subtract and simplify. please help

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:

18

Step-by-step explanation:

if neither wants to go, from 20 it will be 18

Answer:

Permutation count the number of different arrangements pf r out of n items, while combination count  the number of group of r out of n items.

Step-by-step explanation:

Permutation is the different possible arrangements or different possible order taking by the given things, objects ,words and numbers. it is also know rearranging.

Result are vary with different conditions Like Repetition is allowed or Repetition is not allowed

In mathematics we denote permutation by   [tex]{\textup{n}p_{r}}[/tex] no of permutation of n taken r at a  time.

Combination is a selection of some specific item or all items at a time from a collection is known as combination. It is denote by [tex]{\textup{n}c_{r}}[/tex] number of combination of n  different things taken r at a time

Eg. We have to choose 2 boys in group of 5 so, we can choose by many ways

Combination is widely used in lottery system.

So

Permutation count the number of different arrangements pf r out of n items, while combination count  the number of group of r out of n items.

Permutations count arrangements with order, combinations count groups without order.

The main difference between a situation in which the use of the permutations rule is appropriate and one in which the use of the combinations rule is appropriate is:

In summary, permutations focus on arrangements where order matters, while combinations focus on groups where order doesn't matter.

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Answer: you should deposit $236.2 each month.

Step-by-step explanation:

We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as

S = R[{(1 + r)^n - 1)}/r][1 + r]

Where

S represents the future value of the investment.

R represents the regular payments made(could be weekly, monthly)

r = represents interest rate/number of payment intervals.

n represents the total number of payments made.

From the information given,

there are 12months in a year, therefore

r = 0.09/12 = 0.0075

n = 12 × 35 = 420

S = $700000

Therefore,

700000 = R[{(1 + 0.0075)^420 - 1)}/0.0075][1 + 0.0075]

700000 = R[{(1.0075)^420 - 1)}/0.0075][1.0075]

700000 = R[{(23.06 - 1)}/0.0075][1.0075]

700000 = R[{22.06}/0.0075][1.0075]

700000 = R[2941.3][1.0075]

700000 = 2963.36R

R = 700000/2963.36

R = 236.2

Answer:

(a) The mean is 4.31 pounds. The variance is 51.42 pounds.

(b) The average shipping cost of a package is $10.78.

(c) The probability that the weight of a package exceeds 59 pounds is 0.0027.

Step-by-step explanation:

The probability density function of the weight of packages is:

[tex]f(x) = \frac{70}{69x^{2}};\ 1 < x < 70[/tex]

(a)

The formula for expected value (or mean) of X is:

[tex]E(X)=\int\limits^a_b {x\times f(x)} \, dx[/tex]

Compute the expected value of X as follows:

[tex]E(X)=\int\limits^{70}_{1} {x\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {x^{-1}} \, dx=\frac{70}{69} |\ln x|^{70}_{1}\\=\frac{70}{69}\times\ln 70\\=4.31[/tex]

Thus, the mean is 4.31 pounds.

The formula to compute the variance is:

[tex]V(X)=E(X^{2})-[E(X)]^{2}[/tex]

Compute the E (X²) as follows:

[tex]E(X^{2})=\int\limits^{70}_{1} {x^{2}\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x^{2} \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {1} \, dx=\frac{70}{69} | x|^{70}_{1}\\=\frac{70}{69}\times69\\=70[/tex]

The variance is:

[tex]V(X)=E(X^{2})-[E(X)]^{2}\\=70-(4.31)^{2}\\=51.4239\\\approx51.42[/tex]

Thus, the variance is 51.42 pounds.

(b)

It is provided that the shipping cost for per pound is, C = $2.50.

Compute the average shipping cost of a package as follows:

[tex]Average\ cost=Cost\ per\ pound\times E(X)\\=2.50\times4.31\\=10.775\\\approx10.78[/tex]

Thus, the average shipping cost of a package is $10.78.

(c)

Compute the probability that the weight of a package exceeds 59 pounds as follows:

[tex]P(59<X<70)=\int\limits^{70}_{59} {\frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{59} {x^{-2}} \, dx\\=\frac{70}{69} |-\frac{1}{x}|^{70}_{59}=\frac{70}{69} [-\frac{1}{70}+\frac{1}{59}]\\=\frac{70}{69}\times0.0027\\=0.0027[/tex]

Thus, the probability that the weight of a package exceeds 59 pounds is 0.0027.

