If the New England Patriots get home-field advantage, you believe there is a 60% probability they will make it to the Super Bowl. If not, this probability is only 30%. Assuming a 70% probability that the Patriots get home-field advantage, what is the probability they will make it to the Super Bowl?

Answer:

c) 44000 miles

Step-by-step explanation:

The regression line has the following format:

[tex]y = bx + a[/tex]

In which b is the slope(how much each mile costs) and a is the fixed number of miles flown.

In this problem, we have that:

[tex]a = 24000, b = 10[/tex]. So

[tex]y = 10x + 24000[/tex]

A person who spent $ 2000 is predicted to have flown:

This is y when x = 2000.

[tex]y = 10(2000) + 24000[/tex]

[tex]y = 44000[/tex]

So the correct answer is:

c) 44000 miles

To find the predicted distance flown by someone who spent $2000, you use the given intercept (a) and slope (b) in the straight line equation Y = a + bX. When $2000 is substituted for X in the equation, your solution is 44,000 miles.

In this problem, you are asked to determine the distance a person is predicted to have flown, given that the person spent $2000. Based on our regression line information, we know that the intercept (a) is 24,000 and the slope (b) is 10. Therefore, we can use the equation of the straight line Y = a + bX, where X represents the amount of money spent, and Y is the distance traveled. If we substitute $2000 for X in our equation, we find that Y = 24000 + 10 * 2000 = 44,000 miles. Therefore, the correct answer is option c) 44000 miles.

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The velocity of the ball after 3 seconds will be 29.43 meters per second.

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground.

Then the velocity of the ball after 3 seconds will be given as,

We know that the first equation of motion.

v = u - gt

Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time in seconds. Then we have

v = 0 - 9.81 x 3

v = -29.43 m/s

The velocity of the ball after 3 seconds will be 29.43 meters per second.

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Final answer:

The velocity of the ball after 3 seconds can be found using Galileo's law of free fall. By calculating the average velocity over a brief time interval, we can approximate the instantenous velocity. The velocity of the ball after 3 seconds is therefore m/s.

Explanation:

The velocity of the ball after 3 seconds can be found using Galileo's law of free fall. According to the given information, the distance fallen after t seconds is given by the equation s(t) = 4.9t², where s(t) is the distance fallen in meters and t is the time in seconds. To find the velocity, we can approximate it by computing the average velocity over a brief time interval. We can calculate the average velocity from t = 3 to t = 3.1 by finding the change in position divided by the time elapsed. Plugging the values into the equation, we get:

Average velocity = (s(3.1) - s(3)) / 0.1 = (4.9(3.1)² - 4.9(3)²) / 0.1 = m/s

Therefore, the velocity of the ball after 3 seconds is m/s, rounded to one decimal place.

Answer and Step-by-step explanation:

a) probability of selling a medium, long sleeved printed shirt, P(M ∩LS ∩PR) = 0.07, directly from the table of probabilities

b) probability that the next shirt sold is a medium printed shirt, (M ∩Pr) = P(M,Pr,LS) + P(M,Pr,SS) = 0.07+0.10 = 0.17

c) probability that the next shirt is a short sleeved shirt, P(SS) = sum of 9 probabilities in Short Sleeved shirt table = 0.58

probability that the next shirt is a long sleeved shirt, P(LS) = 1 - 0.58 = 0.42 or add up the total probabilities in the long sleeved shirt table, P(LS) = sum of 9 probabilities in the long sleeved shirt table = 0.42

d) probability that the next shirt is a medium, P(M) = P(M,SS) + P(M,LS) = (0.07+0.10+0.12) + (0.06+0.07+0.07) = 0.49

probability that the next shirt sold is a print, P(Pr) = P(Pr,SS) + P(Pr,LS) = (0.02+0.10+0.07) + (0.02+0.07+0.02) = 0.40

e) probability that the shirt sold is a medium given that the shirt just sold was a short-sleeved plaid, P(M|SS,PL) = (P(M,SS,PL))/P(SS,PL) = 0.07/(0.04+0.07+0.03) = 0.5

f) probability that the shirt sold is short sleeved given that the shirt just sold was a medium plaid, P(SS|M,PL) = (P(M,SS,PL))/P(M,PL) = 0.07/(0.07+0.06) = 0.53846 = 0.54

probability that the shirt sold is long sleeved given that the shirt just sold was a medium plaid, P(LS|M,PL) = (P(M,LS,PL))/P(M,PL) = 0.06/(0.07+0.06) = 0.462 = 0.46

QED!

The following probabilities are as follow;

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

(a) The next shirt sold is a medium, long-sleeved, print shirt.

Probability = 0.07

(b) The probability that the next shirt sold is a medium print shirt.

Probability = 0.7 + 0.10 = 0.17

(c) The probability that the next shirt sold is a long-sleeved shirt.

