 0.95.

P(A) is the probability of the alarm sounding. This is 95% of 0.4%(real emergency situation) and 2% of 99.6%(no real emergency situation). So

P(A) = 0.95*0.04 + 0.02*0.996 = 0.05792

Given that the alarm has just sounded, what is the probability that a real emergency situation exists?

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.004*0.95}{0.05792} = 0.0656[/tex]

6.56% probability that a real emergency situation exists.

Answer:

A number to the third power would be a number cubed, so the answer is "Cubed".

Answer:

Option a

Step-by-step explanation:

Given that all of the seniors and none of the freshmen voted for it.

Some of the juniors and some of the sophomores voted for the proposal and some voted against it.

If the proposal was to be adopted then a simple of majority of votes should have been cast.

Since all seniors voted for it, and no freshmen number of freshmen has to be less than seniors then only majority would have voted.

Regarding juniors and sophomores we assume some voted and some against thus approximately nullifying their votes.

So the correct option would be

a. There are more seniors than freshmen on the council.

Option b wrong since only some voted for it.

C is wrong because actual seniors were not given.

D is wrong because while seniors voted for sophomores voted against.

e is wrong since sophomores and freshmen nullified each other

Answer with Step-by-step explanation:

We are given that

[tex]yx+9y=0[/tex]

[tex]-4x+5y=3[/tex]

We have to find the angle between the lines.

[tex]y(x+9)=0[/tex]

[tex]y=0,x+9=0\implies x=-9[/tex]

[tex]y=0[/tex]..(1)

[tex]x=-9[/tex]..(2)

[tex]-4x+5y=3[/tex]..(3)

The angle between two lines

[tex]a_1x+b_1y+c_1=0[/tex]

[tex]a_2x+b_2y+c_2=0[/tex]

[tex]tan\theta=\mid \frac{a_1b_2-b_1a_2}{a_1a_2+b_1b_2}\mid[/tex]

By using the formula the angle between equation (1) and equation (2) is given by

[tex]tan\theta_1=\mid\frac{0\times 0-1\times 1}{0+0}\mid=\infty=90^{\circ}[/tex]degree

[tex]tan90^{\circ}=\infty[/tex]

It is not possible because we are given that the acute angle between intersecting lines that do not cross at right angles is same as the angle determined by vectors that either normal to the lines or parallel to lines.

By using the formula the angle between equation (2) and equation(3)

[tex]tan\theta_2=\mid\frac{1(5)-0(4)}{-4(1)+5(0)}\mid=\frac{5}{4}[/tex]

[tex]\theta_2=tan^{-1}(1.25)[/tex] degree

By using the formula the angle between equation (3) and equation(1)

[tex]tan\theta_3=\mid\frac{-4(1)-5(0)}{-4(0)+5(1)}\mid=\frac{4}{5}[/tex]

[tex]\theta_3=tan^{-1}(\frac{4}{5})[/tex]degree

Answer:

The answer to your question is 3.35 ft

Step-by-step explanation:

Data

height = 19 ft

angle = 80°

Process

1.- It is formed a right triangle so use a trigonometric function that relates the opposite side and the adjacent side. This trigonometric function is tangent.

               tan Ф = Opposite side/adjacent side

               adjacent side = Opposite side / tan Ф

               adjacent side = 19 / tan 80

               adjacent side = 19 / 5.67

               adjacent side = 3.35 ft

Answer: The base of the latter needs to be positioned 3.6 feet out from the building.

Step-by-step explanation:

The ladder forms a right angle triangle with the building and the ground. The length of the ladder represents the hypotenuse of the right angle triangle. The height from the where the top of the ladder touches the window to the base of the building represents the opposite side of the right angle triangle.

The distance from the bottom of the ladder to the base of the building represents the adjacent side of the right angle triangle.

To determine distance,h from the bottom of the ladder to the base of the building, we would apply

the tangent trigonometric ratio.

Tan θ = opposite side/adjacent.

Tan 80 = 19/h

h = 19/Tan 80 = 19/5.6713

h = 3.6 feet

Answer:

Step-by-step explanation:

The formula for finding the sum of the measure of the interior angles in a regular polygon is expressed as (n - 2) × 180. Therefore,

(n - 2) × 180 = 9000

180n - 360 = 9000

180n = 9000 + 360 = 9360

n = 9360/180

n = 52

The regular polygon has 52 sides

7) The sum of the angles in the quadrilateral is 360°. Let x represent the missing angle. Therefore,

64 + 116 + 120 + x = 360

300 + x = 360

x = 360 - 300

x = 60°

8a) let x represent the missing side. Therefore,

24/15 = x/10

Cross multiplying,

15x = 240

x = 240/15 = 16

8b) let x represent the missing side. Therefore,

6/12 = 5/x

6x = 60

x = 60/6 = 10

a. The probability is 0.30. The advertisement seems to have a positive effect on customer purchases.

b. The market share is 0.20. The advertisement could potentially increase the company's market share

c. The conditional purchase probability of 0.333. The second advertisement seems to have had a slightly bigger effect on customer purchases.

a. We are asked to find the probability of an individual purchasing the product given that the individual recalls seeing the advertisement, i.e., P(B|S).

