At most restaurants, the more food you order, the more money you have to pay. This is true whether you go to an overpriced "gourmet" restaurant where the plates are tiny but very expensive, or to a cafeteria where the portions are huge and the prices cheap. If you collect data at any ONE restaurant, the correlation between the amount of food served per person and the price paid for it should be:

A.positive and fairly strong
B.negative and fairly strong
C.fairly weak

Answer:

0.028 Gallons/mile

Step-by-step explanation:

1 Gallon = 3.78541 Litres

1 Mile    = 1.60934 Km

Total Fuel Expenditure (Dec 2019) = PKR 7,000

Fuel cost / Litre = 114Litre/Km

Total Fuel consumption = (7000/114)

                                      = 61.40 Litres or 16.22 Gallons

Total Distance Travelled in Dec 2019 = 921 Km or 572.28 miles

Fuel Economy = 16.22/572.28

                       = 0.028 Gallons/mile

One time tank fill up cost = PKR 1,500

Answer:

We can answer that the product of 9 and 7 1/8 lies between 64 and 65, the two whole numbers asked.

Step-by-step explanation:

Let's find the product of 9 and 7 1/8, this way:

9 * 7 1/8 = 9 * 7.125

9 * 7.125 = (9 * 7) + (9 * .125)

63 + 1.125 = 64.125

We can answer that the product of 9 and 7 1/8 lies between 64 and 65, the two whole numbers asked.

Answer:

64 and 65

Step-by-step explanation:

9 X 7 1/8

=64.125

Answer:

a) The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]

b) The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]

c) The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]                  

Step-by-step explanation:

We are given the following data on oxygen consumption (mL/kg/min):

28.6, 49.4, 30.3, 28.2, 28.9, 26.4, 33.8, 29.9, 23.5, 30.2

a) The sample range

Range = Maximum - Minimum

[tex]\text{Range} = 49.4 - 23.5 = 25.9[/tex]

The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]

b) The sample variance

[tex]\text{Variance} = \displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{309.2}{10} = 30.92[/tex]

Sum of squares of differences =

5.3824 + 341.5104 + 0.3844 + 7.3984 + 4.0804 + 20.4304 + 8.2944 + 1.0404 + 55.0564 + 0.5184 = 444.096

[tex]s^2 = \dfrac{444.096}{9} = 49.344[/tex]

The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]

c)  The sample standard deviation

It is the square root of sample variance.

[tex]s = \sqrt{s^2} = \sqrt{49.344} = 7.0245[/tex]

The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]

Answer:

84

Step-by-step explanation:

21x4=84

Answer:the total number of pieces of equipment in the bus is 84

Step-by-step explanation:

The total number of members of the soccer team on the bus is 21.

if each player carries on 4 pieces of equipment, then the total number of pieces of equipment in the bus would be

21 × 4 = 84

Answer:

The generated random variate is  X = 6.173

Explanation:

To generate an exponential random variate (with parameter λ) from a pseudo-random number U, we use the formula:

X = − ln(U)/λ

Fro the given problem, we have  U = 0.128 and λ = 3.

So, X = -In(0.128) / (1/3) = -In(0.128)/ 0.333 = 2.056/0.333 = 6.173

To generate an Exponential random variate with a given pseudo-random number, we use the formula -ln(U) / λ, where U is the pseudo-random number and λ is the rate parameter. In this case, plugging in U = 0.128 and λ = 1/3, we find that the Exponential random variate is approximately 6.147.

To generate an Exponential random variate, we use the formula:

X = -ln(U) / λ

where U is a pseudo-random number between 0 and 1, and λ is the rate parameter for the Exponential distribution.

In this case, U = 0.128 and λ = 1/3. Plugging in these values, we have:

X = -ln(0.128) / (1/3)

X ≈ -ln(0.128) / (1/3) ≈ -(-2.049) / (1/3) ≈ 2.049 / (1/3) ≈ 6.147

Therefore, the Exponential random variate is approximately 6.147.

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Answer:

The life expectancy for 2100 will be 134 years.

Step-by-step explanation:

Consider the provided information.

In 1900 women life expectancy was 49 years and in 2000 it was 81 years.