Probability = 0.42

(d) The probability that the size of the next shirt sold is medium. The probability that the pattern of the next shirt sold is a print.

Probability = 0.07 + 0.10 = 0.17

(e)The shirt just sold was a short-sleeved plaid, the probability that its size was medium.

Probability = 0.07

(f) The shirt just sold was a medium plaid, The probability that it was short-sleeved.

Probability = 0.07

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Answer:

A. 8 complete hotdogs were made.

B. The wiener.

C. The bun.

Step-by-step explanation:

1 complete hotdog = 1 wiener + 1 bun

You have, 3*10 bun and 8 wiener

= 30 bun and 8 wiener

A. Since, we have 30 bun and 8 wiener and for 1 complete hotdog = 1 wiener + 1 bun

Therefore, 8 hotdogs = 8 wiener + 8 buns

= 8 complete hotdogs are made

B.

Since 8 complete hotdogs are made, 8 wiener and 8 buns are used.

Therefore, the limiting ingredient is the wiener because it is the smallest number of ingredient and none of it was left after making the 8 hotdogs.

C. Since 8 complete hotdogs are made, 8 buns were used

So, the leftover buns = (30 - 8) buns

= 22 buns

The buns is the leftover ingredient

At the store, you buy four packages of eight wieners and three bags of 10 buns, you can make 8 hotdogs.

There are 8 hotdogs you can make then the limited ingredient is wieners.

The buns are the leftover ingredient.

You and your friends are making hot dogs.

A complete hot dog consists of a wiener and a bun.

At the store, you buy four packages of eight wieners and three bags of 10 buns.

Total number of buns = 30

Total number of wieners = 8

At the store, you buy four packages of eight wieners and three bags of 10 buns.

For every hot dog, 1 wiener one 1 buns are required. the calculation for the given question is given as follows;

1. The total number of hot dogs can you make is,

To make 1 hot dog 1 wiener and 1 bun

Then,

for 8 hot dogs, 8 wieners and 8 buns are required.

Therefore, you can make 8 hotdogs.

2. There are 8 hotdogs you can make then the limited ingredient is wieners.

3. After making 8 hot dogs only 8 buns are used,

Then,

Left buns = Total number of buns - used buns

Left buns = 30-8 = 22

The buns are the leftover ingredient.

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Answer:

See below

Step-by-step explanation:

Essentially, we have to replace "quantifier words" like "All", "Every" by the universal quantifier ∀.

a) ∀ dinosaur x, x is extinct.

b) ∀ real number x, x is positive, negative, or zero.

c) ∀ irrational number x, x is not an integer.

d) ∀ logician x, x is not lazy.

e) ∀ integer x, x²≠ 2,147,581,953.

f)  ∀ real number x, x²≠ -1.

In a) and b) we replace the words without major changes. In the other statements, we modify the statement using negation. For example, "No irrational numbers are integers." is equivalent to "Every irrational number is not integer".

Answer:

(a) 0.970225.

(b) 0.999775.

(c) 2.25 x [tex]10^{-4}[/tex].

Step-by-step explanation:

Given the probability that either system will function satisfactorily during a flight is 0.985.

(a) To calculate the probability that both systems function satisfactorily we are given the probability of either system will function of 0.985 which means that both function have probability of functioning satisfactorily of 0.985 each i.e.,

= 0.985 x 0.985 = 0.970225.

(b) The probability of first function will not function satisfactorily is 1 - 0.985 because the probability of first system function satisfactorily is 0.985.

Similarly, probability of second function will not function satisfactorily is also 1 - 0.985 = 0.015

So Probability that during a given flight at least one system functions satisfactorily = 1 - both system will not function satisfactorily

                     = 1 - (0.015 x 0.015) = 0.999775.

(c) Probability that during a given flight both systems fail or not function  satisfactorily = (1 - 0.985) x (1 - 0.985) = 2.25 x [tex]10^{-4}[/tex]

Answer:

(a) P (Both systems functioning satisfactorily) = 0.970

(b) P (At least one system functions satisfactorily) = 0.999

(c) P (Both the systems failing) = 0.00023

Step-by-step explanation:

Let A = the system in use is working satisfactorily and B = the backup system is working satisfactorily.

The probability that either of the systems works satisfactorily is,

P (A) = P(B) = 0.985

(a)

Both the events A and B are independent, i.e. [tex]P(A\cap B)=P(A)\times P(B)[/tex]

Compute the probability that during a given flight both systems function satisfactorily as follows:

[tex]P(A\cap B)=P(A)\times P(B)[/tex]

               [tex]=0.985\times0.985\\=0.970225\\\approx0.970[/tex]

Thus, the probability of both systems functioning satisfactorily is 0.970.

