At a camground, a rectangular fire pit is 6 feet by 5 feet. What is the area of the largest circular fire that can be made in inches

Answer:

24.06% of total sales

Step-by-step explanation:

Total sales, which were $170,000 originally, are expected to grow by 10%. The expected value of total sales this year is:

[tex]S=\$170,000*1.10 = \$187,000[/tex]

If Basil sales remain at the same value of $45,000, the percentage of sales corresponding to basil is:

[tex]B=\frac{\$45,000}{\$187,000}*100\%\\B=24.06\%[/tex]

Therefore, basil will correspond to 24.06% of total sales.

Basil sales will represent approximately 24.06% of Scarborough Faire Herb Farm's projected total sales this year, given that total sales are forecasted to increase by 10% and basil sales remain unchanged.

The student's question concerns the percentage of total sales that basil sales will represent for Scarborough Faire Herb Farm in the current year, assuming a 10% increase in total sales from the previous year and unchanged basil sales. Firstly, we calculate the projected total sales for this year by increasing last year's total sales by 10%. The calculation is as follows:

Since the basil sales are to remain the same as last year ($45,000), we now calculate what percentage this represents of the projected total sales for this year:

Thus, basil sales will account for approximately 24.06% of the Scarborough Faire Herb Farm's total sales this year.

For [tex]0\le t\le1[/tex], we have

[tex]y'+6y=1\implies e^{6t}y'+6e^{6t}=(e^{6t}y)'=e^{6t}\implies y=\dfrac16+Ce^{-6t}[/tex]

Given that [tex]y(0)=0[/tex], we have

[tex]0=\dfrac16+C\implies C=-\dfrac16[/tex]

so that

[tex]y=\dfrac16(1-e^{-6t})[/tex]

For [tex]t>1[/tex] (I think you mistakenly wrote [tex]t>0[/tex], which overlaps with the first definition of [tex]g(t)[/tex]), we have

[tex]y'+6y=0\implies e^{6t}y'+6e^{6t}y=(e^{6t}y)'=0\implies y=Ke^{-6t}[/tex]

We want this to be a continuation of the previously found solution [tex]y[/tex] at [tex]t=1[/tex], which means we need to pick [tex]K[/tex] such that

[tex]\dfrac16(1-e^{-6})=Ke^{-6}\implies K=\dfrac16(e^6-1)[/tex]

Then the solution to the IVP is

[tex]y(t)=\begin{cases}\frac16(1-e^{-6t})&\text{for }0\le t\le1\\\frac{e^6-1}6e^{-6t}&\text{for }t>1\end{cases}[/tex]

Alternatively, we can get around treating [tex]g(t)[/tex] piecemeal and resorting to the Laplace transform. Write

[tex]g(t)=\begin{cases}1&\text{for }0\le t\le1\\0&\text{for }t>1\end{cases}=u(t)-u(t-1)[/tex]

where

[tex]u(t-c)=\begin{cases}0&\text{for }t<c\\1&\text{for }t\ge c\end{cases}[/tex]

is the unit step function.

Take the Laplace transform of both sides of the ODE:

[tex]y'+6y=g(t)\overset{\text{L.T.}}{\implies}(sY-y(0))+6Y=\mathcal L_s\{g(t)\}[/tex]

where [tex]Y=Y(s)[/tex] is the Laplace transform of [tex]y(t)[/tex].

We have

[tex]\mathcal L_s\{g(t)\}=\displaystyle\int_0^\infty g(t)e^{-st}\,\mathrm dt=\int_0^1e^{-st}\,\mathrm dt=\dfrac{1-e^{-s}}s[/tex]

so that

[tex](s+6)Y=\frac{1-e^{-s}}s\implies Y=\dfrac{1-e^{-s}}{s(s+6)}=\dfrac{1-e^{-s}}6\left(\dfrac1s-\dfrac1{s+6}\right)[/tex]

Taking the inverse transform yields

[tex]y=\dfrac{1-u(t-1)}6-\dfrac{e^{-6t}}6(e^tu(t-1)-1)[/tex]

[tex]y=\dfrac{1-e^{-6t}}6+\dfrac{e^{6-6t}-1}6u(t-1)[/tex]

which is equivalent to the same solution found earlier; for [tex]0\le t\le1[/tex], [tex]u(t-1)=0[/tex], so that [tex]y=\frac{1-e^{-6t}}6[/tex]; for [tex]t>1[/tex], [tex]u(t-1)=1[/tex], and [tex]y=\frac{1-e^{-6t}}6+\frac{e^{6-6t}-1}6=\frac{(e^6-1)e^{-6t}}6[/tex].

