how do I solve these three

Answer: C. If he passes the class, she will pay $100.

A. If she ended up paying $100, he must have passed the class.

Step-by-step explanation:

For a statement to be true, one assumption might imply the other assumption. In this case, A and C fulfills it.

Final answer:

The mathematician's statement makes passing the class a necessary condition for her son to receive $100. If he received the money, he must have passed the class, but not receiving the money does not conclusively prove he did not pass. The statement does not make passing the class a sufficient condition, as there could be other requirements.

Explanation:

The statement made by the mathematician to her son is a classic example of a conditional statement in logic, particularly looking at necessary and sufficient conditions. To analyze the truth of the given statements, we must understand the relationship between passing the class and receiving the $100.

Answer:

A) The actual error |f(0.5) - P3(0.5)| is 0.0340735. The error using Taylor's error formula R₃(0.5) is 0.2917.

B) Te error in the interval [0.5,1.5] is 0.0340735.

C) The approximate integral using P₃(x) is 1/12

D) The real error between the two integrals is 4.687·10^(-3) while the error using \int_{0.5}^{1.5}|R_{3}(x)dx| is 0.018595

Step-by-step explanation:

A) To determine the error we first have to write the third grade Taylor polynomial P3(x):

[tex]P_n(x)=\displaystyle\sum_{k=0}^N \frac{f^{k}(a)}{k!}(x-a)[/tex] with N=3

[tex]P_3(x)=\displaystyle\sum_{k=0}^3 \frac{f^{k}(1)}{k!}(x-1)=(x-1)^2-\frac{1}{2} (x-1)^3[/tex]

The error for the third grade Taylor polynomial R₃(x) is represented as the next term non-written in the polynomial expression near to the point x:

[tex]R_3(x)=\displaystyle \frac{f^{4}(x)}{4!}(x-a)=\frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4[/tex]

Therefore the real error is:

[tex]Err(0.5)=|f(0.5)-P_3(0.5)|=0.0340735[/tex]

The error of the Taylor polynomial R₃(0.5) is:

[tex]R_3(0.5)=\displaystyle \frac{1}{24} (\frac{2}{0.5^3}+\frac{6}{0.5^4})(0.5-1)^4=0.2917[/tex]

B) The error in the interval [0.5,1.5] is the maximum error in that interval.

This is found in the extremes of the intervals. We analyze what happens in X=1.5:

[tex]Err(1.5)=|f(1.5)-P_3(1.5)|=0.01523[/tex]

[tex]R_3(1.5)=\displaystyle \frac{1}{24} (\frac{2}{1.5^3}+\frac{6}{1.5^4})(1.5-1)^4=\frac{1}{216}=4.63\cdot 10^{-3}[/tex]

Both of these errors are smaller than Err(0.5) and R₃(0.5). Therefore the error in this interval is Err[0.5,1.5] is Err(0.5).

C) The approximation of the integral and the real integral is:

[tex]I_f=\displaystyle\int_{0.5}^{1.5} f(x)\, dx=\int_{0.5}^{1.5} (x-1)ln(x)\, dx=0.08802039[/tex]

[tex]I_{p3}=\displaystyle\int_{0.5}^{1.5} P_3(x)\, dx=\int_{0.5}^{1.5} (x-1)^2-0.5(x-1)^3\, dx=1/12=0.0833333[/tex]

D) The error in the integrals is:

[tex]Errint=|I_f-I_{p3}|=4.687\cdot10^{-3}[/tex]

[tex]Errint(R_3)=\displaystyle\int_{0.5}^{1.5} R_3(x)\, dx=\int_{0.5}^{1.5} \frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4\, dx=0.018595[/tex]

Answer:

sin theta = opp/hyp = √5/3 = (sq5)/3

csc theta = 1 / sin theta = 1 / (√5/3) =  1 x 3/√5 = 3/√5 = 3/(sq5)

csc theta = 3/(sq5) and tan theta = -(sq5)/2 = B

Step-by-step explanation:

if cos theta = -2/3

the -2 is adjacent

3 is the hypothenus

using pythagoras theorem

(hypothenus)² = (opposite)² + (adjacent)²

3² = opp²+ (-2)² = opp²+ (4)

9-4 = opp² = 5

opp = √5

sin theta = opp/hyp = √5/3 = (sq5)/3

tan theta = opp/adj = √5/-2  = (sq5)/2

csc theta = 1 / sin theta = 1 / (√5/3) =  1 x 3/√5 = 3/√5 = 3/(sq5)

csc theta = 3/(sq5) and tan theta = -(sq5)/2 = b

Answer:

A and B are correct!

