The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 415 grams and a standard deviation of 23 grams. Find the weight that corresponds to each event.

Answer:

1364

Step-by-step explanation:

Answer:

Upper bound: 26.45

Lower bound: 15.55            

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 21

Sample size, n = 350

Alpha, α = 0.05

Population standard deviation, σ = 52

95% Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

[tex]21 \pm 1.96(\dfrac{52}{\sqrt{350}} ) = 21 \pm 5.45 = (15.55,26.45)[/tex]

Upper bound: 26.45

Lower bound: 15.55

Final answer:

The 99% confidence interval for the mean change in score in the population of high school seniors ranges from approximately 13.89 to 28.11 points.

Explanation:

To find the 99% confidence interval for the mean change in score μ in the population of all high school seniors, we'll use the formula for the confidence interval of the mean for a population when the standard deviation is known:

Confidence interval = μ ± (z*·σ/√n)

Where μ is the mean, σ is the standard deviation, n is the sample size, and z* is the z-score associated with the desired confidence level. Here, the sample mean (μ) is 21, the standard deviation (σ) is 52, and the sample size (n) is 350. The z-score for a 99 percent confidence level is approximately 2.576.

Confidence interval = 21 ± (2.576*52/√350)

First, calculate the margin of error:

Margin of error = 2.576 * (52/√350) ≈ 7.11

Then, calculate the confidence interval:

Lower bound = 21 - 7.11 = 13.89 (rounded to two decimal places)

Upper bound = 21 + 7.11 = 28.11 (rounded to two decimal places)

Therefore, the 99 percent confidence interval for the mean change in score μ is approximately from 13.89 to 28.11 points.

Answer:

(a) The Nyquist rate for the given signal is 2 × 5000π = 10000π. Therefore, in order to be able

to recover x(t) from xp(t), the sampling period must at most be Tmax =

2π

10000π = 2 × 10−4

sec. Since the sampling period used is T = 10−4 < Tmax, x(t) can be recovered from xp(t).

(b) The Nyquist rate for the given signal is 2 × 15000π = 30000π. Therefore, in order to be able

to recover x(t) from xp(t), the sampling period must at most be Tmax =

2π

30000π = 0.66×10−4

sec. Since the sampling period used is T = 10−4 > Tmax, x(t) cannot be recovered from

xp(t).

(c) Here, Im{X(jω)} is not specified. Therefore, the Nyquist rate for the signal x(t) is indeterminate. This implies that one cannot guarantee that x(t) would be recoverable from xp(t).

(d) Since x(t) is real, we may conclude that X(jω) = 0 for |ω| > 5000. Therefore, the answer to

this part is identical to that of part (a).

(e) Since x(t) is real, X(jω) = 0 for |ω| > 15000π. Therefore, the answer to this part is identical

to that of part (b).

(f) If X(jω) = 0 for |ω| > ω1, then X(jω) ∗ X(jω) = 0 for |ω| > 2ω1. Therefore, in this

part, X(jω) = 0 for |ω| > 7500π. The Nyquist rate for this signal is 2 × 7500π = 15000π.

Therefore, in order to be able to recover x(t) from xp(t), the sampling period must at most

be Tmax =

2π

15000π = 1.33 × 10−4

sec. Since the sampling period used is T = 10−4 < Tmax,

x(t) can be recovered from xp(t).

(g) If |X(jω)| = 0 for |ω| > 5000π, then X(jω) = 0 for |ω| > 5000π. Therefore, the answer to

this part is identical to the answer of part (a).

Answer:

15 sec

Step-by-step explanation:

Data:

let m = 600 kg

    b  = 50

the differential equation will be:

[tex]m\frac{dv}{dt} = mg - bv(t)[/tex]

The equation of motion of the object can be determined considering the forces acting on it, including the force due to gravity and the force due to air resistance. The equation can be rearranged and solved using the given values in Newton's method to approximate the time when the object strikes the ground.

The equation of motion of the object can be determined by considering the forces acting on it. The main force acting on the object is the force due to gravity, which is given by F = mg, where m is the mass of the object and g is the acceleration due to gravity. However, in this case, we also need to consider the force due to air resistance, which is proportional to the velocity of the object.

Let's denote the velocity of the object as v and the constant of proportionality for the force due to air resistance as b. The equation of motion can then be written as:

m(dv/dt) = mg - bv

This equation can be rearranged to get:

(dv/dt) + (b/m)v = g

Now, we can solve this differential equation to find the velocity as a function of time, and then integrate the velocity to get the position as a function of time. However, since the exponential term is too large to ignore, we can use Newton's method to approximate the time when the object strikes the ground. The equation of motion and the value of g are given, so we can substitute those values into Newton's method to solve for the time when the object strikes the ground.

We can calculate the 92% confidence interval for the mean assembly time to be approximately 15.572 to 16.828 minutes. The information provided, however, is insufficient to answer part b) of the question.

In order to construct a 92% confidence interval for the mean assembly time, we first have to find the z-score corresponding to this level of confidence. The z-score for a 92% confidence interval is approximately 1.75. This value is found using a standard normal distribution table (or z-table).