First calculate the difference of life expectancy between 1900 to 2000.

81-49=32 years

Percentage change is:

[tex]\text{Percentage change}=\frac{32}{49}\times 100\approx65.31[/tex]

It is given that the increase in percentage remains same.

Therefore, next year life expectancy will be:

[tex]\begin{aligned}81+65.31\% \times81&=81(1+\frac{65.31}{100})\\&=81(1.6531)\\&=133.9011\approx134\end{aligned}[/tex]

Hence, the life expectancy for 2100 will be 134 years.

To find the life expectancy for women in 2100 with the same percentage increase as from 1900 to 2000, we calculate the percentage increase between these years and apply it to the year 2000 figure, resulting in an estimated life expectancy of approximately 134 years in 2100.

In 1900, the life expectancy for women was 49 years, and in 2000 it was 81 years. To calculate the life expectancy for women in 2100, assuming it increases by the same percentage as from 1900 to 2000, we must first determine the percentage increase from 1900 to 2000.

The percentage increase is calculated as follows:

Percentage Increase = ((Life Expectancy in 2000 - Life Expectancy in 1900) / Life Expectancy in 1900) * 100

This gives us:

Percentage Increase = ((81 - 49) / 49) * 100 = (32 / 49) * 100 ≈ 65.31%

Now, we apply this percentage increase to the life expectancy in 2000 to estimate the life expectancy for women in 2100:

Estimated Life Expectancy in 2100 = Life Expectancy in 2000 * (1 + Percentage Increase)

Estimated Life Expectancy in 2100 = 81 * (1 + 0.6531) ≈ 133.9

Therefore, if the life expectancy for women continues to increase by the same percentage, it will be approximately 134 years in 2100.

Answer:

The correct answer is B. N = st

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

t = number of toothpicks received by each student

s = number of students

N = total number of students

2. Which equation can be used to find the total number of toothpicks, N, given out by the teacher for the project?

Total number of toothpicks for the project = number of toothpicks received by each student * number of students, replacing with the variables:

N = t * s = s * t

The correct answer is B. N = st

Answer:

B. N =st

Step-by-step explanation:

Answer:

[tex]y=2x+2[/tex]

Step-by-step explanation:

we know that

The equation of a linear function has no exponents higher than 1, and the graph of a linear function is a straight line.

Verify each case

case 1) we have

[tex]y=2x+2[/tex]

Is the equation of a line in slope intercept form

so

Is a straight line and has no exponents higher than 1

therefore

Is a linear equation

case 2) we have

[tex]y=2x^{2}+2[/tex]

Is a quadratic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

case 3) we have

[tex]y=2x^{3}+2[/tex]

Is a cubic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

case 4) we have

[tex]y=2x^{4}+2[/tex]

Is a quartic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

Answer:

The true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22

Step-by-step explanation:

False.

By definition the sample mean is defined as:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

For this case the value for the sample size is n =190 and the calculated sample mean is [tex] \bar X=22[/tex]. This value represent the sample and for this case we can't assume that this value represent at all the population as the population mean [tex]\mu[/tex] since we probably have variability from the data of the students at City College.

So we can conclude that the true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22

The true mean  it probably could be larger, smaller, or equal to 22.

Given that:

Total student, n = 190.

Average age of 190 students, [tex]\bar X = 22\\[/tex].

By definition the sample mean is defined as:

[tex]\bar X =\dfrac{{\sum_{i=1}^nX_i}}{n}[/tex]

For this case, the sample size n =190 and the calculated sample mean is [tex]\bar X = 22\\[/tex]. This value represent the sample and for this case can't assume that this value represent at all the population as the population mean  since probably have variability from the data of the students at city college.