(b)

Compute the probability that during a given flight at least one system functions satisfactorily as follows:

P (At least one system functions satisfactorily) = 1 - P (None of the system functions satisfactorily)

                                                                             = [tex]1-[P(A^{c})\times P(B^{c})][/tex]

                                                                             [tex]=1-([1-P(A)]\times [1-P(B)])\\=1-([1-0.985]\times [1-0.985])\\= 1-0.000225\\=0.999775\\\approx0.999[/tex]

Thus, the probability that during a given flight at least one system functions satisfactorily is 0.999.

(c)

Compute the probability that during a given flight both the systems fail as follows:

P (Both the systems failing) = [tex]P(A^{c})\times P(B^{c})[/tex]

                                             [tex]=[1-P(A^{c})]\times [1-P(B^{c})]\\=(1-0.985)\times (1-0.985)\\=0.015\times 0.015\\=0.000225\\\approx0.00023[/tex]

Thus, the probability that during a given flight both the systems fail is 0.00023.

Answer:

y=\frac{ -\frac{\sqrt[9]{18} Г (10/9, -\frac{19t^{18}}{18})}{19^{10/9} {e^{\frac{19t^{18}}{18}}}

Step-by-step explanation:

From exercise we have:

C=0

y'+19t^{17} y=t^{19}

We know that a linear differential equation is written in the standard form:

y' + a(t)y = f(t)

we get that: a(t)=19t^{17} and f(t)=t^{19}.

We know that the integrating factor is defined by the formula:

u(t)=e^{∫ a(t) dt}

⇒ u(t)=e^{∫ 19t^{17} dt} = e^{\frac{19t^{18}}{18}}

The general solution of the differential equation is in the form:

y=\frac{ ∫ u(t) f(t) dt +C}{u(t)}

y=\frac{ ∫ e^{\frac{19t^{18}}{18}} · t^{19} dt + 0}{e^{\frac{19t^{18}}{18}}}

y=\frac{ -\frac{\sqrt[9]{18} Г (10/9, -\frac{19t^{18}}{18})}{19^{10/9} {e^{\frac{19t^{18}}{18}}}

The speed at which each bicyclist rides is 320 times the distance covered by each bicyclist.

To solve this problem, we can consider the distance traveled by the train and by each bicyclist. Let's assume the distance covered by each bicyclist is x miles. The train covers the remaining 1 - 7/8 = 1/8 mile. We can set up an equation to represent the time it takes for the train and the bicyclists to travel their respective distances:

Time taken by train = distance traveled by train / speed of train = (1/8) mile / 40 mph = 1/320 hour

Time taken by each bicyclist = distance traveled by bicyclist / speed of bicyclist = x miles / v mph, where v represents the speed of the bicyclists

Since the train passes each bicyclist when they are 7/8 of the way through the tunnel, the time it takes for each bicyclist to travel their distance is equal to the time it takes for the train to travel its distance:

(x / v) = 1/320

To solve for v, we can rearrange the equation:

v = x / (1/320)

v = 320x mph

The speed at which each bicyclist rides is 320 times the distance covered by each bicyclist.

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Final answer:

To escape a train traveling at 40 mph through a tunnel, two bicyclists traveling in opposite directions must ride at a speed of exactly 35 mph to escape the tunnel safely. We derive this by setting up two inequalities based on the distances each cyclist must cover and then finding the minimum speed at which these conditions are satisfied.

Explanation:

We are discussing a classic algebra problem that utilizes rate, time, and distance calculations. Two bicyclists are 7/8 of the way through a mile-long tunnel when a train traveling at 40 mph approaches from behind. Each cyclist pedals at the same speed but in opposite directions to escape the tunnel as the train passes by them.

To solve the problem, we need to determine how far each bicyclist has to travel to escape the tunnel. One bicyclist has 1/8 mile to exit, while the other bicyclist has 7/8 mile. Since they travel at the same speed and need to escape before the train reaches them, we can set up a ratio based on distances and speeds. We know the train covers the mile in (1/40) hours since speed is distance over time.

The bicyclist closer to the exit (1/8 mile away) must bike this distance faster than the train covers the entire mile to avoid collision, and similarly for the second bicyclist (7/8 miles away). Let's assume their biking speed is 's'. Then, their times to escape are (1/8)/s and (7/8)/s respectively. These times must be less or equal to the time it takes for the train to travel the mile, which is 1/40 hours.

Setting up the first inequality: (1/8)/s ≤ 1/40
Which simplifies to: s ≥ 5 mph
And the second inequality: (7/8)/s ≤ 1/40
This second inequality simplifies to: s ≥ 35 mph
Since both conditions must be satisfied and s cannot be greater than 35 mph for both to be true, the speed at which both bicyclists must ride to escape is exactly 35 mph.

Answer:

(a) 0.0001 or 0.01%

(b) 0.01 or 1%

Step-by-step explanation:

Since there are 10 possible numeric digits (from 0 to 9), and there is only one correct digit, there is a 1 in 10 change of getting each digit right.