The given differential equation needs to be solved separately for two time ranges because of the piecewise-defined function g(t). Solution for the corresponding equations are founded using the techniques of homogeneous equation solutions and the integrating factor method. These solutions are then matched at the point of continuity.

The given differential equation is a first-order linear differential equation of the form y′+p(t)y=g(t). We need to solve this equation considering two cases due to the piecewise definition of g(t).

Case 1: For 0 ≤ t ≤ 1, g(t) = 1. The corresponding homogeneous equation is y' + 6y = 0, with the solution being y(t) = Ce-6t. We find the particular solution using the integrating factor method, yielding y(t) = t/6 - 1/36 + Ce-6t. Substituting the initial condition y(0) = 0 helps us solve for C, giving the final solution for this range as y(t) = t/6 - 1/36.

Case 2: For t > 1, g(t) = 0. The homogeneous solution is the same as in Case 1, but in this case, no particular solution needs to be added, so the solution is y(t) = Ce-6t. The constant is determined by making the function continuous at t=1. We ultimately get y(t) = (1-e-6(t-1))/36.

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Answer:

The probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E) = 0.76

Step-by-step explanation:

Probability of wearing a seatbelt in an accident = P(B) = 65% = 0.65

Probability of not wearing a seatbelt in an accident = P(B') = 1 - 0.65 = 0.35

Probability of escaping hospitalization and/or death given that one is wearing a seatbelt = P(E|B) = 83% = 0.83

Probability of escaping hospitalization and/or death given that one isn't wearing a seatbelt = P(E|B') = 0.49

Find the probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E)

The probability of P(X|Y) is given mathematically as P(X n Y)/P(Y)

P(B|E) = P(B n E)/P(E)

But P(E) is unknown at the moment.

But P(E) = P(B n E) + P(B' n E) mathematically,.

P(B n E) can be obtained using P(E|B) and P(B)

P(E|B) = P(B n E)/P(B)

P(B n E) = P(E|B) × P(B) = 0.83 × 0.65 = 0.5395

And

P(B' n E) can be obtained using P(E|B') and P(B')

P(E|B') = P(B' n E)/P(B')

P(B' n E) = P(E|B') × P(B') = 0.49 × 0.35 = 0.1715

P(E) = P(B n E) + P(B' n E) = 0.5395 + 0.1715 = 0.711

The probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E)

P(B|E) = P(B n E)/P(E) = 0.5395/0.711 = 0.76

The probability that all three students solve the problem is calculated by multiplying their individual success probabilities together. The total probability in this case is 64.6%.

Your question pertains to probability, a topic in Mathematics. When three students independently attempt to solve a problem, and you have the probabilities of their success, the probability that all three will successfully solve the problem is determined by the product of their respective probabilities.

Therefore, the probability that all three students - the first with a probability of 0.95, the second with 0.85, and the third with 0.80 - will successfully solve the problem is calculated as follows:

0.95 * 0.85 * 0.80 = 0.646

Hence, there is a 64.6% probability that all three students will successfully solve the problem.

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Answer:

33% probability that the company will win at least one of the two contracts

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a company is awarded the first contract.

B is the probability that a company is awarded the second contract.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that a company is awarded the first contract but not the second and [tex]A \cap B[/tex] is the probability that a company is awarded both contract.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

The probability that it will get both contracts is .13.

This means that [tex]A \cap B = 0.13[/tex]

The probability that it will be awarded a second contract is .21

This means that [tex]B = 0.21[/tex]

[tex]B = b + (A \cap B)[/tex]

[tex]0.21 = b + 0.13[/tex]

[tex]b = 0.08[/tex]

The probability that a company will be awarded a certain contract is .25

This means that [tex]A = 0.25[/tex]

[tex]A = a + (A \cap B)[/tex]

[tex]0.25 = a + 0.13[/tex]

[tex]a = 0.12[/tex]

What is the probability that the company will win at least one of the two contracts?

[tex]A \cup B = a + b + A \cap B = 0.12 + 0.08 + 0.13 = 0.33[/tex]

33% probability that the company will win at least one of the two contracts

The probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\) or 0.34.[/tex]

To find the probability that the company will win at least one of the two contracts, we can use the principle of inclusion-exclusion. The principle states that the probability of the union of two events (in this case, winning either contract) is the sum of the probabilities of each event occurring individually, minus the probability of both events occurring together.

