Since the ratios for sides of triangles are the same, when the angles are the same, ? have been developed which show these ratios for every angle encountered in a triangle.

Final answer:

Keith Bowie should deposit $667.45 into an account today to have enough money in 6 years to pay for the engine overhaul.

Explanation:

To determine the amount Keith Bowie should deposit today in an account earning 6% interest compounded annually, we can use the concept of present value. The present value of a future cash flow is calculated by dividing the future cash flow by (1 + interest rate) raised to the power of the number of years. In this case, we want to find the amount needed that would accumulate to cover the engine overhaul cost in 6 years. Let's assume the cost of the overhaul is $390 with a 10% probability, $570 with a 30% probability, $750 with a 50% probability, and $790 with a 10% probability. We can calculate the present value by multiplying each cash flow amount with its respective probability, discounting it back at the interest rate over 6 years, and adding them all up:

Present Value = (390 * 0.10) / (1 + 0.06)6 + (570 * 0.30) / (1 + 0.06)6 + (750 * 0.50) / (1 + 0.06)6 + (790 * 0.10) / (1 + 0.06)6 = $667.45

Therefore, Keith Bowie should deposit $667.45 into the account today to ensure he has enough money on hand in 6 years to pay for the engine overhaul.

Answer:

2 contracts

Step-by-step explanation:

Break even point refers to the number of units or sales that needs to be generated for the company to make neither a profit nor a loss.

This means that at the break even point, sales is equivalent to the cost incurred (both fixed and variable).

Let the number of contracts that must be signed to break even be s

The rent is a fixed cost while the cost associated with each contract is variable.

5687.1s = 7000 + 1260.7s

5687.1s - 1260.7s = 7000

4426.4 s = 7000

s = 1.58

≈ 2

She must facilitate 2 contracts each month to break even.

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

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Step-by-step explanation:

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Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

[tex]\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018[/tex]

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

[tex]z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55[/tex]

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

[tex]P(0.79<p<0.81)=P(-0.55<z<0.55)\\\\P(0.79<p<0.81)=P(z<0.55)-P(z<-0.55)\\\\P(0.79<p<0.81)=0.70884-0.29116=0.41768[/tex]

Following are the solution to the given question:

By using the approximation of normal for the proportions so, the values:  

[tex]\to P(\hat{p} < p ) = P(Z < \frac{(\hat{p} - p)}{\sqrt{\frac{p ( 1 - p)}{n}}} )[/tex]

So,

[tex]\to P(0.79 < \hat{p} < 0.81) = P( \hat{p} < 0.81) - P( \hat{p} < 0.79)[/tex]

                                  [tex]= P(Z<\frac{( 0.81 - 0.80)}{\sqrt{\frac{0.80( 1 - 0.80)}{500}}}) - P(Z < \frac{( 0.79 - 0.80)}{ \sqrt{ \frac{0.80 ( 1 - 0.80)}{500}}}) \\\\[/tex]

                                   [tex]= P(Z < 0.56) - P(Z < -0.56)\\\\= 0.7123 - 0.2877 \ \ \text{(using the Z table)}\\\\= 0.4246[/tex]

Therefore, the final answer is "0.4246".

Learn more:

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Answer:

1) 0.75

2) No, the 2 events are not mutually exclusive because P(A&B) ≠ 0

3)The 2 events are not independent because P(A | B) ≠ P(A).

Step-by-step explanation:

1) P(A) = 0.59 and P(B) = 0.41

Thus; to find the probability that at least one of the stocks will rise in price;

P(A or B) = P(A) + P(B) - P(A&B)

Now, we know that if B rises in price, the probability that A will also rise is 0.61 i.e P(A|B) = 0.61

Hence;P(A & B) = P(A|B) x P(B) =0.61 x 0. 41 = 0.25

So P(A or B) = 0.59 + 0.41 - 0.25 = 0.75

B) we want to find out if events A and B are mutually exclusive ;

P(A|B) = P(A&B)/P(B), For the events to be mutually exclusive, P(A&B) has to be zero. Since P(A&B) ≠ 0,the 2 events are not mutually exclusive.

C) Since P(A|B) = 0.61 and P(A) is 0.59. Thus, P(A | B) ≠ P(A).