Step-by-step explanation:

Opposite side= - or +(sq5)

Adjacent side= -2

Hypotenuse= 3

Answer:

[tex]\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7[/tex]

[tex] Mode= 53987[/tex]

[tex] Median = \frac{52762+53108}{2}=52935[/tex]

[tex] Midrange=\frac{50554+54364}{2}=52459[/tex]

Step-by-step explanation:

We can calculate the mean with the following formula:

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

And if we replace the values given we got:

[tex]\bar X = \frac{52384 +54079 +52762 +54364 +52584+53987 +53108 +51588 +50554 +53987}{10}=52939.7[/tex]

The mode is the most repeated value and for this case with a frequency of 2 the mode is:

[tex] Mode= 53987[/tex]

In order to find the median we need to order the dataset on increasing way like this:

50554, 51588, 52384  ,52584, 52762

53108, 53987 , 53987 , 54079, 54364

Since we have 10 values an even number the median is calculated from the average between positions 5 and 6 on the data ordered, and we got:

[tex] Median = \frac{52762+53108}{2}=52935[/tex]

The mid range is defined like this:

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

And if we replace we got:

[tex] Midrange=\frac{50554+54364}{2}=52459[/tex]

The 'Top 10' list gives insights into the population of college tuitions in a country. The mean, midrange, median, and mode of the data set are $53,138.50, $52,459, $52,584., and  $53,987 respectively.

The "Top 10" list of annual tuition amounts for the most expensive colleges in a country provides insights into the range and distribution of college tuitions within the country's higher education system.

Mean: The mean (average) tuition is calculated by summing up all the tuition amounts and dividing by the total number of colleges (in this case, 10).

Mean = ($52,384 + $54,079 + $52,762 + $54,364 + $52,584 + $53,987 + $53,108 + $51,588 + $50,554 + $53,987) / 10 = $53,138.50 (rounded to two decimal places).

Midrange: The midrange is the average of the minimum and maximum values in the dataset.

Midrange = ($50,554 + $54,364) / 2 = $52,459 (rounded to two decimal places).

Median: The median is the middle value when the data is arranged in ascending or descending order. In this case, the data is already provided in ascending order, so the median is the fifth value.

Median = $52,584.

Mode: The mode is the value that appears most frequently in the dataset.

Mode = $53,987.

In summary, the "Top 10" list of college tuitions in the country indicates that there is a range of tuition amounts, with the mean being around $53,138.50. The midrange falls around $52,459, the median is $52,584, and the mode is $53,987.

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Answer:

f(x) = 0 , when x=(-2) and x=3

Step-by-step explanation:

for a≠0 and b positive then for the function

f(x) = a*(x+2)*(x-3)ᵇ

f(x) = 0 when

1) (x+2)=0 → x=(-2)

2) (x-3)=0 → x=-3

then for x=3 and x=(-2) , f(x)=0

Answer:

A = 3.706 square meters

Step-by-step explanation:

the length of a cube has equal side.

therefore, the volume of a cube is given by S³

V = S³ = 7.14

where S is the length of a side

the surface area of a cube is = 6S²

where the area a aside is calculated, then it is multiply by 6.

if S³ = 7.14

S = ∛(7.14) = 1.925 m

What is the cross-sectional area that is parallel to one of its faces?

this is saying we should calculate for the cross-sectional area of one face. All faces in a cube is equal and parallel to each other.

the crossectional  area of one side of a cube is = S² = 1.925² = 3.706 square meters

A = 3.706 square meters

Answer:

[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]    

[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]    

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

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".