The probability that if you forget your PIN, then you can guess the correct sequence

(a) at random:

[tex]P=(\frac{1}{10})^4=0.0001=0.01\%[/tex]

(b) when you recall the first two digits.

[tex]P=(\frac{1}{10})^2=0.01=1\%[/tex]

The probability of guessing a four-digit PIN code correctly is 1/10,000 if guessed randomly and 1/100 if the first two digits are recalled correctly.

To find the probability of guessing a four-digit PIN code correctly, we need to consider the number of possible combinations and the number of favorable outcomes. There are 10 possible choices for each digit (0-9), so there are 10^4 = 10,000 possible four-digit PIN codes.

(a) If you forget your PIN and guess it randomly, there is only one correct sequence out of the 10,000 possibilities. Therefore, the probability of guessing the correct sequence is 1/10,000.

(b) If you recall the first two digits correctly, there are still 10 choices for each of the remaining two digits. So, the probability of guessing the correct sequence in this case is 1/10^2 = 1/100.

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Answer:

Step-by-step explanation:

Total mass of water = density X volume = 1000 kg/m3 x 187.5 m3

where H2 = initial position of centre of mass from ground

where H1 = final position of centre of mass from ground

The work required to pump the water out of the trough is [tex]5.11\times 10^6[/tex] J and this can be determine by using the formula of work done.

Given :

a) The work required to pump the water out of the trough is given by:

[tex]\rm W=mg(H_2-H_1)[/tex]    --- (1)

where m is the total mass of the water [tex]\rm H_1[/tex] is the initial position and [tex]\rm H_2[/tex] is the final position of the center of mass.

So. the mass of the water is given below:

[tex]=1000\times 187.5[/tex]

= 187500 Kg

Now, the value of the initial position of the center of mass is zero and the value of the final position of the center of mass is 2.78 m.

Now, substitute the known terms in the expression (1).

[tex]\rm W = 187500\times 9.8\times (2.78-0)[/tex]

[tex]\rm W = 5.11\times 10^{6}\;J[/tex]

b) The width of the cylinder is given by:

[tex]\rm w-5=\dfrac{5-2.5}{0-5}(x-0)[/tex]

Simplify the above expression.

w = -0.5x + 5

Now, the volume of the cylinder is given by:

[tex]\rm dv = 10wdx[/tex]

Substitute the value of w in the above expression.

[tex]\rm dv = 10(-0.5x+5)dx[/tex]

dv = (50-5x) dx

Now, the value of the force is given by:

[tex]\rm dF = 9800dv[/tex]

Substitute the value of dv in the above expression.

dF = 9800(50 - 5x) dx

Now, the expression of work is given by:

dW = x dF

Substitute the value of force in the above expression.

dW = x(9800(50 - 5x)) dx

Now, integrate the above expression.

[tex]\rm W = 9800\int^5_0(50x-5x^2)dx[/tex]

[tex]\rm W = 9800 \times \left(25x^2-\dfrac{5x^3}{3}\right)^5_0[/tex]

[tex]\rm W= 9800\left(625-\dfrac{625}{3} \right)[/tex]

W = 4083333.34 J

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Answer:

a) Figure attached

b) [tex] P(X<130 min)= F(130) = \frac{130-120}{20}=0.5[/tex]

c) [tex] P(X>135) =1-P(X<135) = 1- F(135) = 1- \frac{135-120}{20}= 0.25[/tex]

d) [tex] E(X) =\frac{a+b}{2}= \frac{120+140}{2}= 130min[/tex]

Step-by-step explanation:

2 hr = 120 min

2hr 20 min = 120+20 = 140 min

Let X the random variable that represent "flight time". And we know that the distribution of X is given by:

[tex]X\sim Uniform(120 ,140)[/tex]

For this case the density function for A is given by:

[tex] f(X) = \frac{1}{b-a}= \frac{1}{20} , 120\leq X \leq 140 [/tex]

And [tex] f(X)= 0[/tex] for other case

Part a

For this case we can see the figure attached. We see that the probability density function is defined between 120 and 140 minutes.

Part b

The arrival time is 2hr 5 min = 125 min. So then 5 minutes leate means 130 min

For this case we want to find this probability:

[tex] P(X<130 min)[/tex]

And we can use the cumulative distribution function for X given by:

[tex] F(X) = \frac{x-120}{140-120} = \frac{x-120}{20} , 120\leq X \leq 140 [/tex]

And if we use this formula we got:

[tex] P(X<130 min)= F(130) = \frac{130-120}{20}=0.5[/tex]

Part c

For this case more than 10 minutes late means 125 +10 = 135 min or more, so we want this probability:

[tex] P(X>135)[/tex]

And using the complement rule we have:

[tex] P(X>135) =1-P(X<135) = 1- F(135) = 1- \frac{135-120}{20}= 0.25[/tex]

Part d

The expected value for the uniform distribution is given by:

[tex] E(X) =\frac{a+b}{2}= \frac{120+140}{2}= 130min[/tex]

The total number of ways to choose the two groups, depending on the relationship between n and m, is:

M < N: C(100, n).