So, therefore the 2 events are not independent.

Answer:

P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

Assuming the box is of unit length, a = 1,

0.29a = 0.29 and 0.34a = 0.34

P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = 0.167 = 0.17 to 2 s.f

Step-by-step explanation:

ψ(x) = 2a(√sin(πxa)

The probability of finding a particle in a specific position for a given energy level in a one-dimensional box is related to the square of the wavefunction

The probability of finding a particle between two points 0.29a and 0.34a is given mathematically as

P(0.29a < X < 0.34a) = ∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx

That is, integrating from 0.29a to 0.34a

ψ²(x) = [2a(√sin(πxa)]² = 4a² sin(πxa)

∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx = ∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa)) dx

∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa))

= - [(4a²/πa)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ

- [(4a/π)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]

Assuming the box is of unit length, a = 1

P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = (-1.273) [0.482 - 0.613] = 0.167.

Answer:

a) We have the following distribution [tex] X \sim N(\mu =650, \sigma =10)[/tex]

And we want to calculate:

[tex] P(X>654) [/tex]

And in order to calclate this in the ti 84 we can use the following code

2nd> Vars>DISTR

And then we need to use the following code:

1-normalcdf(-1000,654,650,10)

And we got:

[tex] P(X>654)=0.3445783 [/tex]

b) For this case we assume a normal standard distribution and we want to calculate:

[tex] P(z<0.72)[/tex]

And using the following code in the Ti84 or using the normal standard table we got:

normalcdf(-1000,0.72,0,1)

[tex] P(z<0.72)=0.76424[/tex]

So this part was the wrong solution from the solution posted,

Step-by-step explanation:

Part a

We have the following distribution [tex] X \sim N(\mu =650, \sigma =10)[/tex]

And we want to calculate:

[tex] P(X>654) [/tex]

And in order to calclate this in the ti 84 we can use the following code

2nd> Vars>DISTR

And then we need to use the following code:

1-normalcdf(-1000,654,650,10)

And we got:

[tex] P(X>654)=0.3445783 [/tex]

Part b

For this case we assume a normal standard distribution and we want to calculate:

[tex] P(z<0.72)[/tex]

And using the following code in the Ti84 or using the normal standard table we got:

normalcdf(-1000,0.72,0,1)

[tex] P(z<0.72)=0.76424[/tex]

So this part was the wrong solution from the solution posted,

The question discusses two principal sources for conducting learning research: 1. nature of knowledge and diverse learning methods, 2. cognitive processes. The outlined chapters provide a systematic framework to understand personal learning, covering areas from motivation, mindset, individual learning styles to personality types, and practical application of learning knowledge.

The student's question pertains to the two fundamental sources for conducting research on learning. The first source refers to the nature of knowledge and the diversity of learning methods, while the second digs into cognitive processes or what goes on in our minds.

The chapters provided, under the title of 'Knowing Yourself as a Learner', present a systematic examination of personal learning. It starts with 'The Power to Learn', which considers the capacity and eagerness for learning. The 'Motivated Learner' and 'It's All in the Mindset' chapters elaborate on how motivation and mindset affect learning.

The 'Learning Styles' and 'Personality Types and Learning' sections delve into the individual differences that shape how we absorb and process information. Finally, 'Applying What You Know about Learning' encourages students to utilize their understanding of their learning attributes to optimize knowledge acquisition.

https://brainly.com/question/32403971

#SPJ12

Answer:

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

For this case we know this:

[tex] n=37 ,  p=0.2[/tex]

We can find the standard error like this:

[tex] SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658[/tex]

So then our random variable can be described as:

[tex] p \sim N(0.2, 0.0658)[/tex]

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

[tex] P(p>0.4)[/tex]

We can use the z score given by:

[tex] z = \frac{p -\mu_p}{SE_p}[/tex]

And using this we got this:

[tex] P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z<3.04) = 0.0012[/tex]

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

We need to check if we can use the normal approximation:

[tex] np = 37 *0.2 = 7.4 \geq 5[/tex]

[tex] n(1-p) = 37*0.8 = 29.6\geq 5[/tex]

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

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

For this case we know this:

[tex] n=37 ,  p=0.2[/tex]

We can find the standard error like this:

[tex] SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658[/tex]

So then our random variable can be described as:

[tex] p \sim N(0.2, 0.0658)[/tex]

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

[tex] P(p>0.4)[/tex]

We can use the z score given by:

[tex] z = \frac{p -\mu_p}{SE_p}[/tex]

And using this we got this:

[tex] P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z<3.04) = 0.0012[/tex]

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Answer:

0.0497 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean number of calls = 3 calls per minute

The number of calls coming per minute is following a Poisson distribution.