[tex]\bar X= 0.82[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=0.048 represent the sample standard deviation

n=15 represent the sample size  

Solution to the problem

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

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=15-1=14[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that [tex]t_{\alpha/2}=2.977[/tex]

Now we have everything in order to replace into formula (1):

[tex]0.82-2.977\frac{0.048}{\sqrt{15}}=0.783[/tex]    

[tex]0.82+2.977\frac{0.048}{\sqrt{15}}=0.857[/tex]    

So on this case the 99% confidence interval would be given by (0.783;0.857)

So then we can conclude that at 99% of confidence the true mean is between (0.783;0.857) and the specification is satisfied since the value of 0.8 is on the confidence interval

To find a 99% confidence interval for the true average concentration of the active ingredient, we can use the formula: CI = (x - z*(s/sqrt(n)), x + z*(s/sqrt(n))). Plugging in the given values and using the critical value for a 99% confidence level, we get the confidence interval to be (0.799, 0.841) grams per liter.

To find a 99% confidence interval for the true average concentration of the active ingredient, we can use the formula:

CI = (x - z*(s/sqrt(n)), x + z*(s/sqrt(n)))

Where CI is the confidence interval, x is the sample mean, z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.

Plugging in the given values: x = 0.82, s = 0.048, and n = 15, and using the critical value for a 99% confidence level (z = 2.576), we get:

CI = (0.82 - 2.576*(0.048/sqrt(15)), 0.82 + 2.576*(0.048/sqrt(15)))

Simplifying the expression, we get the confidence interval to be (0.799, 0.841) grams per liter.

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Answer:

[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.

Step-by-step explanation:

The daily demand for milk containers has a Normal (or Gaussian) distribution, and we can use values from the cumulative distribution function and z-scores to solve the question.

We know from the question that the mean of the distribution is:

[tex] \\ \mu = 55 [/tex]

And a standard deviation of:

[tex] \\ \sigma = 6 [/tex]

The z-scores permit calculates the probabilities for any case whose values have a Normal o Gaussian distribution. Then, for this, we need to calculate the z-scores for 49 containers and 61 containers to establish the corresponding probabilities, as well as the differences between these two values to determine the probability between them.

These z-scores are given by:

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

Thus,

The z-scores for 49 and 61 containers are:

[tex] \\ z_{49} = \frac{49 - 55}{6} = \frac{-6}{6} = -1 [/tex] [1]

[tex] \\ z_{61} = \frac{61 - 55}{6} = \frac{6}{6} = 1 [/tex] [2]

Well, this is a special case when in both cases the values are one standard deviation from the mean, but in one case ([tex] \\ z_{49} = -1 [/tex]) the values are smaller than the mean and in the other case ([tex] \\ z_{61} = 1 [/tex]) the values are greater than the mean.

In other words, the cumulative probability for ([tex] \\ z_{61} = 1 [/tex]), obtained from any Table of the Normal Distribution available on the Web, is: 0.8413 (or 84.13%) and the cumulative probability for ([tex] \\ z_{49} = -1 [/tex]) is: 1 - 0.8413 = 0.1587 (or 15.87%), because of the symmetry of the Normal Distribution.

Then, the probability of expecting to sell between 49 and 61 containers in a day is the difference of both obtained probabilities:

[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.

See the graph below.

Final answer:

Approximately 68% of the time, the number of milk containers sold in a day at the supermarket is expected to be between 49 and 61 containers, following the Empirical Rule for normal distribution.

Explanation:

The question asks us about the probability of daily sales being between certain values when they follow a bell-shaped distribution, specifically between 49 and 61 containers. Since we know the distribution is approximately normal with a mean of 55 containers and a standard deviation of six containers, we can use the Empirical Rule or the Standard Normal Distribution to find the probability.

In the context of a normal distribution, we know that:

About 68% of the data falls within one standard deviation of the mean. (Mean ± 1SD)

About 95% falls within two standard deviations. (Mean ± 2SD)

For this supermarket, one standard deviation from the mean (55 ± 6) gives us a range from 49 to 61 containers. Therefore, approximately 68% of the time, we can expect the number of milk containers sold to be between 49 and 61 containers in a day, based on the Empirical Rule.

answer and step by step in the attachment

The value of PR is 28.

PQ = x + 10

QR = 4x - 2

To solve the question, we'll equate both equations which will be:

PQ = QR.

x + 10 = 4x - 2

Collect like terms

4x - x = 10 + 2.

