If a = 6 and c = 15, what is the measure of ∠A? (round to the nearest tenth of a degree) Q: A: A) 21.8° B) 22.7° C) 23.6° D) 66.4°

Answer:

A:   d. Yes, the program is statistically significant because the results are likely to occur by chance.

B: d. Yes, the program is practically significant because the amount of lost weight is large enough to be considered practically significant.

Step-by-step explanation:

A likely event is the one which is likely to occur or show the same results.

An unlikely event is one in which there's a chance of not getting the desired result or may not show the same results.

In the given scenario the program is statistically significant because even if the 28 % results occur likely by chance the rest  72 % will show some desired results.

In part B the program is statistically practical as the remaining 72 % represent a significant stats.

Answer:

The correct answer is 36.2

Step-by-step explanation:

Hello!

Given the data:

Rainfall(inch) x: 10.5, 8.8, 13.4, 12.5, 18.8, 10.3, 7.0, 15.6, 16.0

Yield (bushels per acre) y: 50.5, 46.2, 58.8, 59.0, 82.4, 49.2, 31.9, 76.0, 78.8

The response variable is

Y: Yield of wheat (bushels per acre)

Explanatory variable:

X: Rainfall in a certain period (inch)

I've calculated the equation of regression using a statistics software, the estimated equation is:

^Y= 4.27 + 4.38X

To calculate the value that will take the response variable for a given value of X:

^Y= 4.27 + 4.38*7.3= 36.24 bushels per acre.

I hope it helps!

Without the necessary data such as a regression equation or the correlation coefficient, we are unable to predict the yield of wheat when rainfall is 7.3 inches. A regression equation is a statistical tool that helps understand the relationship between predictor and response variables, in this case, rainfall and yield of wheat.

In order to predict the value of yield of wheat (y) given the rainfall (x = 7.3), we would need to understand the relationship between x and y, typically via a regression equation. A regression equation is a statistical tool used to understand the relationship between predictor variables (in this case, rainfall) and response variables (in this case, yield of wheat).

However, from the details provided, it seems we do not have the necessary data to conduct a regression analysis or connect x and y values in a meaningful way. Therefore, we're unable to directly predict the specific yield of wheat when rainfall is 7.3 inches over this data gap.

An accurate value can only be predicted if we are provided with more data or the specific correlation coefficient between the two variables and the regression equation (y = ax + b).

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Answer:

The probability is 0.27

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

Odds of a win are 27 to 73.

This means that for each 27 games that you are expcted to win, you are also expected to lose 73.

So

Desired outcomes:

27 wins

Total outcomes:

27 + 73 = 100 games

Probability

[tex]P = \frac{27}{100} = 0.27[/tex]

Answer:

The probability is 0.27

Answer:

Step-by-step explanation:

This is a circle with radius 2 and z = y

All points on or within the circle x2 + y2 +4 and in the plane z = y

Answer:

The answer to your question is 146 adult tickets

Step-by-step explanation:

Data

Total number of tickets = 584

student ticket = s

adult ticket = a

Condition

                      3a = s

Equation

                     s + a = 584

Substitution

                     3a + a = 584

Simplification

                            4a = 584

Solve for a

                              a = 584/4

                              a = 146

Find s

                             s = 3(146)

                             s = 438

Answer:146 adult tickets were sold.

Step-by-step explanation:

Let x represent the number of adult tickets that were sold.

Let y represent the number of student tickets that were sold.

A total of 584 tickets were sold for the school play. This means that

x + y = 584 - - - - - - - - - - - - - - 1

The number of student tickets sold was three times the number of adult tickets sold. This means that

y = 3x

Substituting y = 3x into equation 1, it becomes

x + 3x = 584

4x = 584

x = 584/4 = 146

y = 3x = 3 × 146

y = 438

Answer: Lateral area of the pyramid is 144

Step-by-step explanation:

The formula for determining the lateral surface area of a regular pyramid is expressed as

Lateral area = (perimeter of base x slant height of pyramid) /2

From the information given,

Length of base = 12

Since the base is a square, the perimeter of the base is

4L = 4 × 12 = 48

Height of pyramid = 6

Therefore, lateral surface area of the pyramid is

(48 × 6)/2 = 288/2 = 144

Lateral area of square pyramid with height 6 and base side 12 is 144 square units.