So, the 92% confidence interval for the mean assembly time of 16.2 minutes can be computed as follows: 16.2 ± 1.75*(3.6/sqrt(120)). This gives a confidence interval of approximately 15.572 to 16.828 minutes.

Regarding the second part of your question (b), it seems unfinished and lacks enough information for me to provide a complete answer. Typically, a desired margin of error or a certain level of confidence would need to be specified to determine the sample size.

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a) The 92% confidence interval for the mean assembly time is (15.75 minutes, 16.65 minutes).  b) About 993 workers are needed for the study to estimate the mean assembly time with a 0.2-minute margin of error.

To construct the confidence interval, use the formula for the confidence interval of the mean for a normal distribution: [tex]\[ \text{CI} = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}} \][/tex]

1. Calculate the standard error of the mean: [tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{120}} \approx 0.329 \][/tex]

2. Find the critical z-value for a 92% confidence interval (using a z-table or calculator): [tex]\( z = 1.75 \) (approximately).[/tex]

3. Substitute the values into the formula: [tex]\[ \text{CI} = 16.2 \pm 1.75 \times 0.329 \][/tex]

4. Calculate the confidence interval: [tex]\[ \text{CI} = (16.2 - 0.575, 16.2 + 0.575) \][/tex]

[tex]\[ \text{CI} = (15.625, 16.775) \][/tex]

So, the 92% confidence interval for the mean assembly time is approximately (15.625 minutes, 16.775 minutes).

b) To estimate the mean assembly time with a desired margin of error, use the formula for the required sample size: [tex]\[ n = \left( \frac{z \times \sigma}{\text{ME}} \right)^2 \][/tex]

Given the desired margin of error (ME), let's say 0.2 minutes, and using \( z = 1.75 \) from the 92% confidence level:

[tex]\[ n = \left( \frac{1.75 \times 3.6}{0.2} \right)^2 \]\[ n = (31.5)^2 \]\[ n = 992.25 \][/tex]

So, approximately 993 workers should be involved in the study to estimate the mean assembly time with a margin of error of 0.2 minutes.

Answer:

a. H + K = {u + v}

b. See explanation below

Step-by-step explanation:

Given

H+K={w:w=u+v for some u in H and some v in K}

a.

From the (given) above,

We have that

0 = v1 + 0.u

v1 represents the zero vector in K

We also have that

0 = v + u1

u1 represents the zero vector in H

Up this point, we've shown that both H and K have vector factors, V

We'll use the available data to solve for the zero vector of H + K

This is given by

u1 + v1 + u2 + v2 ---- Rearrange

u1 + u2 + v1 + v2 ---- Let u3 = u1 + u2 and v3 = v1 + v2.

So, we have

u3 + v3 as the zero vector of H + K

H + K = {u + v}

So, H + K is a subspace of V

b

Since H, K are subspace of V, then they are

1. Both close under scalar multiplication

2. Both closed under vector addition

1 and 2 gives

c(u + v) where c is a constant

= c.u + c.v

We can then conclude that H is a subspace of H+K and K is a subspace of H+K because H + K is closed under scalar multiplication and vector addition

The sum of two subspaces H and K is a subspace of V as it contains the zero vector, is closed under addition, and under scalar multiplication. Additionally, H and K themselves also qualify as subspaces of this resultant sum subspace, H+K.

To show that H+K is a subspace of V, we need to prove that it satisfies three properties: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication.

1. Zero Vector: Since H and K are subspaces, they both contain the zero vector. So, 0 = 0 + 0 represents a vector in H+K (with the first 0 from H and the second from K).

2. Closure under Addition: If w1 and w2 are vectors in H+K, w1 can be written as h1 + k1 (h1 in H, k1 in K) and w2 as h2 + k2 (h2 in H, k2 in K). Adding w1 and w2 gives: (h1 + k1) + (h2 + k2) = (h1 + h2) + (k1 + k2), where h1+h2 is in H and k1+k2 is in K, hence, the result is in H+K.

3. Closure under Scalar Multiplication: Let w be in H+K and c be any scalar. Then w = h + k, so cw = c(h + k) = ch + ck, where ch is in H and ck is in K, hence, cw is in H+K.

To show that H is a subspace of H+K and K is a subspace of H+K, note that for any vector h in H, it can be written as h+0 (with 0 as the zero vector of K), and similarly for any vector k in K as 0+k. Thus, H and K are subspaces of H+K.

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The probability question from part (a) requires calculating the chance of getting all heads or all tails on multiple days in a year, which involves complex probability distributions. For part (b), using a Poisson distribution could be appropriate due to the rarity of the event and the high number of trials involved.

The question pertains to the field of probability theory and involves calculating the probability of specific outcomes when flipping a fair coin. For part (a), Jack flips a coin ten times each morning for a year, counting the days (X) when all flips are identical (all heads or all tails). The exact expression for P(X > 1), the probability of more than one such day, requires several steps. First, we find the probability of a single day having all heads or all tails, then use that to calculate the probability for multiple days within the year. For part (b), whether it is appropriate to approximate X by a Poisson distribution depends on the rarity of the event in question and the number of trials. A Poisson distribution is typically used for rare events over many trials, which may apply here.