Formula:

[tex]P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}[/tex]

We have to evaluate

[tex]P( x =0) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\P(x = 0) = \frac{3^0e^{-3}}{0!} = 0.0497[/tex]

0.0497 is the probability that no calls are received in a given one-minute period.

Answer:

B. approximately normal with a mean of 2.02 dollars and a standard error of 0.45 dollars

Step-by-step explanation:

We use the Central Limit Theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 2.02, \sigma = 3, n = 45, s = \frac{3}{\sqrt{45}} = 0.45[/tex]

So the correct answer is:

B. approximately normal with a mean of 2.02 dollars and a standard error of 0.45 dollars

Answer:

There is enough evidence to support the claim that students study more than two hours per weeknight on the average.      

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 2 hours = 120 minutes

Sample mean, [tex]\bar{x}[/tex] = 148 minutes = 2.467 hours

Sample size, n = 255

Alpha, α = 0.05

Population standard deviation, σ = 65 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 120\text{ minutes}\\H_A: \mu > 120\text{ minutes}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{148 - 120}{\frac{65}{\sqrt{255}} } = 6.8788[/tex]

Now, we calculate the p-value from the normal standard table.

P-value = 0.00001

Since the p-value is smaller than the significance level, we fail accept the null hypothesis and reject it, We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that students study more than two hours per weeknight on the average.

Answer:

There is enough evidence to support the claim that students study more than two hours per weeknight on the average.    

Step-by-step explanation:

Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d) [tex]\mu_{X}=6[/tex]

(e) [tex]\sigma_{X}=2.45[/tex]

Step-by-step explanation:

Let X = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, λ = 6.

The random variable X follows a Poisson distribution with parameter λ = 6.

The probability mass function of a Poisson distribution is:

[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...[/tex]

(a)

Compute the value of P (X = 7) as follows:

[tex]P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388[/tex]

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2)

              [tex]=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380[/tex]

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

                   [tex]=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443[/tex]

Thus, the value of P (2 < X < 7) is 0.5433.

(d)

The mean of the Poisson distribution is:

[tex]\mu_{X}=\lambda=6[/tex]

(e)

The standard deviation of the Poisson distribution is:

[tex]\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45[/tex]

Answer:

"half of the 12-inch quesadilla" is larger

Step-by-step explanation:

The "inch size" is the diameter of the circular quesadilla.

The area of circle is given by  [tex]\pi r^2[/tex]

We also know radius is half of diameter.

HALF OF 12-INCH QUESADILLA:

12in diameter = 6 inch radius, the area is:

[tex]\pi(6)^2=36\pi[/tex]

Half of this is:

[tex]\frac{36\pi}{2}=18\pi[/tex]

ENTIRE 6-INCH QUESADILLA:

6 inch diameter means 6/2 = 3 inch radius

The area of this is:

[tex]\pi(3)^2 = 9\pi[/tex]

So, we can say that "half of the 12-inch quesadilla" is larger.

Answer:

lateral​ surface area of the cone =49π

Step-by-step explanation:

y= x/4

dx/dy = 4

Given x range as  from 0 ≤ x ≤ 7 ⇒ x ranges between (a,b)

a= x/4 = 0/4 = 0

b = x/4= 7/4

lateral​ surface area of the cone

[tex]\int\limits^b_a {2\pi x\sqrt{1 + (\frac{dx}{dy})^2 } } \, dx \\\\= \int\limits^{\frac{7}{4}}_0 {2\pi (4y)\sqrt{1 + (4)^2 } } \, dx \\\\= 32\pi \int\limits^{\frac{7}{4}}_0 { y } \, dx \\\\= 32\pi [\frac{y^2}{2}]|^\frac{7}{4}_0[/tex]

=32π * 49/(16 *2)

=49π

Final answer:

To find the lateral surface area of the cone with the given line segment, use the formula [tex]L = \(\ r l\)[/tex], where r is the base radius, calculated from the given equation, and l is the slant height determined by the Pythagorean theorem.

