(19.-2).(-11, 10) find the slope

(a) Assuming that Q satisfies the differential equation Q' = -rQ, determine the decay constant r for carbon-14. (b) Find an expression for Q(t) at any time t, if Q(0) = Qo. (c) Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains.

Answer:

a) r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121/year

b) Q(t) = Q₀ (e^-rt)

c) Are of the 20% remnant of Carbon-14 = 13301.14 years.

Step-by-step explanation:

Q' = -rQ

Q' = dQ/dt

dQ/dt = -rQ

dQ/Q = -rdt

Integrating the left hand side from Q₀ to Q₀/2 and the right hand side from 0 to t1/2 (half life, t1/2 = 5730 years)

In ((Q₀/2)/Q₀) = -r(t1/2)

In (1/2) = -r(t1/2)

In 2 = r(t1/2)

r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121 /year

b) Q' = -rQ

Q' = dQ/dt

dQ/dt = -rQ

dQ/Q = -rdt

Integrating the left hand side from Q₀ to Q(t) and the right hand side from 0 to t.

In (Q(t)/Q₀) = -rt

Q(t)/Q₀ = e^(-rt)

Q(t) = Q₀ (e^-rt)

c) Q(t) = Q₀ (e^-rt)

Q(t) = 0.2Q₀, t = ? and r = 0.000121/year

0.2Q₀ = Q₀ (e^-rt)

0.2 = e^-rt

In 0.2 = -rt

-1.6094 = - 0.000121 × t

t = 1.6094/0.000121 = 13301.14 years.

Hope this Helps!

Radiocarbon dating is a method used in archeological research to determine the age of artifacts based on the ratio of carbon-14 to the original amount. This technique is limited to time periods of 50,000 years or more.

Radiocarbon dating is an important tool in archeological research that allows scientists to determine the age of wood, plant remains, and other artifacts. This method is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. By measuring the ratio of carbon-14 to the original amount, scientists can accurately determine the age of these remains. Radiocarbon dating is limited to time periods of 50,000 years or more.

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Answer:

[tex] P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]

Solution to the problem

For this case the time between breakdowns representing our random variable T is exponentially distirbuted [tex] T \sim Exp (\mu = 10)[/tex]

So on this case we can find the value of [tex]\lambda[/tex] like this:

[tex] \lambda = \frac{1}{\mu} = \frac{1}{10}[/tex]

So then our density function would be given by:

[tex]P(T)=\lambda e^{-\frac{t}{10}}[/tex]

The exponential distribution is useful when we want to describe the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time between two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

[tex]P(T>t)= e^{-\lambda t}[/tex]

And on this case we are looking for this probability:

[tex] P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048[/tex]

Answer:

10

Step-by-step explanation:

The number of tiles in the design is 1 + 2 + 3 + ...

We can model this as an arithmetic series, where the first term is 1 and the common difference is 1.  The sum of the first n terms of an arithmetic series is:

S = n/2 (2a₁ + d (n − 1))

Given that a₁ = 1 and d = 1:

S = n/2 (2(1) + n − 1)

S = n/2 (n + 1)

Since S ≤ 60:

n/2 (n + 1) ≤ 60

n (n + 1) ≤ 120

n must be an integer, so from trial and error:

n ≤ 10

Mr. Tong should use 10 tiles in the final row to use the most tiles possible.

Answer: A. - 1/a

Step-by-step explanation:

Two lines are said to be perpendicular if the product of their slope equals -1 . That is , if the slope of the first line is [tex]m_{1}[/tex] and the slope of the second line is [tex]m_{2}[/tex] , if they are perpendicular , then :

[tex]m_{1}[/tex] x [tex]m_{2}[/tex] = -1.

Therefore : the perpendicular slope to m is [tex]\frac{-1}{a}[/tex]

Answer:

The confidence level that was used is 0.25% .

Step-by-step explanation:

We are given that mean of the selected sample of 16 accounts is $5,000 and a standard deviation of $400.

It has also been reported that the sample information indicated the mean of the population ranges from $4,739.80 to $5,260.20. This represents the Confidence Interval for population mean.

But we have to find that at what confidence level this information about range of population men has been stated.

Since we know that Confidence Interval for population mean is given by :

   C.I. for population mean = Sample mean(xbar) [tex]\pm[/tex] z value * [tex]\frac{Standard deviation}{\sqrt{n} }[/tex]

i.e.,if we have 95% C.I. = xbar [tex]\pm[/tex] 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] .

