8. The distribution for the time it takes a student to complete the fall class registration has mean of 94 minutes and standard deviation of 10 minutes. For a random sample of 80 students, determine the mean and standard deviation (standard error) of the sample mean. What can you say about the sampling distribution of the sample mean and why

Answer:

Step-by-step explanation:

We need to observe total 9 days.First we need to choose any 4 days from the 9 days. From these total 9 days, 4 days can be chosen in [tex]^9C_4 = \frac{9!}{4!\times5!} = \frac{6\times7\times8\times9}{4!} = 126[/tex] ways.

The probability of occurring an auto related accident is [tex]\frac{52}{100}[/tex].

Similarly, the probability of not occurring an accident is [tex]\frac{100 - 52}{100} = \frac{48}{100}[/tex].

Hence, the required probability is [tex]126\times(\frac{52}{100} )^4(\frac{48}{100} )^5[/tex].

Answer:

Width = 25

Length = 25

Area = 625

Step-by-step explanation:

The perimeter of a rectangle is given by the sum of its four sides (2L+2W) while the area is given by the product of the its length by its width (LW). It is possible to write the area as a function of width as follows:

[tex]100 = 2L+2W\\L = 50-W\\A=LW=W*(50-W)\\A=50W - W^2[/tex]

The value of W for which the derivate of the area function is zero is the width that yields the maximum area:

[tex]A=50W - W^2\\\frac{dA}{dW}=0=50 - 2W\\ W=25[/tex]

With the value of the width, the length (L) and the area (A) can be also be found:

[tex]L=50-25 = 25\\A=W*L=25*25\\A=625[/tex]

Since the values satisfy the condition W≤L, the answer is:

Width = 25

Length = 25

Area = 625

Answer:

The sequence {an} is A. unbounded

Step-by-step explanation:

The sequence {an} is unbounded.

we say a sequence is bounded if and only it is bounded both above and it is also bounded below. clearly the sequence {an} is an unbounded sequence.

The sequence {an} is bounded if there is a number M>0 such that |an|≤M for every positive n.Every unbounded sequence is divergent

The sequence{an}. is an unbounded sequence, because it has no a finite upper bound

Answer:

Yes, Tom must be admitted to this university.

Step-by-step explanation:

We are given that the scores on national test are normally distributed with a mean of 500 and a standard deviation of 100.

Also, we are provided with the condition that Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test.

Let, X = score in national test, so X ~ N([tex]\mu=500 , \sigma^{2} = 100^{2}[/tex])

The standard normal z distribution is given by;

              Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

Now, z score of probability that tom scores 585 is;

             Z = [tex]\frac{585-500}{100}[/tex] = 0.85

Now, proportion of students scoring below 85% marks is given by;

P(Z < 0.85) = 0.80234

This shows that Tom scored 80.23% of the students who took test while he just have to score more than 70%.

So, it means that Tom must be admitted to this university.

Answer:

a) [tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332[/tex]

b) [tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)][/tex]

And we can find the individual probabilities like this:

[tex]P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649[/tex]

[tex]P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099[/tex]

[tex]P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040[/tex]

[tex]P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465[/tex]

[tex]P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732[/tex]

[tex]P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228[/tex]

[tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033[/tex]

And replacing we got:

[tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)] = 1-0.825=0.175 [/tex]

c) [tex] P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)[/tex]

And replacing we got 0.825

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=17, p=0.7)[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Part a

And we want to find this probability:

[tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332[/tex]

Part b fewer than 10 will check bags

We want this probability:

[tex] P(X<10) [/tex]

We can use the complement rule and we have:

[tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)][/tex]

And we can find the individual probabilities like this:

[tex]P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649[/tex]

[tex]P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099[/tex]

[tex]P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040[/tex]

[tex]P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465[/tex]

[tex]P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732[/tex]

[tex]P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228[/tex]

[tex]P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033[/tex]

And replacing we got:

[tex]P(X<10) = 1-P(X\geq 10) = 1- [P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)] = 1-0.825=0.175 [/tex]

Part c at least 10 bags

We can find this probability like this:

[tex] P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)[/tex]

And replacing we got 0.825

The probability that all 16 passengers will check their bags given a 70% chance for each passenger is found using a binomial distribution, resulting in a probability of 0.0047 when rounded to four decimal places.

The question involves calculating the probability of passengers checking bags in a binomial distribution context. Since each passenger is an independent trial and has a 70 percent chance or probability (p=0.70) of checking bags, this indeed constitutes a binomial problem. Here, success is defined as a passenger checking their bags.

To find the probability that all 16 passengers will check their bags, we need to consider the binomial probability formula for exactly 'k' successes in 'n' trials: P(X = k) = (n choose k) * p^k * (1-p)^(n-k). For this case, every passenger checks their bags, meaning 'k' is 16 and 'n' is also 16.

The formula simplifies in this scenario to P(X = 16) = 0.70^16, as the combination of 16 choose 16 is 1. Following the calculation:

P(X = 16) = 0.70^16 = 0.0047 (rounded to four decimal places).