To find the lateral area of a square pyramid, you need to calculate the area of each triangular face and then sum them up.

1. Find the slant height (l):

In a regular square pyramid, the slant height (l) can be found using the Pythagorean theorem. Each triangular face is an isosceles right triangle. So,

[tex]\[ l = \sqrt{(s/2)^2 + h^2} \][/tex]

Where:

[tex]- \( s \) is the length of one side of the square base (given as 12 in this case).- \( h \) is the height of the pyramid (given as 6).[/tex]

[tex]\[ l = \sqrt{(12/2)^2 + 6^2} \]\[ l = \sqrt{(6)^2 + 6^2} \]\[ l = \sqrt{36 + 36} \]\[ l = \sqrt{72} \]\[ l = 6\sqrt{2} \][/tex]

2. **Calculate the lateral area (A):**

Each triangular face has a base of length 12 and a height of 6 (same as the height of the pyramid). So, the area of each triangular face (A) can be calculated as:

Since there are four triangular faces in a square pyramid, the total lateral area (LA) is:

[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]\[ A = \frac{1}{2} \times 12 \times 6 \]\[ A = 36 \][/tex] [tex]\[ LA = 4 \times A \]\[ LA = 4 \times 36 \]\[ LA = 144 \][/tex]

So, the lateral area of the square pyramid is 144 square units.

Answer:

The probability that the sixth customer buys sour milk is [tex]\frac{1}{5}[/tex].

Step-by-step explanation:

The grocery store has a total of 10 cartons of milk.

The number of cartons of milk that are sour is, 2.

       P (Both the sour cartons are available to be sold to the sixth customer)

        = [tex]P(2\ sour\ cartons)=\frac{2}{5}[/tex]

        P (Only one sour cartons is available to be sold to the sixth customer)

            = [tex]P(1\ sour\ cartons)=\frac{1}{5}[/tex]

        P (None of the sour cartons is available to be sold to the sixth customer)

         = [tex]P(0\ sour\ cartons)=\frac{0}{5}[/tex]

Compute the probability that the sixth customer buys sour milk:

= P (Both sour milk is available for the 6th customer) +  

      P (Only one sour milk is available for the 6th customer) +  

          P (None of the sour milk is available for the 6th customer)

[tex]=\frac{{8\choose 5}{2\choose 0}}{{10\choose 5}} \times\frac{2}{5} +\frac{{8\choose 4}{2\choose 1}}{{10\choose 5}} \times\frac{1}{5} +\frac{{8\choose 3}{2\choose 2}}{{10\choose 5}} \times\frac{0}{5} \\=\frac{56\times2}{252\times5} +\frac{140\times1}{252\times5} +0\\=\frac{1}{5}[/tex]

Thus, the probability that the sixth customer buys sour milk is [tex]\frac{1}{5}[/tex].

Answer:

(a) 0.0152%

(b) 1.1663%

(c) 19.4297%

(d) 79.3787%

Step-by-step explanation:

Tickets bought by organizers = 4

Number of tickets = 55

Prizes = 3

(a) The probability that the four organizers win all of the prizes is:

[tex]P = \frac{4}{55}*\frac{3}{54}*\frac{2}{53}\\P=0.0152\%[/tex]

(b) The probability that the four organizers win exactly two of the prizes is:

[tex]P = \frac{4}{55}*\frac{3}{54}*\frac{51}{53}+\frac{4}{55}*\frac{51}{54}*\frac{3}{53}+\frac{51}{55}*\frac{4}{54}*\frac{3}{53}\\P=1.1663\%[/tex]

(c) The probability that the four organizers win exactly one of the prizes is:

[tex]P = \frac{4}{55}*\frac{51}{54}*\frac{50}{53}+\frac{51}{55}*\frac{4}{54}*\frac{50}{53}+\frac{51}{55}*\frac{50}{54}*\frac{4}{53}\\P=19.4397\%[/tex]

(d) The probability that the four organizers win none of the prizes is:

[tex]P = \frac{51}{55}*\frac{50}{54}*\frac{49}{53}}\\P=79.3787\%[/tex]

Answer:

[tex]\frac{3}{14}[/tex]

Step-by-step explanation:

There are 252 (=31+54+42+20+47+58) employees in total. 54 of those are in research. So the chances that one check that gets lost belongs to a research employee can be calculated as follows:

[tex]P=\frac{54}{252}= \frac{3}{14}[/tex]

Answer:

(a) 0.25

(b) 0.5

(c) 0.5256

(d) 0.025

Step-by-step explanation:

(a) There are 20 even numbers out of 40 the probability that both numbers are even is:

[tex]P=\frac{20}{40} *\frac{20}{40} =\frac{1}{4}=0.25[/tex]

(b) The events for which one number is even and one number is odd are:

- First is odd, second is even

- First is even, second is odd.