Explanation:

The student has asked for help in finding the lateral surface area of a cone generated by revolving a line segment around the x-axis.

The specific line segment is given by the equation [tex]y = \frac{1}{4} x[/tex], for 0 \<= x \<= 7. To find this surface area, we can use the formulas for the surface area of a cone, which is \rl\, where \ is the radius of the base of the cone and \ is the slant height.

In this case, when x=7, y = [tex]y = \frac{1}{4} x[/tex] = 7/4, which gives us the radius of the base of the cone. The slant height can be calculated using the Pythagorean theorem, since we have a right triangle with the slant height as the hypotenuse, and the radius and height of the cone as the other two sides. Thus, [tex]l = {7^2+(7/4)^2}[/tex]

Finally, the formula for the lateral surface area of the cone is L = rl, which, after substituting the given values, results in the exact answer needed. The calculation will result in the following:

[tex]L = \frac{7}{4}\times\sqrt{7^2+(7/4)^2}[/tex]

Answer:

The distribution of scores on this final exam is left-skewed.

Step-by-step explanation:

We use the Pearson Mode Skewness to solve this question. It states that:

If the median is higher than the mean, the distribution is left-skewed.

If the median is lower than the mean, the distribution is right-skewed.

If the median is the same as the mean, the distribution is symmetric.

In this problem, we have that:

Median = 74

Mean = 70

Median higher than the mean

So the distribution of scores on this final exam is left-skewed.

Answer:

Step-by-step explanation:

median 74

mean 70

this is the answer

Answer:

a) 0.85

b) 0.15

c) 0.40

d) 0.70

Step-by-step explanation:

P(A) = 0.55

P(B) = 0.45

P(A n B) = 0.15

a) Probability of A or B occurring = P(A u B) = P(A) + P(B) - P(A n B) = 0.55 + 0.45 - 0.15 = 0.85

b) Probability of A and B occurring = P(A n B) = 0.15

c) Probability of just A occurring = P(A n B') = P(A) - P(A n B) = 0.55 - 0.15 = 0.40

d) Probability of just A or just B occurring = P(A n B') + P(A' n B) = 0.4 + (0.45 - 0.15) = 0.4 + 0.3 = 0.70

Answer:

dx/dt  =  0,04 m/sec

Step-by-step explanation:

Area of the circle is:

A(c) =π*x²      where   x is a radius of the circle

Applying differentiation in relation to time we get:

dA(c)/dt   =  π*2*x* dx/dt    

In this equation we know:

dA(c)/dt  = 0,5 m²/sec

And are looking for dx/dt then

0,5  = 2*π*x*dx/dt    when the area of the sheet is 12 m²  (1)

When  A(c) = 12 m²      x = ??

A(c)  =  12  =  π*x²      ⇒    12  =  3.14* x²    ⇒  12/3.14  =  x²

x²  = 3,82     ⇒   x  = √3,82    ⇒  x = 1,954 m

Finally plugging ths value in equation (1)

0,5  = 6,28*1,954*dx/dt

dx/dt  =  0,5 /12.28

dx/dt  =  0,04 m/sec

The rate at which the radius is decreasing when the area of the sheet is 12 m² is; dr/dt = 0.041 m/s

We are given;

Area of sheet; A = 12 m²

Rate of change of area; dA/dt = 0.5 m²/s

Now, formula for area of the circular sheet is given as;

A = πr²

Thus; 12 = πr²

r = √(12/π)

r = 1.9554 m

Now, we want to find the rate at which the radius is decreasing and so we differentiate both sides of the area formula with respect to t;

dA/dt = 2πr(dr/dt)

Thus;

0.5 = 2π × 1.9554(dr/dt)

dr/dt = 0.5/(2π × 1.9554)

dr/dt = 0.041 m/s

Read more at; https://brainly.in/question/42255951

Answer:

57.93% probability that a trip will take at least 35 minutes.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 36, \sigma = 4.9[/tex]

What is the probability that a trip will take at least 35 minutes

This probability is 1 subtracted by the pvalue of Z when X = 35. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{35 - 36}{4.9}[/tex]

[tex]Z = -0.2[/tex]

[tex]Z = -0.2[/tex] has a pvalue of 0.4207

1 - 0.4207 = 0.5793

57.93% probability that a trip will take at least 35 minutes.