Using the formula for conditional probability:

P(B|S) = P(B∩S) / P(S)

Given:

P(B∩S) = 0.12

P(S) = 0.40

So, P(B|S)

= 0.12 / 0.40

= 0.30

Seeing the advertisement increases the probability that an individual will purchase the product from 0.20 (P(B)) to 0.30 (P(B|S)). Therefore, the advertisement seems to have a positive effect on customer purchases.

As a decision maker, if the cost of the advertisement is reasonable, it would be recommended to continue the advertisement since it increases the likelihood of product purchases.

b. The company's market share can be estimated by considering the probability of individuals purchasing the company's soap product and the probability of individuals purchasing from competitors.

The probability of an individual purchasing from competitors

= 1 - P(B)

= 1 - 0.20

= 0.80

Therefore, the company's market share is:

Market Share = P(B)

                      = 0.20

Continuing the advertisement could potentially increase the company's market share since the advertisement has a positive effect on customer purchases, as shown in part (a).

c. For the other advertisement:

Given:

P(S) = 0.30

P(B∩S) = 0.10

Using the formula for conditional probability:

P(B|S) = P(B∩S) / P(S)

So, P(B|S)

= 0.10 / 0.30

= 0.333

Comparing the two advertisements, the first advertisement had a conditional purchase probability of 0.30, while the second advertisement had a conditional purchase probability of 0.333. Therefore, the second advertisement seems to have had a slightly bigger effect on customer purchases.

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The advertisements increase the chance for an individual to purchase the product. The second advertisement yields a higher likelihood of purchase, implying it's more effective.

The problem is related to conditional probability. To solve the question:

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Answer:

a.) 0.4565

b.) 0.8656

c.) 0.4615

Step-by-step explanation:

We solve this using the probability distribution formula of combination.

nCr * p^r * q^n-r

Where

n = number of trials

r = successful trials

probability of success = p = 77% =0.77

Probability of failure= q = 1-0.77 = 0.23

a.) When exactly 3 opponents are defeated, When n = 3 and r = 3, probability becomes:

= 3C3 * 0.77³ * 0.23^0

= 1 * 0.456533 * 1

= 0.456533 = 0.4565 (4.d.p)

b.) When at least 2 opponents are defeated, that is when r = 2 and when r = 3,

When r = 2, probability becomes:

= 3C2 * 0.77² * 0.23¹

= 3 * 0.5929 * 0.23

= 0.409101

When 3 opponents are defeated, we calculated it earlier to be 0.456533

Hence, probability that at least 2 opponents are defeated

= 0.409101 + 0.456533

= 0.865634 = 0.8656(2.d.p)

c.) If 2 games are played, probability he defeat all 3 at least once in the game will be the sum (probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game) + (probability of defeating all three opponents in both games)

Probability of defeating all three opponents in the first game = 0.456533

Probability of not defeating all three opponents in the second game = 1 - 0.456533 = 0.543467

Hence ,

probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game = 0.465633 * 0.543467 = 0.253056

probability of defeating all three opponents in both games

= 0.456533 * 0.456533

=0.208422

Probability he defeats all three opponents at least once in 2games

= 0.253056 + 0.208422

=0.461478 = 0.4615(4.d.p)

To find the angle of intersection (θ) between the curves r1(t) and r2(t) at the origin, we calculate the dot product of their tangent vectors and use the arccosine formula. θ ≈ 79 degrees.

To find the angle of intersection (θ) between the curves r1(t) = (3t, [tex]t^2[/tex], [tex]t^4[/tex]) and r2(t) = (sin(t), sin(2t), 5t) at the origin, we can use the dot product formula for angles between vectors.

First, we need to calculate the tangent vectors at the origin for both curves. The tangent vector for r1(t) is (3, 2t, [tex]4t^3[/tex]), and for r2(t), it is (cos(t), 2cos(2t), 5).

Next, evaluate these vectors at t = 0 (the origin) to get the tangent vectors at the point of intersection: r1'(0) = (3, 0, 0) and r2'(0) = (1, 2, 5).

Now, calculate the dot product of these vectors:

r1'(0) · r2'(0) = (3 × 1) + (0 × 2) + (0 × 5) = 3.

The magnitude of r1'(0) is [tex]\sqrt{ (3^2 + 0^2 + 0^2)[/tex] = 3, and the magnitude of r2'(0) is [tex]\sqrt{(1^2 + 2^2 + 5^2[/tex]) = [tex]\sqrt{(1 + 4 + 25)[/tex] = √30.