The probability is:

[tex]P = \frac{20}{40}*\frac{20}{40}+\frac{20}{40}*\frac{20}{40}=\frac{1}{2}=0.5[/tex]

(c) There are 29 numbers that are less than 30, the probability that both numbers are less than 30 is:

[tex]P=\frac{29}{40}*\frac{29}{40}=\frac{841}{1600}=0.5256[/tex]

(d) If any number from 1 to 40 is selected in the first pick, the probability that the same number is selected again is:

[tex]P=\frac{1}{40} =0.025[/tex]

The manufacturer's profit is P(p) = 120p^2 - 6000p. The average profit for producing 0 to 15 recorders is 15 * 120. At 15 units, the rate of profit change is -3000.

The function for the manufacturer's profit P is given by P(p) = pq - 50q where q=120p. Thus, P(p) = 120p * p - 50 * 120p, simplifying to P(p) = 120p^2 - 6000p.

To express the average rate of profit from q = 0 to q=15, we need to calculate P(15)-P(0) and divide it by 15 - 0. Thus, ((120 * 15^2 - 6000 * 15) - (120 * 0^2 - 6000 * 0)) / (15 - 0) = 15*120.

The rate at which profit is changing when q = 15 is the derivative of the profit function evaluated at q = 15. We obtain P'(q) = 240p - 6000, and by inserting 15 for p, we obtain -3000 (at dollars per unit).

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a) The manufacturer's profit can be expressed as P = (p - 50)q. b) The average rate of profit obtained as the level of production increases from q=0 to q=15 is p - 50 dollars per unit. c) When q=15 recorders are produced, the rate at which profit is changing is p - 50 dollars per unit.

a) To express the manufacturer's profit P as a function of q, we can use the equation: P = (p - 50)q. This equation represents the total revenue minus the total cost, where the total revenue is the selling price p multiplied by the quantity q, and the total cost is the cost of production per unit (50 dollars) multiplied by the quantity q.

b) The average rate of profit obtained as the level of production increases from q=0 to q=15 can be found by calculating the change in profit divided by the change in quantity. In this case, the change in profit is (p - 50)(15) - (p - 50)(0) = 15(p - 50) dollars, and the change in quantity is 15 units. Therefore, the average rate of profit is (15(p - 50))/15 = p - 50 dollars per unit.

c) To find the rate at which profit is changing when q=15 recorders are produced, we can calculate the derivative of the profit function with respect to q. Taking the derivative of P = (p - 50)q with respect to q gives us dP/dq = p - 50 dollars per unit. So, when q=15, the rate at which profit is changing is p - 50 dollars per unit.

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Answer:

a) [tex]\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96[/tex]

[tex] Median = 25[/tex]

b) [tex] Mode = 25, 35[/tex]

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) [tex] Midrange = \frac{70+13}{3}=41.5[/tex]

d) [tex] Q_1 = \frac{20+21}{2} =20.5[/tex]

[tex] Q_3 =\frac{35+35}{2}=35[/tex]

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96[/tex]

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

[tex] Median = 25[/tex]

Part b

The mode is the most repeated value on the dataset on this case would be:

[tex] Mode = 25, 35[/tex]

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

[tex] Midrange = \frac{Max+Min}{2}[/tex]

And if we replace we got:

[tex] Midrange = \frac{70+13}{3}=41.5[/tex]

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

[tex] Q_1 = \frac{20+21}{2} =20.5[/tex]

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

[tex] Q_3 =\frac{35+35}{2}=35[/tex]

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Answer:

a) For this case the population of study is " the 12,304 students who live in off-campus housing and have not yet graduated " since the objective of the study is know how much students pay per month for rent in off-campus housing. So thn th epopulation size is N = 12304.

b) They select 200 students who live in off-campus housing and have not yet graduated selected using random sampling from the total population of 12304. So the sample size is n=200

As we can see in the info they got response just for 78 questionairres, so then the response rate is (78/200)*100 =39% and the non response rate is 100-39= 61.

