A Chinese restaurant offers 10 different lunch specials. Each weekday for one week, Fiona goes to the restaurant and selects a lunch special. How many different ways are there for her to select her lunches for the week

Answer:

5

Step-by-step explanation:

numerator degrees of freedom=[tex]Treatments-1[/tex]

[tex]N_d_f=5-1=4[/tex]

Total degrees of freedom=[tex]Treatments\times \ Observations Recorded-1[/tex]

[tex]T_d_f=5\times13-1=64[/tex]

Denominator Degrees of freedom=[tex]T_d_f-N_d_f=64-4=60[/tex]

Therefore, to calculate the degrees of freedom corressponding to treatments:

[tex]F=\frac{MSR}{MSE}\\MSR=200\div4=50\\MSE=(800-200)\div 60=10\\F=50\div10=5[/tex]

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:

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:

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:

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:

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:

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:

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:

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.

Step-by-step explanation:

Let x be line xy.

y represents line zm.

z is line zx or line zy

The area of one of the smaller triangles is x/2* y.

same for the other triangle.

The perimeter for either one of the triangles is x/2+y+z.

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:

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:

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:

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:

81.2 is the 62th percentile

Step-by-step explanation:

What is the interpretation for a percentile?

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

300 observations, there are 186 points less than 81.2. Find the percentile for 81.2.

So

[tex]p = \frac{186}{300} = 0.62[/tex]

81.2 is the 62th percentile

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:

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 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:

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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:

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

Answer:

The relation represents y as a function of x, because each value of x is associated with a single value of y.

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:

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:

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: there were 40 questions on her math final.

Step-by-step explanation:

Let x represent the total number of questions in the math test.

Mary got 85% correct on her math test. This means that the number of questions that she got right is

85/100 × x = 0.85 × x = 0.85x

This also means that the number of questions that she missed would be

x - 0.85x = 0.15x

Therefore, if Mary missed 6 questions, it means that

0.15x = 6

Dividing both sides of the equation by 0.15, it becomes

0.15x/0.15 = 6/0.15

x = 40

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:

0.6

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

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Answer:

Mean = 94

Standard deviation = 1.12

The sampling distribution of the sample mean is going to be normally distributed, beause the size of the samples are 80, which is larger than 30.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 94, \sigma = 10[/tex]

By the Central Limit Theorem

The sampling distribution of the sample mean is going to be normally distributed, beause the size of the samples are 80, which is larger than 30.

Mean = 94

Standard deviation:

[tex]s = \frac{10}{\sqrt{80}} = 1.12[/tex]