Hence, conclude that the true mean  it probably could be larger, smaller, or equal to 22

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Answer:

unit normal vector n will be n=(a,b,c) = (4/√171,11/√171,6/√171)

Step-by-step explanation:

There are several ways to solve this problem

1) build 2 vectors AB and BC such that the vectorial product ABxBC is the orthogonal vector to the plane , then find unit vector

2) since the 3 points belongs to the plane solve a linear system of 4 equation with 4 variables

for the second solution , the equation of the plane with normal vector n=(a,b,c) and containing the point (x₀,y₀,z₀) is

a*(x-x₀)+b*(y-y₀)+c*(z-z₀) =0

and

a²+b²+c² = 1 (unit vector)

then choosing A=(x₀,y₀,z₀)=(1,0,0)

a*(x-1)+b*(y-0)+c*(z-0) =0

for B

a*(3-1)+b*(-1-0)+c*(-3-0) =0

1) 2*a - b - 3*c =0

for C

a*(1-1)+b*(3-0)+c*(-2-0) =0

2) 3*b - 2*c=0 → b= 2/3*c

replacing in 1)

2*a -  2/3*c - 3*c =0

2*a-11/3*c=0 → a=11/6*c

thus

a²+b²+c² = 1

(11/6*c)²+(2/3*c)²+c² = 1

(121/36+4/9+1)*c² = 1

171/36*c²=1 → c= 6/√171

therefore

a=11/6*c = 11/6*6/√171= 11/√171

b=2/3*c= 2/3*6/√171= 4/√171

then the unit normal vector n will be

n=(a,b,c) = (4/√171,11/√171,6/√171)

Answer:

a)  10 pieces of luggage to be lost is not likely because its z-score is -2.29

b) mean=20 and standard deviation=4.359

c) 40 pieces of luggage lost is very extreme given that in average 5% of the luggage is lost.

d) The probability that less than 6 pieces of luggage are lost or stolen is 0.0007

Step-by-step explanation:

Let p be the proportion of stolen or damaged luggages of all the luggage handled by the airlines at JFK in the last 10 years.

p=0.05 (5%)

b) Mean and standard deviation of the binomial distribution is given as:

Thus,

Mean=400×0.05=20 and

Standard Deviation=[tex]\sqrt{400*0.05*0.95}[/tex] ≈ 4.359

a) We should calculate z-score of 10 pieces of luggage lost to decide if it is unusual.

Z-score can be calculated as follows:

[tex]z= \frac{X-M}{s}}[/tex] where

Thus, [tex]z= \frac{10-20}{4.359}[/tex] ≈ -2.29

This is unusual because it is in the 1st percentile and 99% of the possible number of lost luggage are higher than this score.

c) 40 pieces of lost luggage is also unusual because its z score is:

[tex]z= \frac{40-20}{4.359}[/tex] ≈ 4.59, in the 100th percentile, that is very extreme.

d) the probability that less than 6 pieces of luggage are lost or stolen

= P(z<z*) where z* is the z-score of 6 pieces of luggage are lost or stolen

That is  [tex]z= \frac{6-20}{4.359}[/tex] ≈ -3.21

And P(z<3.21)= 0.0007

Answer:

a) H0: μm = μf versus Ha: μm < μf

b) [tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]

c) [tex]p_v =P(t_{133}<-0.200)=0.421[/tex]  

d) Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

Step-by-step explanation:

Data given and notation  

[tex]\bar X_{m}=2.68[/tex] represent the mean for the sample male

[tex]\bar X_{f}=2.7[/tex] represent the mean for the sample female

[tex]s_{m}=0.65[/tex] represent the sample standard deviation for the males

[tex]s_{f}=0.5[/tex] represent the sample standard deviation for the females  

[tex]n_{m}=66[/tex] sample size for the group male  

[tex]n_{f}=69[/tex] sample size for the group female  

t would represent the statistic (variable of interest)  

Part a

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the average male soldier spends less time in the shower than the average female soldier, the system of hypothesis would be:  

Null hypothesis:[tex]\mu_{m}-\mu_{f}\geq 0[/tex]  

Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]  

Or equivalently:

Null hypothesis:[tex]\mu_{m}-\mu_{f}= 0[/tex]  

Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]  

And the best option is:

H0: μm = μf versus Ha: μm < μf

Part b

We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  

[tex]t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{f}}{n_{f}}}}[/tex] (1)

And the degrees of freedom are given by [tex]df=n_m +n_f -2=66+69-2=133[/tex]  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

What is the test statistic?

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

[tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]

Part c What is the p-value?