So at the end the final sample consist on 39% of the original sample with 78 responses available for the analysis.

For this case we can redefine the sample as "The k students selected at random who live in off-campus housing and have not yet graduated  and answer the mail questionairries ". And as we can see we have a better definition fo the real sample for this situation.

Step-by-step explanation:

Previous concepts

The term population represent to the total set of observations in a sample space S defined, and with an specified characteristics.

The term sample represent a set of individuals or objects who are a subsample of the population who are "collected or selected from a statistical population by a defined procedure".

Part a

For this case the population of study is " the 12,304 students who live in off-campus housing and have not yet graduated " since the objective of the study is know how much students pay per month for rent in off-campus housing. So thn th epopulation size is N = 12304.

Part b

They select 200 students who live in off-campus housing and have not yet graduated selected using random sampling from the total population of 12304. So the sample size is n=200

As we can see in the info they got response just for 78 questionairres, so then the response rate is (78/200)*100 =39% and the non response rate is 100-39= 61.

So at the end the final sample consist on 39% of the original sample with 78 responses available for the analysis.

For this case we can redefine the sample as "The k students selected at random who live in off-campus housing and have not yet graduated  and answer the mail questionairries ". And as we can see we have a better definition fo the real sample for this situation.

The population in this study is the 12,304 students who live in off-campus housing and have not yet graduated. The sample is the 200 students selected at random who received the questionnaire. The sample provides information about the population of students living in off-campus housing.

(a) Population: The population in this study refers to all the 12,304 students who live in off-campus housing and have not yet graduated. This is the group about which the university's housing and residence office wants information.

(b) Sample: The sample in this study refers to the 200 students selected at random who received the questionnaire. This is the group from which the office obtains information. The sample can be seen as a representation of the population.

The important message in this problem is that the sample can help provide information about the population of students living in off-campus housing.

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Answer:

[tex] P(A \cap B) = P(X=9) =\frac{16}{78} \neq 0[/tex]

The correct answer would be:

NO

Step-by-step explanation:

For this case we have the following dataset given

Hours    Number of students (f)

_______________________________

   4                     15

   5                     11

    6                     19

   7                      6

   8                      9

    9                     16

    10                     2

______________________________

Total                  78

For this case we have defined the following events:

A = event the student took at most 9 hours

B = event the student took at least 9 hours

And we can find the empirical probability for both elements like this:

[tex] P(A) = \frac{78-2}{78}= \frac{76}{78}[/tex]

[tex] P(B) = \frac{16+2}{78}= \frac{18}{78}[/tex]

And for this case we want to see if A and B are disjoint

From definition two events X and Y are disjoint if the two sets not have a common elements, and we satisfy that:

[tex] P(X \cap Y) =0[/tex]

So this case the intersection for the events A and B is X=9, because at most 9 means [tex] X \leq 9[/tex] and at least 9 means [tex] X \geq 9[/tex] and the intersection between [tex] X \leq 9[/tex]  and [tex] X \geq 9[/tex]  is X=9

So then the probability:

[tex] P(A \cap B) = P(X=9) =\frac{16}{78} \neq 0[/tex]

So then we can conclude that the two events not are disjoint

The correct answer would be:

NO

No, the events A and B are not disjoint.

If two events have no outcomes in common, then they are called disjoint.

We have data of the number of hours sixth grade students took to complete a research project as:

For this case we have the following dataset given

Hours    Number of students (f)

 4                     15

5                     11

6                     19

7                      6

8                      9

9                     16

10                     2

Total                  78

Two events are:

A = event the student took at most 9 hours

B = event the student took at least 9 hours

Now, the number of students who took at most 9 hours

= 78 - 2

= 76

So, [tex]P(A)=\frac{76}{78}[/tex]

The number of students who took at least 9 hours

=16 +2

=18

So, [tex]P(B)=\frac{16}{78}[/tex]

Number of students who read exactly 9 hours

P(A n B)[tex]=\frac{16}{78}[/tex][tex]\neq 0[/tex]

Therefore the events A and B disjoint are not disjoint.