Answer:

0.717 or 71.7%

Step-by-step explanation:

P(M) = 0.852

P(D) = 0.759

P(M or D) = 0.894

The probability that a randomly selected American has both medical and dental insurance is given by the probability of having medical insurance, added to the probability of having dental insurance, minus the probability of having either insurance:

[tex]P(M\ and\ D) = P(M)+P(D)-P(M\ or\ D)\\P(M\ and\ D) =0.852+0.759-0.894\\P(M\ and\ D) =0.717=71.7\%[/tex]

The probability is 0.717 or 71.7%.

The result is a 71.7% probability of an adult American having both insurances.

The question revolves around finding the probability of a randomly selected adult American having both medical and dental insurance. To calculate this, we use the principle of inclusion-exclusion. According to the problem statement, the percentage of adults with medical insurance is 85.2%, with dental insurance is 75.9%, and with either of the two is 89.4%. The principle of inclusion-exclusion states that the probability of the union of two events (medical or dental insurance) is equal to the sum of the probabilities of each event minus the probability of their intersection (both insurances).

Let's denote the following:

Using the principle of inclusion-exclusion, we find P(Medical and Dental) as follows:

P(Medical and Dental) = P(Medical) + P(Dental) - P(Medical or Dental)

P(Medical and Dental) = 85.2% + 75.9% - 89.4%

P(Medical and Dental) = 160.1% - 89.4%

P(Medical and Dental) = 71.7%

Therefore, the probability that a randomly selected adult American will have both medical and dental insurance is 71.7%.

Answer:

The critical value is 6.314

Step-by-step explanation:

Null hypothesis: The proportion of businesses that earn a profit within the first two years of operation is 80%

Alternate hypothesis: The proportion of businesses e earn a profit within the first two years of operation is greater than 18%

The hypothesis test has one critical value because it is a one-tailed test. It is a one-tailed test because the alternate hypothesis is expressed using the inequality, greater than.

n = 2

degree of freedom = n - 1 = 2 - 1 = 1

significance level = 10%

Using the t-distribution table, critical value corresponding to 1 degree of freedom and 10% significance level is 6.314.

Answer:

To determine the critical value or values for a one-proportion z-test at the 10% significance level when the hypothesis test is left- or right-tailed, we must use the look-up table for zz0.01. Since this is a right-tailed test, our critical value is positive 1.282.

Answer:

a) 0.9961

b) 0.9886          

Step-by-step explanation:

We are given the following information:

We treat passenger not showing up as a success.

P(passenger not showing up) = 0.10

Then the number of passengers follows a binomial distribution, where

[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 = 125

a) probability that every passenger who shows up can take the flight

[tex]P(x \geq 5) = 1- P(x = 0) - P(x = 1)-P(x = 2) - P(x = 3) - P(x = 4) \\= 1-\binom{125}{0}(0.10)^0(1-0.10)^{125} -...-\binom{125}{4}(0.10)^4(1-0.10)^{121}\\=0.9961[/tex]

b)  probability that the flight departs with empty seats

[tex]P(x > 5) =P(x\geq 5) - P(x = 5) \\= 0.9961 -\binom{125}{5}(0.10)^5(1-0.10)^{120}\\=0.9961-0.0075\\ = 0.9886[/tex]

Answer:

0.6

Step-by-step explanation:

Data:

Let W be the event the order is not ready in 30 mins.

Then, let A be the event the first worker got the order. So, we want the probability, P(A/W). Using formula for conditional probability gives:

[tex]P (A/W) = \frac{P(AnW)}{P(W)}[/tex]

In this case, we need to calculate a couple of probabilities.

The event that W can happen can be like this:

1. the first worker got the order

2. the second worker got the order and it is not ready.

the probability for both events are disjoint, so:

for item (1) the probability will be 0.6 times the probability that the first worker takes 30 mins to do the job.

Calculating:

The probability that an exponential with parameter is 2 is given by:

[tex]P = \frac{1}{2}e^-{\frac{2}2} }[/tex]

so, the probability is [tex]0.6e^{-1}[/tex]

thus, the probability is > t is [tex]e^{-\lambda t }[/tex]

That's the same as 0.6

The probability that the first specialist is working on it is 0.4764.

P(not ready in 30 minutes(0.5 hrs) will be:

=P(specialist 1) × P(not ready in 30 minutes |specialist 1) + P(specialist 2) × P(not ready in 30 minutes specialist 2)

= 0.6 × (e-0.5*3) + 0.4 × (e-0.5*2)

= 0.6 × 0.2231 + 0.4 × 0.3679

= 0.2810

Therefore P(first specialist given not ready in 30 minutes(0.5 hrs))

=P(specialist 1) × P(not ready in 30 minutes |specialist 1)/P(not ready in 30 minutes(0.5 hrs))

=0.6 × (e-0.5*3)/0.2810

=0.6 × 0.2231/0.2810

= 0.4764

Learn more about probability on:

https://brainly.com/question/24756209

Answer:

See explanation to get answer.