3x = 12

x = 12/3

x = 4

Therefore,

PQ = x + 10 = 4 + 10 = 14

QR = 4x - 2 = 4(4) - 2 = 16 - 2 = 14

Therefore, PR = 14 + 14 = 28

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Answer:

the consecutive integars is 11,12,13,14,15

Step-by-step explanation:

then it says the sum of the first and 4 times the third = 60 less than 3 times the sum of the second, fourth and fifth

60 less than 3 times the sum of the second, fourth and fifth means that 3 times the sum of the second, fourth and fifth minus 60

y+ 4(y+2) = 3{ (y +1) +(y+3) + (y+4)} -60

y + 4y + 8 =3[ y +1 + y + 3 +y+4} - 60

5y +8=3(y+y+y+1+3+4)-60

5y+8=3(3y+8)-60

5y+8=9y+24-60

5y+8=9y-36

5y+8-8=9y-36-8

5y=9y-44

5y-9y=9y-9y-44

-4y=-44

-4y/-4=-44/-4 ( minus divided by minus is plus

y = 11

Answer:

Step-by-step explanation:

given are four statements and we have to find whether true or false.

.1 If two matrices are equivalent, then one can be transformed into the other with a sequence of elementary row operations.

True  

2.Different sequences of row operations can lead to different echelon forms for the same matrix.

True in whatever way we do the reduced form would be equivalent matrices

3.Different sequences of row operations can lead to different reduced echelon forms for the same matrix.

False the resulting matrices would be equivalent.

4.If a linear system has four equations and seven variables, then it must have infinitely many solutions.

True, because variables are more than equations.  So parametric solutions infinite only is possible

Statements 1 and 2 are true, and statements 3 and 4 are false. While equivalent matrices can be transformed into each other and different row operations can yield different echelon forms, the reduced echelon form is unique, and a linear system with more variables than equations does not necessarily have infinitely many solutions.

1. The statement is true. If two matrices are equivalent, one can indeed be transformed into the other through a sequence of elementary row operations. These operations include swapping two rows, multiplying a row by a non-zero scalar, or adding a multiple of one row to another row.

2. The statement is true. Different sequences of row operations can yield different echelon forms of the same matrix as the operations can redistribute information about the system of equations in different ways.

3. The statement is false. Regardless of the sequence of row operations performed, the reduced echelon form of a matrix is unique. This is because the reduced echelon form is a canonical form, meaning there's only one possible reduced echelon form for a given matrix.

4. The statement is false. Even though a linear system has more variables than equations, this does not guarantee that it will have infinitely many solutions. It could have no solutions or, under certain conditions, even a unique solution. Further analysis is required in these cases.

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Answer:

There is a 3.95% probability that none are mechanical engineering majors.

Step-by-step explanation:

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

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

In this problem, we have that:

There are 125 students.

51 of them are mechanical engineering majors and 125-51 = 74 are not mechanical engineering majors.

Suppose six students from the class are chosen at random what is the probability none are mechanical engineering majors?

The total number of students is 125. So total number of 6 student groups is

[tex]C_{125,6} = \frac{125!}{6!119!}[/tex]

The total number of non mechanical engineering students is 74. So the total number of 6 non mechanical engineering students is.

[tex]C_{74,6} = \frac{74!}{6!68!}[/tex]

The probability is:

[tex]\frac{C_{74,6}}{C_{125,6}} = 74/125 * 73/124 * 72/123 * 71/122 * 70/121 * 69/120 = 0.0395[/tex]

The 68/119 ends up simplified in this exercise, this is your mistake.

There is a 3.95% probability that none are mechanical engineering majors.

Answer: False.

Step-by-step explanation:

The probability for any event = [tex]\dfrac{\text{favorable outcomes}}{\text{Total outcomes}}[/tex]

Definition : A simple event is an event that has only one possible favorable outcomes.

If we toss a coin and roll a die .

The possible outcomes for event​ "tossing tails and rolling a 4 or 6​" = {(T, 4) , (T,6)}

Since this event has 2 possible favorable  outcomes , Therefore it is not an simple event.

Hence, the given statement is "false".

The statement 'tossing tails and rolling a 4 or 6' is not a simple event. A simple event only consists of a single outcome, whereas this case includes multiple outcomes: getting tails in the coin toss and either a 4 or 6 in the die roll. As such, the statement is false.