For part (a), the probability on any given day is the sum of the probabilities of all heads or all tails: 2*(0.5^10). Over a year (365 days), we need to calculate the probability distribution for this outcome occurring on multiple days. To find P(X > 1), we would need to use the binomial distribution and subtract the probability of the event not occurring at all (P(X=0)) and occurring exactly once (P(X=1)) from 1. However, this calculation can become quite complex due to the large number of trials.

For part (b), given the low probability of the event (all heads or all tails) and the high number of trials (365), a Poisson distribution may be an appropriate approximation. The mean (λ) for the Poisson distribution would be the expected number of times the event occurs in a year. Since the probability of all heads or all tails is low, it can be considered a rare event, and the Poisson distribution is often used for modeling such scenarios.

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(a) Probability of upgrading in less than 12 minutes is practically 0. (b) Approximately 66 new files are required for 95% completion in less than 10 minutes.

Part (a):

First, let's calculate the mean and standard deviation of the time it takes to install all 68 files.

Given:

Mean time to install one file: [tex]$\mu = 15 , \text{sec}$[/tex]

Variance: [tex]$\sigma^2 = 11 , \text{sec}^2$[/tex]

The total time T to install all 68 files is the sum of 68 random variables, each with mean [tex]\mu$ and variance $\sigma^2$[/tex]. Since each installation time is independent, the mean of the total time is the sum of the individual means, and the variance of the total time is the sum of the individual variances.

Mean of total time:

[tex]\mu_{\text {total }}=68 \times \mu=68 \times 15 \mathrm{sec}=1020 \mathrm{sec}[/tex]

Variance of total time:

[tex]\sigma_{\text {total }}^2=68 \times \sigma^2=68 \times 11 \mathrm{sec}^2=748 \mathrm{sec}^2[/tex]

Standard deviation of total time:

[tex]\sigma_{\text {total }}=\sqrt{\sigma_{\text {total }}^2}=\sqrt{748} \mathrm{sec} \approx 27.34 \mathrm{sec}[/tex]

Now, to find the probability that the whole package is upgraded in less than 12 minutes (720 seconds), we convert this time into seconds and then use the cumulative distribution function of the normal distribution.

[tex]Z=\frac{X-\mu_{\text {total }}}{\sigma_{\text {total }}}[/tex]

Where:

X=720 (time in seconds)

[tex]\mu_{\text {total }}[/tex] =1020 (mean time in seconds)

[tex]\sigma_{\text {total }}[/tex] = [tex]\sqrt{748[/tex] (standard deviation in seconds)

Substituting the values:

[tex]\begin{aligned}& Z=\frac{720-1020}{27.34} \\& Z \approx-7.316\end{aligned}[/tex]

Now, we look up this Z value in a standard normal distribution table or use software to find the corresponding probability. The probability we seek is P(Z<−7.316).

Using a calculator or statistical software, this probability is extremely close to 0 (practically 0). Hence, the probability that the whole package is upgraded in less than 12 minutes is very close to 0.

Part (b):

Given that 95% of the time upgrading takes less than 10 minutes (600 seconds), we can use the same approach as above to find the corresponding Z value.

[tex]\begin{aligned}&Z=\frac{600-1020}{27.34}\\&Z \approx-14.468\end{aligned}[/tex]

To find the corresponding percentile from the standard normal distribution, we look up Z = -14.468 and find the corresponding percentile, which should be close to 0.05.

Now, let's find the number of files N required to ensure that 95% of the time the upgrading takes less than 10 minutes. We want the probability of completing the installation in less than 10 minutes to be 0.95, which corresponds to the Z value of approximately -1.645 (from standard normal distribution tables).

[tex]Z=\frac{X-\mu_{\text {total }}}{\sigma_{\text {total }}}=-1.645[/tex]

Solving for X:

[tex]\begin{aligned}& -1.645=\frac{X-1020}{27.34} \\& X=-1.645 \times 27.34+1020 \\& X \approx 978.52\end{aligned}[/tex]

So, the total time to install N files is approximately 978.52 seconds. Since we know the mean time to install one file is 15 seconds, we can find N by:

[tex]\begin{aligned}& N=\frac{978.52}{15} \\& N \approx 65.235\end{aligned}[/tex]

So, to ensure that 95% of the time upgrading takes less than 10 minutes, we would need to install approximately 65 new files. Since we can't install a fraction of a file, we would round up to the nearest whole number, giving us N = 66.

Complete Question:

Upgrading a certain software package requires installation of 68 new files. Files are installed consecutively. The installation time is random, but on the average, it takes 15 sec to install one file, with a variance of 11 square sec.

(a) What is the probability that the whole package is upgraded in less than 12 minutes?

(b) A new version of the package is released. It requires only N new files to be installed, and it is promised that 95% of the time upgrading takes less than 10 minutes. Given this information, compute N.