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Answer:

This approach to (0,0) also gives the value 0

Step-by-step explanation:

Probably, you are trying to decide whether this limit exists or not. If you approach through the parabola y=x², you get

[tex]\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{\sqrt{x^2+y^2}}=\lim_{(x,x^2)\rightarrow (0,0)}\frac{xx^2}{\sqrt{x^2+(x^2)^2}}=\lim_{x\rightarrow 0}\frac{x^3}{|x|\sqrt{1+x^2}}=0[/tex]

It does not matter if x>0 or x<0, the |x| on the denominator will cancel out with an x on the numerator, and you will get the term x²/(√(1+x²) which tends to 0.

If you want to prove that the limit doesn't exist, you have to approach through another curve and get a value different from zero.

However, in this case, the limit exists and its equal to zero. One way of doing this is to change to polar coordinates and doing a calculation similar to this one. Polar coordinates x=rcosФ, y=rsinФ work because the limit will only depend on r, no matter the approach curve.

Focus on the sub-triangle on the left. It is a right triangle with legs 9 and 6, so its hypothenuse is

[tex]\sqrt{9^2+6^2}=\sqrt{81+36}=\sqrt{117}[/tex]

Now focus on the sub-triangle on the right. It is a right triangle with legs 6 and x, so its hypothenuse is

[tex]\sqrt{6^2+x^2}=\sqrt{x^2+36}[/tex]

Now, the entire triangle has legs [tex]\sqrt{117}[/tex] and [tex]\sqrt{x^2+36}[/tex], and its hypothenuse is [tex]9+x[/tex]. Write the Pytagorean theorem one last time to get

[tex]117+(x^2+36)=(9+x)^2\iff x^2+153=81+18x+x^2 \iff 18x+81=153[/tex]

Subtract 81 from both sides to get

[tex]18x=72 \iff x=\dfrac{72}{18}=4[/tex]

Answer: x = 4

Step-by-step explanation:

The attached photo shows a clearer illustration of the given triangle.

Looking at the photo, assuming ∆BCD is a right angle triangle. To determine BC, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

BC² = 9² + 6²

BC² = 81 + 36 = 117

BC = √117

To determine θ, we would apply the tangent trigonometric ratio.

Tan θ opposite side/adjacent side

Tan θ = 6/9 = 0.6667

θ = 33.6914

Considering ∆ABC,

Hypotenuse = x + 9

Adjacent = √117

Cos θ = adjacent side/ hypotenuse

Cos 33.6914 = √117/(x + 9)

Cross multiplying, it becomes

0.8320 = √117/(x + 9)

x + 9 = √117/0.8320

x + 9 = 13

x = 13 - 9 = 4

Answer:

11,713 bacteria

Step-by-step explanation:

Integrating the growth rate function gives us the population of bacteria at any given moment 't', in hours:

[tex]r(t) = (450.263)e^{1.12567t} \\\int\ {r(t)} \, dt =P(t) = \frac{450.263}{1.12567}*e^{1.12567t} +C[/tex]

Since at t=0, P(t) = 400, the value of C is:

[tex]P(0) = \frac{450.263}{1.12567}*e^{1.12567*0} +C\\400 = 400*1+C\\C=0[/tex]

The number of bacteria after 3 hours is:

[tex]P(3) = 400*e^{1.12567*3}\\P(3) =11,713[/tex]

Answer:

5.70456 slug

Step-by-step explanation:

Data provided in the question:

Dimensions of the room = 20 ft long, 15 ft wide, and 8 ft high

Now,

Volume of the room = 20 × 15 × 8

or

Volume of the room = 2400 ft³

we know,          

Density of air = 0.0023769 slug/ft³

Therefore,

Mass of air in the room = Volume × Density

= 0.0023769 × 2400

= 5.70456 slug

The mass of air in the room is approximately 1268 slugs and the weight is approximately 40825.6 pounds.

To calculate the mass of air in a room, we first need to find the volume of the room. The volume of a rectangular room can be calculated by multiplying its length, width, and height. So, the volume of the room is 20 ft x 15 ft x 8 ft = 2400 ft³.

Next, we need to convert the volume from cubic feet to cubic meters. Since 1 ft³ is approximately equal to 0.0283 m³, we can multiply the volume in cubic feet by 0.0283 to get the volume in cubic meters. Therefore, the volume of the room is 2400 ft³ x 0.0283 m³/ft³ = 67.92 m³.