The weights of the package follows a probability density function

The probability density function is given as:

[tex]\mathbf{f(x) = \frac{70}{69x^2},\ 1 < x < 70}[/tex]

(a) The mean and the variance

The mean is calculated as:

[tex]\mathbf{E(x) = \int\limits^a_b {x \cdot f(x)} \, dx }[/tex]

So, we have:

[tex]\mathbf{E(x) = \int\limits^{70}_1 {x \cdot \frac{70}{69x^2} } \, dx }[/tex]

[tex]\mathbf{E(x) = \int\limits^{70}_1 {\frac{70}{69x} } \, dx }[/tex]

Rewrite as:

[tex]\mathbf{E(x) = \frac{70}{69}\int\limits^{70}_1 {x^{-1} } \, dx }[/tex]

Integrate

[tex]\mathbf{E(x) = \frac{70}{69} {ln(x)}|\limits^{70}_1 } }[/tex]

Expand

[tex]\mathbf{E(x) = \frac{70}{69} \cdot {(ln(70) - ln(1)) }}[/tex]

[tex]\mathbf{E(x) = 4.31 }}[/tex]

The variance is calculated as:

[tex]\mathbf{Var(x) = E(x^2) - (E(x))^2}[/tex]

Where:

[tex]\mathbf{E(x^2) = \int\limits^a_b {x^2 \cdot f(x)} \, dx }[/tex]

So, we have:

[tex]\mathbf{E(x^2) = \int\limits^{70}_1 {x^2 \cdot \frac{70}{69x^2} } \, dx }[/tex]

[tex]\mathbf{E(x^2) = \int\limits^{70}_1 {\frac{70}{69} } \, dx }[/tex]

Rewrite as:

[tex]\mathbf{E(x^2) = \frac{70}{69}\int\limits^{70}_1 {1 } \, dx }[/tex]

Integrate

[tex]\mathbf{E(x^2) = \frac{70}{69} x|\limits^{70}_1 } }[/tex]

Expand

[tex]\mathbf{E(x^2) = \frac{70}{69} \cdot {(70 - 1) }}[/tex]

[tex]\mathbf{E(x^2) = 70 }}[/tex]

So, we have:

[tex]\mathbf{Var(x) = E(x^2) - (E(x))^2}[/tex]

[tex]\mathbf{Var(x) = 70 - 4.31^2}[/tex]

[tex]\mathbf{Var(x) = 51.42}[/tex]

Hence, the mean is 4.31 and the variance is 51.42, respectively.

(b) The average cost of shipping a package

In (a), we have:

[tex]\mathbf{E(x) = 4.31 }}[/tex] ---- Mean

So, the average cost of shipping a package is:

[tex]\mathbf{Average =4.31 \times 2.50}[/tex]

[tex]\mathbf{Average =10.78}[/tex]

Hence, the average cost of shipping a package is $10.78

(c) The probability a package weighs over 59 pounds

This is represented as: P(x > 59)

So, we have:

[tex]\mathbf{P(x > 59) = P(59 < x < 70)}[/tex]

So, we have:

[tex]\mathbf{P(x > 59) = \int\limits^{70}_{59} { \frac{70}{69x^2}} \, dx }[/tex]

Rewrite as:

[tex]\mathbf{P(x > 59) = \frac{70}{69}\int\limits^{70}_{59} { x^{-2}} \, dx }[/tex]

Integrate

[tex]\mathbf{P(x > 59) = \frac{70}{69} \cdot { -\frac 1x}|\limits^{70}_{59}}[/tex]

Expand

[tex]\mathbf{P(x > 59) = \frac{70}{69} \cdot (-\frac{1}{70} + \frac{1}{59})}[/tex]

[tex]\mathbf{P(x > 59) = 0.0027}[/tex]

Hence, the probability a package weighs over 59 pounds is 0.0027

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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:

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:

Step-by-step explanation:

answer is attached below

Answer:

Have a broad mix of organizational levels in the JAD session

Explanation:

It is not possible for Hamid to include every employee in the JAD session, what Hamid needed to do is to select participants from the different departments and other key important people to ensure every one is well represented at the JAD session. Selecting lower-level and mid-level managers from the affected departments as well as the some senior staff and the project sponsor for the JAD session will ensure everyone's interest is well represented at the session.

Answer:

The probability mass function is expressed as:

P(x) = [(x+r-1)C(r-1)]*[p^r]*[(1-p)^x]

Step-by-step explanation:

This is not a binomial distribution. It is actually a negative binomial distribution. The probability mass function is expressed below:

P(x) = [(x+r-1)C(r-1)]*[p^r]*[(1-p)^x]

where:

x = number of failures

r-1 = number of successes (10 in this scenario)

p = probability of a success

nCr = n!/[r!(n-r)!]

The main formula difference in the positive binomial versus negative binomial is this: With respect to the negative binomial, it is obviously  known that the last event will be: when we reach our 10th "head", we stop .

Thus, the last flip will ALWAYS be a "head".

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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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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Find the total of everything he is using:

1 1/2 + 1/4 + 1/4 = 2 total cups.

For every two cups of trail mix he uses 1/4 cup of pretzels.

15 cups/ 2 cups = 7.5

7.5 x 1/4 cup of pretzels = 1 7/8 cups of pretzels.