Step-by-step explanation:

Solution

As per the data and information given in the question, there are three Bidders, one each on Monday, Tuesday and Wednesday.

Bidder 1 may bid on Monday either for $2,000,000 or $3,000,000 with probabilities 0.5 each

Therefore expected pay-off for Bidder 1 is $2,500,000 (2,000,000*.52 + 3,000,000*.5)

Bidder 2 may bid on Tuesday either for $2,000,000 with probability 0.9 or $4,000,000 with probability 0.1

Therefore expected pay-off for Bidder 2 is $2,200,000 (2,000,000*.9 + 4,000,000*.1)

Bidder 3 may bid on Wednesday either for $1,000,000 with probability 0.7 or for $4,000,000 with probability 0.3

Therefore expected pay-off for Bidder 3 is $1,900,000 (1,000,000*.7 + 4,000,000*.3)

Based on the comparison of the above mentioned calculations for the expected pay-off for the bidders, it is recommended that the optimal decision strategy among the bidders is to go for Bidder 1 with highest expected pay-off of $2,500,000 and accept the bid of Bidder 1.

Risk profile for the optimal solution is $2,000,000 with probability 0.5 and $3,000,000 with probability 0.5

Bid may be for $4,000,000 with probability of 0.1 by Bidder 2 or with probability 0.3 by Bidder 3

Answer:

0.3075 = 30.75% probability that a person will wait for more than 7 minutes.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The standard deviation is the square root of the variance.

In this problem, we have that:

[tex]\mu = 6, \sigma = \sqrt{4} = 2[/tex]

Find the probability that a person will wait for more than 7 minutes.

This is 1 subtracted by the pvalue of Z when X = 7. So

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

[tex]Z = \frac{7 - 6}{2}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915

1 - 0.6915 = 0.3075

0.3075 = 30.75% probability that a person will wait for more than 7 minutes.

The probability that a person will wait for more than 7 minutes is 30.85%.

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given;  mean of 6 minutes and variance = 4 minutes, hence:

Standard deviation = √variance = √4 = 2 minutes

For > 7 minutes:

z = (7 - 6)/2 = 0.5

P(z > 0.5) = 1 - P(z < 0.5) = 1 - 0.6915 = 0.3085

The probability that a person will wait for more than 7 minutes is 30.85%.

Find out more on z score at: https://brainly.com/question/25638875

Answer:

[tex] y'' + \frac{7}{t} y = 1[/tex]

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

[tex] y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o[/tex]

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and [tex] q(t) = \frac{7}{t}[/tex] and [tex] g(t) =1[/tex]

We see that [tex] q(t)[/tex] is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by [tex] (0, \infty)[/tex]

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

[tex] t y'' + 7y = t[/tex]

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

[tex] y'' + \frac{7}{t} y = 1[/tex]

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

[tex] y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o[/tex]

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and [tex] q(t) = \frac{7}{t}[/tex] and [tex] g(t) =1[/tex]

We see that [tex] q(t)[/tex] is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by [tex] (0, \infty)[/tex]

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Answer:

The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (0,∞)

Step-by-step explanation:

Given the differential equation:

ty'' + 7y = t .................................(1)

Together with the initial conditions:

y(1) = 1, y'(1) = 7

We want to determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution.

First, let us have the differential equation (1) in the form:

y'' + p(t)y' + q(t)y = r(t) ..................(2)

We do that by dividing (1) by t

So that

y''+ (7/t)y = 1 ....................................(3)

Comparing (3) with (2)

p(t) = 0

q(t) = 7/t

r(t) = 1

For t = 0, p(t) and r(t) are continuous, but q(t) = 7/0, which is undefined. Zero is certainly out of the required points.

In fact (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous. But t = 1, which is contained in the initial conditions is found in (0,∞), and that makes it the correct interval.

So the largest interval containing 1 on which p(t), q(t) and r(t) are defined and continuous is (0,∞)

Answer:

The value of test statistic is 1.338

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 16.000

Sample mean, [tex]\bar{x}[/tex] = 16.218

Sample size, n = 22

Alpha, α = 0.05

Sample standard deviation, s = 0.764

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 16.000\text{ cm}\\H_A: \mu < 16.000\text{ cm}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{16.218 - 16.000}{\frac{0.764}{\sqrt{22}} } = 1.338[/tex]

Thus, the value of test statistic is 1.338

Answer: US predicted population in 2020 and 2030 will be 333 million and 353 million, respectively.