So, our Confidence Interval for population is written as :

 [$4,739.80 , $5,260.20] = $5000 [tex]\pm[/tex] z value * [tex]\frac{400}{\sqrt{16} }[/tex]

 $5000 - z value * 100 = $4739.80    { Solving these we get Z value = 2.602}

 $5000 + z value * 100 = $5260.20

In z table we find that at value of 2.60 the probability is 0.99534 so subtracting this from 1 we get confidence level for one tail i.e.0.5%(approx).

Therefore, for two tail Confidence level will be 0.5%/2 = 0.25% .

Using the t-distribution, it is found that a confidence level of 98% was used.

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The information given is:

The margin of error is of:

[tex]M = t\frac{s}{\sqrt{n}}[/tex]

In which t is the critical value.

For this problem, the margin of error is:

[tex]M = \frac{5260.2 - 4739.8}{2} = 260.2[/tex]

Hence, the critical value is found solving the equation of the margin of error for t.

[tex]M = t\frac{s}{\sqrt{n}}[/tex]

[tex]260.2 = t\frac{400}{\sqrt{16}}[/tex]

[tex]100t = 260.2[/tex]

[tex]t = \frac{260.2}{100}[/tex]

[tex]t = 2.602[/tex]

Looking at the t-table, with 16 - 1 = 15 df, t = 2.602 is associated with a confidence level of 98%.

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Answer:

Mean = 0.3026

Median = 0.308

Mode = 0.327                    

Step-by-step explanation:

We are given the following data set:

0.327, 0.295, 0.318, 0.310, 0.285, 0.280, 0.314, 0.323, 0.310, 0.295, 0.283, 0.279, 0.277, 0.313, 0.327, 0.306, 0.327, 0.317, 0.292, 0.275

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{6.053}{20} = 0.3026[/tex]

Sorted data: 0.275, 0.277, 0.279, 0.280, 0.283, 0.285, 0.292, 0.295, 0.295, 0.306, 0.310, 0.310, 0.313, 0.314, 0.317, 0.318, 0.323, 0.327, 0.327, 0.327

[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]

[tex]\text{Median} = \dfrac{10^{th} + 11^{th}}{2} = \dfrac{0.306 + 0.310}{2} = 0.308[/tex]

Mode is the mos frequent observation in the data.

Mode = 0.327

It appeared 3 times.

The mean of the MLB Batting Averages is around 0.306, the median is 0.308, and the mode is 0.327.

In the field of mathematics, especially in statistics, mean, median and mode are measures of central tendency that provide an overview of the data set. Given the MLB Batting Averages, we start by arranging the data in ascending order. After that, we calculate the mean by adding up all the values and dividing by the number of values. The median is the middle value in an ordered dataset, and the mode is the value that appears most often in a dataset.

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Answer:Yes (A can occur when B occurs)

Step-by-step explanation: Independent events are events that are not determined by the occurrence of another. In this case EVENT A CAN OCCUR AS EVENT B OCCURS.

Justification: Landing on tails after tossing a coin and landing on a Five (5) after rolling the DIE Landing on Six (6) after rolling a DIE and landing on tails after tossing a coin.

Independent probabilities can occur together,they don't depend on each other.

Answer:

The chart A is correct

Pareto Chart

Step-by-step explanation:

Given chart is missing (Attached)

Find:

- Which chart represents the correct data.

- What other chart can be used to express the given data

Solution:

- Use the given values for each color and compare with the three charts A,B and C given.

                For Blue =    A (12)    ,    B(12)    ,    C(11)

                For Green = A(7)       ,    B(13)    ,    C(12)

- Hence,  The chart A is correct.

- Any other chart which can correctly express the information given should be a chart that uses bars or frequency to expresses the percentages. Pareto Chart expresses both bars and line chart(curve) to express the frequency of the data.

Answer:

a) [tex](0.67-0.56) - 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.054[/tex]  

[tex](0.67-0.56) + 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.166[/tex]

And the 90% confidence interval would be given (0.054;0.166).  

b) [tex](0.31-0.25) - 1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=-0.0122[/tex]  

[tex](0.31-0.25) +1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=0.132[/tex]  

And the 90% confidence interval would be given (-0.0122;0.132).

c) [tex](0.46-0.61) - 1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.260[/tex]  

[tex](0.46-0.61) +1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.0404[/tex]  

And the 90% confidence interval would be given (-0.260;-0.0404).