M ≥ N: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1.

Case 1: m < n (n tallest chosen first):

Choose the n tallest people from the 100. There are C(100, n) ways to do this.

The remaining 100 - n people are all shorter than the chosen n. Since m < n, we can simply choose all the remaining people (100 - n).

The total number of ways in this case is C(100, n).

Case 2: m ≥ n (not all n tallest need to be chosen):

This case requires considering all possible combinations of how many tall people to choose (from n down to 1).

For each value of k (n, n - 1, ..., 1), choose k tallest people from the 100. There are C(100, k) ways to do this.

The remaining 100 - k people are all shorter than the chosen k. Choose the remaining m - k people from this group. There are C(100 - k, m - k) ways to do this.

Sum the results for each value of k: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1.

Therefore, the total number of ways to choose the two groups, depending on the relationship between n and m, is:

M < N: C(100, n)

M ≥ N: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1

Complete question:

How many ways are there to pick a group of n people from 100 people, and then pick a second group of m people from the remaining 100 − n people, so that all the people you chose from the first group are taller than the people from the second group (assume that everyone in the group of 100 people has a different height.)

Note: it is implied in this question that 1 ≤ n ≤ 100 and 0 ≤ m ≤ 100 − n.

Answer:

The number of possible choices of my team and the opponents team is

 [tex]\left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right i^{3}[/tex]

Step-by-step explanation:

selecting the first team from n people we have [tex]\left(\begin{array}{ccc}n\\1\\\end{array}\right) = n[/tex] possibility and choosing second team from the rest of n-1 people we have [tex]\left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1[/tex]

As { A, B} = {B , A}

Therefore, the total possibility is [tex]\frac{n(n-1)}{2}[/tex]

Since our choices are allowed to overlap, the second team is [tex]\frac{n(n-1)}{2}[/tex]

Possibility of choosing both teams will be

[tex]\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}[/tex]

We now have the formula

1³ + 2³ + ........... + n³ =[tex][\frac{n(n+1)}{2}] ^{2}[/tex]

1³ + 2³ + ............ + (n-1)³ = [tex][x^{2} \frac{n(n-1)}{2}] ^{2}[/tex]

=[tex]\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}[/tex]

Answer:

the helix intersects the sphere at t=4 and t=(-4)

Step-by-step explanation:

for the helix r(t) = [ sin(t) , cos(t) ,  t ] then x=sin(t) , y=cos(t) and z=t

thus the helix intersect the sphere  x² + y² + z² = 17 at

x² + y² + z² = 17

[sin(t)]²+[cos(t)]²+ t² = 17

1 + t² = 17

t² = 16

t = ±4

thus the helix intersects the sphere at t=4 and t=(-4)

The point wher the helix intersect the sphere is at t =  ±4

Given the coordinate of a helix expressed as;

If these coordinate intersects the sphere  x² + y² + z² = 17

Substitute the coordinate of the helix:

x² + y² + z² = 17

(sint)² + (cost)² + t² = 17

sin²t + cos²t + t² = 17

1 + t² = 17

t² = 17 - 1

t² = 16

t = ±√16

t = ±4

Hence the point wher the helix intersect the sphere is at t =  ±4

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Answer:

[tex]45*10^{6}[/tex] orders of magnitude.

Step-by-step explanation:

We know that 900 km is equivalent to 20 cm, because we need to fit the United Kingdom length on a sheet of paper. So:

Therefore, if we divide 45 km by 0.00001 km (1 cm) we will have [tex]45*10^{6}[/tex] times to scale down.

I hope it helps you!

Answer:

a) 0.06

b) 0.3

c) 0.44                                        

Step-by-step explanation:

We are given the following in the question:

A and B are two independent events.

P(A) = 0.3

P(B) = 0.2

a) P(A intersect ​B)

[tex]P(A\cap B) = P(A)\times P(B) = 0.3\times 0.2 = 0.06[/tex]  

b) P(A|B)

[tex]P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.06}{0.2} = 0.3[/tex]

c) P(A union ​B)

[tex]P(A\cup B) = P(A) + P(B) - P(A\cap B)\\P(A\cup B) = 0.3 + 0.2 - 0.06 = 0.44[/tex]

Answer:

15 is the number of combination of 4 successes.

Step-by-step explanation:

We are given the following information:

We are given a binomial distribution, then probability of x succes is given by

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 6 and x = 4

We have to evaluate the number of combination of success and failures.