Let [tex]\(P(A)\)[/tex] be the probability that the company will win the first contract, [tex]\(P(B)\)[/tex] be the probability that the company will win the second contract, and [tex]\(P(A \cap B)\)[/tex] be the probability that the company will win both contracts. We are given:

[tex]\(P(A) = 0.25\), \(P(B) = 0.21\), \(P(A \cap B) = 0.13\).[/tex]

The probability that the company will win at least one contract, [tex]\(P(A \cup B)\)[/tex], is given by:

[tex]\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\][/tex]

Substituting the given probabilities:

[tex]\[P(A \cup B) = 0.25 + 0.21 - 0.13\] \[P(A \cup B) = 0.46 - 0.13\] \[P(A \cup B) = 0.33\][/tex]

To express this probability as a fraction, we can write 0.33 as [tex]\(\frac{33}{100}\)[/tex], which simplifies to[tex]\(\frac{17}{50}\)[/tex] when reduced to its simplest form.

Therefore, the probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\)[/tex] or 0.34

Answer:

Null hypothesis: [tex]p \geq 0.7[/tex]

Alternative hypothesis: [tex]p<0.7[/tex]

A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.

So the correct option for this case would be:

c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.  

Solution to the problem

On this case we want to test if the sporting goods store claims that at least 70^ of its customers, so the system of hypothesis would be:

Null hypothesis: [tex]p \geq 0.7[/tex]

Alternative hypothesis: [tex]p<0.7[/tex]

A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.7 when actually is not true.

So the correct option for this case would be:

c. The sporting goods store thinks that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores.

Answer:

(a) The number of bulbs often replaces is 66.67.

(b) The fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].

Step-by-step explanation:

Let X = lifetime of a bulb and Y = time after which the bulb is replaced.

It is provided that X follows Exponential distribution with mean lifetime of a bulb is, 200 days.

And the rate at which the bulb is replaced is, 0.01 also following an Exponential distribution.

(a)

A bulb is replaced only after it burns out or a handyman comes at times of a Poisson process and replaces it.

Then min (X, Y) follows an Exponential distribution with parameter [tex](\frac{1}{200}+0.01)[/tex].

The mean of an Exponential distribution with parameter θ is:

[tex]Mean=\frac{1}{\theta}[/tex]

Compute the mean of min (X, Y) as follows:

[tex]Mean =\frac{1}{(\frac{1}{200}+0.01)} =\frac{1}{0.015}= 66.67[/tex]

Thus, the number of bulbs often replaces is 66.67.

(b)

Compute the probability of the event (X < Y) as follows:

[tex]P(X<Y)=\frac{0.005}{0.015} =\frac{1}{3}[/tex]

Thus, the fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].

Answer:

(a) The number of bulbs often replaces is 66.67.

(b) The fraction of the replacements that are due to failure, in the long run, is .

Step-by-step explanation:

The point between (4,16) and (16,16) is (10,16) as calculated using the midpoint formula. This point is exactly halfway between the given points.

To determine a point between the two points (4,16) and (16,16), we need to calculate the midpoint.

[tex]M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)[/tex]

[tex]M = \left(\frac{4 + 16}{2}, \frac{16 + 16}{2}\right) = (10, 16)[/tex]

Therefore, the point that lies between (4,16) and (16,16) is (10,16).

Answer:

Franklin Freightways experienced $66 million.

Step-by-step explanation:

= ( Pretax accounting income + Overweight fines - Temporary difference: Depreciation) * tax rate

($200 + 5 - $40) × 40%

=Tax liability of $66.

Answer:

The regression equation is:

Final Grade = 96.14 - 2.76 Number of absence

A student who was absent for 14 days received a final grade of 58.

Step-by-step explanation:

The general form a regression equation is:

[tex]y=\alpha +\beta x[/tex]

Here,

y = dependent variable = Final grade

x = independent variable = Number of absence

α = intercept

β = slope

The formula to compute the intercept and slope are:

[tex]\alpha =\frac{\sum Y. \sum X^{2}-\sum X.\sum XY}{n.\sum X^{2}-(\sum X)^{2}}[/tex]

[tex]\beta =\frac{n.\sum XY-\sum X.\sum Y}{n.\sum X^{2}-(\sum X)^{2}}[/tex]

The value of α and β are computed as follows:

[tex]\alpha =\frac{\sum Y. \sum X^{2}-\sum X.\sum XY}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(722\times460-(52\times3732)}{(9\times460)-(52)^{2}} =96.139\approx96.14[/tex]

[tex]\beta =\frac{n.\sum XY-\sum X.\sum Y}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(9\times3732-(52\times722)}{(9\times460)-(52)^{2}} =-2.755\approx-2.76[/tex]

The regression equation is:

Final Grade = 96.14 - 2.76 Number of absence

For the value of Number of absence = 14 compute the value of Final grade as follows:

[tex]Final\ Grade = 96.14 - 2.76\ Number\ of\ absence\\=96.14-(2.76\times14)\\=57.5\\\approx58[/tex]

Thus, a student who was absent for 14 days received a final grade of 58.