Now, use the dot product formula for angles:

cos(θ) = (r1'(0) · r2'(0)) / (|r1'(0)| ×|r2'(0)|)

cos(θ) = 3 / (3 × [tex]\sqrt{30}[/tex]) = 1 / [tex]\sqrt30}[/tex]

Now, find θ:

θ = arc cos(1 / [tex]\sqrt{30[/tex])

Using a calculator, θ ≈ 79 degrees (rounded to the nearest degree).

So, the angle of intersection θ is approximately 79 degrees.

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Answer:

[tex]t=\frac{11-8}{\frac{6}{\sqrt{9}}}=1.5[/tex]    

[tex]p_v =P(t_{(8)}>1.5)=0.086[/tex]  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, we can conclude that the mean is higher than 8 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=11[/tex] represent the mean height for the sample  

[tex]s=6[/tex] represent the sample standard deviation for the sample  

[tex]n=9[/tex] sample size  

[tex]\mu_o =8[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.   (assumed)

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 8, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 8[/tex]  

Alternative hypothesis:[tex]\mu > 8[/tex]  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{11-8}{\frac{6}{\sqrt{9}}}=1.5[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=9-1=8[/tex]  

Since is a one side upper test the p value would be:  

[tex]p_v =P(t_{(8)}>1.5)=0.086[/tex]  

Conclusion  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, we can conclude that the mean is higher than 8 at 5% of signficance.  

To determine if the population standard deviation is known, we can use the formula for the standard error of the mean (SEM) and use the t-distribution for a sample size of 9.

To determine if the population standard deviation LaTeX: \sigma\sigma is known, we can use the formula for the standard error of the mean (SEM):

SE = \frac{s}{\sqrt{n}}

If the sample size is less than or equal to 30, we can use the t-distribution to find the critical value for a given level of significance. If the sample size is greater than 30, we can use the z-distribution. In this case, since the sample size is 9, the t-distribution should be used.

Thus, with a sample size of 9 and the population standard deviation not known, \sigma\sigma is not known.

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Answer:

a(n)=14+6(n-1)

Step-by-step explanation:

The first term is 14. This would be an arithmetic sequence, so you will add 6 to every term: the common difference is 6.

14+6= 20

20+6=26

You have the formula a(n)= a(1)+d(n-1)

a(n)= 14+6(n-1)

14 for the first term, 6 for the common difference.

Answer:

I think 25000

Step-by-step explanation:

Answer:

Therefore the measurement of EF,

[tex]EF=1.98\ units[/tex]

Step-by-step explanation:

Given:

In Right Angle Triangle DEF,

m∠E=90°

m∠D=26°

∴sin 26 ≈ 0.44

DF = Hypotenuse = 4.5    

To Find:

EF = ? (Opposite Side to angle D)

Solution:

In Right Angle Triangle DEF, Sine Identity,

[tex]\sin D= \dfrac{\textrm{side opposite to angle D}}{Hypotenuse}\\[/tex]

Substituting the values we get

[tex]\sin 26= \dfrac{EF}{DF}=\dfrac{EF}{4.5}[/tex]

Also,  sin 26 ≈ 0.44 .....Given

[tex]EF = 4.5\times 0.44=1.98\ units[/tex]

Therefore the measurement of EF,

[tex]EF=1.98\ units[/tex]

Answer:

a)  D_u .  ∀ f ( 2 , 1 ) = 3 , Q = 72 degrees

b) D_u .  ∀ f ( 2 , 1 ) = 4*sqrt(2) , Q = 80 degrees

c) D_u .  ∀ f ( 2 , 1 ) = - sqrt(2) , Q = -54.74 degrees

d) u = 1 / sqrt(34) *(-3i +5j) ,  D_u .  ∀ f ( 2 , 1 )  = +/- sqrt (34) , Q = 80.27 degrees

Step-by-step explanation:

Given:

- The terrain is modeled as a surface:

                                   f(x , y) = x*y + y^3 - x^2

At point P( 2 , 1 , -1 ) is the current position:

Find:

(a) Determine the slope you would encounter if you headed due West from your position. What angle of inclination does this correspond to?

(b) Determine the slope you would encounter if you headed due North-West from your position. What angle of inclination does this correspond to?

(c) Determine the slope you would encounter if you headed due South-West from your position. What angle of inclination does this correspond to?

(d) Determine the steepest slope you could encounter from your position and the direction of that slope (as a unit vector).

Solution:

- We will compute the gradient of the vector ∀ f @ point P( 2 , 1 , -1 ) as follows:

                                ∀ f ( x , y ) = (y - 2x ) i + ( x + 3y^2) j

- Evaluate at point P ( 2 , 1 ):

                               ∀ f ( 2 , 1 ) = (1 - 2*2 ) i + ( 2 + 3*1^2) j    

                               ∀ f ( 2 , 1 ) = -3 i + 5 j        

- We will use the result of ∀ f @ point P( 2 , 1 , -1 ) for the all the parts.

a)

- The direction due west can be written as a unit vector u_w = - i .