Answer:

Yes, this a biased sampling method.

Step-by-step explanation:

The sampling done is a biased sampling method.

This is because usually people who are upset or dissatisfied with the service are the ones who write online reviews, as there are no other way to release their frustration with the service provided.

The bias is likely to be directed towards the proportion of people who are not satisfied with service.

The corrected parts of the question has been attached to this answer.

Answer:

A) Probability that the error is less than 0.2 mm; P(X < 0.2) = 0.0272

B) Mean Error (E(X)) = 0.6

C) Variance Error (V(X)) = 0.04

D) Answer properly written in attachment (Page 2)

E) P(0<X<0.8) = 0.8192

Step-by-step explanation:

The probability density function of X is;

f(x) = { 12(x^(2) −x^(3) ; 0<x<1

So, due to the integral symbol and for clarity sake, i have attached all the explanations for answers A - D.

E) The probability that the specification for the error to be between 0 to 0.8 mm is met will be;

P(0<X<0.8) = F(0.8) − F(0) =12([(0.8)^(3)] /3] −[(0.8)^(4)]/4]

= 0.8192

So, the probability is 0.8192.

The second and third function are missing and they are;

The second function; ((2^n) + n^(2))(n^(3) + 3^(n))

The third function is;

(n^(n) + n^(2n) + 5^(n))(n! + 5^(n))

Answer:

A) O((n^(3)) logn)

B) O(6^(n))

C) O(n^(n)(n!))

Step-by-step explanation:

I've attached explanation of the 3 answers.

The big-O estimate for the function f(n) = [tex]f(n) = (n log n + n^2)(n^3 + 2)[/tex] is [tex]O(n^5)[/tex] , as the product of the highest order terms [tex]n^2[/tex] and [tex]n^3[/tex] is [tex]n^5[/tex], which dominates the function's growth.

To find a big-O estimate for the function [tex]f(n) = (n log n + n^2)(n^3 + 2)[/tex], we must look for the term with the highest growth rate as n goes to infinity. This function is the product of two terms, n log [tex]n log n + n^2[/tex] and [tex]n^3 + 2[/tex]. The term with the highest growth rate in the first part is n2, and in the second part, it is n3. Therefore, the product of the two highest order terms will give us the term that grows the fastest: [tex]n^2 \times n^3 = n^5[/tex].

Thus, the given function is in big-O of [tex]n^5[/tex], which means [tex]f(n) \in O(n^5)[/tex]. The higher order term [tex]n^5[/tex]completely dominates the growth of the function, making the other terms and constants irrelevant for big-O notation. Remember that big-O notation provides an upper bound and is used to describe the worst-case scenario for the growth rate of the function.

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Answer:

6 extra hot dogs.

Step-by-step explanation:

A dependent variable is the variable being tested and measured in a scientific experiment. The dependent variable is 'dependent' on the independent variable

The 6 hot dogs are dependent on whether the teammates will bring friends. In other words, if there are no friends, there are no hot dogs. Note that the teammates are the independent variable that is they are not dependent on any variable.

Answer:

Step-by-step explanation:

a. The frequency and percent frequency of complaints by industry is shown below:

Industry            Frequency    Percentage Frequency

Bank                     26                    13%

Cable                    44                    22%

Car                        42                    21%

Cell                        60                   30%

Collection              48                   13%

b. Cell has the highest complaints (60).

c. The decreasing order of complaints goes thus: Cell (60%) , Cable (22%), Car(21%).

Both Collection and Bank, together account nearly up to 27% of the total complaints.

Answer:A researcher uses probability to decide whether the sample she obtained is likely to be a sample from a particular population.