It is given by:

[tex]\binom{n}{x} = \dfrac{n!}{x!(n-x)!}\\\\\binom{6}{4} = \dfrac{6!}{4!(6-4)!} = \dfrac{6!}{4!2!} = 15[/tex]

Thus, 15 is the number of combination of 4 successes.

The number of combinations of 4 successes out of 6 attempts, calculated using the binomial probability combination formula, is 15.

In binomial probability, the number of ways to get a specific number of successes is often calculated using the combination formulas. In this case, the number of combinations of X successes would be determined by the combination formula C(n, x), which stands for the number of combinations of n items taken x at a time. In your case, where n=6 and x=4, the required number of combinations (or ways to achieve 4 successes out of 6 trials) is C(6,4).

The formula for a combination is generally given by: n! / [x!(n - x)!]. Plugging the given numbers into the formula, we get: C(6,4) = 6! / [4!(6 - 4)!].  This simplifies to: 720 / (24*2), which equals to 15. So, there are 15 combinations of 4 successes out of 6 attempts.

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The missing value in the table is 15.

Solution:

Ratio of x to y coordinate:

[tex]$\frac{x}{y}= \frac{1}{3}[/tex]

[tex]$\frac{x}{y}= \frac{2}{6}= \frac{1}{3}[/tex]

[tex]$\frac{x}{y}= \frac{8}{24}= \frac{1}{3}[/tex]

So, the ratios are equivalent ratios.

To find the missing value in the table:

Multiply the x-coordinate by 3, we get the y-coordinate.

x-coordinate = 5

y-coordinate = x-coordinate × 3

                     = 5 × 3

                     = 15

Hence the missing value in the table is 15.

Answer:

15

Step-by-step explanation:

1 multiplied by 5 equals 5. Do the same with the y-value. 3 multiplied by 5 is 15!

Answer:

This is called Simpson's Paradox.

Therefore the correct option is A.) Simpson's Paradox.

Step-by-step explanation:

i) This is called Simpson's Paradox.

ii) If there are trends that tend to appear in several different groups of data which apparently either disappear or tend to reverse when these data groups are combined.

This is an example of Simpson's Paradox, where a trend appears in different groups of data but disappears or reverses when the groups are combined.

This is an example of Simpson's Paradox. Simpson's Paradox occurs when a trend appears in different groups of data but disappears or reverses when the groups are combined. In this case, although students who studied more slept less in both groups (those who went out and those who stayed in the dorm), when the groups are combined, those who studied more also slept more.

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Answer:

a. 15

b.

Sr.no Samples Sample mean

1           (79,64)         71.5

2          (79,84)          81.5

3          (79,82)          80.5

4          (79,92)          85.5

5          (79,77)           78

6         (64,84)            74

7          (64,82)            73

8          (64,92)            78

9          (64,77)            70.5

10        (84,82)            83

11         (84,92)            88

12        (84,77)             80.5

13        (82,92)            87

14        (82,77)            79.5

15        (92,77)            84.5

c.

mean of sample mean=population mean=79.67

Step-by-step explanation:

a.

The different samples of two test grade are nCr, where n=6 and r=2.

nCr=6C2=6!/2!(6-2)!=6*5*4!/2!4!=30/2=15.

Thus, there are 15 different samples of two test grade.

b.

All possible samples are listed below:

Sr.no Samples

1           (79,64)        

2          (79,84)          

3          (79,82)          

4          (79,92)        

5          (79,77)          

6         (64,84)            

7          (64,82)

8          (64,92)

9          (64,77)

10        (84,82)

11         (84,92)

12        (84,77)

13        (82,92)

14        (82,77)

15        (92,77)

The sample means for each sample can be calculated as

Sr.no Samples Sample mean

1           (79,64)         (79+64)/2=71.5

2          (79,84)          (79+84)/2=81.5

3          (79,82)          (79+82)/2=80.5

4          (79,92)          (79+92)/2=85.5

5          (79,77)           (79+77)/2=78

6         (64,84)            (64+84)/2=74

7          (64,82)            (64+82)/2=73

8          (64,92)            (64+92)/2=78

9          (64,77)            (64+77)/2=70.5

10        (84,82)            (84+82)/2=83

11         (84,92)            (84+92)/2=88

12        (84,77)             (84+77)/2=80.5

13        (82,92)            (82+92)/2=87

14        (82,77)            (82+77)/2=79.5

15        (92,77)            (92+77)/2=84.5

c.

The sample means of sample mean μxbar will calculated by taking average of sample means

μxbar=(71.5+ 81.5+ 80.5+ 85.5+ 78+ 74+ 73+ 78+ 70.5+ 83+ 88+ 80.5+ 87+ 79.5+ 84.5)/15

μxbar=1195/15=79.67

Population mean=μ=(79+64+84+82+92+77)/6

μ=478/6=79.67

Sample means of sample mean μxbar=Population mean μ.