To find the predicted final grade for 14 absences, calculate the slope and y-intercept of the regression line for the given data set to form the equation y=mx+b. With x as 14, solve the equation.

To answer this question, we first need to find the equation of the regression line using the given number of absences (x) and final grades (y). This is achieved by calculating the slope and y-intercept of the best fit line for the data set. The formula for the slope (m) is given by the expression [N(Σxy) - (Σx)(Σy)] / [N(Σx^2) - (Σx)^2] and the y-intercept (b) by (Σy - m(Σx)) / N. After calculating these values, you can form the equation y = mx + b. Using the equation, input the absent times (14) into the x-variable to predict the final grade.

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Answer:

The following option is not correct:

(a) If 100 numbers are generated comma each integer between 0 and 99 must occur exactly once.

Step-by-step explanation:

This is not correct, as there is a possibility of an integer being generated twice when 100 numbers are generated.

This can be explained with an example such as stated below:

3 numbers are to be generated.

The number generated can either be 1, 2, or 3.

The probability for all three numbers to be generated once when the generator is run 3 times is:

(1/3)*(1/3)*(1/3) * Number of ways to arrange the three numbers

Thus this probability will be:

Probability = (1/3)^3 * 3!

Probability = 0.222

Since the probability here is not equal to 1, the probability for the same thing happening at a larger scale will also not be 1.

Final answer:

In a random number generator for numbers between 0 and 99, each integer should occur exactly once when 100 numbers are generated.

Explanation:

The correct statement among the options is:

Let's break it down:

Answer:

w = $1 or $10 or $25

Probability distribution of w = ($1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $10,$1  $10,$1  $10,$10  $10,$25  $10,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25)

Step-by-step explanation:

Since the winner gets the larger amount of the two slips picked, random variable 'w' which is the amount awarded could be $1 or $10 or $25.

The probability distribution of 'w' is the values that the statistic takes on. Which could be: $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $1,$1  $1,$1  $1,$10  $1,$25  $1,$25  $10,$1  $10,$1  $10,$10  $10,$25  $10,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25  $25,$1  $25,$1  $25,$10  $25,$25  $25,$25

Answer:

Discrete variable        

Step-by-step explanation:

We are given the following in the question:

Variable:

Total number of points scored in a football game

Discrete and continuous data:

Since, total number of points score in a foot game are expressed in whole numbers and cannot be expressed in decimals, they are discrete variable. They cannot take all the values within an interval and they are usually counted.

Thus,

Total number of points scored in a football game is a discrete variable.

Answer:

Answer:

a).

The amount spent on school materials for each term of all ST201students

b).

a).

It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.

b).

The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.

c).

mean: $617

z critical value at 95%: 1.96

standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017

lower limit= mean-1.96*se = 617-1.96*55.9017=507.43

upper limit= mean+1.96*se = 617+1.96*55.9017=726.57

d).

The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).

Step-by-step explanation:

Answer:

a). 0.03593, 0.27425

(bl 0.0703

(c) the lifespan of such bulb is less than 210 days

(d) 0.0298.

Step-by-step explanation:

Given that U = 300, sd= 50

Z = X-U/sd

(a) We find the probability

Pr(X<210) = Pr(Z<(210-300)/50)

= Pr(Z<-1.8)

= 0.03593

Pr(X >330) = Pr(Z >(330-300)/50))

= Pr(Z> 0.6)

= 1- PR(Z<0.6) = 1 - 0.72575

Pr(X>330)= 0.27425

(b) Pr(280< X< 330) = Pr( 280 < Z< 380)

= Pr([280-300/50] < Z < [380-300/50])

= Pr(0.4 < Z < 1.6)

= Pr(Z< 1.6) - Pr(Z < 0.4)

=0.72575 - 0.65542

= 0.07033

= 0.0703

(c) the lifespan of such bulbs is less than 210 days

(d) given n = 6, x = 4 ,

Pr(X>330) = 0.27425 = p

q = 1-p = 0.72575

Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²

Pr(X=4) = 10 × 0.005657 × 0.52671

Pr(X= 4) = 0.029795

Pr(X= 4) = 0.0298

mADC=360-mABC=360-184=176⁰

So, measure of angle ABC= mADC/2=176/2=88⁰

The required measure of angle ABC is 88° for the given figure. The correct answer is option A.