Lastly, we need to find the mass of air in the room. The average molar weight of air is approximately 28.8 g/mol. Since the mass of one cubic meter of air is 1.28 kg, the mass of air in the room is 67.92 m³ x 1.28 kg/m³ = 86.86 kg. To convert the mass from kg to slugs, we divide it by the conversion factor of 0.0685218 slugs/kg. Therefore, the mass of air in the room is 86.86 kg / 0.0685218 slugs/kg ≈ 1268 slugs.

To calculate the weight in pounds, we multiply the mass in slugs by the acceleration due to gravity. The acceleration due to gravity is approximately 32.2 ft/s². Therefore, the weight of the air in the room is 1268 slugs x 32.2 ft/s² = 40825.6 lb.

The total number of students in the class is 50, which is calculated based on the given that 20 women make up 40% of the class.

The question deals with the concept of percentages. In this case, the number of women in the class represents 40% of the total number of students. We have been told that there are 20 women in the class.

Here's how we can solve it step by step:

Therefore, the total number of students in the class is 50.

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Answer:

P(T) = 1/20 = 0.05

The probability of randomly selecting an umbrella and a shaving kit in that order is 0.05

Step-by-step explanation:

The probability of randomly selecting an umbrella and a shaving kit in that order.

P(T) = Probability of selecting umbrella first P(U) × probability of selecting shaving kit second P(S)

P(U) = 1/5 (1 umbrella out of five possible gifts)

P(S) = 1/4 (1 shaving kit out of four remaining possible gifts)

P(T) = 1/5 × 1/4

P(T) = 1/20 = 0.05

The probability that a person will select an umbrella and a shaving kit in that order is 1/20.

The following can be deduced from the information given:

P(U) = 1/5 (1 umbrella out of five possible gifts)

P(S) = 1/4 (1 shaving kit out of four remaining possible gifts)

Therefore, the probability that a person will select an umbrella and a shaving kit in that order will be:

P(T) = 1/5 × 1/4

P(T) = 1/20 = 0.05

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Answer:

...the size tolerance controls both the measurements or dimensions of a piece, and the size.

Step-by-step explanation:

However, dimensional tolerance controls neither the shape, nor the position, nor the orientation of the elements to which said tolerance applies. In manufacturing, geometric irregularities occur that can affect the shape, position or orientation of the different elements of the pieces. An applied dimensional tolerance, for example, has an effect on the parallelism and flatness of that piece.

Answer:

The equation relating j to d can be written as:

If J = the amount of juice in fluid ounces and D = its diameter, then:

J=kD³ where k = the constant of proportionality.

The equation that represents the relationship between the amount of juice in a grapefruit (j) and its diameter (d) with constant of proportionality (k), given that they are directly proportional, is j = kd^3.

The question given is a classic example of direct variation, more specifically, cubic direct variation since the amount of juice in a grapefruit varies with the cube of its diameter. When two quantities are directly proportional or vary directly, they form a relationship that can be described by an equation of the form y = kx^n, where k is the constant of proportionality, x is one quantity and y is the other. Here, the amount of juice in a grapefruit (j) is directly proportional to the cube of the diameter (d^3), with k being the constant of proportionality. Therefore, the equation that represents this relationship is j = kd^3.

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Answer:

B. 72

Step-by-step explanation:

The total number of different systems that can be bundled together is the product of the possible number of ways to select 1 out of 4 stereo receivers, by 1 out of 3 CD decks and by 1 out of 6 speakers. Assuming that the order at which products are picked does not matter, the number of different systems is:

[tex]n=\frac{4!}{(4-1)!1!}*\frac{3!}{(3-1)!1!}*\frac{6!}{(6-1)!1!}\\n=4*3*6\\n=72\ systems[/tex]

The electronics firm can offer 72 different systems.

The electronics firm can offer 72 different systems by combining 4 models of stereo receivers, 3 CD decks, and 6 speaker brands together.

To find the number of different systems the electronics firm can offer, we use the combinations formula. Since there are 4 models of stereo receivers, 3 CD decks, and 6 speaker brands, the total number of different systems can be calculated as:

4 models * 3 CD decks * 6 speaker brands = 72 different systems

Therefore, the answer to the question is 72 different systems (option B).