Answer: Ted needs 1.875 cups of pretzels to make 15 cups of trail mix

Step-by-step explanation:

He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels. This means that the ratio of ratio of the number of cups of nuts used to the number of cups of raisins used to the number of cups of pretzels used is

1.5 : 0.25 : 0.25

The total ratio is

1.5 + 0.25 + 0.25 = 2

Ted need to make 15 cups of trail mix. Therefore, the number of cups of pretzels that he needs to use is

0.25/2 × 15 = 1.875 cups of pretzels

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:

60

Step-by-step explanation:

, $&7%"""%- &xgsfx,, 77$""$66"'++

Answer:

95% Confidence interval: (39.43, 61.58)

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = $50.50

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation = 20

95% Confidence interval:

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.05} = \pm 2.1447[/tex]  

[tex]=50.50 \pm 2.1447(\dfrac{20}{\sqrt{15}} ) \\\\= 50.50 \pm 11.0751 \\= (39.4249,61.5751)\\\approx (39.43, 61.58)[/tex]  

95% Confidence interval: (39.43, 61.58)

The 95% confidence interval for the mean amount spent on their first visit to the chain's new store in the mall by credit card customers is approximately $41.99 to $59.01.

We can construct the 95% confidence interval using the sample mean ( X = $50.50) and the standard deviation ( S = $20). Since we know that the distribution is normal, we can use the z-score for a 95% confidence level, which is approximately 1.96.

The formula for a confidence interval is given by: X ± Z*(S/√n). By substituting the given values into the formula, we get: 50.5 ± 1.96*(20/√15).

After calculating, we find that the 95% confidence interval is approximately $41.99 - $59.01. Thus, we are 95% confident that the true mean amount spent by the department store's credit card customers on their first visit is between $41.99 and $59.01.

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Answer:

XYZ is NOT a good buy.

Step-by-step explanation:

Calculate the market price of stock:

[tex]\frac{Next year's Dividend}{Reqd.return - Growth rate}[/tex]

[tex]= \frac{(2.4)(1.065)}{0.115-0.065}[/tex]

[tex]= 51.12[/tex]

The Market price of the stock is $51. Therefore, buying the stock at $60 is overpriced and is NOT a good buy.

Answer: the length of the altitude is 10 mm

Step-by-step explanation:

The formula for determining the volume of a square base pyramid is expressed as

Volume = Ah

Where

A represents the area of the square base.

h represents the height or altitude of the pyramid.

From the information given,

Length of each side of the square base = 2 millimeters

Volume of the square pyramid is 40 cubic.

Area of square base = 2² = 4 mm²

Therefore,

40 = 4h

h = 40/4

h = 10 mm

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.

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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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).

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:

The population of interest to the researcher were the 250 first-year students that were monitored.

Step-by-step explanation:

In descriptive statistics, the portion of the cost of college education to be determined and has been selected for analysis is calle d "sample", the sample the researcher is interested in, considers the textbooks cost of first-year students, therefore the 250 first-year students is the researcher´s population of interest. This method involved the collection, presentation, and characterization.

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

Answer:

Option B) 19.1 years

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 9

Sample mean, [tex]\bar{x}[/tex] = 19.1 years

Alpha, α = 0.05

Population standard deviation, σ =  1.5 years

We have to approximate best point estimate for population mean.

The best point estimate for population mean is the sample mean.

Thus, we can write

[tex]\mu = \bar{x} = 19.1[/tex]

Thus, the correct answer is

Option B) 19.1 years

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:

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.

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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)

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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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Product A performed worse in the study because 22% failed to get relief with it, whereas only 14% failed to get relief with Product B.

In the given study of pain relievers, we need to determine which product performed worse based on the percentage of people who did not experience relief. For Product A, 50 people were given the product and all but 11 experienced relief. This means that 11 out of 50 people did not experience relief, so we calculate the failure rate as follows: (11/50) * 100 = 22%. For Product B, 100 people were given the product and all but 14 experienced relief, therefore the failure rate is: (14/100) * 100 = 14%.

With these failure rates, we can now fill in the blanks of the statement:

Product A performed worse in the study because 22% failed to get relief with this product, whereas only 14% failed to get relief with Product B.

Answer:

(A) 0.999996

(B) 0.11680

Step-by-step explanation:

We are given that a Randstad Harris interactive survey reported that 25% of employees said their company is loyal to them.

And 9 employees are selected randomly and interviewed about company loyalty.

The Binomial probability distribution is given by;

[tex]P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]

where, n = number of trials (samples) taken

             r = number of success

             p = probability of success

In our question; n = 9 , p = 0.25 (as employees saying their company is loyal to them is success to us)

(A) Probability that none of the 9 employees will say their company is loyal to them = 1 - Probability that all 9 employees will say their company is loyal to them

= 1 - P(X = 9)  { As here number of success is 9 }

= 1 - [tex]\binom{9}{9}0.25^{9}(1-0.25)^{9-9}[/tex] = 1 - [tex]0.25^{9}[/tex] = 0.999996

(B) Probability that 4 of the 9 employees will say their company is loyal to them = P(X = 4)

    P(X = 4) = [tex]\binom{9}{4}0.25^{4}(1-0.25)^{9-4}[/tex]

                  = [tex]126*0.25^{4}*0.75^{5}[/tex] = 0.11680

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: