Suppose the exchange rate of US dollar to Japanese yen exchange rate is $1 for every 107.35 yen, and the Japanese yen to Bitcoin exchange rate is 1,086,300 yen for every 1 Bitcoin. If someone traded $83,000 US dollars for Japanese yen, then traded the yen for Bitcoin, how many Bitcoin would that person end up with? Round your answer to the nearest whole Bitcoin.

Answers

Answer 1

Answer:

The person would end up with 8 Bitcoins.

Step-by-step explanation:

This question can be solved by consecutive rules of three.

If someone traded $83,000 US dollars for Japanese yen, then traded the yen for Bitcoin, how many Bitcoin would that person end up with?

Each US dollar is worth 107.35 yen. So how many yens are $83,000 US dollars worth?

$1 - 107.35 yen

$83,000 - x yen

[tex]x = 83000*107.35[/tex]

[tex]x = 8,910,050[/tex]

The person has 8,910,050 yens. Each bitcoin is worth 1,086,300 yens. How many bitcoins are worth 8,910,050 yens?

1 bitcoin - 1,086,300 yens

x bitcoins - 8,910,050 yens

[tex]1086300x = 8910050[/tex]

[tex]x = \frac{8910050}{1086300}[/tex]

[tex]x = 8.2[/tex]

Rouded to the nearest whole Bitcoin, is 8.

So the person would end up with 8 Bitcoins.

Answer 2
Final answer:

By first converting the US dollars to yen and then trading the yen for Bitcoin, using the provided exchange rates, we determine that the person would end up with roughly 8 Bitcoin.

Explanation:

To answer this exchange rate problem, we must first convert the US dollars to yen, then convert the yen to Bitcoin.

First, we multiply the amount of US dollars, $83,000 by the US dollar to yen exchange rate, which is 107.35 yen for every 1 US dollar. This gives us:

$83,000 * 107.35 yen/US dollar = 8,910,050 yen

Next, we trade the yen for Bitcoin by dividing by the yen to Bitcoin exchange rate. Our yen to Bitcoin rate is 1,086,300 yen for 1 Bitcoin:

8,910,050 yen ÷ 1,086,300 yen/Bitcoin ≈ 8.2 Bitcoin.

Rounding this to the nearest whole number, we find that the person ends up with approximately 8 Bitcoin.

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Related Questions

How many ways are there to pick a group of n people from 100 people (each of a different height) and then pick a second group of m other people such that all people in the first group are taller than the people in the second group

Answers

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.

Two bicyclists are 7/8 of the way through a mile long tunnel when a train approaches the closer end at 40 mph. The riders take off at the same speed in opposite directions and each escapes the tunnel as the train passes them. How fast did they ride?

Answers

Final answer:

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

Explanation:

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.

The binomial formula has two parts. The first part of the binomial formula calculates the number of combinations of X successes. The second part of the binomial formula calculates the probability associated with the combination of success and failures. If N=6 and X=4, what is the number of combinations of X successes?156486!

Answers

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.

Final answer:

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

Explanation:

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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For two independent​ events, A and​ B, ​P(A)equals . 3 and​P(B)equals . 2 . a. Find ​P(Aintersect ​B). b. Find​ P(A|B). c. Find ​P(Aunion ​B). a.​ P(Aintersect ​B)equalsnothing b.​P(A|B)equals nothing c.​ P(Aunion ​B)equalsnothing. Any expert yet

Answers

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]

Suppose that prices of a gallon of milk at various stores in one town have a mean of $3.94 with a standard deviation of $0.11. Using Chebyshev's Theorem, state the range in which at least 88.9% of the data will reside. Please do not round your answers.

Answers

Answer:

($3.61, $4.27) is the range in which at least 88.9% of the data will reside.

Step-by-step explanation:

Chebyshev's Theorem

This theorem states that at least [tex]1 - \dfrac{1}{k^2}[/tex]  percentage of data lies within k standard deviation from the mean.For k = 2

[tex]\mu \pm 2\sigma\\\\1 - \dfrac{1}{(2)^2}\% = 75\%[/tex]

Atleast 75% of data lies within two standard deviation of the mean for a non-normal data.

For k = 3

[tex]\mu \pm 3\sigma\\\\1 - \dfrac{1}{(3)^2}\% = 88.9\%[/tex]

Thus, range of data for which 88.9% of data will reside is

[tex]\mu \pm 3\sigma\\= (\mu - 3\sigma , \mu + \3\sigma)[/tex]

Now,

Mean = $3.94

Standard Deviation = $0.11

Putting the values, we get,

Range =

[tex]3.94 \pm 30.11\\= (3.94 - 3(0.11) , 3.94 + 3(0.11))\\=(3.61,4.27)[/tex]

($3.61, $4.27) is the range in which at least 88.9% of the data will reside.

Answer:

Using Chebyshev's Theorem, the range in which at least 88.9% of the data reside is [$3.82, $4.06].

Step-by-step explanation:

Given:

The mean of the prices of a gallon of milk is, [tex]\mu=\$3.94[/tex]

The standard deviation of the prices of a gallon of milk is, [tex]\sigma=\$0.11[/tex]

The Chebyshev's Theorem states that for a random variable X with finite mean μ and finite standard deviation σ and for a positive constant k the provided inequality exists,

                                      [tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}[/tex]

The value of [tex]\frac{\sigma^{2}}{k^{2}} = 0.889[/tex]

Then solve for k as follows:

[tex]\frac{\sigma^{2}}{k^{2}} = 0.889\\\frac{(0.11)^{2}}{k^{2}}=0.889\\k^{2}=\frac{(0.11)^{2}}{0.889}\\k=\sqrt{0.0136108} \\\approx0.1167[/tex]

The range in which at least 88.9% of the data will reside is:

[tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}\\P(\mu-k\leq X\leq \mu+k)\leq 0.889\\P(3.94-0.1167\leq \leq X\leq 3.94+0.1167)\leq 0.889\\P(3.8233\leq X\leq 4.0567)\leq 0.889\\\approxP(3.82\leq X\leq 4.06)\leq 0.889[/tex]

Thus, the probability of prices of a gallon of milk between $3.82 and $4.06 is 0.889.

Rewrite each of the following statements in the form "∀ _____ x, _____." (a) All dinosaurs are extinct. ∀ x, . (b) Every real number is positive, negative, or zero. ∀ x, . (c) No irrational numbers are integers. ∀ x, .(d) No logicians are lazy. ∀ x, .(e) The number 2,147,581,953 is not equal to the square of any integer. ∀ x, .(f) The number −1 is not equal to the square of any real number.

Answers

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".

A sociologist studying freshmen carried out a survey, asking, among other questions, how often students went out per week, how many hours they studied per day, and how many hours they slept at night.

The tables provide the answers on the time slept and the time spent studying by whether or not students went out.

Go out Hours/day studied Hours slept less than 6 Hours slept 6 or more Total Stay in dorm Hours/day studied Hours slept less than 6 Hours slept 6 or more Total Less than 2 3 9 12 Less than 2 40 120 160 2 or more 14 14 28 2 or more 20 20 40 Total 17 23 40 Total 60 140 200

When we examine the data, we find that students who studied more slept less, both among those who go out and among those who stay in the dorm.

When we combine both groups of students, we find that those who studied more also slept more.

This is an example of:

a. Simpson's Paradox.

b. Andersen's Paradox.

c. the Probability Paradox.

d. the Gaussian Paradox.

Answers

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.

Final answer:

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.

Explanation:

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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The standard deviation of pulse rates of adult males is more than 11 bpm. For the random sample of 153 adult males the pulse rate have a standard deviation of 11 .3 bpm. Find the value of the test statistic.

Answers

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

In the 2012 National Football League (NFL) season, the first three weeks’ games were played with replacement referees because of a labor dispute between the NFL and its regular referees. Many fans and players were concerned with the quality of the replacement referees’ performance. We could examine whether data might reveal any differences between the three weeks’ games played with replacement referees and the next three weeks’ games that were played with regular referees. For example, did games generally take less or more time to play with replacement referees than with regular referees? The pair of dotplots below display data about the duration of games (in minutes), separated by the type of referees officiating the game.​

Answers

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.

Consider this scatter plot.


(a) How would you characterize the relationship between the hours spent on homework and the test

scores? Explain.


(b) Paul uses the function y = 8x + 42 to model the situation. What score does the model predict for 3 h

of homework?

Answers

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

A roofing contractor purchases a shingle delivery truck with a shingle elevator for $42,000. The vehicle requires an average expenditure of $9.50 per hour for fuel and maintenance, and the operator is paid $11.50 per hour. Write a linear equation giving the total cost C of operating this equipment for t hours. (Include the purchase cost of the equipment.)

Answers

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]

Bob can go to work by one of four modes of transportation: 1.) bicycle, 2.) car, 3.) bus and 4.) train. If he takes his bicycle, Bob has a 25% chance of getting to work on time. Because of heavy traffic, there is a 43% chance of being late if he takes his car. If he goes by bus, which has special reserve lanes to help keep schedule, but can be over crowded forcing him to wait for the next bus, he has a 15% chance of being late. The train runs on a very specific schedule, so there is only 5% probability of being late if he decides to take the train. Suppose that Bob is late and his boss see's Bob coming in late. His boss tries to estimate the probability Bob took his car in. Not knowing Bob's travel habbits, what does the boss estimate Bob's probability of taking the car given that he was late.

Answers

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

Final answer:

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%.

Explanation:

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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anju plots points showing equivalent ratios on the coordinate plane below​

Answers

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!

ATM personal identification number (PIN) codes typically consist of four-digit sequences of numbers. Find the probability that if you forget your PIN, then you can guess the correct sequence (a) at random and (b) when you recall the first two digits.

Answers

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]

Final answer:

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.

Explanation:

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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A frequent flyer was interested in the relationship between dollars spent on flying and the distance flown. She sampled 20 frequent flyers of a certain airline. She collected the number of miles flown in the previous year and the total amount of money the flyer spent. A regression line of distance flown on money spent was fit to the data, and the intercept and slope were calculated to be a = 24,000 and b = 10.
A person who spent $ 2000 is predicted to have flown:

a) 34000 miles. b) 54000 miles. c) 44000 miles. d) 24000 miles.

Answers

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

Final answer:

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.

Explanation:

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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Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct): 79, 64, 84, 82,92, and 77. Instead of averaging the six scores, the instructor indicated he would randomly select two grades and report that grade to the student records office.
a. How many different samples of two test grades are possible?
b. List all possible samples of size two and compute the mean of each.
c. Compute the mean of the sample means and compare it to the population mean.

Answers

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.

The United Kingdom is about 900km long from north to south. By how many orders of magnitude (i.e. how many times) would you have to scale down a map of the United Kingdom so that it can fit on a sheet of paper 20cm long

Answers

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:

 1 cm in a paper will be 45 km in the real ground.

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!

Western Athletic Club International (WACI) owns and operates a chain of fitness clubs. Currently WACI has 675 members at its Collegetown location, a 13% increase over the previous year. Next year, WACI hopes to grow by the same number of members. 66% of its members are women. What percent increase will next year's growth represent?

Answers

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%

Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allowed to overlap). Show that the number of possible choices of your team and the opponent’s team equals Pn−1 i=1 i 3 .

Answers

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]

The relationship between the number of games won by a minor league baseball team and the average attendance at their home games is analyzed. A regression to predict the average attendance from the number of games won has an r = 0.73. Interpret this statistic.

Answers

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 .

It tells the direction and the strength of the relation.When r is negative it means there is a negative relationship between the variables .When r is positive it means there is a positive relationship between the variables .When |r|=1 , strong correlation , When r=0 , there is no correlation.If 0.70<|r|<1 , there is a strong correlation.If 0.50<|r|<0.70 , there is a moderate correlation.If 0.30<|r|<0.50 , there is a low correlation.

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.

A trough has a trapezoidal cross section with a height of 5 m and horizontal sides of width 5/2 m and 5 m. Assume the length of the trough is 10 m. Complete parts? (a) and? (b) below.

a) How much work is required to pump the water out of the trough? (to the level of the top of the? trough) when it is? full? Use 1000 kg/m3 for the density of water and9.8 m/s2 for the acceleration due to gravity. Draw a? y-axis in the vertical direction? (parallel to? gravity) and use the midpoint of the bottom of one edge of the trough as the location of the origin. For 0?y?5?, find the? cross-sectional area? A(y) in terms of y.

b) Set Up the integral that gives the work required to pump the water out of the tank.

Answers

Answer:

Step-by-step explanation:

Considering Volume of water inside this = Base area x HeightBase is trapezoid = 0.5 (2.5 + 5) X 5 = 18.75 m2Height = 10 mVolume = 10 * 18.75 = 187.5 m3

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

= 187500kgWork done to pump out the water = change in potential energy of system of waterWork = mg(H2 - H1)

where H2 = initial position of centre of mass from ground

where H1 = final position of centre of mass from ground

here H2 = 2.78 m (from bottom) H1 = 0Work = 187500kg X 9.8 X (2.78 - 0) = 5.11 x 106 JThe detailed analysis of the (b) part is as shown in the attachment

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 trough has a trapezoidal cross section with a height of 5 m and horizontal sides of width 5/2 m and 5 m. Assume the length of the trough is 10 m.

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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Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.46cm and a standard deviation of 0.39cm. Using the empirical rule, what percentage of the apples have diameters that are greater than 7.07cm? Please do not round your answer.

Answers

Final answer:

Using the empirical rule, approximately 84% of apple diameters are greater than 7.07 cm, since this value is within one standard deviation below the mean diameter of 7.46 cm.

Explanation:

The empirical rule (also known as the 68-95-99.7 rule) is a statistical rule which states that for a normally distributed set of data, approximately 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

In this particular problem, we know the mean (7.46cm) and standard deviation (0.39cm) of the apple diameters. We are asked to find the percentage of apple diameters that are greater than 7.07cm. Since 7.07cm is less than one standard deviation below the mean (7.46 - 0.39 = 7.07), from the normal distribution we can say that approximately 84% (50% of the data is above the mean and 34% is between the mean and one standard deviation below the mean) of the apple diameters are larger than 7.07cm according to the Empirical rule.

Thus, using Empirical rule and given statistical mean and standard deviation, we can estimate that about 84% of the apples from this species have a diameter greater than 7.07 cm.

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Using the Empirical Rule, approximately 84% of the apples have diameters greater than 7.07cm since 7.07cm is exactly one standard deviation below the mean of 7.46cm in a normal distribution.

To answer the student's question, we need to apply the Empirical Rule which is used in statistics to describe the distribution of data in a bell-shaped curve, also known as a normal distribution. This rule states that for a data set with a normal distribution, approximately 68% of the data falls within one standard deviation from the mean, 95% falls within two standard deviations, and over 99% falls within three standard deviations.

Given that the mean diameter of the apples is 7.46cm and the standard deviation is 0.39cm, first we need to determine how many standard deviations 7.07cm is from the mean. This is calculated as:

(7.07cm - 7.46cm) / 0.39cm = -1 standard deviation.