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

[tex]p_1[/tex] represent the real population proportion for sample 1

[tex]\hat p_1 =0.67[/tex] represent the estimated proportion for sample 1

[tex]n_A=400[/tex] is the sample size required for sample 1

[tex]p_2[/tex] represent the real population proportion for sample 2

[tex]\hat p_2 =0.56[/tex] represent the estimated proportion for sample 2

[tex]n_2=400[/tex] is the sample size required for sample 2

[tex]z[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

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

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]  

For the 90% confidence interval the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.67-0.56) - 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.054[/tex]  

[tex](0.67-0.56) + 1.64 \sqrt{\frac{0.67(1-0.67)}{400} +\frac{0.56(1-0.56)}{400}}=0.166[/tex]

And the 90% confidence interval would be given (0.054;0.166).  

Part b

[tex](0.31-0.25) - 1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=-0.0122[/tex]  

[tex](0.31-0.25) +1.64 \sqrt{\frac{0.31(1-0.31)}{180} +\frac{0.25(1-0.25)}{250}}=0.132[/tex]  

And the 90% confidence interval would be given (-0.0122;0.132).

Part c

[tex](0.46-0.61) - 1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.260[/tex]  

[tex](0.46-0.61) +1.64 \sqrt{\frac{0.46(1-0.46)}{100} +\frac{0.61(1-0.61)}{120}}=-0.0404[/tex]  

And the 90% confidence interval would be given (-0.260;-0.0404).

Answer:

(-2.95,0)

x = -2.95 is the zero of the function.                                                      

Step-by-step explanation:

We are given the function:

[tex]f(x) = 1.5x + 3(x-3) + 5.5(x+7)[/tex]

We have to find the zero of the function.

Zero of function:

We can find zero of the function in the following manner:

[tex]f(x) = 1.5x + 3(x-3) + 5.5(x+7) = 0\\1.5x + 3x -9 + 5.5x + 38.5 = 0\\(1.5x + 3x + 5.5x) + (38.5-9) = 0\\10x + 29.5 = 0\\10x = -29.5\\x = -2.95[/tex]            

Thus, x = -2.95 is the zero of the function.        

Answer:

The diameter was [tex]d=\frac{45}{\pi }[/tex] in.

Step-by-step explanation:

The arc of a circle is given by

[tex]s=r\theta[/tex]

where

s = arc length

r = radius of the circle

θ = measure of the central angle in radians.

From the information given

s = 4 in

θ = 32º

To find the diameter of the original​ pizza, we use the formula of the diameter of a circle

[tex]d=2r[/tex]

First, we need to convert the angle to radians

[tex]\theta=32\º \cdot \frac{\pi}{180\º} =\frac{8\pi }{45}[/tex]

Next, solve for r from the arc formula

[tex]r=\frac{s}{\theta} =\frac{4}{\frac{8\pi }{45}} =\frac{45}{2\pi }[/tex]

Then, we use the diameter of a circle formula

[tex]d=2r=2(\frac{45}{2\pi })=\frac{45}{\pi }[/tex]

Answer:

(a) P (Both vehicles are available at a given time) = 0.81

(b) P (Neither vehicles are available at a given time) = 0.01

(c) P (At least one vehicle is available at a given time) = 0.99

Step-by-step explanation:

Let A = Vehicle 1 is available when needed and B = Vehicle 2 is available when needed.

Given:

The availability of one vehicle is independent of the availability of the other, i.e. P (A ∩ B) = P (A) × P (B)

P (A) = P (B) = 0.90

(a)

Compute the probability that both vehicles are available at a given time as follows:

P (Both vehicles are available) = P (Vehicle 1 is available) ×

                                                              P (Vehicle 2 is available)

                                  [tex]P(A\cap B)=P(A)\times P(B)[/tex]

                                                  [tex]=0.90\times0.90\\=0.81[/tex]

Thus, the probability that both vehicles are available at a given time is 0.81.