Since is a left tailed test the p value would be:  

[tex]p_v =P(t_{133}<-0.200)=0.421[/tex]  

Part d

The significance level given is [tex] \alpha =0.1[/tex] since the p value is higher than the significance level we can conclude:

Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

The question relates to hypothesis testing in statistics, insights into average shower times for male and female soldiers. After formulating the hypotheses, we calculate the test statistic using the provided sample data, and then find the corresponding P-value. If the P-value is less than our significance level, we reject the null hypothesis and side with the alternative hypothesis.

The subject of this question falls under Mathematics, specifically it deals with hypothesis testing statistics.

a) The appropriate hypotheses for the engineer to consider would be: H0: μm = μf versus Ha: μm > μf

b) To calculate the test statistic, we use the formula for the test statistic in a independent two-sample t-test which incorporates the sample sizes, means, and standard deviations from the two groups. The formula is (avg(male soldiers)-avg(female soldiers))/sqrt(((sd(male soldiers))^2/number(male soldiers))+((sd(female soldiers))^2/number(female soldiers))). Plug in given values, we can obtain the test statistic.

c) The P-value can be obtained by looking up the test statistic in the T distribution table.

d) If the P-value is greater than the level of significance (0.1), we would fail to reject the null hypothesis. If the P-value is less than the level of significance, we would reject the null hypothesis. The conclusion is based on the specific P-value we computed.

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Answer:

a) [tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]

[tex]P(z<-0.882)=0.189[/tex]

b) [tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]

[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]

c)  [tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]

[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(4.5,1.7)[/tex]  

Where [tex]\mu=4.5[/tex] and [tex]\sigma=1.7[/tex]

We are interested on this probability

[tex]P(X<3.0)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]

And we can find this probability using excel or the normal standard table:

[tex]P(z<-0.882)=0.189[/tex]

Part b

We are interested on this probability

[tex]P(X>3.0)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]

And we can find this probability using excel or the normal standard table:

[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]

Part c

[tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]

And we can find this probability using excel or the normal standard table liek this:

[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]

Answer:

D. $140.

Step-by-step explanation:

Given:

Cost of clothes = $76.17

Cost of books = $44.98

Cost of meal = $19.15

We need to find the best estimate of total cost of her shopping trip.

Solution:

First we will find the total cost of her shopping trip.

total cost of her shopping trip is equal to sum of Cost of clothes, Cost of books and Cost of meal.

framing in equation form we get;

total cost of her shopping trip = [tex]76.17+44.98+19.15 = \$140.3[/tex]

Now we can say that;

140.3 is close to 140

Hence Best estimate of total cost of shopping trip is $140.

Answer: a. Parameter

Step-by-step explanation:

Parameter can be defined as a fact/characteristic about a whole population.

For example:

i. 20% of my class are boys

ii. 40% of Canadian senators are women.

Parameter usually deal with a small measurable population.

While statistic is a characteristic/fact about the sample.

Therefore the case above is a parameter because the students in the class is the population and the stated (Two thirds of all the students in this class are women) is a fact about the population.

Answer:

Parameter

Step-by-step explanation:

The given value is Parameter because it contains the measurement of population. In short, the given value 2/3 is the estimated value of population.

The population consists of all the students in the class. If the ratio of women from all the students in the class is calculated, then it is a measure of population. Thus, the given value is parameter.

Answer:  y=x+2

Step-by-step explanation:

Final answer:

The angle -6 radians places the terminal side in the third quadrant with a reference angle of 0.28 radians when expressed in the positive acute form and rounded to four decimal places.

Explanation:

The question involves determining the quadrant of an angle measured in radians and finding its corresponding reference angle, which is a common task in Mathematics, specifically in the study of trigonometry.

To find the quadrant in which the terminal side of the angle -6 radians lies, we need to recall that one full revolution around the unit circle is 2π or approximately 6.28 radians. Since the given angle is negative, we move in the clockwise direction from the positive x-axis. Dividing -6 by 6.28, we realize that it is just shy of a full negative revolution, thereby placing the terminal side in the third quadrant.