Answer:

Mean ;

Median;

Step-by-step explanation:

The mean is used when the data under consideration is more of quantitative and in which the data is devoid of outliers as such the values are assumed to follow a normal distribution.

The median on the other hand is considered when the data are more of qualitative and usually contain outliers. Median on the other hand is best used when there is a skewed symmetry in the values given.

Mean ;

Median;

Whether mean or median provides a better description of a data set depends on the skewness and outliers in the data. Generally, the mean is more sensitive to outliers whereas the median can better represent the central tendency of skewed distributions.

In statistics, mean and median are two measures of central tendency. The mean is the average of the data points, while the median is the middle value. Whether the mean or median provides a better description depends on the distribution of the data.

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Answer:

4.23% probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill.

Step-by-step explanation:

For each installation, there are only two possible outcomes. Either it reduces the utility bill, or it does not. The probabilities for each installation reducing the utility bill are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

In this problem we have that:

Solar-heat installations successfully reduce the utility bill 60% of the time, which means that [tex]p = 0.6[/tex]

What is the probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill?

This is [tex]P(X \geq 9)[/tex] when [tex]n = 10[/tex]. So

[tex]P(X \geq 9) = P(X = 9) + P(X = 10)[/tex]

In which

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

[tex]P(X = 9) = C_{10,9}.(0.6)^{9}.(0.4)^{1} = 0.0363[/tex]

[tex]P(X = 10) = C_{10,10}.(0.6)^{10}.(0.4)^{0} = 0.0060[/tex]

So

[tex]P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0363 + 0.0060 = 0.0423[/tex]

4.23% probability that at least 9/10 solar-heat installations are successful and will reduce the utility bill.

The probability of at least 90% success in solar-heat installations is found by using the binomial probability formula to calculate and add together the probabilities of exactly 9 and 10 successful installations out of 10.

The problem in question is a classic scenario of binomial probability. Here, each solar-heat installation attempt is independent and each attempt is a success (reduces the utility bill) 60% of the time. We are interested in the probability of having 90% or more success in ten attempts.

In a binomial distribution, the formula for calculating the probability of k successes in n attempts is:

P(X=k) = C(n, k) * (p^k) * (1-p)^(n-k)

where C(n, k) is the binomial coefficient ('n choose k'), p is the probability of success on an individual trial, n is the number of trials, and k is the number of successes.

To calculate the probability that at least 9 out of 10 solar-heat installations are successful, we need to calculate P(X=9) and P(X=10) and add these probabilities together.

Calculations like these help inform decisions in a range of fields - from individual choices about energy saving at homes to policy and planning decisions at the level of energy utilization for entire nations.

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The value of B is determined by the equation 2π / 12.7, which corresponds to the tide's period. The variable C represents the time delay from midnight to the first high tide, which is a phase shift in the function.

The Bay of Fundy tidal pattern can be modeled using a cosine function. Since tides go through a complete cycle (360 degrees or 2π radians) every 12.7 hours, the value of B, the frequency, can be determined by dividing 2π by the period of the tide in hours.

Therefore, B = 2π / 12.7.

The variable C in the equation represents a phase shift. In this context, a phase shift refers to a horizontal shift of the cosine function, which corresponds to a time delay or advance of the tides. The meaning of C is the time delay between midnight and the first high tide of the day.

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Answer: t = - 5 ∈ [tex]I_{1}[/tex] = ( -∞ , -4 )

Step-by-step explanation:

The standard form of O.D.E is written as :

[tex]y^{1}[/tex] + [tex]p(t) = g(t)[/tex]

Equation given :

[tex](16-t^{2} )y^{1}[/tex] + [tex]9ty[/tex] = [tex]5t^{2}[/tex] ,       [tex]y(-5) = 1[/tex]

The first thing to do is to write the O.D.E in standard form , that is we will divide through by [tex]16 - t^{2}[/tex] , so we have

[tex]y^{1} + \frac{9ty}{16-t^{2}}=\frac{5t^{2}}{16-t^{2}}[/tex]

With this , we can see that [tex]p(t)[/tex] and [tex]g(t)[/tex] are both continuous in the same domain. Therefore , the intervals are :

[tex]I_{1}[/tex] = ( -∞ , -4 )