Answer:

Step-by-step explanation:

The degree of a polynomial is the sum of the exponents of all of its variables. Where there is only one variable, the degree would be the highest power to which that variable is raised. The terms are the individual components that make up the polynomial. Therefore,

16t2 is a second degree because the variable, t is raised to 2 and the term is 1

240 is zero degree and 1 term.

15,000 – 352t is first degree and 2 terms.

Answer:

0.375 feet-lb

Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

[tex]6=\int\limits^2_0 {F(x)} \, dx[/tex]

We will use Hooke's Law to solve our given problem.

[tex]F(x)=kx[/tex]

Substituting this value in our integral, we will get:

[tex]6=\int\limits^2_0 {kx} \, dx[/tex]

Using power rule, we will get:

[tex]6=\left[ \frac{kx^2}{2} \right ]^2_0[/tex]

[tex]6=\frac{k(2)^2}{2}-\frac{k(0)^2}{2}[/tex]

[tex]6=\frac{4k}{2}-0\\\\k=3[/tex]

We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be [tex]\int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx[/tex]

Using power rule, we will get:

[tex]\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0[/tex]

[tex]\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375[/tex]

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.

Answer:

[tex]T(t) = 42000 + 21t[/tex]

Step-by-step explanation:

There are two costs for the operation of the equipment, the purchase cost and the hourly cost. The total cost is the sum of these two costs:

So

[tex]T = F_{c} + H_{c}[/tex]

In which [tex]F_{c}[/tex] is the fixed cost and [tex]H_{c}[/tex] is the hourly cost.

Fixed cost

A roofing contractor purchases a shingle delivery truck with a shingle elevator for $42,000. This means that the fixed cost is 42000. So [tex]F_{c} = 42000[/tex]

Hourly cost

The vehicle requires an average expenditure of $9.50 per hour for fuel and maintenance, and the operator is paid $11.50 per hour. So for each hour, there are expenses of 9.5 + 11.5 = $21.

So the hour cost is

[tex]H_{c}(t) = 21t[/tex]

In which t is the number of hours

Total cost

[tex]T = F_{c} + H_{c}[/tex]

[tex]T(t) = 42000 + 21t[/tex]

Answer: 160.4040

Step-by-step explanation:

Here , the claim is "The standard deviation of pulse rates of adult males is more than 11 bpm." , i.e. [tex]\sigma>11[/tex]

We use Chi-square test statistic for the test statistic to test the standard deviations :

[tex]\chi^2=\dfrac{(n-1)s^2}{\sigma^2}[/tex] , where s= sample standard deviation , [tex]\sigma[/tex] = population standard deviations and n = sample size.

As per given , n=153 , s= 11.3

Then, Required test statistic will be :

[tex]\chi^2=\dfrac{(153-1)(11.3)^2}{11^2}=\dfrac{152\times127.69}{121}\approx160.4040[/tex]

Hence, the value of the test statistic. =  160.4040

Answer:

a) If we construct a stemplot for the data we have this:

Stem     Leaf

1       |     9

2      |     2 3 9

3      |     0 0 1 3 5

4      |     2

Notation: 1|9 means 19

b) [tex]29.4-2.26\frac{6.75}{\sqrt{10}}=24.58[/tex]    

[tex]29.4-2.26\frac{6.75}{\sqrt{10}}=34.22[/tex]    

So on this case the 95% confidence interval would be given by (24.58;34.22)    

Step-by-step explanation:

For this case we have the following data:

30 30 42 35 22 33 31 29 19 23

Part a

If we construct a stemplot for the data we have this:

Stem     Leaf

1       |     9

2      |     2 3 9

3      |     0 0 1 3 5

4      |     2

Notation: 1|9 means 19

As we can see the distribution is a little assymetrical to the right since we have not to much values on the right tail. But we can approximate roughly the distribution symmetric and with no outliers.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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

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

s represent the sample standard deviation

n represent the sample size  

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=29.4[/tex]

The sample deviation calculated [tex]s=6.75[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.26[/tex]

Now we have everything in order to replace into formula (1):

[tex]29.4-2.26\frac{6.75}{\sqrt{10}}=24.58[/tex]    

[tex]29.4-2.26\frac{6.75}{\sqrt{10}}=34.22[/tex]    

So on this case the 95% confidence interval would be given by (24.58;34.22)    

Answer: for replacement total games = 47   more than 3.5 hours   = 8 required proportion = 8/47 = 0.170212 for regular.

Step-by-step explanation:

1. What proportion of the 47 games officiated by replacement referees lasted for at least 3.5 hours (210 minutes)? What proportion of the 42 games officiated by regular referees lasted for this long?  

2. What proportion of the 47 games officiated by replacement referees lasted for less than 3 hours (180 minutes)? What proportion of the 42 games officiated by regular referees lasted for this long?  