To find the reference angle ¯θ, which is the positive acute angle the terminal side makes with the x-axis, we need to subtract the given angle from one full revolution (if necessary) and find the absolute value. Hence, ¯θ = |2×π - (-6)| = |6.28 - (-6)| = |0.28|, which is the same as 0.28 radians when rounded to four decimal places.

Answer: 0.4667

Step-by-step explanation:

According to 68–95–99.7 rule , About  99.7% of all data values lies with in 3 standard deviations from population mean ([tex]\mu[/tex]).

Here , margin of error = 3s , where s is standard deviation.

As per given , we have want our sample mean [tex]\overline{x}[/tex] to estimate μ μ with an error of no more than 1.4 point in either direction.

If 99.7% of all samples give an [tex]\overline{x}[/tex] within 1.4 , it means that

[tex]3s=1.4[/tex]

Divide boths ides by 3 , we get

[tex]s=0.466666666667\approx0.4667[/tex]

Hence, So [tex]\overline{x}[/tex] must have 0.4667 as standard deviation so that 99.7 % 99.7% of all samples give an [tex]\overline{x}[/tex] within 1.4 point of μ .

To estimate the mean score of an MCAT with an error of no more than 1.4 points, one needs to calculate the standard deviation for the sample mean that allows 99.7% of samples to fall within this range. After determining this standard deviation, apply the Central Limit Theorem to increase sample size and approach the population mean.

To estimate the mean score μ of those who took the Medical College Admission Test on your campus with an error of no more than 1.4 points in either direction, you are seeking a standard deviation for the sample mean ¯x that will allow for 99.7% of samples to fall within this range. This relates to the 68 – 95 – 99.7 rule, which in this context will be interpreted as within three standard deviations of the mean, hence, you need ¯x to have a standard deviation of 1.4/3.

Using the formula for ¯x standard deviation which is σ/√n (where σ is population standard deviation and n is the sample size), you can rewrite the formula as 1.4/3 = 6.2/√n and solve for n to get the desired sample size.

This use of the empirical rule and standard deviations is part of a bigger concept known as the Central Limit Theorem, which states that as sample size increases, the sample mean gets nearer to the population mean.

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Answer:

11.46°

Step-by-step explanation:

Let x be the angle that yields an arc length of 1 feet, if r= 5 feet, applying the circumference length equation and assuming that a full circumference has 360 degrees:

[tex]1=\frac{x}{360}*2\pi r \\1=\frac{x}{360}*2\pi 5\\x=\frac{360}{2 \pi 5}=11.46^o[/tex]

Rounding to the nearest hundredth, the angle should be 11.46°

The differential equation y' = y(y - 3) satisfies all the given conditions. It has equilibrium solutions at y = 0 and y = 3, y' > 0 for 0 < y < 3, and y' < 0 for -inf < y < 0 and 3 < y < inf.

The autonomous differential equation that meets the given conditions is

y' = y(y-3). To confirm this, we can check each condition:

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Answer:

a) N(P) = -6P + 16000

b) slope = -6 computers per dollar

That means the number of computer sold reduce by 6 per dollar increase in price.

c) ∆N = -660 computers

Step-by-step explanation:

Since N(P) is a linear function

N(P) = mP + C

Where m is the slope and C is the intercept.

Case 1

N(1000) = 10000

10000 = 1000m + C ....1

Case 2

N(1700) = 5800

5800 = 1700m + C ....2

Subtracting equation 1 from 2

700m = 5800 - 10000

m = -4200/700

m = -6

Substituting m = -6 into eqn 1

10000 = (-6)1000 + C

C = 10000+ 6000 = 16000

N(P) = -6P + 16000

b) slope = -6 computers per dollar

That means the number of computer sold reduce by 6 per dollar increase in price.

Slope is the change in number of computer sold per unit Change in price.

c) since slope m = -6 computers per dollar

∆P = 110 dollars

∆N = m × ∆P

Substituting the values,

∆N = -6 computers/dollar × 110 dollars

∆N = -660 computers.