Answer:

-0.51, 5/3, sqrt (2), and 1.37

Step-by-step explanation:

As we know probability must be between 0 and 1. It cannot be greater than 1 and cannot be less than 0 or negative

1,  can be the probability as it lies between 0 and 1

-0.51,  cannot be the value of probability as it is negative

5/3 = 1.66 which is greater than 1, so cannot be the value of probability

[tex]\sqrt{2}[/tex] = 1.41 which is greater than 1, so cannot be the value of probability

3/5 = 0.6 can be the probability as it lies between 0 and 1

0.03, can be the probability as it lies between 0 and 1

1.37 which is greater than 1, so cannot be the value of probability

Answer:

[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]

Step-by-step explanation:

It is given that

[tex]\sinh x=\dfrac{e^x-e^{-x}}{2}[/tex]

[tex]\cosh x=\dfrac{e^x+e^{-x}}{2}[/tex]

[tex]f(x)=\sinh x\cosh x=[/tex]

Using the given hyperbolic functions, we get

[tex]f(x)=\dfrac{e^x-e^{-x}}{2}\times \dfrac{e^x+e^{-x}}{2}[/tex]

[tex]f(x)=\dfrac{(e^x)^2-(e^{-x})^2}{4}[/tex]

[tex]f(x)=\dfrac{e^{2x}-e^{-2x}}{4}[/tex]

Differentiate both sides with respect to x.

[tex]\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left(\dfrac{e^{2x}-e^{-2x}}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2e^{2x}-(-2)e^{-2x}}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\left(\dfrac{2(e^{2x}+e^{-2x})}{4}\right )[/tex]

[tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex]

Hence, [tex]\dfrac{d}{dx}f(x)=\dfrac{e^{2x}+e^{-2x}}{2}[/tex].

Answer:

[tex]10\sqrt 2 iN[/tex] and [tex]10\sqrt 2 jN[/tex]

Step-by-step explanation:

We are given that

Force =F=20 N

[tex]\theta=\frac{\pi}{4}[/tex]

We have to find the horizontal and vertical component of the force.

Horizontal component of the force =[tex]F_x=Fcos\theta i[/tex]

Horizontal component of the force=[tex]F_x=20cos\frac{\pi}{4}i=10\sqrt 2 iN[/tex]

Vertical component of the force,[tex]F_y=Fsin\theta j[/tex]

Vertical component of the force,[tex]F_y=20sin\frac{\pi}{4}j=10\sqrt 2 jN[/tex]

Force,[tex]F=F_x+F_y[/tex]

[tex]F=10\sqrt 2i N+10\sqrt 2j N[/tex]

a. Null hypothesis (H0): Mean cost for the new production method is $215 per hour.

Alternative hypothesis (H1): Mean cost for the new production method is less than $215 per hour.

b. Type I error: Rejecting the null hypothesis when it is true; consequence: Implementing the new method incorrectly and incurring unnecessary costs.

c. Type II error: Failing to reject the null hypothesis when it is false; consequence: Missing the opportunity to adopt a cost-effective production method and continuing with higher operating costs.

We have,

a. The appropriate null and alternative hypotheses would be:

Null Hypothesis (H0): The mean cost for the new production method is equal to or greater than $215 per hour.

Alternative Hypothesis (H1): The mean cost for the new production method is less than $215 per hour.

b. The Type I error in this situation would be rejecting the null hypothesis when it is actually true.

In other words, it would mean concluding that the new production method reduces the mean operating cost per hour when it actually doesn't.

The consequence of making this error is that resources, time, and effort might be invested in implementing the new method based on incorrect information, potentially leading to unnecessary costs and inefficiencies.

c. The Type II error in this situation would be failing to reject the null hypothesis when it is actually false.

In other words, it would mean failing to conclude that the new production method reduces the mean operating cost per hour when it actually does.

The consequence of making this error is that the opportunity to adopt a more cost-effective production method would be missed, potentially leading to continued higher operating costs and missed efficiency gains.

Thus,

a. Null hypothesis (H0): Mean cost for the new production method is $215 per hour.

Alternative hypothesis (H1): Mean cost for the new production method is less than $215 per hour.

b. Type I error: Rejecting the null hypothesis when it is true; consequence: Implementing the new method incorrectly and incurring unnecessary costs.

c. Type II error: Failing to reject the null hypothesis when it is false; consequence: Missing the opportunity to adopt a cost-effective production method and continuing with higher operating costs.

Learn more about hypothesis testing here:

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The null and alternative hypotheses are set as μ ≥ 215 and μ < 215 respectively. A Type I error, rejecting the null hypothesis when it's true, could lead to unnecessary changes. A Type II error, failing to reject the null hypothesis when it's false, could result in missed cost savings.

a. In this situation, the null hypothesis (H0) would state that the mean cost of the new production method is equal to or greater than 215. We can denote this as: H0: μ ≥ 215. The alternative hypothesis (Ha) posits that the mean cost of the new production method is less than 215, denoted as: Ha: μ < 215.

b. A Type I error would occur if we incorrectly reject the null hypothesis, concluding that the new method reduces the cost when, in fact, it does not. The consequence could be implementing a production method that doesn't actually provide the desired savings, which could lead to unnecessary changes and associated costs.

c. A Type II error would happen if we fail to reject the null hypothesis when it's actually false, meaning the new method does reduce the cost but we fail to recognize it. The consequence of this error could be missing out on an opportunity to reduce operating costs, potentially leading to higher than necessary expenses.