- Now compute the directional derivative in the direction of u_w = - i

                                D_u .  ∀ f ( 2 , 1 ) =-3 i + 5 j . - i

                                D_u .  ∀ f ( 2 , 1 ) = 3

- Now compute the angle of inclination Q for the following direction:

                                Q = arctan(3) = 71.57 = 72 degrees

 Hence, along the direction due west from position we ascend with an inclination of approximately 72 degrees.

b)

- The direction due north-west can be written as a unit vector u_w = - i + j .

- Now compute the directional derivative in the direction of u_w = - i + j

                                D_u .  ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i + j) / sqrt(2)

                                D_u .  ∀ f ( 2 , 1 ) = 8 / sqrt(2) = 4*sqrt(2)

- Now compute the angle of inclination Q for the following direction:

                                Q = arctan(4*sqrt(2)) = 79.98 = 80 degrees

 Hence, along the direction due north-west from position we ascend with an inclination of approximately 80 degrees.    

c)

- The direction due south-west can be written as a unit vector u_w = - i - j .

- Now compute the directional derivative in the direction of u_w = - i - j

                                D_u .  ∀ f ( 2 , 1 ) =-3 i + 5 j . (- i - j) / sqrt(2)

                                D_u .  ∀ f ( 2 , 1 ) = -2 / sqrt(2) = - sqrt(2)

- Now compute the angle of inclination Q for the following direction:

                                Q = arctan(-sqrt(2)) = -54.74 degrees

 Hence, along the direction due south-west from position we descend with an inclination of approximately 55 degrees.    

d)

- We know that ∀ f ( 2 , 1 ) gradient points in the direction greatest increase, hence, So from P the direction of greatest increase is  ∇f(P) = -3i +5j. The unit vector pointing in this direction is:

                                  u = 1 / sqrt(34) *(-3i +5j)

- so we have:

                     D_u .  ∀ f ( 2 , 1 ) =  u . ∀ f ( 2 , 1 ) = (-3i +5j) . (-3i +5j) / sqrt(34)

                     = +/- sqrt (34)

- Hence, the steepest ascent or decent is of :

                     Q = arctan ( sqrt(34)) = 80.27 degrees

Answer:

0.9688

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either it is heads, or it is tails. The probabilities for each coin toss are independent from each other. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

For each time the coin is tossed, heads or tails are equally as likely, since the coin is fair. So [tex]p = \frac{1}{2} = 0.5[/tex]

You toss a fair coin 5 times. What is the probability of at least one head?

Either there are no heads, or there is at least one head. The sum of the probabilities of these events is decimal 1. So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

We want to find [tex]P(X \geq 1)[/tex], when [tex]n = 5[/tex].

So

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

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

[tex]P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.03125 = 0.96875[/tex]

Rounding to the nearest ten-thousandth(four decimal places), this probability is 0.9688.

The probability of getting at least one head when tossing a fair coin 5 times is 31/32 or 0.96875 when rounded to the nearest ten-thousandth.

The subject of this question is probability, a field within Mathematics. To answer your question: the probability of getting at least one head when tossing a fair coin 5 times can be found by calculating the probability of not getting a head (which is tossing tails 5 times in a row) and then subtracting that from 1 (representing certainty). Each toss of the fair coin has two outcomes, heads or tails, with equal probability of 1/2. If you toss the coin 5 times, the total number outs comes is 2^5, or 32. The chance of getting all tails is (1/2)^5, which is 1/32.

So, the probability of not getting a head (only tails) in 5 tosses is 1/32. Subtract this from 1 to find the probability of getting at least one head: This equals 1 - 1/32 = 31/32 = 0.96875, when rounded to the nearest ten-thousandth.

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Answer:

a) F(t) = R[P(t)]

b) the output units of the new function = F(t) in dollars per gallon

Step-by-step explanation:

a) There are two function R(t) which shows the average price in dollars of a gallon of regular unleaded gasoline and P(t) which shows the purchasing power of the dollar as measured by consumer prices based on 2010 dollars.

To write the function which gives the rice of gasoline in constant 2010 dollars ;

From the analysis , this is an example of a composition of function as such the relationship =

F(t) = R[P(t)]

b) the output units of the new function = F(t) in dollars per gallon

This shows that the value of F(t) is the dependent variable

Final answer:

To determine the real price of gasoline in constant 2010 dollars, combine the functions R(t) and P(t) using the formula R(t) ÷ P(t). This calculation adjusts the nominal price of gasoline for inflation, resulting in the price of gasoline in terms of 2010 dollars, with the output units being 2010 dollars per gallon.