The arc is a portion of the circumference of a circle. The circumference of a circle is the distance or perimeter around a circle

The measure of the arc is given as follows:

mABC = 184°

According to the given figure, mADC is the part of the full circle which is complete arc that measure of angle is 360°.

mADC = 360° - mABC

mADC = 360° - 184°

mADC = 176°

As we know that the measure of angle ABC is equal to half of mADC.

The measure of angle ABC = mADC/2

The measure of angle ABC = 176/2

The measure of angle ABC = 88°

Therefore, the required measure of angle ABC is 88°.

Learn more about the arc of the circle here:

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Using the limit laws of addition and multiplication, the limit of the given equation is solved by combining the calculated limits of the function and the constant, giving an answer of approximately 13.78.

To solve this problem, we can use the limit laws of addition and multiplication. The limit laws state that the limit of the sum of two functions is equal to the sum of their individual limits, and the limit of the product of two functions is equal to the product of their individual limits.

So, based on these laws, we first separate the function into two parts: the limit of 3√(f(x).g(x)) as x approaches 3 and the limit of 10 as x approaches 3.

Given lim f(x)=>6 and lim g(x)=>9 when x approaches 3, we multiply these values together to get a product of 54. The cubed root of 54 is approximately 3.78.

The constant 10 has a limit of 10, as constants maintain their value irrespective of the limit.

Adding these values together, 3.78 + 10, gives us a limit of approximately 13.78.

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Answer:

a) [tex]P(5<X<6)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5-7.37}{1.25}<Z<\frac{6-7.37}{1.26})=P(-1.90<z<-1.10)[/tex]

And we can find this probability with this difference:

[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)[/tex]

And using the norma standard distribution or excel we got:

[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)=0.136-0.029=0.107[/tex]

b) [tex]P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)[/tex]

And using the complement rule we got:

[tex]P(Z>-1.096) =1-P(Z<-1.096) = 1-0.137= 0.863[/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 variable of interest of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(7.37,1.25)[/tex]  

Where [tex]\mu=7.37[/tex] and [tex]\sigma=1.25[/tex]

We are interested on this probability

[tex]P(5<X<6)[/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(5<X<6)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5-7.37}{1.25}<Z<\frac{6-7.37}{1.26})=P(-1.90<z<-1.10)[/tex]

And we can find this probability with this difference:

[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)[/tex]

And using the norma standard distribution or excel we got:

[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)=0.136-0.029=0.107[/tex]

Part b

For this case we want this probability:

[tex] P(X>6)[/tex]

And we can use the z score and we got:

[tex]P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)[/tex]

And using the complement rule we got:

[tex]P(Z>-1.096) =1-P(Z<-1.096) = 1-0.137= 0.863[/tex]

To find the probability that the hospital stay is from 5 to 6 days, we need to standardize the values using the z-score formula. The probability that their hospital stay is from 5 to 6 days is approximately 0.10756. The probability that their hospital stay is greater than 6 days is approximately 0.86301.

To find the probability that the hospital stay is from 5 to 6 days, we first need to standardize the values using the z-score formula.

z = (x - µ) / σ

Let's calculate the z-scores for x = 5 and x = 6.

For x = 5:

z = (5 - 7.37) / 1.25 = -1.896

For x = 6:

z = (6 - 7.37) / 1.25 = -1.096

Next, we can use the standard normal distribution table or a calculator to find the probabilities associated with these z-scores:

P(z < -1.896) = 0.02999

P(z < -1.096) = 0.13699

To find the probability that the hospital stay is from 5 to 6 days, we subtract P(z < -1.096) from P(z < -1.896):

P(5 < x < 6) = P(z < -1.896) - P(z < -1.096) = 0.02999 - 0.13699 = 0.10756

Therefore, the probability that their hospital stay is from 5 to 6 days is approximately 0.10756, rounded to five decimal places.

To find the probability that their hospital stay is greater than 6 days, we can use the standard normal distribution table or a calculator to find the probability associated with the z-score for x = 6:

P(z > -1.096) = 1 - P(z < -1.096) = 1 - 0.13699 = 0.86301

Therefore, the probability that their hospital stay is greater than 6 days is approximately 0.86301, rounded to five decimal places.