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Answer:

T= A person selected in the poll travel to Europe

NT= A person selected in the poll NOT travel to Europe

For this case we have the following respondents for each event

n(T)= 68

n(NT) = 124

So then the total of people for the poll are:

[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]

And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:

[tex] p =\frac{Possible}{Total}[/tex]

And if we replace we got:

[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]

So then the probability of getting someone who has traveled to Europe is 0.354

Step-by-step explanation:

For this case we define the following events:

T= A person selected in the poll travel to Europe

NT= A person selected in the poll NOT travel to Europe

For this case we have the following respondents for each event

n(T)= 68

n(NT) = 124

So then the total of people for the poll are:

[tex] n = n(T) + n(NT)= 68 +124= 192[/tex]

And we are interested on the probability of getting someone who has traveled to Europe, and we can use the empirical definition of probability given by:

[tex] p =\frac{Possible}{Total}[/tex]

And if we replace we got:

[tex] p = \frac{n(T)}{n}= \frac{68}{192}= 0.354[/tex]

So then the probability of getting someone who has traveled to Europe is 0.354

Answer:

1120 Eagles

Step-by-step explanation:

If the population of Eagles decreases by 4% percent each year, then in 2030 (which is 15 years later), the population would decrease by (4 × 15)%

that is 60%

This indicates that the population of bald eagles in the year 2030 would be 60% less of what it was in the year 2015.

so we have, [tex]\frac{60}{100}[/tex] × 2800

which gives us 1680

this means that the number of bald eagles in the year 2030 would be 1680 less than the number of bald eagles in the year 2015

hence 2800 - 1680

then the answer is 1120 Bald Eagles

Answer: the approximate population of bald eagles be in the year 2030 is 15178

Step-by-step explanation:

The population of bald eagles is decreasing 4% each year. This means that the rate is exponential.

We would apply the formula for exponential decay which is expressed as

A = P(1 - r)^t

Where

A represents the population after t years.

t represents the number of years.

P represents the initial population.

r represents rate of growth.

From the information given,

P = 2800

r = 4% = 4/100 = 0.04

t = 2030 - 2015 = 15 years

Therefore

A = 2800(1 - 0.04)^ 15

A = 2800(0.96)^ 15

A = 15178

Answer:

4.45 ccs will remain after 5 hours.

Step-by-step explanation:

The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.

(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)

(dQ/dt) = -kQ

(dQ/Q) = -kdt

 ∫ (dQ/Q) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.

We obtain

In (Q/Q₀) = -kt

(Q/Q₀) = e⁻ᵏᵗ

Q(t) = Q₀ e⁻ᵏᵗ

the initial injection was 8 ccs and 5 ccs remain after 4 hours

Q₀ = 8 ccs,

At t = 4 hours, Q = 5 ccs

5 = 8 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.625

-kt = In (0.625) = -0.47

-4k = 0.47

k = 0.1175 /hour

Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ

At t = 5 hours, Q = ?

Q = 8 e⁻⁰•¹¹⁷⁵ᵗ

0.1175 × 5 = 0.5875

Q = 8 e(^-0.5875)

Q = 4.45 ccs

Hope this Helps!!!

Answer:

$12,884.08

Step-by-step explanation:

Assuming that interest is compounded annually, the future value of an invested amount 'P', at an interest rate 'r' for a period of 'n' years is given by the following equation:

[tex]FV = P*(1+r)^n[/tex]

Therefore, an investment of $8,000 at a rate of 10% per year for 5 years has a future value of:

[tex]FV = 8,000*(1+0.1)^5\\FV=\$12,884.08[/tex]

There will be $12,884.08 in the account after 5 years.

Answer:

Step-by-step explanation:

Assuming the interest was compounded annually, we would apply would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 8000

r = 10% = 10/100 = 0.1

n = 1 because it was compounded ince in a year.

t = 5 years

Therefore,

A = 8000(1+0.1/1)^1 × 5

A = 8000(1.1)^5

A = $12884.1

Answer:

[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]

[tex]E(T) =900+900+900+900 =3600[/tex]

And the mean would be:

[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]

And the standard deviation of total time would be:

[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]

[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/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".  

Solution to the problem

Let total life time of t four transistors T = X1 + X2 + X3 + X4 (where X1,X2,X3 and X4 are life time of individual transistors

For this case the mean length of time that four transistors will last

[tex]E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4) [/tex]

[tex]E(T) =900+900+900+900 =3600[/tex]

And the mean would be:

[tex] \bar X = \frac{T}{4} = \frac{3600}{4}= 900[/tex]

And the standard deviation of total time would be:

[tex]SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))[/tex]

[tex]\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60[/tex]

The mean lifetime of four transistors is 3600 hours and the standard deviation of the lifetime is 60 hours.

The mean and standard deviation are statistical concepts used to describe a dataset. The mean, or average, is the sum of all data points divided by the number of data points. In this case, the mean lifetime of a single transistor is 900 hours. If you have four transistors, their total lifetime would be four times the mean of a single transistor. So, the mean lifetime for four transistors is 4 * 900 = 3600 hours.