The statement 'tossing tails and rolling a 4 or 6' is false. This is because it is not a simple event. In probability theory, a simple event is an event that consists of a single outcome. However, in this case, there are multiple outcomes - getting a tails on the coin toss, and getting either a 4 or 6 on the die roll.

For example, the possible sets are: {Tails, 4} and {Tails, 6}. This indicates that there are multiple outcomes within this event, thus it cannot be considered a simple event.

Hence, the given statement is false because a single outcome in this scenario could only be {Tails, 4} or {Tails, 6}, not both of them combined.

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Answer:

D

Step-by-step explanation:

The regression equation is

y=a+bx

Now, we have to estimate a and b.

We know that slope can be determine as

b=byx=r*(sy/sx)

=0.5*(8/40)

b=0.1

So, we have the slope of regression equation is 0.1.

Now the intercept "a" can be estimated as

y=a+bx

a=ybar-bxbar

a=75-0.1*280

a=47

So, we have the intercept of regression equation is 47.

y=a+bx

y=47+0.1x

Final answer:

The least-squares regression line of final-exam scores on pre-exam total scores, given the statistics provided, is y = 47 + 0.1x, which corresponds to option (D).

Explanation:

The question involves identifying the least-squares regression line of final-exam scores on pre-exam total scores in an economics course. Given the correlation coefficient (r = 0.5), the mean and standard deviation of both the pre-exam totals and the final-exam scores, we can use the formula for the regression line, which is:

y = b0 + b1*x

where y is the predicted final exam score, x is the pre-exam total score, b0 is the y-intercept, and b1 is the slope of the line. The slope (b1) is calculated by:

b1 = r * (standard deviation of y / standard deviation of x) = 0.5 * (8 / 40) = 0.1

And the y-intercept (b0) is found using the means of x and y:

b0 = mean of y - b1 * mean of x = 75 - 0.1 * 280 = 47

So, the regression line is y = 47 + 0.1x, which corresponds to option (D).

The following statement

'In general, men who took drug x maintained or increased the number of visible scalp hairs; while scalp hairs counts in men who took the placebo continued to decrease'

can be referred as descriptive statistics, since we are saying that the msot of the men who took the drug gain scalp hairs while the contrary happened from those men who took placebo.

On the other hand, the statement

'drug x is effective in maintaining or increasing the amount of scalp hair in men'

Is a deduction you make, so it can be referred as inferential statistics

Answer:

Step-by-step explanation:

The first step is to rearrange the given quadratic equation so that it will take the form of the standard quadratic equation which is expressed as

ax² + bx + c = 0

The given quadratic equation is expressed as

2 = - x + x² - 4

Rearranging it, we would subtract 2 from the left hand side and the right hand side of the equation. It becomes

2 - 2 = - x + x² - 4 - 2

- x + x² - 6 = 0

x² - x - 6

a = 1

b = - 1

c = - 6

Substituting into the quadratic formula, it becomes

x = [- (- 1) ± √(- 1)² - 4(1)(- 6)] /2(1)

Answer:

The answer to your question is 2.6 in

Step-by-step explanation:

Data

width = ? in

width = 65 mm

Process

1.- To solve this problem, use proportions and cross multiplication

                   1 in ---------------------- 25.4 mm

                   x    ----------------------  65 mm

                   x = (65 x 1) / 25.4

2.- Simplification

                   x = 65 / 25.4

3.- Result

                   x = 2.6 in      

Answer:

Option A.

Step-by-step explanation:

Let M be the milk per gallon.

C be the cookies per dozen.

B be the bundle (one gallon of milk and a dozen cookies ).

Milk can be sold by itself for a profit of $1.50 per gallon. Cookies can likewise be sold at a profit of $2.50 per dozen and bundle is sold for a profit of $3.00 per bundle.

[tex]Profit= 1.5M + 2.5C + 3B[/tex]

We need to maximize the profits. So, our objective function is

Therefore, the correct option is A.

The probability of drawing an ace or a heart from a full deck of cards is approximately 5.88%. The probability of drawing a card that is neither an ace nor a heart is approximately 45.89%.

This question is about probability in Mathematics. A full deck has 52 cards, 4 of which are aces, and 13 of which are hearts. The total number of favorable outcomes are the scenarios where we draw an ace or a heart.

For part a) There are 16 favorable outcomes (4 aces + 12 remaining hearts). For 2 draws, that's 16/52 * 15/51 = 0.0588 or 5.88% chance of drawing an ace or a heart.