Step-by-step explanation:

Three different points are required to determine the coefficients of correspondent second-order polynomial. Three linear equations are form after substituting the variables associated with those points. [tex]t^{*}[/tex] is the year and [tex]p[/tex] is the population according to US census, measured in millions. That is to say:

[tex]a_{2}\cdot 1990^{2} + a_{1}\cdot 1990 + a_{0} = 249\\a_{2}\cdot 2000^{2} + a_{1}\cdot 2000 + a_{0} = 281\\a_{2}\cdot 2010^{2} + a_{1}\cdot 2010 + a_{0} = 309[/tex]

There are different approaches to solve linear equation systems. In this problem, a matrix-based approach will be used and a solver will be applied in order to minimize the effort and time required to make the need operations. The solution of the 3 x 3 linear system is shown as following:

[tex]a_{2} = -\frac{1}{50},a_{1}=83,a_o=-85719[/tex]

Now, the second-order polynomial is:

[tex]p(t)=-\frac{1}{50}\cdot (t+1990)^{2}+83\cdot(t+1990)-85719[/tex], where [tex]p(t) = 249[/tex] when [tex]t=0[/tex].

The predicted populations are:

[tex]p(30) = 333, p(40) = 353[/tex]

US predicted population in 2020 and 2030 will be 333 million and 353 million, respectively.

Answer:

Yes, Hypothesis test represents a statistically significant difference from 50% .

Step-by-step explanation:

We are given that 400 students were randomly sampled from a large university, and 289 said they did not get enough sleep.

Let Null Hypothesis, [tex]H_0[/tex] : p = 0.50 {means from the students sampled from the university, proportion of them who did not get enough sleep is 50%}

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.50 {means from the students sampled from the university, proportion of them who did not get enough sleep is different from 50%}

The test statistics we will use here is;

        T.S. = [tex]\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion = 289/400 = 0.7225

            n = sample of students = 400

So, test statistics = [tex]\frac{0.7225 - 0.50}{\sqrt{\frac{0.7225(1-0.7225)}{400} } }[/tex] = 3.143

Now, at 1% level of significance, z table gives critical value of 2.5758. Since our test statistics is more than the critical value which means test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that from the students sampled from the university, proportion of them who did not get enough sleep is different from 50%.

The question asks to perform a hypothesis test to determine if there is a significant difference between the proportion of university students who do not get enough sleep and the assumed 50%. The steps include stating hypotheses, calculating the test statistic, and comparing it with a critical value at a 0.01 significance level. A decision to reject or not reject the null hypothesis will be made based on this comparison.

The student's question involves conducting a hypothesis test to determine if there is a statistically significant difference between the proportion of university students who do not get enough sleep and the assumed proportion of 50%. To perform this test at the given significance level of 0.01, the following steps need to be taken:

For our example, with 289 out of 400 students not getting enough sleep, the sample proportion p-hat is 289/400 = 0.7225. Using the formula, the test statistic is then calculated and compared with the critical z-value for a 0.01 significance level. If the absolute value of the test statistic is greater than the critical z-value, the null hypothesis is rejected, suggesting that there is a statistically significant difference from 50%.

If the test statistic does not exceed the critical value, we fail to reject the null hypothesis, indicating that there is not enough evidence to suggest a significant deviation from 50%. Based on the results of the hypothesis test, conclusions about the sleep habits of university students can be drawn.

Answer:

This isn't an answer but it might help

for exponential decay use a(1-r)^t

exponential growth uses a(1+r)^t

compound interest uses p(1+r/n)^(t x n)  

and for interest compounded continuously use Pe^(rt)

a= initial amount, r= rate(%), t= time, P= principle (practically the same as a), and n= number of times compounded

for the different # of times compounded it goes like this: yearly n=1, daily n=365, monthly n=12, weekly n=52, quarterly n=4, semi-annually n=2, and bi-monthly (this is the tricky one) can be either n=6 or n= 24.

also as long as you have a calculator that can do semi-complex things like a Ti 30xs multiview or a ti 36x pro, I have both of those and they can do it as long as you input the formula, hope all this advice helps because I saw you post many of the same types of question. good luck.

Step-by-step explanation:

Answer: she would need $123775.5 in her account when she retires.

Step-by-step explanation:

We would apply the formula for determining future value involving constant deposits at constant intervals. It is expressed as

S = R[{(1 + r)^n - 1)}/r][1 + r]

Where

S represents the future value of the investment.

R represents the regular payments made(could be weekly, monthly)

r = represents interest rate/number of payment intervals

n represents the total number of payments made.

From the information given,

Since she would be taking $1500 four times in a year, then

R = 1500

r = 0.04/4 = 0.01

n = 3 × 20 = 60 times in 20 years

Therefore,

S = 1500[{(1 + 0.01)^60 - 1)}/0.01][1 + 0.01]

S = 1500[{(1.01)^60 - 1)}/0.01][1.01]

S = 1500[{(1.817 - 1)}/0.01][1.01]

S = 1500[0.817/0.01][1.01]

S = 1500[81.7][1.01]

S = 1500 × 82.517

S = 123775.5

Answer:

Option D) The data are continuous because the data can take on any value in an interval          

Step-by-step explanation:

Discrete Data:

Continuous data:

The volumes (in cubic feet) of different rooms

Since the volume can be expressed in decimals, it can take all the values within an interval. Also volume is measured not counted. Hence, it is a continuous variable.