The number of computer sold reduce by 660 when the price increase by 110 dollars

Answer:

110

Step-by-step explanation:

Answer:

For this case the wisth of the interval represent 2ME and we have this:

[tex] 0.432-0.408 = 2ME[/tex]

And if we solve for ME we got:

[tex] ME = \frac{0.432-0.408}{2}=0.012[/tex]

So then the correct answer would be:

A. 0.012

Step-by-step explanation:

Previous concepts

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

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

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

The margin of error is given by:

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

For this case the wisth of the interval represent 2ME and we have this:

[tex] 0.432-0.408 = 2ME[/tex]

And if we solve for ME we got:

[tex] ME = \frac{0.432-0.408}{2}=0.012[/tex]

So then the correct answer would be:

A. 0.012

Answer:

p = -0.004n+34

Step-by-step explanation:

The slope of the linear equation can be found by using the two given points (30, 1000) and (22, 3000):

[tex]m = \frac{p_1-p_0}{n_1-n_0}= \frac{30-22}{1000-3000}\\m=-0.004[/tex]

Applying the point (30, 1000) to the general form of a linear equation with m =-0.004, gives us the linear relationship between price (p) and number of shirts (n):

[tex](p-p_0) = m(n-n_0)\\p-30=-0.004(n-1000)\\p=-0.004n+34[/tex]

Answer: p = -0.004n+34

Step-by-step explanation:

Given that we need to derived the linear equation of the form;

p = mn+b ....1

Where p is the p is the price and n is the number of shirts they can sell

We need to substitute the values of p and n for the two cases to determine the slope m and constant b.

Case 1:

n = 1000 and p = $30

Substituting into the equation 1, we have:

30 = 1000m +b .....2

Case 2:

n = 3000 and p = $22

Substituting into the equation 1

22 = 3000m + b ......3

Substracting equation 2 from 3, we have

22-30 = 3000m-1000m +b-b

-8 = 2000m

m = -8/2000

m = -0.004

Substituting the value of m into equation 2

30 = 1000(-0.004) + b

b = 30 + 1000(0.004)

b = 30 + 4 = 34

b = 34

Therefore substituting the values of m and b into equation 1, we have our linear equation:

p = -0.004n+34

Answer:

The question is not clear, but here is a similar question with the same approach.

Integral from 0 to 1 Integral from 1 to 2 Integral from 2 to 3, { (x+y+z) } dx dy dz by changing the order of integration in an appropriate way.

Step-by-step explanation:

The approach is that of multiple integral where changing the order of integration is done appropriately

The step by step with detailed workings are shown in the attachment below.

Answer:

a) [tex] S= [x \in x \geq 0][/tex]

Because the weigth can't be negative.

b) AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.

c) A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .

d) A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A

e)  A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"

f) (A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"

g) A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time

h) B' ∩ C = The complement of B are the [tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].

i) A ∪ (B ∩ C) = For this case (B ∩ C)  represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]

Step-by-step explanation:

For this case we have defined the following events, assuming that X represent the weight:

A= "a weight exceeds 11 grams" [tex]X>11[/tex]

B= " a weight is less than or equal to 15 grams" [tex] X \leq 15[/tex]

C= "a weight is greater than or equal to 8 grams and less than 12 grams" [tex] 8 \leq X < 12[/tex]

Part a

The sample space is given by:

[tex] S= [x \in x \geq 0][/tex]

Because the weigth can't be negative.

Part b

AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.

Part c

A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .

Part d

A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A

Part e

A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"

Part f

(A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"

Part g

A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time

Part h

B' ∩ C = The complement of B are the values in[tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].

Part i

A ∪ (B ∩ C) = For this case (B ∩ C)  represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]

The question refers to the mathematical concept of events and their union and intersection. In this context, A, B, and C represent specific weight ranges. Various combinations of these, such as A ∪ B or A ∩ B ∩ C, refer to different sets of possible weights.

Sample space in this experiment would basically consist of all possible weights that the digital scale can measure to the nearest gram. In Principle, it's unlimited but practically it will depend on the maximum limit of the scale.

A is the event that a weight exceeds 11 grams. B is the event that a weight is less than or equal to 15 grams. C is the event that a weight is greater than or equal to 8 grams and less than 12 grams.

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Answer:

a) 0.913

b) 0.397

c) 0.087

Step-by-step explanation:

We are given the following information:

We treat wearing tie too tight as a success.