Answer:

a) 0.00563

b) 1

c) 0

Step-by-step explanation:

Total = 40+30+20 =90

a) (20C8×30C6×40C6)/90C20

= 0.00563

b) 1 - (no chocolate + 1 chocolate)

1 - [(50C20) + (40C1×50C19)]/90C20

1 - 0.00002478

= 0.9999752187

c) [40C20+20C20+30C20]/90C20

= 0.0000000027045

This is about permutations and combinations.

a) Probability = 0.00563

b) Probability = 0.99997522

c) Probability = 0.0000000027045

Chocolate = 40

Coconut = 30

Banana = 20

Total number of chocolates = 40 + 30 + 20

Total number of chocolates = 90

a) Probability that she will choose 8 banana and 6 chocolate cakes if she chooses 20 cupcakes at random will be;

(20C₈ × 30C₆ × 40C₆)/90C₂₀

(125970 × 593775 × 3838380)/50980740277700939310

This gives us   0.00563

b) Probability of at least 2 chocolate cupcakes is;

1 - [P(no chocolate) + P(1 chocolate)]

P(no chocolate) = (50C₂₀)/90C₂₀

P(1 chocolate) = (40C₁ × 50C₁₉)/90C₂₀

Thus;

1 - [P(no chocolate) + P(1 chocolate)] = 1 - [(40C₁ × 50C₁₉) + 50C₂₀]/90C₂₀

This gives us;  0.99997522

c) Probability of getting all cupcakes of same kind is;

(40C₂₀ + 20C₂₀ + 30C₂₀)/90C₂₀

⇒ 0.0000000027045

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Answer:

The average rate of change as follows of I is -0.185.

Step-by-step explanation:

The relation between voltage V, current I, and resistance R in a circuit, according to the Ohm's law is:

[tex]V = IR[/tex]

It is provided that:

V = 20 V

The interval between which R varies is, R = 8 to R = 8.1.

Compute the value of I as follows:

[tex]V=IR\Rightarrow I=\frac{I}{R}\Rightarrow I=\frac{20}{R}[/tex]

Compute the average rate of change as follows of I as follows:

[tex]\frac{\delta I}{\delta R}=\frac{(I\times8.1)-(I\times8)}{8.1-8}[/tex]

    [tex]=\frac{1}{0.1}[\frac{12}{8.1}-\frac{12}{8}][/tex]

    [tex]=\frac{12}{0.1}[\frac{8-8.1}{64.8}][/tex]

    [tex]=-0.1852[/tex]

Thus, the average rate of change as follows of I is -0.185.

Answer:

The parking rate at the hospital a parking garage are 50 cents for the 1st hour and 25 cents for each additional hour if guan part in the hospital parking garage for 8 hours how much will the total of charge for parking be

Answer:

poalo is correct

Step-by-step explanation

Answer:

Neither Marco nor Paolo is correct, because neither one used the rational exponent property correctly.

Step-by-step explanation:

I just did the test and it told me that I got it right.

Hope that helps! Please mark me brainliest.

Answer:10 gallons of 30% solution and 30 gallons of 70% solution should be mixed.

Step-by-step explanation:

Let x represent the number of gallons of the 30% solution that should be mixed.

Let y represent the number of gallons of the 70% solution that should be mixed.

The total number of gallons of the mixture to be made is 40. This means that

x + y = 40

The 30% solution of fertilizer is to be mixed with a 70% solution of fertilizer in order to get 40 gallons of a 60% solution. This means that

0.3x + 0.7y = 0.6 × 40

0.3x + 0.7y = 24- - - - - - - - - - - - - -1

Substituting x = 40 - y into equation 1, it becomes

0.3(40 - y) + 0.7y = 24

12 - 0.3y + 0.7y = 24

- 0.3y + 0.7y = 24 - 12

0.4y = 12

y = 12/0.4

y = 30

x = 40 - y = 40 - 30

x = 10

Answer:

The Expected cosy of the builder is $433.3

Step-by-step explanation:

$400 is the fixed cost due to delay.

Given Y ~ U(1,4).