Explanation:

To combine the two functions representing the average price of a gallon of regular unleaded gasoline, R(t), and the purchasing power of the dollar as measured by consumer prices based on 2010 dollars, P(t), into a new function giving the price of gasoline in constant 2010 dollars, we use the formula:

R(t) ÷ P(t)

This formula represents the price of gasoline adjusted for inflation, giving us the real price of gasoline in terms of 2010 dollars. Here, R(t) gives the average price of gasoline in year t, and P(t) gives the purchasing power of the dollar in year t, compared to 2010 dollars. By dividing R(t) by P(t), we adjust the nominal price of gasoline to reflect its real value, accounting for changes in the purchasing power of the dollar over time.

The output units of this new function would be 2010 dollars per gallon. This metric allows economists and analysts to compare the price of gasoline across different years on a level playing field, eliminating the effects of inflation.

Answer:

1) 41.67% probability that dice A is larger than dice B.

2) Given hat the roll resulted in a sum of 5 or less, there is a 20% conditional probability that the two dice were equal.

3) Given that the two dice land on different numbers there is a 26.67% conditional probability that the two dice differed by 2.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have these possible outcomes:

Format(Dice A, Dice B)

(1,1), (1,2), (1,3), (1,4), (1,5),(1,6)

(2,1), (2,2), (2,3), (2,4), (2,5),(2,6)

(3,1), (3,2), (3,3), (3,4), (3,5),(3,6)

(4,1), (4,2), (4,3), (4,4), (4,5),(4,6)

(5,1), (5,2), (5,3), (5,4), (5,5),(5,6)

(6,1), (6,2), (6,3), (6,4), (6,5),(6,6)

There are 36 possible outcomes.

1) Find the probability that dice A is larger than dice B.

Desired outcomes:

(2,1)

(3,1), (3,2)

(4,1), (4,2), (4,3)

(5,1), (5,2), (5,3), (5,4)

(6,1), (6,2), (6,3), (6,4), (6,5)

There are 15 outcomes in which dice A is larger than dice B.

There are 36 total outcomes.

So there is a 15/36 = 0.4167 = 41.67% probability that dice A is larger than dice B.

2) Given that the roll resulted in a sum of 5 or less, find the conditional probability that the two dice were equal.

Desired outcomes:

Sum of 5 or less and equal

(1,1), (2,2)

There are 2 desired outcomes

Total outcomes:

Sum of 5 or less

(1,1), (1,2), (1,3), (1,4)

(2,1), (2,2), (2,3)

(3,1), (3,2)

(4,1)

There are 10 total outcomes.

So given hat the roll resulted in a sum of 5 or less, there is a 2/10 = 20% conditional probability that the two dice were equal.

3) Given that the two dice land on different numbers, find the conditional probability that the two dice differed by 2.

Desired outcomes

Differed by 2

(1,3), (2,4), (3,1), (3,5),(4,2),(4,6), (5,3), (6,4).

There are 8 total outcomes in which the dices differ by 2.

Total outcomes:

There are 30 outcomes in which the two dice land of different numbers.

So given that the two dice land on different numbers there is a 8/30 = 0.2667 = 26.67% conditional probability that the two dice differed by 2.

Answer:

See explanation below.

Step-by-step explanation:

When we want to fit a linear model given by:

[tex] y = \beta_0 + \beta_1 x[/tex]

Where y is a vector with the observations of the dependent variable, [tex]\beta_0 , \beta_1 [/tex] the parameters of the model and x the vector with the observations of the independent variable.

For this case this population regression function represent the conditional mean of the variable Y with values of X constant. And since is a population regression the parameters are not known, for this reason we use the sample data to obtain the sample regression in order to estimate the parameters of interest [tex] \beta_0, \beta_1[/tex]

We can use any method in order to estimate the parameters for example least squares minimizing the difference between the fitted and the real observations for the dependenet variable.  After we find the estimators for the regression model then we have a model with the estimated parameters like this one:

[tex] \hat y = \hat b_0 +\hat b_1 x[/tex]

With [tex] \hat \beta_0 = b_o , \hat \beta_1 = b_1[/tex]

And this model represent the sample regression function, and this equation shows to use the estimated relation between the dependent and the independent variable.

Answer:

6 fl. oz

Step-by-step explanation:

The most accurate measurement for the volume a teacup can hold is 6 fluid ounces.

To determine the most accurate measurement for the amount that a teacup can hold, we need to consider the typical volume of a teacup and understand the relationship between different units of capacity. A standard teacup holds about 6 fluid ounces. Given the unit conversion factors, we know that:

1 cup = 8 fluid ounces

1 pint = 2 cups = 16 fluid ounces

1 quart = 2 pints = 32 fluid ounces

1.5 quarts = 48 fluid ounces

Considering these units:

6 fluid ounces is less than 1 cup (8 fl oz).

2 cups equal 16 fluid ounces, which is more than a typical teacup can hold.

1 pint (16 fluid ounces) is also more than what a teacup can hold.

1.5 quarts (48 fluid ounces) is significantly more than a teacup's capacity.