P(Tight tie) = 15% = 0.15

Then the number of businessmen 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 = 15

We have to evaluate:

a) at least one tie is too tight

[tex]P(x \geq 1) = P(x = 1) +....+ P(x = 15)\\=1 - P(x = 0)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}\\=1 - 0.087\\= 0.913[/tex]

b) more than two ties are too tight

[tex]P(x > 2) = P(x = 3) +....+ P(x = 15)\\=1 - P(x = 0) - P(x=1) - P(x=2)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}-\binom{15}{1}(0.15)^1(1-0.15)^{14}-\binom{15}{0}(0.15)^2(1-0.15)^{13}\\=1 - 0.087 - 0.231 - 0.285\\= 0.397[/tex]

c) no tie is too tight

[tex]P(x = 0)\\=\binom{15}{0}(0.15)^0(1-0.15)^{15}\\=0.087[/tex]

d) at least 18 ties are not too tight

This probability cannot be evaluated as the number of success or the failures exceeds the number of trials given which is 15.

The probability is asked for 18 failures which cannot be evaluated.

To calculate the probabilities, we can use the concept of binomial probability. Using appropriate formulas, we can find the probabilities (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight (d) at least 18 ties are not too tight.

To calculate the probabilities, we can use the concept of binomial probability. In this case, the probability that a businessman wears the tie too tightly is 15% or 0.15. The total number of businessmen at the board meeting is 15.

(a) To calculate the probability that at least one tie is too tight, we need to find the complement of the probability that no ties are too tight. Using the formula 1 - P(no tight ties), we calculate 1 - (0.85^15) = 0.999 or 99.9%.

(b) To calculate the probability that more than two ties are too tight, we need to find the sum of probabilities of having 3 or more ties too tight. Using the binomial probability formula, we calculate P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) = 1 - (0.85^15 + 15*0.15*(0.85^14) + 105*(0.15^2)*(0.85^13)) = 0.014 or 1.4%.

(c) To calculate the probability that no ties are too tight, we use the probability P(X=0) = 0.85^15 = 0.203 or 20.3%.

(d) To calculate the probability that at least 18 ties are not too tight, we need to find the sum of probabilities of having 18 or more ties not too tight. Using the binomial probability formula, we calculate P(X >= 18) = P(X=18) + P(X=19) + ... + P(X=15) = (15 choose 18)*(0.85^18)*(0.15^2) + (15 choose 19)*(0.85^19)*(0.15^1) + (15 choose 20)*(0.85^20)*(0.15^0) = 0.999 or 99.9%.

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Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

[tex]\frac{dV}{dt} = -kA[/tex]

Using Euler's method

[tex]\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i}    \\[/tex]

Where initial droplet volume is:

[tex]V(0) = \frac{4pi}{3} * r(0)^3 =  \frac{4pi}{3} * 2.5^3 = 65.45 mm^3[/tex]

Hence, the iterative solution will be as next:

[tex]V'_{i}  = -k *4pi*(\frac{3*65.45}{4pi})^(2/3)  = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88[/tex]

[tex]V'_{i}  = -k *4pi*(\frac{3*63.88}{4pi})^(2/3)  = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33[/tex]

[tex]V'_{i}  = -k *4pi*(\frac{3*62.33}{4pi})^(2/3)  = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813[/tex]

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

[tex]r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\[/tex]

The average change of droplet radius with time is:

Δr/Δt = [tex]\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min[/tex]

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

Using Euler's method, we approximated the volume and surface area of the object over a period of 10 minutes. Starting with an initial radius of 2.5 mm, and given a decay constant of 0.08 mm/min, we computed the final volume to be approximately 26.19 mm³ and the final surface area to be about 31.70 mm². This resulted in a final radius of approximately 2.06 mm.