Calculating the Variable Cost, V

V = $0 if Y≤ 2

V = 50(Y-2) if Y > 2

This can be summarised to

V = 50 max(0,Y)

Cost = 400 + 50 max(0, Y-2)

Expected Value is then calculated by;

Waiting day =2

Additional day = at least 1

Total = 3

E(max,{0, Y - 2}) = integral of Max {0, y - 2} * ⅓ Lower bound = 1; Upper bound = 4, (4,1)

Reducing the integration to lowest term

E(max,{0, Y - 2}) = integral of (y - 2) * ⅓ dy Lower bound = 2; Upper bound = 4 (4,2)

E(max,{0, Y - 2}) = integral of (y) * ⅓ dy Lower bound = 0; Upper bound = 2 (2,0)

Integrating, we have

y²/6 (2,0)

= (2²-0²)/6

= 4/6 = ⅔

Cost = 400 + 50 max(0, Y-2)

Cost = 400 + 50 * ⅔

Cost = 400 + 33.3

Cost = 433.3

Answer:

the expected value of the builder’s cost due to waiting for supplies is $433.3

Step-by-step explanation:

Due to the integration symbol and also for ease of understanding, i have attached the explanation as an attachment.

The maximum height reached by the shell is approximately 18255.79 meters.

To find when the shell reaches its maximum height and the value of the maximum height, we need to consider the forces acting on the shell and analyze its motion.

1. Force due to gravity:

The force due to gravity is given by the formula:

Force_gravity = mass × acceleration_due_to_gravity

Here, the mass of the shell is 2 kg, and the acceleration due to gravity is 9.81 m/s².

2. Force due to air resistance:

The force due to air resistance is given by the formula:

Force_air_resistance = (0.1) × velocity²

Here, the velocity of the shell is given as 600 m/s.

Using Newton's second law, we can calculate the net force acting on the shell:

Net force = Force_gravity - Force_air_resistance

When the shell reaches its maximum height, the net force is equal to zero because there is no acceleration at that point. Therefore, we can set the net force equation to zero and solve for the time:

0 = Force_gravity - Force_air_resistance

mass × acceleration_due_to_gravity = (0.1) × velocity²

2 kg × 9.81 m/s² = (0.1) × (600 m/s)²

Simplifying further, we find:

19.62 = 0.1 × 360,000

Time = 600 m/s / 9.81 m/s²

Time ≈ 61.15 seconds

Therefore, the shell will reach its maximum height approximately 61.15 seconds after being shot upward.

To find the maximum height, we can use the kinematic equation:

h = v₀t - (1/2)gt²

Substituting the given values into the equation, we find:

h = (600 m/s) × (61.15 s) - (1/2) × (9.81 m/s²) × (61.15 s)²

h ≈ 18255.79 meters

Therefore, the maximum height reached by the shell is approximately 18255.79 meters.

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The shell will reach its maximum height after approximately 4.51 seconds and this maximum height is around 100 meters.

This problem involves the physics of motion under the influence of gravity and air resistance. An important concept here is the 'terminal velocity', which is the velocity at which the upward force (due to the initial impulse provided to the shell) balances out the downward force (due to gravity and air resistance).

First, we must establish that the terminal velocity 'v' of the shell upwards will be achieved when the sum of the forces acting on it are zero. This gives us:

0 = -0.1v^2 + 2 * 9.81

. Solving this quadratic equation reveals v = sqrt(2*9.81/0.1) m/s ≈ 44.3 m/s.

After achieving this terminal velocity, the shell will start decelerating at a rate of 9.81 m/s^2 (the gravity acceleration). The time 't' that takes for the shell to stop moving upwards (so to reach its maximum height) can be calculated using the formula:

t = v / gravity = 44.3/9.81 ≈ 4.51 seconds

.

As for the maximum height 'h', it can be calculated using this formula:

h = v * t + 0.5 * (-gravity) * t^2

By inserting the values of v, gravity, and t in this equation, we find h ≈ 100 m.

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There is a useful linear relationship between rainfall and runoff and the relationship is y = 0.85x - 2.65

Deciding whether there is a useful linear relationship between rainfall and runoff

From the question, we have the following parameters that can be used in our computation:

x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127

y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105

Using a graphing tool, we have the following summary

The regression equation can be represented as

y = bx + a

Where

b = SP/SSX = 17313.93/20086.93 = 0.86

a = MY - bMX = 43.27 - (0.86*53.27) = -2.65

So, we have

y = 0.85x - 2.65

Hence, there is a useful linear relationship between rainfall and runoff

Answer:

The correct answers are

option(1), option (4),option (5)

Step-by-step explanation:

One to one: Every image has exactly one unique pre-image in domain.

(1)

f(a) = b, f(b)=a, f(c)=c, f(d)=d

b has a pre-image in domain i.e a

a has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is one to one.