[tex]I_{2}[/tex] = ( - 4 , 4 )

[tex]I_{3}[/tex] = ( 4 , -∞ )

recall that y(−5) = 1 , then t = -5

This means that :

t = - 5 ∈ [tex]I_{1}[/tex] = ( -∞ , -4 )

Answer:

Step-by-step explanation:

Since each trial is independent of the other

no of mistakes he does is binomial with p = 1/3

a) the probability that he makes no mistakes on his first 10 orders but the 11th order is a mistake

= [tex](\frac{2}{3}) ^{10} *\frac{1}{3}\\=\frac{2^{10} }{3^{11} }[/tex]

b) Prob that shanker quits = P(Shankar does I one mistake and Fran does not do the first one)+Prob (Shanker does mistake in the II one while Fran does both right)

= [tex]\frac{1*5}{3*6} +\frac{2}{3} \frac{1}{3}(\frac{5}{6})^2 =\frac{5}{18} +\frac{50}{216} \\=\frac{55}{108}[/tex]

Final answer:

Calculating probabilities in scenarios involving making mistakes in orders in a coffee shop setting. he probability that Shankar quits and goes to grad school before Fran is approximately 0.651.

Explanation:

a. For Shankar to make no mistakes on his first 10 orders but the 11th order is a mistake, the probability of making no mistakes on the first 10 orders and then making a mistake on the 11th order can be calculated as follows:

[tex]\[ P(\text{No mistakes on first 10 orders}) \times P(\text{Mistake on 11th order}) \][/tex]

[tex]\[ = \left(\frac{2}{3}\right)^{10} \times \frac{1}{3} \][/tex]

[tex]\[ = \left(\frac{1024}{59049}\right) \times \frac{1}{3} \][/tex]

[tex]\[ \approx 0.0173 \][/tex]

b. For Shankar to quit and go to grad school before Fran, Shankar must make a mistake before Fran does. The probability of Shankar quitting and going to grad school can be calculated as follows:

[tex]\[ P(\text{Shankar quits}) = 1 - P(\text{Fran quits first}) \][/tex]

Since Fran's probability of making a mistake on an order is [tex]\( \frac{1}{6} \)[/tex], the probability of Fran making no mistakes on an order is [tex]\( \frac{5}{6} \).[/tex] Thus, the probability of Fran not making a mistake before Shankar is:

[tex]\[ P(\text{Fran makes no mistakes before Shankar}) = \left(\frac{5}{6}\right)^{10} \][/tex]

Therefore,

[tex]\[ P(\text{Shankar quits}) = 1 - \left(\frac{5}{6}\right)^{10} \][/tex]

[tex]\[ \approx 0.651 \][/tex]

So, the probability that Shankar quits and goes to grad school before Fran is approximately \(0.651\).

Answer:

24, 204

Step-by-step explanation:

to find the height

we would use the heron formula  when all the sides are given

S which is equal to half of the perimeter of the triangle which is (a+b+c)/2.

S = (17+ 25+ 26)/2 = 68/2 = 34

Area = √S(S-A)(S-B)(S-C)

Area= (1/2)bh

we equate it back to the formula Area = √S(S-A)(S-B)(S-C)

it becomes

(1/2)bh  = √S(S-A)(S-B)(S-C)

A = IABI= 25

B= IBCIM = 26

C= IACI = 17

b = base = IACI = 17

S = 34

(1/2)bh  = √S(S-A)(S-B)(S-C)

(1/2)17h  = √34(34-25)(34-26)(34-17)

(17/2)h = √34(9)(8)(17) = √34 x 9 x 8 x 17

(17/2)h = √41616 = 204

17h/2 = 204

17h = 204 x 2 = 408

h = 408/17 = 24 inch

height = h = IBDI = 24 in

Area = (1/2)bh

      = (1/2) x 17 x 24

    =  12 x 17 = 204  or we use the heron formula just like the above which we get 204 before multiplication by 2.

Answer:

2.2 times

Step-by-step explanation:

Given that,

Sales = $4,400,000

Average total assets (excluding long-term investments) = 2,000,000

Therefore, it is as follows;

Asset turnover ratio:

= Sales ÷ Average total assets (excluding long-term investments)

= $4,400,000 ÷ 2,000,000

= 2.2 times

Hence, the asset turnover ratio of this company is 2.2 times.