To solve this problem using Euler's method, we'll use the following formulas:

dV/dt = -kA

A = 4πr²

V = (4/3)πr³

Given that the initial radius r₀ = 2.5 mm, we can compute the initial volume V₀ and surface area A₀. Then, we'll iterate through time steps using Euler's method:

Vₙ₊₁ = Vₙ - kA * Δt

Aₙ₊₁ = 4π(rₙ - (k/3) * Δt)²

Using k = 0.08 mm/min and a step size of Δt = 0.25 min, we perform the calculations:

V₀ = (4/3)π(2.5)³ ≈ 65.45 mm³

A₀ = 4π(2.5)² ≈ 98.17 mm²

Iterating:

t = 0.25 min:

V₁ ≈ 65.45 - 0.08 * 98.17 * 0.25 ≈ 61.44 mm³

A₁ ≈ 4π(2.44)² ≈ 59.17 mm²

t = 0.5 min:

V₂ ≈ 61.44 - 0.08 * 59.17 * 0.25 ≈ 59.58 mm³

A₂ ≈ 4π(2.42)² ≈ 58.01 mm²

Continuing this process until t = 10 min, we obtain:

V₄₀ ≈ 26.19 mm³

A₄₀ ≈ 31.70 mm²

Finally, calculating the radius r₄₀ corresponding to V₄₀:

r₄₀ = ((3V₄₀)/(4π))^(1/3) ≈ 2.06 mm

To learn more about Euler's method

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Answer:

13112221

Step-by-step explanation:

Each sequence of numbers is a verbal representation of the sequence before it. Thus, starting with 1, the next sequence would be "one one," or "11." That sequence is followed by "two one," or "21," and so on and so forth.

This may also be a good explanation:

The first number is just ONE (amount) "1" (0-9 numeral). So if you say there's ONE "1" (seriously just say it aloud) the next number would be an 11. Then there are TWO "1's", creating 21. Then ONE "2" and ONE "1" which creates 1,211. Then ONE "1", ONE "2", and TWO "1's" creating 111,221 ... and so on.

The first number 1 is read as one one, so the second number is written as 11, this is read as two ones, so the next number is written as 21 ( two ones)

This continues throughout the sequence.

The last number written is 312211 which is read as one three, one one, two twos, two ones

This gets written as 13112221

Answer:the company bought 12 computers and 4 printers.

Step-by-step explanation:

Let x represent the number of computers that the company bought.

Let y represent the number of printers that the company bought.

The company buys a total of 16 machines. It means that

x + y = 16

Each computer costs $550 and each printer costs $390. If the company spends $8160 for all the computers and printers that was bought, it means that

550x + 390y = 8160 - - - - - - - - - - 1

Substituting x = 16 - y into equation 1, it becomes

550(16 - y) + 390y = 8160

8800 - 550y + 390y = 8160

- 550y + 390y = 8160 - 8800

- 160y = - 640

y = - 640/ - 160

y = 4

Substituting y = 4 into x = 16 - y, it becomes

x = 16 - 4

x = 12

The company bought 12 computers and 4 printers.

To solve this problem, we can set up a system of equations. Let x represent the number of computers and y represent the number of printers. We have two equations: x + y = 16 and 550x + 390y = 8160. We can solve this system by substitution or elimination. Let's use elimination.

Multiply the first equation by 390: 390x + 390y = 6240. Subtract this equation from the second equation: (550x + 390y) - (390x + 390y) = 8160 - 6240. Simplify: 160x = 1920. Divide both sides by 160: x = 12.

Substitute this value into the first equation: 12 + y = 16. Solve for y: y = 4. Therefore, the company bought 12 computers and 4 printers.

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Answer:

The required probability is 0.0391.

Step-by-step explanation:

Consider the provided information.

The percent of people with blue eyes is 32%

Thus, p = 32% = 0.32

The percent of people with non blue eyes is 100%-32%=68% = 0.68

q = 68% = 0.68

We need to determine the probability that exactly 9 of them will have blue eyes if 17 are selected.

Thus, n=17 and r=9

Use the formula of binomial distribution: [tex]P(r) = ^nC_r p^r q^{n-r}[/tex]

Substitute the respective values in the above formula.

[tex]P(r=9) = ^{17}C_9 \times(0.32)^9 \times 0.68^{17-9}[/tex]

[tex]P(r=9) = \frac{17!}{9!8!} \times(0.32)^9 \times 0.68^{8}[/tex]

[tex]P(r=9) \approx0.0391[/tex]

Hence, the required probability is 0.0391.