(2)

f(a) = b, f(b)=a, f(c)=c, f(d)=d

b has a pre-image in domain i.e a

a has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is one to one.

(3)

f(a) = b, f(b)=b, f(c)=d, f(d)=c

b has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

b has two pre image.

Here all elements have not a unique pre- image in domain.

Therefore it is not a one to one mapping.

(4)

f(a) = b, f(b)=b, f(c)=d, f(d)=c

b has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

b has two pre image.

Here all elements have not a unique pre- image in domain.

Therefore it is not a one to one mapping.

(5)

f(a) = d, f(b)=b, f(c)=d, f(d)=c

d has a pre-image in domain i.e a

b has a pre-image in domain i.e b

d has a pre-image in domain i.e c

c has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is a one to one mapping.

(6)

f(a)=d, f(b)=b,f(c)=c,f(d)=d

d has a pre-image in domain i.e a

b has a pre-image in domain i.e b

c has a pre-image in domain i.e c

d has a pre-image in domain i.e d

Here all elements have a unique pre- image in domain.

Therefore it is a one to one mapping.

f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

One-to-one - every image has exactly one unique pre-image in the domain.

1. f(a) = b, f(b) = a, f(c) = c, f(d) = d

b has a pre image in a domian that is a

a has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

2. f(a) = b, f(b) = a, f(c) = c, f(d) = d

b has a pre image in a domian that is a

a has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

3. f(a) = b, f(b) = b, f(c) = d, f(d) = c

b has a pre image in a domian that is a

b has a pre image in a domian that is b

d has a pre image in a domian that is c

c has a pre image in a domian that is d

Here all elements do have not a unique pre-image in a domain.

Therefore it is not one-to-one.

4. f(a) = b, f(b) = b, f(c) = d, f(d) = c

b has a pre image in a domian that is a

b has a pre image in a domian that is b

d has a pre image in a domian that is c

c has a pre image in a domian that is d

Here all elements do have not a unique pre-image in a domain.

Therefore it is not one-to-one.

5. f(a) = d, f(b) = b, f(c) = c, f(d) = d

d has a pre image in a domian that is a

b has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

6. f(a) = d, f(b) = b, f(c) = c, f(d) = d

d has a pre image in a domian that is a

b has a pre image in a domian that is b

c has a pre image in a domian that is c

d has a pre image in a domian that is d

Here all elements have a unique pre-image in a domain.

Therefore it is one-to-one.

More about the function link is given below.

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Answer:

Step-by-step explanation:

A. Given

  T = 0.05x +35 . . . . Terri's cost of operating a craft booth

  D = 0.03x +55 . . . . Donna's cost of operating a craft boot

  T = D

where x is the dollar amount of sales.

__

B. Solution

Subtracting the equation for D from that of T, we get ...

  T - D = 0

  (0.05x +35) -(0.03x +55) = 0 = 0.02x -20

  0 = x -1000 . . . . . divide by 0.02

  x = 1000

  T = D = 0.05(1000) +35 = 85

  (x, T) = (x, D) = (1000, 85)

__

C. Meaning

According to the given definitions of the variables, each booth pays a total of $85 in fees for sales of $1000.

The system of equations is T = 0.05x + 35 and D = 0.03x + 55. The solution is (1000, 85), meaning Terri and Donna each earned $1000 from sales and paid $85 total in booth and commission fees.

The system of equations to model the amount of money Terri and Donna pay each weekend at the craft shows can be written as follows: T = 0.05x + 35 and D = 0.03x + 55.

Since we know that Terri and Donna both paid the same total amount of money for their booths and commission, it means T = D. Or, we can equate the two equations: 0.05x + 35 = 0.03x + 55. Solving for x gives us x = 1000. Substituting x = 1000 into T = 0.05x + 35 equation, we get T (and D) = 85. So, the solution to the system of equations is (1000, 85).

In the context of this situation, the solution means that Terri and Donna both earned $1000 from sales, and each paid $85 total for their booth and commission fees.

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Answer:

At least 421 units of boards need to be sold to limit the cost per board to $288

Step-by-step explanation:

Let the number of surfboards made or sold be x

Total cost = fixed cost + variable cost

Fixed Cost = $61466

Variable Cost = 142 × x = $142x

Total cost = 61466 + 142x

Revenue = unit price × quantity = 288×x = 288x

The number of boards that needs to be sold to limit the cost off a board to $288 is the number of units at the point where the total cost matches the revenue.

61466 + 142x = 288x

288x - 142x = 61466

146x = 61466

x = 421 units.