Thus, the correct answer is:

Option D) The data are continuous because the data can take on any value in an interval

Final answer:

The correct answer is D: The data are continuous because the data can take on any value in an interval.

Explanation:

The volumes (in cubic feet) of different rooms would be considered continuous data because they can take on any value within an interval.

Unlike discrete data, which can only take on specific counting numbers, continuous data includes an infinite number of potential values.

For instance, a room can have a volume of 500.5 cubic feet, 500.55 cubic feet, 500.555 cubic feet, and so forth.

This level of precision is due to the fact that volume is a measurable attribute that does not have to be a whole number and can include fractions or decimals as well.

Therefore, the correct answer is D: The data are continuous because the data can take on any value in an interval.

Answer:

The service line is J9263 x 300 to report a unit of 150(300x 0.5 mg = 150).                    

Step-by-step explanation:

The drug J9263 Eloxatin contains 0.5 mg oxaliplatain.

For a infusion of 50 mg the unit reported for service line information is:

- Service line: J9263 x 100

- Unit reported for service line information: 50 = 100 x 0.5 mg

Hence, for a infusion of 150 mg, the unit reported for service line information is:

- Unit reported: 150(300 x 0.5 mg = 150)

- Service line information: J9263 x 300

Therefore, if the physician provided 150 mg infusion of the drug instead of an injection the service line is J9263 x 300 to report a unit of 150(300x 0.5 mg = 150).

I hope it helps you!                          

The unit reported for service line information for 150 mg infusion based on the injection description is :

Given the injection description :

For 150 mg infusion :

Unit reported would be:

Therefore, the service line information and the unit reported would be:

Learn more : https://brainly.com/question/11277823

Answer:

See below

Step-by-step explanation:

Recall first:

Given a vector  

[tex](x_1,x_2,...,x_n)[/tex]

Euclidean norm

[tex]\sqrt{x_1^2+x_2^2+...+x_n^2}[/tex]

Sum norm

[tex]|x_1|+|x_2|+...+|x_n|[/tex]

Max norm

[tex]Max\{|x_1|,|x_2|,...|x_n|\}[/tex]

Now let us apply these definitions to our vectors

Vector (1,1,1)

Euclidean norm

[tex]\sqrt{1^2+1^2+1^2}=\sqrt{3}[/tex]

Sum norm

|1|+|1|+|1| = 3

Max norm

Max{|1|, |1|, |1|} = |1| = 1

Vector (3,0,0)

Euclidean norm

[tex]\sqrt{3^2+0^2+0^2}=\sqrt{3^2}=3[/tex]

Sum norm

|3|+|0|+|0| = 3

Max norm

Max{|3|, |0|, |0|} = |3| = 3

Vector (-1,1,4)

Euclidean norm

[tex]\sqrt{(-1)^2+1^2+4^2}=\sqrt{18}[/tex]

Sum norm

|-1|+|1|+|4| = 1+1+4 =6

Max norm

Max{|-1|, |1|, |4|} = |4| = 4

Vector (-1.4, 3)

Euclidean norm

[tex]\sqrt{(-1.4)^2+3^2}=\sqrt{10.96}[/tex]

Sum norm

|-1.4|+|3| = 1.4+3 = 4.4

Max norm

Max{|-1.4|, |3|} = |3| = 3

Vector (4,4,4,4)

Euclidean norm

[tex]\sqrt{4^2+4^2+4^2+4^2}=\sqrt{4*4^2}=8[/tex]

Sum norm

|4|+|4|+|4|+|4| = 16

Max norm

Max{|4|, |4|, |4|, |4|} = |4| = 4

Final Answer:

- Vector (a) has a Euclidean norm of approximately 1.732, a sum norm of 3, and a max norm of 1.
- Vector (b) has a Euclidean norm of 3, a sum norm of 3, and a max norm of 3.
- Vector (c) has a Euclidean norm of approximately 4.243, a sum norm of 6, and a max norm of 4.
- Vector (d) has a Euclidean norm of approximately 4.123, a sum norm of 5, and a max norm of 4.
- Vector (e) has a Euclidean norm of 8, a sum norm of 16, and a max norm of 4.

Explanation:

Sure, let's calculate the norms for each of the given vectors:

(a) For the vector (1, 1, 1):
- The Euclidean norm (also known as the L2 norm) is the square root of the sum of the squares of the components. For this vector, it's [tex]\(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \approx 1.732\)[/tex].
- The sum norm (or L1 norm) is the sum of the absolute values of the components. For this vector, it's |1| + |1| + |1| = 3.
- The max norm (also known as the infinity norm) is the maximum absolute value component of the vector. For this vector, it's [tex]\(\max(|1|, |1|, |1|) = 1\)[/tex].