Therefore, the most accurate measurement to describe the amount a teacup can hold is 6 fluid ounces.

Answer:

a) [tex] \hat p =\frac{X}{n}=\frac{1760}{2000}=0.88[/tex]

b) [tex]ME=1.64* \sqrt{\frac{0.88*(1-0.88)}{2000}}=0.0119[/tex]

c) [tex]0.88 - 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.868[/tex]

[tex]0.88 + 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.892[/tex]

And the 90% confidence interval would be given (0.868;0.892).

d) [tex]0.88 - 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8658[/tex]

[tex]0.88 + 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8942[/tex]

And the 90% confidence interval would be given (0.8658;0.8942).

Step-by-step explanation:

Part a

For this case the point of estimate for the population proportion is given by:

[tex] \hat p =\frac{X}{n}=\frac{1760}{2000}=0.88[/tex]

Part b

The confidence interval would be given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.64[/tex]

The margin of error is given by:

[tex] ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

And if we replace we got:

[tex]ME=1.64* \sqrt{\frac{0.88*(1-0.88)}{2000}}=0.0119[/tex]

Part c

And replacing into the confidence interval formula we got:

[tex]0.88 - 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.868[/tex]

[tex]0.88 + 1.64 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.892[/tex]

And the 90% confidence interval would be given (0.868;0.892).

Part d

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.96[/tex]

[tex]0.88 - 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8658[/tex]

[tex]0.88 + 1.96 \sqrt{\frac{0.88(1-0.88)}{2000}}=0.8942[/tex]

And the 90% confidence interval would be given (0.8658;0.8942).

The Main Answer for:

a. The point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern is approximately [tex]0.88[/tex].

b. The margin of error at [tex]90[/tex]% confidence is approximately [tex]0.0120[/tex].

c. The [tex]90[/tex]% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is approximately ([tex]0.868, 0.892[/tex]).

d. The [tex]95[/tex]% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is approximately ([tex]0.8657, 0.8943[/tex]).

a. Point Estimate: The point estimate of the population proportion can be calculated by dividing the number of respondents who think the future health of Social Security is a major economic concern by the total number of respondents surveyed.

Point Estimate = Number of respondents concerned about Social Security / Total number of respondents

Given: Number of respondents concerned about Social Security = [tex]1760[/tex]

Total number of respondents surveyed = [tex]2000[/tex]

Point Estimate = [tex]1760 / 2000 =0.88[/tex]

So, the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern is approximately [tex]0.88[/tex].

b. Margin of Error: The margin of error (E) can be calculated using the formula:

[tex]E=z*\sqrt{p(1-p)/n}[/tex]

where:

• z is the z-score corresponding to the desired confidence level

• p is the point estimate of the population proportion

• n is the sample size

Since we are aiming for a 90% confidence interval, we find the z-score corresponding to a 90% confidence level, which is approximately 1.645 (you can find this value in a standard normal distribution table).

[tex]E=1.645*\sqrt{0.88(1-0.88)/ 2000} \\E=1.645*\sqrt{0.88*0.12/2000} \\E=1.645*\sqrt{0.1056/2000} \\E=1.645*\sqrt{0.0000528} \\E=1.645*0.00727\\E=0.01196[/tex]

So, the margin of error at [tex]90[/tex]% confidence is approximately [tex]0.0120[/tex].

c. Confidence Interval (90%): The confidence interval can be calculated using the point estimate and the margin of error.

[tex]CI=(p-E,p+E)\\CI=(0.88-0.0120,0.88+0.0120)\\CI=(0.868,0.892)[/tex]

So, the [tex]90[/tex]% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is approximately ([tex]0.868, 0.892[/tex]).

d. Confidence Interval (95%): To calculate the 95% confidence interval, we use the same formula but with a different z-score. For a 95% confidence level, the z-score is approximately 1.96.

[tex]E=1.96*\sqrt{0.88(1-0.88)/ 2000} \\E=1.96*\sqrt{0.0000528} \\E=1.96*0.00727\\E=0.01427\\CI=(0.88-0.0143,0.88+0.0143)\\CI=(0.8657,0.8943)\\[/tex]

So, the [tex]95[/tex]% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is approximately ([tex]0.8657, 0.8943[/tex]).

COMPLETE QUESTION:

The Consumer Reports National Research Center conducted a telephone survey of [tex]2000[/tex] adults to learn about the major economic concerns for the future (Consumer Reports, January [tex]2009[/tex]). The survey results showed that [tex]1760[/tex] of the respondents think the future health of Social Security is a major economic concern.

a. What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern (to 2 decimals)?

b. At [tex]90[/tex]% confidence, what is the margin of error (to 4 decimals)?

c. Develop a [tex]90[/tex]% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern (to 3 decimals).(, )

d. Develop a [tex]95[/tex]% confidence interval for this population proportion (to 4 decimals).