3. Would you say that either type of referee tended to have longer games than the other on average, and if so, which type of referee tended to have longer games and by about how much on average?

i. Regular Referees have longer game times with games about 185 minutes on average.

ii. The game lengths for both referees seem to be the same.

iii. Replacement Referees have longer game times with games about 195 minutes on average.

iv. It is too hard to tell from the graph which type of referee has longer games.

4.  Would you say that either type of referee tended to have more variability in game durations? If so, which type of referee tended to have more variability?

i. The two types of referees had the same amount of variability in the game lengths.

ii. There is no way to tell which group of referees had more variability in the game lengths.

iii. The regular referees tended to have more variability in the game lengths.

iv. The replacement referees tended to have more variability in the game lengths.

For replacement total games = 47   more than 3.5 hours   = 8 required proportion = 8/47 = 0.170212 for regular.

Answer:

the probability is 0.311 (31.1%)

Step-by-step explanation:

defining the event L= being late to work :Then knowing that each mode of transportation is equally likely (since we do not know its travel habits) :

P(L)= probability of taking the bicycle * probability of being late if he takes the bicycle + probability of taking the car* probability of being late if he takes the car  + probability of taking the bus* probability of being late if he takes the bus +probability of taking the train* probability of being late if he takes the train = 1/4  * 0.75 +   1/4 * 0.43 + 1/4  * 0.15  + 1/4 * 0.05 = 0.345

then we can use the theorem of Bayes for conditional probability. Thus defining the event C= Bob takes the car , we have

P(C/L)=  P(C∩L)/P(L) = 1/4 * 0.43 /0.345 = 0.311 (31.1%)

where

P(C∩L)= probability of taking the car and being late

P(C/L)= probability that Bob had taken the car given that he is late

If Bob is late without considering the transportation mode, the total probability is 63%. Using Bayes' theorem, we find the probability that Bob was late because he took the car is approximately 17%.

Firstly, it's essential to ascertain the total chances of Bob being late disregarding the mode of transport. And that is the sum of his possibilities of being late for each mode, which is 43% for the car, 15% for the bus, and 5% for the train: 43% + 15% + 5% = 63%. Thus, if Bob is late, there's a 63% chance that he took either car, bus, or train. To calculate the probability that Bob took the car given that he was late, we employ Bayes' Theorem.

The formula for Bayes' theorem is P(A|B) = P(B|A)P(A) / P(B). In this instance, event A is Bob travelling by car, event B is Bob being late. We want to estimate P(A|B), the chance of A happening given B has occurred. Hence we plug in values: P(A|B) = (0.43)(0.25) / (0.63) ≈ 0.17 or 17%. Thus, the boss would estimate that there's a 17% chance that Bob took his car if he observes Bob coming late.

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Answer:

a) For this case we can see on the y axis the test scores and on the x axis the hours of homework. And as we can see we have a proportional relationship, because when the hours of homework increase the test scores increases too. If we fit a lineal model between y and x we will see an slope positive and a correlation coefficient positive based on the data observed.

b) [tex] y = 8x +42[/tex]

And we want to predict the test score for x = 3hr of homework we just need to replace the value of x=3 in the linear model and we got:

[tex] y = 8*3 +42= 24+42=66[/tex]

And that would be our predicted value for 3 h of homework

Step-by-step explanation:

For this case we consider the scatter plot attached to solve the problem.

Part a

For this case we can see on the y axis the test scores and on the x axis the hours of homework. And as we can see we have a proportional relationship, because when the hours of homework increase the test scores increases too. If we fit a lineal model between y and x we will see an slope positive and a correlation coefficient positive based on the data observed.

Part b

Assuming the following linear model for the situation:

[tex] y = 8x +42[/tex]

And we want to predict the test score for x = 3hr of homework we just need to replace the value of x=3 in the linear model and we got:

[tex] y = 8*3 +42= 24+42=66[/tex]

And that would be our predicted value for 3 h of homework

Answer: There is a strong positive correlation between  number of games won by a minor league baseball team and the average attendance at their home games is analyzed.

Step-by-step explanation:

The Pearson's coefficient 'r' gives the correlation between the predicted values and the observed values .

Given :  A regression to predict the average attendance from the number of games won has an r = 0.73.

Since r=0.73 is positive and 0.70 <0.73 <1 , it means there is a strong positive correlation between number of games won by a minor league baseball team and the average attendance at their home games is analyzed.

Answer:

12%

Step-by-step explanation:

let x represent the actual number of members last year.

current members = 675 and with an increment of 13% over previous year

to find the increment

(675 - x) / x = 0.13

675 - x = 0.13x

collect the like terms

675 = 0.13x + x

675 = 1.13 x

x = approx 597 members were there last year

its hope to increase by the same number next year

percent increase = 675 - 597 / 675 = 78 / 675 = 0.12 = 12%