Step-by-step explanation: Inferential statistics is a Statistical process used to compare two or more samples or treatments.

Probability helps in inferential statistics to decide whether the sample obtained is likely from the population of interest.

Inferential statistics use data obtained from the sample of interest in a research to compare the treatment or samples.Through Inferential statistics researchers make conclusions about the entire population.

Probability in inferential statistics is used to make inferences about a population based on sample data. It provides the foundation for statistical methods such as confidence intervals and hypothesis testing to evaluate the accuracy of the sample in representing the population.

Probability in inferential statistics is critical in helping researchers make inferences about a population from a sample. When researchers collect data from a sample, they use probability theory to deduce how likely it is that their observations are reflective of the entire population or occured by chance. Inferential statistical methods, such as confidence intervals and hypothesis testing, leverage probability to make these determinations.

Probability enables statisticians to evaluate the accuracy of the sample data in representing the population, decide how confident they can be about their inferences, and test the validity of existing hypotheses about the population parameters based on sample data.

For instance, if an inferential statistical test indicates that the likelihood of obtaining the observed sample results by chance is only 5%, researchers can infer there is a 95% probability that the sample accurately reflects the population, supporting the hypothesis being tested. Therefore, probability is used to determine how much confidence researchers can have in their sample data when making generalizations about a larger group.

a) The marginal cost function is [tex]\[C'(x) = 12 - 0.2x + 0.0015x^2\][/tex].

b) The value C'(500) = 287 predicts that for every additional yard of fabric produced at this level (500 yards), the cost will increase by $287.

c) The cost of manufacturing the 501st yard of fabric is approximately $45,000, which is approximately equal to the marginal cost C'(500)

a)

To find the marginal cost function, we need to take the derivative of the cost function C(x) with respect to x:

[tex]\[C(x) = 1500 + 12x - 0.1x^2 + 0.0005x^3\][/tex]

Taking the derivative:

[tex]\[C'(x) = \frac{d}{dx}(1500 + 12x - 0.1x^2 + 0.0005x^3)\][/tex]

[tex]\[C'(x) = 12 - 0.2x + 0.0015x^2\][/tex]

b)

To find [tex]\(C'(500)\)[/tex] substitute x = 500 into the marginal cost function:

[tex]\[C'(500) = 12 - 0.2(500) + 0.0015(500^2)\][/tex]

[tex]\[C'(500) = 12 - 100 + 375\][/tex]

[tex]\[C'(500) = 287\][/tex]

The value C'(500) = 287 represents the rate at which costs are increasing with respect to the production level when x = 500. It predicts that for every additional yard of fabric produced at this level (500 yards), the cost will increase by $287.

c)

To compare C'(500) with the cost of manufacturing the 501st yard of fabric, calculate C(501) - C(500):

[tex]\[C(501) - C(500) = (1500 + 12(501) - 0.1(501)^2 + 0.0005(501)^3) - (1500 + 12(500) - 0.1(500)^2 + 0.0005(500)^3)\][/tex]

Calculate the difference:

[tex]\[C(501) - C(500) = -45,000\][/tex]

Therefore, the cost of manufacturing the 501st yard of fabric is approximately $45,000, which is approximately equal to the marginal cost C'(500) when rounded to four decimal places.

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(a) Marginal cost function: C'(x) = 12 - 0.2x + 0.0015x².

(b) C'(500) predicts a $287 cost increase for the 500th yard.

(c) Cost for the 501st yard is approximately $512, slightly higher than C'(500).

(a) To find the marginal cost function, we need to calculate the derivative of the cost function C(x) with respect to x:

C(x) = 1500 + 12x - 0.1x² + 0.0005x³

C'(x) = d/dx [1500 + 12x - 0.1x² + 0.0005x³]

C'(x) = 12 - 0.2x + 0.0015x²

So, the marginal cost function is C'(x) = 12 - 0.2x + 0.0015x².

(b) To find C'(500), we plug in x = 500 into the marginal cost function:

C'(500) = 12 - 0.2(500) + 0.0015(500)²

C'(500) = 12 - 100 + 375

C'(500) = 287

C'(500) represents the rate at which costs are increasing with respect to the production level when 500 yards of fabric are produced. In this case, it predicts that the cost is increasing at a rate of $287 per yard for the 500th yard produced.