(b) For the vector (3, 0, 0):
- The Euclidean norm of this vector is [tex]\(\sqrt{3^2 + 0^2 + 0^2} = \sqrt{9} = 3\)[/tex].
- The sum norm of this vector is |3| + |0| + |0| = 3.
- The max norm of this vector is max(|3|, |0|, |0|) = 3.

(c) For the vector (-1, 1, 4):
- The Euclidean norm of this vector is [tex]\(\sqrt{(-1)^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} \approx 4.243\)[/tex].
- The sum norm is (|-1| + |1| + |4| = 1 + 1 + 4 = 6).
- The max norm is max(|-1|, |1|, |4|) = 4.

(d) For the vector (-1, 4):
- The Euclidean norm is [tex]\(\sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.123\)[/tex].
- The sum norm is |-1| + |4| = 1 + 4 = 5.
- The max norm is max(|-1|, |4|) = 4.

(e) For the vector (4, 4, 4, 4):
- The Euclidean norm is [tex]\(\sqrt{4^2 + 4^2 + 4^2 + 4^2} = \sqrt{16 + 16 + 16 + 16} = \sqrt{64} = 8\)[/tex].
- The sum norm is |4| + |4| + |4| + |4| = 4 + 4 + 4 + 4 = 16.
- The max norm is max(|4|, |4|, |4|, |4|) = 4.

Now we have calculated the three types of norms for each of the vectors:
- Vector (a) has a Euclidean norm of approximately 1.732, a sum norm of 3, and a max norm of 1.
- Vector (b) has a Euclidean norm of 3, a sum norm of 3, and a max norm of 3.
- Vector (c) has a Euclidean norm of approximately 4.243, a sum norm of 6, and a max norm of 4.
- Vector (d) has a Euclidean norm of approximately 4.123, a sum norm of 5, and a max norm of 4.
- Vector (e) has a Euclidean norm of 8, a sum norm of 16, and a max norm of 4.

Answer:

2,075,673,600 batting orders may occur.

Step-by-step explanation:

The order of the first eight batters in the batting order is important. For example, if we exchange Jonathan Schoop with Adam Jones in the lineup, that is a different lineup. So we use the permutations formula to solve this problem.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

[tex]P_{(n,x)} = \frac{n!}{(n-x)!}[/tex]

First 8 batters

8 players from a set of 16. So

[tex]P_{(16,8)} = \frac{16!}{(16 - 8)!} = 518918400[/tex]

Last batter:

Any of the four pitchers.

How many different batting orders may​ occur?

4*518918400 = 2,075,673,600

2,075,673,600 batting orders may occur.

Final answer:

20,922,789,888,000 different ways.

Explanation:

To find out the number of different batting orders that may occur with the given constraints, we consider that there are 16 players who can play any position other than pitcher, and since the pitcher bats last, we only need to determine the batting order for the first 8 positions which any of the 16 players can fill. Since the order matters, we use permutations to calculate the number of ways to arrange these players. After the 8 positions are filled, the pitcher is automatically assigned the last batting slot.

The formula for permutations is:

P(n, k) = n! / (n-k)!,

where n is the total number of items to choose from, k is the number of items to arrange, and '!' denotes factorial which is the product of an integer and all the integers below it.

Therefore, the number of different batting orders is P(16, 8) which equals:

16! / (16-8)! = 16! / 8! = 20,922,789,888,000 possible ways.

This includes 1 of the 4 pitchers, who will be the last to bat in each permutation.

Answer:

The driving distance for meeting at the three centres can be written as

d1=columbus

d2=Des moines

d3=Boise

The distance from des moines to columbus to 650 miles

from Boise to des moines is 1350

Therefore the distance from columbus to Boise

if the meeting is holding at columbus

0+650+1350=2000

Pr(columbus)=1/3

Pr(Des moines)=1/3

Pr(Boise)=1/3

a. Probability distribution is

capital       c            DM         B

di               0            650        2000

Pr(di)           1/3           1/3         1/3

b. Expected value is multiplication of the probability of d1 and the outcome

E(x)=0*1/3=0

c. find the variance of d1 first

Var=(x-E(x))^2*Pr(d1)

Var=(0-0)^2*1/3

Var=0

the square root of var=standard deviation

S.D=0

d. probability distribution of d2 and d3 is equal to the probability distribution of d1 , because they all have a probability of 1/3(the likelihood that an event will occur is 1/3 for the meeting \location

e. d1=0

d2=650

d1+d2=650

pr(d1+d2)=1/3+1/3=2/3

Pr(d1+d2) will be on the vertical axis, while d1+d2 will be plotted on the horizontal axis of the probability distribution graph

Step-by-step explanation:

The driving distance for meeting at the three centres can be written as

d1=columbus

d2=Des moines

d3=Boise

The distance from des moines to columbus to 650 miles

from Boise to des moines is 1350

Therefore the distance from columbus to Boise

if the meeting is holding at columbus

0+650+1350=2000

Pr(columbus)=1/3

Pr(Des moines)=1/3

Pr(Boise)=1/3

a. Probability distribution is

capital       c            DM         B

di               0            650        2000

Pr(di)           1/3           1/3         1/3

b. Expected value is multiplication of the probability of d1 and the outcome

E(x)=0*1/3=0

c. find the variance of d1 first

Var=(x-E(x))^2*Pr(d1)

Var=(0-0)^2*1/3

Var=0

the square root of var=standard deviation

S.D=0

d. probability distribution of d2 and d3 is equal to the probability distribution of d1 , because they all have a probability of 1/3(the likelihood that an event will occur is 1/3 for the meeting \location

e. d1=0

d2=650

d1+d2=650

pr(d1+d2)=1/3+1/3=2/3

Pr(d1+d2) will be on the vertical axis, while d1+d2 will be plotted on the horizontal axis of the probability distribution graph

Answer:

The 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).

It is better to take a larger sample to derive conclusion about the true parameter value.

Step-by-step explanation:

The (1 - α) % confidence interval for proportion is:

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

Given:

n = 2000

X = 200

The sample proportion is:

[tex]\hat p=\frac{X}{n}=\frac{200}{2000}=0.10[/tex]

The critical value of z for 95% confidence interval is:

[tex]z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

Compute the 95% confidence interval as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.10\pm1.96\times\sqrt{\frac{0.10(1-0.10)}{2000}}\\=0.10\pm0.0034\\=(0.0966, 0.1034)[/tex]

Thus, the 95% confidence interval for the true proportion of defective batteries is (0.0966, 0.1034).

Now if in a sample of 100 batteries there are 15 defectives, the the 95% confidence interval for this sample is:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.15\pm1.96\times\sqrt{\frac{0.15(1-0.15)}{100}}\\=0.15\pm0.0706\\=(0.0794, 0.2206)[/tex]

It can be observed that as the sample size was decreased the width of the confidence interval was increased.

Thus, it can be concluded that it is better to take a larger sample to derive conclusion about the true parameter value.

Answer:

C. 0.896

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 80, p = 0.76[/tex]

So

[tex]E(X) = np = 80*0.76 = 60.8[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{80*0.76*0.24} = 3.82[/tex]

What is the probability that at least seventy percent of the sample prefers Coke to Pepsi?

0.7*80 = 56.

This probability is 1 subtracted by the pvalue of Z when X = 56. So

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

[tex]Z = \frac{56 - 60.8}{3.82}[/tex]

[tex]Z = -1.26[/tex]

[tex]Z = -1.26[/tex] has a pvalue of 0.1040.

1 - 0.1040 = 0.8960

So the correct answer is:

C. 0.896

Answer:

0.435 s

Step-by-step explanation:

Weight of car=m=4000 pounds

Spring constant for each shock absorber=k=540lbs/in

Effective spring constant=4k=4(540)=2160 lbs/in

We have to find the period of oscillation of the vertical motion by assuming no damping.

Time period, T=[tex]2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{w}{gk}[/tex]

Where g=[tex]386 in/s^2[/tex]

[tex]\pi=3.14[/tex]

Using the formula

[tex]T=2\times 3.14\sqrt{\frac{4000}{386\times 2160}}=0.435 s[/tex]

Hence,the period of oscillation of the vertical motion of the car=0.435 s

Answer:

0.1222 or 12.22%

Yes, it is unlikely.

Step-by-step explanation:

If the probability of someone being infected is 0.005, then the probability of someone not being infected is 0.995. In order for the combined sample to test negative (N), all of the 26 people must test negative. Thus, the probability that the combined sample tests positive is:

[tex]P(Sample = P) = 1 -P(N=26)\\P(Sample = P) = 1 -0.995^{26}\\P(Sample = P) = 0.1222=12.22\%[/tex]

There is a 0.1222 or 12.22% probability that the combined sample tests positive, which is unlikely to occur.

Final answer:

To calculate the magnitude of the torque exerted on a bolt, use the formula τ = rFsinθ with the given values: lever arm length of 6 inches, force of 16 lbs, and an assumed angle of 90 degrees. The calculated torque is 96 in.lb

Explanation:

The magnitude of torque (τ) can be calculated using the formula τ = rFsinθ, where r is the distance from the pivot to the point where the force is applied, F is the force applied, and θ is the angle between the force and the arm of the lever—in this case, assuming the force is perpendicular (90 degrees) to the lever arm for maximum torque.

Given that |vector PQ| = 6 in. as the lever arm (r) and |F| = 16 lb. as the force applied, and assuming the angle θ is 90 degrees (since it's not provided but necessary for calculating maximum torque):

τ = rFsinθ = (6 in.)(16 lb.)(sin90°) = 96 in.lb. because sin90° = 1.

Therefore, the magnitude of the torque exerted by the force on the bolt at point P is 96 in.lb.

The rabbit population doubles every 6 weeks. After converting 112 days to 16 weeks, we calculate the growth over 2.67 six-week periods to find that there will be approximately 61 rabbits after 112 days.

The rabbits population grows exponentially and doubles every 6 weeks. To determine the population after a certain time, we would typically use a formula like P(t) = P0 (2^(t/T)), where P(t) is the population at time t, P0 is the initial population, 2 represents the doubling factor, and T is the time period for doubling in the same units as t. However, we need to convert days to weeks since the doubling time is given in weeks. There are 7 days in a week, so 112 days is equal to 112/7 = 16 weeks.

Following this, we calculate the number of periods in 16 weeks by dividing 16 weeks by 6 (the number of weeks it takes for the population to double). This gives us 16/6 = approximately 2.67. The population after 16 weeks would be P(16) = 9 (2^(2.67)). Simplifying with a calculator, we find the population to be about 61 rabbits, rounding to the nearest whole number.

Answer:

0.06597

Step-by-step explanation:

Given that thirty-seven percent of the American population has blood type O+

Five Americans are tested for blood group.

Assuming these five Americans are not related, we can say that each person is independent of the other to have O+ blood group.

Also probability of any one having this blood group = p = 0.37

So X no of Americans out of five who were having this blood group is binomial with p =0.37 and n =5

Required probability

=The probability that at least four of the next five Americans tested will have blood type O+

= [tex]P(X\geq 4)\\= P(X=4)+P(x=5)\\= 5C4 (0.37)^4 (1-0.37) + 5C5 (0.37)^5\\= 0.06597[/tex]

Answer:

Required probability = 0.066

Step-by-step explanation:

We are given that Thirty-seven percent of the American population has blood type O+.

Firstly, the binomial probability is given by;

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

where, n = number of trails(samples) taken = 5 Americans

           r = number of successes = at least four

           p = probability of success and success in our question is % of

                 the American population having blood type O+ , i.e. 37%.

Let X = Number of people tested having blood type O+

So, X ~ [tex]Binom(n=5,p=0.37)[/tex]

So, probability that at least four of the next five Americans tested will have blood type O+ = P(X >= 4)

P(X >= 4) = P(X = 4) + P(X = 5)

                = [tex]\binom{5}{4}0.37^{4}(1-0.37)^{5-4} + \binom{5}{5}0.37^{5}(1-0.37)^{5-5}[/tex]

                = [tex]5*0.37^{4}*0.63^{1} +1*0.37^{5}*1[/tex] = 0.066.

Answer:

Step-by-step explanation:

Hello!

The population of the study is the Toco Toucan, the largest member of the toucan family. It is believed that the length and size of the beak are due to its function of dissipating heat (a cooling mechanism).

You have two variables of interest.

X: Outdoor temperature.

Temperature (ºC) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Y: Total beak heat loss, as a percent of the total body heat loss of a toucan.

Percent heat loss from beak 33 34 33 36 36 47 52 51 41 50 49 50 55 60 60 62

The objective is to construct a least-squares regression line to predict the body heat loss of the toucans given the outdoor temperature.

Using a statistical software I estimated the regression model:

^Yi= 2.71 + 1.96Xi

1. To predict what will be the value of the beak heat loss as a percent of the total body heat loss at a temperature of 25ºC you have to simply replace the value of X in the equation:

^Yi= 2.71 + 1.96 (25)= 51.71ºC

The expected beak heat loss given an outdoor temperature of 25ºC is 51.71ºC.

2. To know what percent of the variation of the beak heat loss is explained by the outdoor temperature you have to calculate the coefficient of determination.

R²= 0.85

This means that 85% of the variability of the beak heat loss as a percent of the total body heat loss of the Toco Toucans is explained by the outdoor Temperature under the estimated model ^Yi= 2.71 + 1.96Xi

3. The correlation coefficient between these two variables is r= 0.92, this means that there is a strong positive linear correlation between the beak heat loss and the outdoor temperature. This means that when the outdoor temperature rises, the beak heat loss increases.

I hope it helps!

Answer:

a. x= $6, y= $8

b. z= $2

Step-by-step explanation:

at equilibrium, the marginal utility per dollar spent will be equal i.e

zA=zB

20-3x=18-2y

upon simplification, we arrive at

[tex]x=\frac{2}{3} (1+y)\\[/tex]......equation 1

since the total amount to spend is $14, then

x+y=14

x=14-y..............equation 2

if we solve equation 1 and equation 2 simultaneously

[tex]14-y=\frac{2}{3} (1+y)\\42-3y=2+2y\\5y=40\\y=8[/tex]

Hence for y=8

x=14-y=14-8

x=6

Hence the amount spent on Product A is $6 and the amount spent on product B is $8

b. to determine the amount of utility the marginal dollar will yield, we substitute the values of y and x into the the given equation,

zA=20-3x=20-3(6)

zA=20-18=2

zB=18-2y=18-2(8)

zB=18-16=2

hence the amount spent on the marginal utility is $2