For part b) There are 36 non-favorable outcomes (52 total - 16 favorable). The chance of drawing a card that is neither ace nor heart is 36/52 * 35/51 = 0.4589 or 45.89%.

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(a) Growth Rate Constants: 6

(b) Carrying Capacity Constants: 0

Interaction Constants: 3

The species compete since the interaction constant is positive, indicating competition for resources between the two species.

For the given system:

[tex]\[ x' = 6x \][/tex]

[tex]\[ S_{xy} = 32y^2 - 8y - 8 + 3xy - 31 \][/tex]

(a) Growth Rate Constants: The growth rate constant for species [tex]\(x\)[/tex] is [tex]\(6\)[/tex].

(b) Carrying Capacity Constants: There is no explicit term indicating a carrying capacity for either species in the given equations, so the carrying capacity constants are both [tex]\(0\)[/tex].

Interaction Constants: The interaction constant between the two species can be identified from the term [tex]\(S_{xy}\)[/tex], which is [tex]\(3\)[/tex] in this case.

So, the answers are:

(a) Growth Rate Constants: [tex]\(6\)[/tex]

(b) Carrying Capacity Constants: [tex]\(0\)[/tex]

Interaction Constants: [tex]\(3\)[/tex]

The species compete since the interaction constant is positive.

Complete Question :

Population models for two species can report either competition for resources (an rise in one species recede the growth rate in the other) or cooperation for asset (an rise in one species rise the growth rate in the other). For the systems below, identify the boundary as growth rates, carrying capacities, and measures of interconnection between species. Determine if the species compete or cooperate. x' = 6x Sxy = 32y^2 - 8y - 8 + 3xy - 31 (a) Growth Rate Constants: (b) Carrying Capacity Constants: interconnection Constants: Note: Your answers must be numbers (which may include 0). If there is more than one answer, different your answers with commas. Choose One: - The Species Do Not Interact - One species take part, the other Cooperates - Help Both Species Cooperate - Both Species Compete.

Answer:

a) [tex]P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 2.2)=1- P(z<2.2)=1-0.986=0.0139 [/tex]

b) [tex]P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 6.6)=1- P(z<6.6)=1-0.999999=2.06x10^{-11} [/tex]

c) No. The mean can be lower than 131.5 if we find the probability:

[tex]P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)[/tex]

[tex]P(Z\leq 6.3) \approx 1 [/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".  

A.    If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

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

[tex]X \sim N(100,15)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

We are interested on this probability

[tex]P(X\geq 133)[/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\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 2.2)=1- P(z<2.2)=1-0.986=0.0139 [/tex]

And the probability is calculated from the normal standard table or with excel.

B.     If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

[tex] \bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)[/tex]

The new z score is defined as

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

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

[tex]P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z\geq 6.6)=1- P(z<6.6)=1-0.999999=2.06x10^{-11} [/tex]

And the probability is calculated from the normal standard table or with excel.

C.     Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

[tex]P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)[/tex]

[tex]P(Z\leq 6.3) \approx 1 [/tex]

Answer:

0.0139

Step-by-step explanation:

We have to find P(X≥133)

IQ scores normally distributed with mean=100 and standard deviation=sd=15

[tex]P(X\geq 133)=P(\frac{x-mean}{sd} \geq \frac{133-100}{15}[/tex]

[tex]P(Z\geq 2.2)=P(0<Z<infinity)-P(0<z<2.2)[/tex]

[tex]P(Z\geq 2.2)=0.5-0.4861[/tex]

[tex]P(Z\geq 2.2)=0.0139[/tex]

P(X≥133)=0.0139.

So, if one person is selected at random the probability of getting someone with at least 133 IQ score is 1.39%

Answer:

cc

Step-by-step explanation:

The additive inverse of the complex number -8 + 3i is 8 - 3i.

The complex number is basically the combination of a real number and an imaginary number.

The additive inverse of a complex number is the number that, when added to the original number, results in zero.

Or it is the negative of the original number.

To find the additive inverse of the complex number -8 + 3i

we need to negate both the real and imaginary parts. So:

Additive inverse = -( -8 + 3i)

= 8 - 3i

Therefore, the additive inverse of the complex number -8 + 3i is 8 - 3i.