According to the Empirical Rule, 68% of apples are within one standard deviation above and below the mean. Since 7.07cm falls exactly at one standard deviation below the mean, we have:

50% of the apples have diameters greater than the mean.Another 34% (half of 68%) have diameters that fall between the mean and one standard deviation below the mean.

Therefore, to find the percentage of apples that have diameters greater than 7.07cm, we would add these two percentages together:

50% + 34% = 84%

Thus, we conclude that approximately 84% of the apples have diameters that are greater than 7.07cm.

Find the general solution to y' + 19t1$ y = t19 Use the variable I=∫et17dt where it occurs in your answer, since this integral is not easily computable. Note that the arbitrary constant C would come from actually computing the integral I, so you do not need to write it.

Answers

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}}}

A space vehicle has an independent backup system for one of its communication networks. The probability that either system will function satisfactorily during a flight is 0.985. What is the probability that during a given flight (a) both systems function satisfactorily, (b) at least one system functions satisfactorily, and (c) both systems fail?

Answers

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.

Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Imagine 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. 1. How many total hot dogs can you make? ________________________________________ 2. Which ingredient limited the number of hot dogs you could make? ____________________ 3. Which ingredient will you have leftovers of? ______________________________________

Answers

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.

Given that,

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.

According to the question,

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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A department store sells sport shirts in three sizes (small, medium, and large), three patterns (plaid, print, and stripe), and two sleeve lengths (long and short). The accompanying tables give the proportions of shirts sold in the various category combinationsShort-sleeved Pattern Size Pl Pr StS 0.04 0.02 0.05M 0.07 0.10 0.12L 0.03 0.07 0.08Long-sleeved Pattern Size Pl Pr StS 0.03 0.02 0.03M 0.06 0.07 0.07L 0.04 0.02 0.08(a) What is the probability that the next shirt sold is a medium, long-sleeved, print shirt?(b) What is the probability that the next shirt sold is a medium print shirt?(c) What is the probability that the next shirt sold is a short-sleeved shirt?What is the probability that the next shirt sold is a long-sleeved shirt?(d) What is the probability that the size of the next shirt sold is a medium?What is the probability that the pattern of the next shirt sold is a print?(e) Given that the shirt just sold was a short-sleeved plaid, what is the probability that its size was medium?(f) Given that the shirt just sold was a medium plaid, what is the probability that it was short-sleeved? What is the probability that it was long-sleeved?

Answers

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;

The probability of a medium, long-sleeved, print shirt is 0.07.The probability of a medium print shirt is 0.17.The probability of a long-sleeved shirt is 0.42.The probability of a medium print shirt is 0.17The probability of a short-sleeved plaid medium is 0.07.The probability of a medium plaid short-sleeved is 0.07.

What is probability?

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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The expression 16t2 approximates a skydiver’s distance, in feet, when in free- a skydiver at 15,000 ft needs to speed up 240 mph (352 ft/s) to join a formation, the expression 15,000 − 352t represents the skydiver’s height above ground after t seconds. Classify each polynomial by its degree and number of terms.
16t2
240
15,000 – 352t
t

Answers

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.

At what points does the helix r(t) = sin(t), cos(t), t intersect the sphere x2 + y2 + z2 = 17? (Round your answers to three decimal places. If an answer does not exist, enter DNE.)

Answers

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

The coordinate of a helix

Given the coordinate of a helix expressed as;

r(t) = [sin(t), cos(t), t]

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

Learn more on helix and sphere here: https://brainly.com/question/1767456

Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine (μg/l). The untrained noses of consumers may be less sensitive, how ever. Here are the DMS odor thresholds for 10 untrained f students:
30 30 42 35 22 33 31 29 19 23
(a) Assume that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 μg/l. Briefly discuss the other two "simple conditions, " using a stemplot to verify that the distribution is roughly symmetric with no outliers.
(b) Give a 95% confidence interval for the mean DMS odor threshold among all students.

Answers

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)    

If the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 6 in. beyond its natural length?

Answers

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.

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