(c) To compare C'(500) with the cost of manufacturing the 501st yard of fabric, we need to calculate C(501) - C(500):

C(501) - C(500) = [1500 + 12(501) - 0.1(501)² + 0.0005(501)³] - [1500 + 12(500) - 0.1(500)² + 0.0005(500)³]

C(501) - C(500) = [1500 + 6012 - 25005 + 627507] - [1500 + 6000 - 25000 + 625000]

C(501) - C(500) = [627012] - [626500]

C(501) - C(500) = 512

So, the cost of manufacturing the 501st yard of fabric is $512, which is approximately equal to C'(500), which was calculated as $287. This means that the cost increased by approximately $287 for the 500th yard and by $512 for the 501st yard.

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Answer:4y^2-10y-1

Step-by-step explanation:

Answer:

Each player gets 6 tennis balls.

4 × t = 24

Step-by-step explanation:

The total number of tennis balls in the basket is, 24 balls.

The number of tennis players is, 4 players.

On equally dividing all the balls among the 4 players, each player gets,

[tex]No.\ of\ balls\ each\ player\ gets=\frac{No.\ of\ balls}{No.\ of\ players} \\=\frac{24}{4} \\=6[/tex]

Thus, each player gets 6 tennis balls.

If we want to represent the equation of the number of players and number of balls, the equation will be,

4 × t = 24

t = number of tennis balls each player gets.

Final answer:

To divide 24 tennis balls evenly among four players, we use the equation 4t = 24, where 't' is the number of balls per player. By dividing 24 by 4, we find that each player receives 6 tennis balls.

Explanation:

The student is asking how to divide 24 tennis balls evenly among four players, which is a basic division problem in mathematics. The correct mathematical equation to represent this situation is C. 4t = 24, where 't' represents the number of tennis balls each player gets. To find the value of 't', we divide the total number of balls, 24, by the number of players, 4.

We perform the division as follows:

Write down the equation: 4t = 24.

Divide both sides of the equation by 4 to solve for 't': t = 24 / 4.

Calculate the result: t = 6.

Each player gets 6 tennis balls.

Answer:

32.33% probability of having at least 3 erros in an hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

The mean number of errors is 2 per hour.

This means that [tex]\mu = 2[/tex]

(a) What is the probability of having at least 3 errors in an hour?

Either you have 2 or less errors in an hour, or we have at least 3 errors. The sum of the probabilities of these events is decimal 1. So

[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]

We want [tex]P(X \geq 3)[/tex]

So

[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]

In which

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353[/tex]

[tex]P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707[/tex]

[tex]P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767[/tex]

[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6767 = 0.3233[/tex]

32.33% probability of having at least 3 erros in an hour.

Final answer:

The question involves calculating the probability of having at least 3 transmission errors in an hour for a long-range radio using the Poisson distribution. Given the mean number of errors is 2 per hour, the resulting probability is found to be 32.33%.

Explanation:

The question involves understanding the Poisson distribution, which is a probability distribution that is used to describe the number of events occurring within a fixed interval of time or space given a known average rate. In this question, we are dealing with a long-range radio transmits across the Atlantic Ocean, where the average number of transmission errors per hour follows a Poisson distribution with a mean (λ) of 2 errors per hour. To find the probability of having at least 3 errors in any given hour, we can use the formula P(X ≥ k) = 1 - [P(X=0) + P(X=1) + P(X=2)], where X is the number of transmission errors.

To calculate P(X=k) for k=0,1,2, we use the Poisson probability formula P(X=k) = (e^{-λ} * λ^k) / k!, where e is the base of the natural logarithm (approximately 2.71828). Thus:

P(X=0) = (e^{-2} x 2⁰) / 0! = 0.1353

P(X=1) = (e^{-2} x 2¹) / 1! = 0.2707

P(X=2) = (e^{-2} x 2²) / 2! = 0.2707

Adding up these probabilities gives us 0.6767. Therefore, the probability of having at least 3 errors in any given hour is 1 - 0.6767 = 0.3233 or 32.33%.