Answer:

813 will exceed 9 months.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they will exceed the gestation period, or they will not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

The expected value of the binomial distribution is:

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

In this problem, we have that:

[tex]n = 3000, p = 0.271[/tex]

In 3000 human gestation​ periods, roughly how many will exceed 9​ months?

[tex]E(X) = np = 3000*0.271 = 813[/tex]

813 will exceed 9 months.

The gestation period should be exceed 9 month is 813.

Given that,

Based on the above information, the calculation is as follows:

[tex]= 0.271 \times 3,000[/tex]

= 813

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"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?

well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.

well, we can check by simply getting the distance from the center to the point (4,-1).

[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}[/tex]

Answer:

Part a: The two events are termed as disjoint because the event B clearly rules out the obese person

Part b: In the plain language, the event "A or B" means that the person is  overweight or obese. Its probability is 0.74.

Part c: If C is the event that the person chosen has normal weight or less, its probability is 0.26.

Step-by-step explanation:

As per the question obtained from the google search, the question has 3 parts as follows:

Part a

Explain why events A and B are disjoint.

Solution

The two events are termed as disjoint because the event B clearly rules out the obese person so the events are disjoint. so the correct option as given in the complete question is A.

Part b

Say in plain language what the event "A or B" is.

What is P(A or B)? (Enter your answer to two decimal places.)

Solution

In the plain language, the event "A or B" means that the person is  overweight or obese. The correction option as given in the complete question is a.

P(A or B) is given as

P(A or B)=P(AUB)=P(A)+P(B)-P(A∩B)

Here from the data of

P(A)=0.41

P(B)=0.33

P(A∩B)=0 (As the events are disjoint)

P(A or B)=P(AUB)=0.41+0.33-0

P(AUB)=0.74

So the probability of A or B is 0.74.

Part c

If C is the event that the person chosen has normal weight or less, what is

P(C)? (Enter your answer to two decimal places.)

Solution

P(C) is given as

P(C)=1-P(AUB)

P(C)=1-0.74

P(C)=0.26

So the probability of event C is 0.26.

Answer:

a. 59049 ways

b. [tex]1.69\times10^{-5}[/tex]

Step-by-step explanation:

If the multiple choice test has 10 questions with 3 choices each, for the each question there are 3 ways to answer the test. Then for n question there are [tex]3^n[/tex] ways to answer the test

The total number of ways to answer 10 questions at 3 choices each is

[tex]3^{10} = 59049[/tex]ways

b. There are 59049 ways to answer the test but if one test must match one other test then the probability for that to happen is

[tex]\frac{1}{59049} = 1.69\times10^{-5}[/tex]

There are 59,049 ways to answer a 10-question test with 3 choices per question. The probability of two tests having the same answers, with all answer patterns being equally likely, is 1/59,049.

This problem belongs to the domain of combinatorics. (a) As there are 3 choices for each question and 10 questions, there are 3^10 or 59,049 ways to answer the test. This assumes you are answering every question. (b) This is essentially the question of the probability that two randomly selected tests are identically answered. Assuming all ways to answer a test are equally likely, there are 59,049 different ways to answer, so the probability that two randomly answered tests are the same is 1/59,049 or approximately 0.000017.

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Answer: It will take 11.56 hours .

Step-by-step explanation:

Exponential growth in population or size formula :

[tex]P(t)=P_0e^{rt}[/tex]

, where [tex]P_0[/tex] = initial size

r= rate of growth

t= time period

As per given , we have

[tex]P_0=10[/tex] grams

At t= 1 , P(t)= 11 grams

Then,

[tex]11=10e^{r(1)}\\\\ 1.1= e^r\\\\\text{Taking natural log on both sides , we get} \\\\\ln (1.1)=r\ln (e)\\\\ r=\ln (1.1)\\\\ r=0.0953101798043\approx0.095[/tex]

When, the  bacteria have tripled in size , P(t) = 3 x10 = 30

Then,

[tex]30=10e^{0.095t}\\\\ 3=e^{0.095t}[/tex]

[tex]\text{Taking natural log on both sides , we get}\\\\ \ln 3=0.095t\\\\ t=\dfrac{\ln3}{0.095}\\\\ t=\dfrac{1.09861228867}{0.095}\approx11.56[/tex]

Hence, it will take 11.56 hours .

A bacteria culture is initially 10 grams at t=0 hours & grows at a rate proportional to its size , After an hour the bacteria culture weighs 11 grams , The bacteria takes 11.56 hours to have tripled in size.

To find the time of bacteria when increasing the growth to tripled.

Given :    when time=0 hours , weight=10 grams.

               when time=1  hours , weight=11 grams.

To find:   when time= ? hours , weight=30grams.

Here according to question, initial size = 10 grams we have asked for tripled in size i.e. 30 grams.