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Answer: The change in the value of x would be : [tex]\Delta x=4.5 [/tex]    .

Step-by-step explanation:

We know that if y represents the value of some varying quantity, then as the value of y varies from p to q.

The change in the value of y is given by :-

[tex]\Delta y=q-p[/tex]     [Next value - previous value]

Given : x represents the value of some varying quantity.

So , As the value of x  varies from 3.5 to 8.

The change in the value of x is given by :-

[tex]\Delta x=8-3.5=4.5 [/tex]  

Hence, the change in the value of x would be : [tex]\Delta x=4.5 [/tex]  

Final answer:

The variance for the given data is 1.69 hours squared.

Explanation:

To find the variance, we need to square the standard deviation. The standard deviation is given as 1.3 hours, so when we square it, we get 1.69 hours squared. Therefore, the variance for the given data is 1.69 hours squared.

By using the distance formula, we calculate the distances between the house, library, and post office, finding the total minimum distance to be approximately 14.8 kilometers.

This question requires the use of the distance formula in mathematics, which is derived from the Pythagorean theorem. The formula is √[(x₂ - x₁)² + (y₂ - y₁)²]. To find the total minimum distance you ride your bike, you calculate the distance from your house to the library, then the library to the post office, and finally, the post office back to your house.

First, calculate the distance from your house at (-3, 2) to the library at (-2, -2): √[(-2 - (-3))² + (-2 - 2)²] = √[1² + -4²] = √[1 + 16] = √17 ≈ 4.1 kilometers.

Second, calculate the distance from the library at (-2, -2) to the post office at (2, 2): √[(2 - (-2))² + (2 - (-2))²] = √[4² + 4²] = √[16 + 16] = √32 ≈ 5.7 kilometers.

Finally, calculate the distance from the post office at (2, 2) back to your house at (-3, 2): √[(-3 - 2)² + (2 - 2)²] = √[-5² + 0²] = √[25 + 0] = √25 = 5 kilometers.

Adding these distances together, 4.1 km + 5.7 km + 5 km = 14.8 kilometers.

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Answer:

3) (98.08, 124.16)

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 111.12

Standard deviation = 13.04

Calculate an interval that is symmetric around the mean such that it contains approximately 68% of fines.

68% of the fines are within 1 standard deviation of the mean speed. So

From 111.12 - 13.04 = 98.08 to 111.12 + 13.04 = 124.16

The interval notation in the smallest value before the highest value.

So the correct answer is:

3) (98.08, 124.16)

Answer:

b) 690 - 7.5*t

c) 0 < t < 92s time (t) is independent quantity

d) 0 < s < 690ft distance from bus stop (s) is dependent quantity

e) f(0) = 690 ft away from bus stop , f(60.25) = 238.125 ft away from bus stop

Step-by-step explanation:

Part a - see diagram

part b

initial distance from bus stop s0 = 690 ft

distance covered = 7.5*t

s = s0 - distance covered

s = 690 - 7.5*t = f(t)

part c

s = 0 or s = 690

0 = 690 -7.5*t

t = 92 s

Hence domain : 0 < t < 92s time (t) is independent quantity

part d

s = 0 or s = 690

Hence range : 0 < s < 690ft distance from bus stop (s) is dependent quantity because it depends on time (t)

part e

f(0) is s @t = 0

f(0) = 690 ft away from bus stop

f(60.25) is s @t = 60.25

f(60.25) = 690 - 7.5*60.25 = 238.125 ft away from bus stop.

Answer:

2.44491234

Step-by-step explanation:

1 gallon = 3785 cm3

Unit conversion is a way of converting some common units into another without changing their real value. The volume of  9255 cm³ in gallons is equal to 2.4451 gallons.

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Since 1 gallon is equal to 4546.09 cubic centimeters, therefore, the volume of 9255 cm³ in gallons can be written as,

1 gallons = 3785 cm³

1 cm³ = 1/3785 gallons

The volume of 9255 cm³ in gallons is,

Volume = 9255 × (1/3785 ) gallons

= 2.4451 gallons

Hence, the volume of  9255 cm³ in gallons is equal to 2.4451 gallons.

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Answer:

68,000

Step-by-step explanation:

$50,000 + $18,000 = $68,000