The standard deviation measures the amount of variation in a set of values. When you are considering the lifetime of multiple transistors, you add the variances - not the standard deviations - and then take the square root. The variance is the square of the standard deviation. So, the standard deviation for the lifetime of four transistors is sqrt(4) * 30 = 60 hours.

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The correct statement is that the probability of both the students who have volunteered for community service the absolute tolerance will be 0.1056.

The calculation of the probability of both the students who have volunteered for the community service getting selected is shown by doing multiple calculations as under.

Hence, correct statement is that the probability of both the students who have volunteered for community service the absolute tolerance will be 0.1056.

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Answer:

The answer is 1/n

Step-by-step explanation:

First number = Alice's number

Second Number = Bob's Number

There are already 2 good answers if the first number is AT LEAST 1 greater than the second.

However, what if the first number is EXACTLY 1 greater than the second?

I will assume the first number is removed from the array after being selected.

If the first number is 1, then it is impossible, because the second number cannot be less than 1.

If the first number is 2, then the second number could be 1, which has a probability of 1/(n-1)

If the first number is 3, then the second number could be 2, which has a probability of 1/(n-1)

etc…

There is a 1/n probability of each first number selected, so the answer would be:

0*1/n+1/n*1/(n−1)+1/n*1/(n−1)+⋯+1/n*1/(n−1)

There are n terms in that series.

That gives us:

0+1/n(n−1)+1/n(n−1)+⋯+1/n(n−1)

There are n-1 of those identical terms, because 1 of the terms is 0.

(n−1)/n(n−1)=1/n

So the answer is 1/n.

Answer:

It increases

Step-by-step explanation:

(5/x) + 5

As x decreases, 5/x increases.  So the expression increases.

Answer:

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

[tex]X \sim N(2.514 cm,0.008 cm)[/tex]  

Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]

b) For this case we can see the interval required on the figure attached,

c) [tex]P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/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 diameters of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(2.514 cm,0.008 cm)[/tex]  

Where [tex]\mu=2.514[/tex] and [tex]\sigma=0.008[/tex]

Part b

For this case we can see the interval required on the figure attached,

Part c

We are interested on this probability

[tex]P(2.49<X<2.51)[/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(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)[/tex]

And we can find this probability with this difference:

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)[/tex]

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

[tex]P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307[/tex]

Answer:

Step-by-step explanation:

The number of observations in model 1 are more compared with model 2 and model 3. And, the coefficient values increased from model 1 to model 3. Moreover, the standard error values are increased. Hence, the test statistic value increases. Therefore, the variable “difference in match win probabilities” is significant.

State the null and alternative Hypotheses:

Null Hypothesis:

H0: The difference in match win probabilities is not significant.

Alternative Hypothesis:

H1: The difference in match win probabilities is significant.

From the given information (model 3), the coefficient of “The difference in match win probabilities” is -0.550 and the corresponding standard error is 0.240. The sample size is 1973.

The test statistic is -2.2917 and it is calculated below: on the picture attached.

Statistics homework question answer, step 1, image 1

The number of sample observations is n=1973 and consider the level of significance as 0.05.

The degree of freedom is, n – 1 = 1973–1 = 1972.

Answer:

(a) P(x=0) = 0.1422

(b) P(x=1) = 0.3012

(c) P(x≥2) = 0.5566

Step-by-step explanation:

The probability that x drivers will have an invalid license follows a binomial distributions, so it is calculated as:

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Where n is equal to the 12 drivers stopped and p is the probability that a driver stopped for speeding have invalid licenses. Then, replacing values, we get:

[tex]P(x)=\frac{12!}{x!(12-x)!}*0.15^{x}*(1-0.15)^{12-x}[/tex]

Now, the probability that none will have an invalid licenses is calculated as:

[tex]P(x=0)=\frac{12!}{0!(12-0)!}*0.15^{0}*(1-0.15)^{12-0}\\P(x=0)=0.1422[/tex]

At the same way, the probability that exactly one will have an invalid license  is calculated as:

[tex]P(x=1)=\frac{12!}{1!(12-1)!}*0.15^{1}*(1-0.15)^{12-1}\\P(x=1)=0.3012[/tex]

Finally, the probability that at least 2 will have invalid licenses is calculated as:

[tex]P(x\geq 2)=1-P(x<2)\\P(x\geq 2)=1-(P(x=0)+P(x=1))\\P(x\geq 2)=1-(0.1422+0.3012)\\P(x\geq 2)=1-0.4434\\P(x\geq 2)=0.5566[/tex]

Answer:

a) [tex] P(X=0) = (12C0) (0.15)^0 (1-0.15)^{12-0} =0.1422[/tex]

b) [tex] P(X=1) = (12C1) (0.15)^1 (1-0.15)^{12-1} =0.3012[/tex]

c) [tex] P(X \geq 2) = 1-P(X<2) = 1-P(X\leq 1)= 1-[P(X=0)+P(X=1)][/tex]

And replacing we got:

[tex]P(X \geq 2)= 1- [0.1422+0.3012]= 0.5566[/tex]

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

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

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=12, p=0.15)[/tex]  

Part a

We want this probability:

[tex] P(X=0) = (12C0) (0.15)^0 (1-0.15)^{12-0} =0.1422[/tex]

Part b

We want this probability:

[tex] P(X=1) = (12C1) (0.15)^1 (1-0.15)^{12-1} =0.3012[/tex]

Part c

We want this probability:

[tex]P(X \geq 2)[/tex]

And we can use the complement rule:

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

And replacing we got:

[tex]P(X \geq 2)= 1- [0.1422+0.3012]= 0.5566[/tex]

Answer:

(1) He would sell 600 pizzas if he cuts his price by 10%

(2) His revenue would increase by 8% ($800)

Step-by-step explanation:

(1) Price elasticity of demand = % change in quantity demanded /% change in price

price elasticity of demand = -2

Initial price = $20

New price = $20 - (10% × $20) = $20 - $2 = $18

Change in price = new price - initial price = 18 - 20 = -2

% change in price = -2/20 × 100 = -10%

% change in quantity demanded = price elasticity of demand × % change in price = -2 × -10% = 20%

Let the new quantity demanded be y

% change in quantity demanded = (y-500)/500

0.2 = (y-500)/500

y-500 = 0.2×500

y-500 = 100

y = 100+500 = 600

If he reduces his price by 10%, he would sell 600 pizzas.

(2) If he sells each pizza $20, quantity demanded is 500

Revenue = 500 × $20 = $10,000

If he cuts his price by 10%, his new price would be $180 and new quantity demanded is 600

Revenue = 600 × $18 = $10,800

If he cuts his price by 10% his revenue would increase by $800 ($10,800 - $10,000 = $800)

Final answer:

By decreasing the price by 10%, the pizzeria can expect a 20% increase in quantity demanded, resulting in 600 pizzas sold and an increase in revenue to $10,800 from the original $10,000, showing the price decrease to be beneficial.

Explanation:

The student's question involves calculating the impact of a price change on demand and revenue for a pizzeria, given a price elasticity of demand of -2. To address this, we can apply the formula for price elasticity of demand, which is the percentage change in quantity demanded divided by the percentage change in price. Knowing that the demand elasticity is -2 and the owner plans to reduce the price by 10%, we can calculate the expected change in quantity demanded and how revenue will be affected.

First, a 10% decrease in the price from $20 to $18 results in a 20% increase in quantity demanded (since the elasticity is -2, the absolute value indicates the proportionate change in quantity for a 1% change in price, thus 10% price change × 2 = 20% quantity change). Hence, the new quantity demanded will be 500 (original) × 1.20 (20% increase) = 600 pizzas. Second, the new revenue will be 600 (quantity) × $18 (new price) = $10,800, compared to the original revenue of 500 × $20 = $10,000. The pizzeria's revenue increases by $800, indicating that the price decrease was beneficial in terms of revenue.

Answer:

A fuel cell converts hydrogen into electricity. This hydrogen is stored in a tank on board the FCEV. Hydrogen and energy components are not synonymous; they can be conveyed in mix or independently. Energy units can work on petroleum gas, which maintains a strategic distance from burning and in this way 90% of airborne toxins. Hydrogen can be scorched in motors and boilers with no immediate CO2 and close to zero NO2 discharges. At the point when utilized together, hydrogen power modules are zero-emanation at the purpose of utilization, with by and large outflows reliant on the fuel creation technique.Because fuel cells allow the regulated reaction of hydrogen (in a tank) and oxygen (from the air) to produce electricity, they are often perceived as non-polluting.Hydrogen (like electricity) is not a primary source of energy but rather an energy carrier. There are no natural reservoirs of pure hydrogen. To produce hydrogen, water can be electrolyzed in the reverse of the fuel-cell reaction. To force electrolysis to occur, considerable amounts of electricity are required, much of which is currently generated by burning fossil fuels. Alternatively, in commercial applications, the vast preponderance of hydrogen generated in the United States is extracted from a fossil fuel: natural gas. When natural gas (basically methane, a lightweight molecule made of carbon and hydrogen) is exposed to steam under high temperatures in the presence of a catalyst, it frees the hydrogen. This is called “reforming,” and the process produces carbon dioxide (CO2).

Step-by-step explanation:

Hydrogen and fuel cells are essential for achieving zero emissions and sustainable energy due to efficiency, renewable sourcing, and zero emissions.

Hydrogen and fuel cells are crucial for achieving 100% zero emissions and sustainable energy conversion due to several reasons:

Efficiency: Fuel cells offer high efficiency, up to 50% or more, compared to traditional combustion engines, contributing to reduced energy waste.

Renewable sourcing: Hydrogen can be produced from renewable sources like solar and wind, ensuring sustainable energy production.

Zero emissions: Hydrogen-powered vehicles emit only water and heat, making them environmentally friendly and reducing greenhouse gas emissions.

Answer:

There is a 99% probability that \mu μ is between 3 and 9.

Step-by-step explanation:

For a confidence level of x% for a mean being found between a and b, it means that we are x% sure, that is, there is a x% probability that the true mean for the population is between a and b.