Now we knows that,

The formula for exponential growth in population or size is

              [tex]\rm (P)=P_0e^{rt}[/tex]  where,

               [tex]\rm P_0=initial\;size\\\\r= rate\;of\;growth\\\\t= time \;period[/tex]

Now, we put the value in formula we get,

[tex]\rm P_0=10\;grams \\\\when ,\\\;\;t=1\;hour P(t)=11 grams\\Then,\\11=10e^{r(1)\\1.1 =e^r\\\\\rm Taking \;log(natural)\;both\;the\; side \;on \;solving\;we\;get,\\ln(1.1)=r\;ln(e)\\r=ln(1.1)\\r=0.953101798043\approx0.095[/tex]

Now when the bacteria increase its size to triple

[tex]\rm P(t) = 3 \times 10 = 30[/tex]

Then, according to the formula we substitute values in the formula,

[tex]\rm 30=10e^{0.095t}\\\\3=e^{0.095t}\\\\Again \;we \;take\;natural\;log\;on \;both\;the\;sides, we\;get\\ln\;3=0.095t\\\\t=\dfrac{\rm ln\;3}{0.095}\\\\\\\\\rm t= \dfrac{1.09861228867}{0.095} \\\\\ t=approx \; 11.56[/tex]

Therefore, The bacteria takes 11.56 hours to have tripled in size.

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Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

[tex]0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0[/tex]

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

                             [tex]\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\[/tex]

Shear Strain Components

                             [tex]\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0[/tex]

Part c: Find the volume change

[tex]\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\[/tex]

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

             [tex]\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\[/tex]

As the strain values are small second and higher order values are ignored so

                                      [tex]\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\[/tex]

As the initial volume of cube is unitary so this result can be proved.

Final answer:

The probability that at least two bills must be selected to obtain the first $10 bill is approximately 24.7%.

Explanation:

To find the probability that at least two bills must be selected to obtain the first $10 bill, we need to calculate the probability of not drawing a $10 bill on the first draw and then drawing a $10 bill on the second draw. In total, there are 5 + 3 + 6 = 14 bills in the wallet.

On the first draw, the probability of not getting a $10 bill is the number of non-$10 bills over the total number of bills, which is (3 $5 bills + 6 $1 bills) / 14 total bills = 9/14.

Assuming a non-$10 bill was drawn first, there are now 13 bills left in the wallet. The probability of drawing a $10 bill on the second draw is now the number of $10 bills remaining over the total number of bills left, which is 5/13.

The combined probability of these two events happening in sequence (not drawing a $10 bill first and then drawing a $10 bill) is the product of their probabilities: (9/14) * (5/13).

Thus, the total probability is (9/14) * (5/13) = 45/182, which simplifies to approximately 0.247 or 24.7%.

Answer:

By looking at the histogram, we can conclude that the distribution is unimodal and skewed to the right. The modal value lies around 30 to 31 minutes and most of the running times range between 29 and 32 minutes.

The histogram is skewed to the right with most of the outliers present at the higher running times because practically, it is most likely possible for a person to run slow and take more time to run 4 miles rather than run fast and take less time.

Step-by-step explanation:

We say that the distribution is unimodal because there is only one peak which is the highest.

The distribution is skewed to the right because most of the outliers are present at the left side of the peak.

The student is undertaking an English language assignment designed to strengthen their understanding of vocabulary, word formation, and spelling through a variety of exercises including word scrambles, pattern recognition, spelling reviews, and word categorization.

The student appears to be working on a language arts activity related to vocabulary and word structure. The question likely requires them to engage with various linguistic exercises such as word scrambles, identifying patterns in spelling or pronunciation, reviewing correct spellings, and categorizing words. Such tasks are designed to help students learn about word formation, synonyms, antonyms, and the nuances of English spelling.

Word Scrambles and Patterns

For word scrambles, students are expected to rearrange the letters to form meaningful words. In doing this, they might uncover a hidden word that pertains to the lesson's focus. Identifying patterns in words might involve recognizing prefixes, suffixes, or roots that appear consistently across different words.

Reviewing Spelling

The student is also asked to choose the word with the correct spelling. This likely involves comparing similar words and identifying the correctly spelled one, perhaps with the aid of a dictionary for verification.

Categorizing Words

In the task of sorting words into groups, students might have to classify words based on different criteria such as part of speech, phonetic features, or spelling patterns. This reinforces their understanding of language structure and proper spelling conventions.

The histogram is shown below.

You'll have 10 bars. Under each bar is a label from 0 to 9. The height of each bar represents the frequency of each units digit.

The histogram shows two bars that are relatively large compared to the rest. These bars have frequency of 16 and 14 (for units digits of 0 and 5 respectively). The distribution is nearly bimodal. If the two frequencies mentioned were the same, say both 16, then it would be exactly bimodal. As you can probably guess, bimodal means "two modes".

The rest of the bar heights are nearly the same, so the remaining portion of the distribution is nearly uniform. A uniform distribution has every bar the same height. Collectively, all the bars of a uniform distribution forms a larger rectangle.