A 99% confidence interval for the mean μ of a population is computed from a random sample and found to be 6 ± 3.

99% probability that the true mean of the population is between 3 and 9.

So the correct answer is:

There is a 99% probability that \mu μ is between 3 and 9.

A 99% confidence interval of 6 ± 3 implies that we are 99% confident that the population mean is between 3 and 9. If further samples were taken, both the mean and margin of error could vary, depending on the specific data points and variability of each sample.

A 99% confidence interval refers to the range of values within which we are 99% confident that the population mean (μ) resides. In this case, the confidence interval is 6 ± 3, which implies that we are 99% confident that the mean is anywhere between 3 and 9. It does not mean that there is a 99% probability that the true mean is 6, or that the true margin of error is 3. The margin of error is a function of your data's variability and your sample size, not a fixed number.

If we take more random samples, the mean could be different because each sample may contain different individuals or data points with different distributions. However, if the sampling process is properly randomized and unbiased, on average, the mean of the samples should be close to the population mean.

While the margin of error would be expected to remain around 3 for similar sample sizes and variability, it is not guaranteed to always be 3, as it depends on the specific data in each sample.

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Answer:

The worth of the company in 2000 is $56,000.

Step-by-step explanation:

The growth rate of the company is:

[tex]f'(t)=2000-12t^{2}[/tex]

To determine the worth of the company in 2000, first compute the change in the net worth during the period 1990 (t = 0) to 2000 (t = 10) as follows:

[tex]\int\limits^{10}_{0} {2000-12t^{2}} \, dt =\int\limits^{10}_{0} {2000} \, dt-12\int\limits^{10}_{0} {t^{2}} \, dt=2000 |t|^{10}_{0}-12|\frac{t^{3}}{3}|^{10}_{0}\\=(2000\times10)-(4\times10^{3})\\=20000-4000\\=16000[/tex]

The increase in the company's net worth from 1990 to 2000 is $16,000.

If the company's worth was $40,000 in 1990 then the worth of the company in 2000 is:

Worth in 2000 = Worth in 1990 + Net increase in company's worth

                        [tex]=40000+16000\\=56000[/tex]

Thus, the worth of the company in 2000 is $56,000.

Answer:

a. (1.811, 1.929)

Step-by-step explanation:

We have to find find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which, z is Z(a/2), [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample. So

[tex]M = 2.58*\frac{0.125}{\sqrt{30}} = 0.059[/tex]

The lower end of the interval is the mean subtracted by M. So it is 1.87 - 0.059 = 1.811.

The upper end of the interval is the mean added to M. So it is 1.87 + 0.059 = 1.929

So the correct answer is:

a. (1.811, 1.929)

Answer:

Tuesday's Z-score is 7.56

Step-by-step explanation:

We are given that a department store, on average, has daily sales of 28,651.79 and the standard deviation of sales is $1000.

Also, it is given that on Tuesday, the store sold $36,211.08 worth of goods.

Let X = Daily sales of goods

So, X ~ N([tex]\mu = 28,651.79, \sigma^{2} =1000^{2}[/tex])

The z-score probability distribution is given by;

        Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ standard normal N(0,1)

Now, Tuesday,s Z-score is given by;

      Z = [tex]\frac{36,211.08-28,651.79}{1000}[/tex] = 7.56

Yes, Tuesday was an unusually good day as on this day more worth of sales takes place as compared to the average daily sales of $28,651.79 .

Answer:

[tex]A_h=150\sqrt{3}\ m^2[/tex]

Step-by-step explanation:

Regular Hexagon

For the explanation of the answer, please refer to the image below. Let's analyze the triangle shown inside of the hexagon. It's a right triangle with sides x,y, and z.

We know that x is half the length of the side length of the hexagon. Thus

[tex]x=5 m[/tex]

Note that this triangle repeats itself 12 times into the shape of the hexagon. The internal angle of the triangle is one-twelfth of the complete rotation angle, i.e.

[tex]\theta=360/12=30^o[/tex]

Now we have [tex]\theta[/tex], the height of the triangle y is easily found by

[tex]\displaystyle tan30^o=\frac{x}{y}[/tex]

Solving for y

[tex]\displaystyle y=\frac{x}{tan30^o}=\frac{5}{ \frac{1} {\sqrt{3} }}=5\sqrt{3}[/tex]

The value of z can be found by using

[tex]\displaystyle sin30^o=\frac{x}{z}[/tex]

[tex]\displaystyle z=\frac{x}{sin30^o}=\frac{5}{\frac{1}{2}}=10[/tex]

The area of the triangle is

[tex]\displaystyle A_t=\frac{xy}{2}=\frac{5\cdot 5\sqrt{3}}{2}=\frac{25\sqrt{3}}{2}[/tex]

The area of the hexagon is 12 times the area of the triangle, thus

[tex]\displaystyle A_h=12\cdot A_t=12\cdot \frac{25\sqrt{3}}{2}=150\sqrt{3}[/tex]

[tex]\boxed{A_h=150\sqrt{3}\ m^2}[/tex]