(b)

Compute the probability that neither vehicles are available at a given time as follows:

P (Neither vehicles are available) = [1 - P (Vehicle 1 is available)] ×

                                                                   [1 - P (Vehicle 2 is available)]

                                    [tex]P(A^{c}\cap B^{c})=[1-P(A)]\times [1-P(B)]\\[/tex]

                                                       [tex]=(1-0.90)\times (1-0.90)\\=0.10\times0.10\\=0.01[/tex]

Thus, the probability that neither vehicles are available at a given time is 0.01.

(c)

Compute the probability that at least one vehicle is available at a given time as follows:

P (At least one vehicle is available) = 1 - P (None of the vehicles are available)

                                                          [tex]=1-[P(A^{c})\times P(B^{c})]\\=1-0.01.....(from\ part\ (b))\\ =0.99[/tex]

Thus, the probability that at least one vehicle is available at a given time is 0.99.

Answer:

[tex]\frac{(x^2\times x^4)^3}{x^8}[/tex] [tex]=x^{10}[/tex]

Step-by-step explanation:

1.

Power of the same base multiply by adding their exponent.

Example: [tex]a^m\times a^n = a^{(m+n)}[/tex]

2.

Power of the same base divide by subtracting their exponent.

Example:[tex]a^m \div a^n = a^{(m-n)}[/tex]

3.

Find power of a power we may multiple the exponent.

Example:[tex](a^m)^n = a^{mn}[/tex]

4.

[tex]\frac{(x^2\times x^4)^3}{x^8}[/tex]

[tex]=\frac{(x^6)^3}{x^8}[/tex]   [ using multiplication rule]

[tex]=\frac{(x^{18})}{x^8}[/tex]  [ using power of power rule]

[tex]=x^{18-8}[/tex]  [ Using division rule]

[tex]=x^{10}[/tex]

Answer:

P(A)= 0.606 and P(B)= 0.237

Step-by-step explanation:

Since A and B are independent

P(A ∩ B) = P(A) *  P(B) → P(B) = P(A ∩ B) / P(A)

and also

P(A ∪ B)=  P(A) + P(B) -  P(A ∩ B)

P(A ∪ B) =  P(A) + P(A ∩ B) / P(A) -  P(A ∩ B)

[P(A ∪ B) +  P(A ∩ B) ]* P(A) =  P(A)² + P(A ∩ B)

P(A)² - [P(A ∪ B) +  P(A ∩ B) ]* P(A) + P(A ∩ B) = 0

P(A)² - [ 0.626+0.144] * P(A) + 0.144 =0

P(A)² - 0.77* P(A) + 0.144 =0

thus

P(A)₁= 0.606 or P(A)₂= 0.1647

for P(A)₁→ P(B)₁ = P(A ∩ B) / P(A)₁ = 0.144/0.606 = 0.237

thus  P(A)₁ > P(B)₁ → correct

for P(A)₂→ P(B)₂ = P(A ∩ B) / P(A)₂ = 0.144/0.1647= 0.8743

thus  P(A)₂ < P(B)₂ → incorrect

therefore

P(A)= 0.606 and P(B)= 0.237

Answer:

Mass of the liquid in the tank will be [tex]1.45\times 10^8kg[/tex]

Step-by-step explanation:

We have given volume of tar [tex]V=3.20\times 10^5gallon[/tex]

1 gallon = 3.785 liter

So [tex]3.20\times 10^5gallon=3.20\times 10^{5}\times 3.785=12.112\times 10^5liter[/tex]

Specific gravity = 1.2

Density of water [tex]=1000kg/m^3[/tex]

We know that specific gravity [tex]=\frac{density\ of\ liquid}{density\ of\ water}[/tex]

[tex]1.2=\frac{density\ of\ liquid}{density\ of\ water}[/tex]

Density of liquid [tex]=1200kg/m^3[/tex]

So mass of liquid = volume × density

= [tex]12.112\times 10^5\times 1200=1.45\times 10^8kg[/tex]

So mass of the liquid in the tank will be [tex]1.45\times 10^8kg[/tex]

Answer:

Step-by-step explanation:

First let's identify decision variables:

X1 - acres of corn

X2 - acres of tobacco

Bradley needs to maximize the profit, MAX = 300X1 + 520X2

The Bradley family owns 410 acres, X1+X2≤410

Each acre of corn costs $105,  each acre of tobacco costs $210

The Bradleys have a budget of $52,500

So 105X1 +210X2≤52,500

There is a restriction on planting the tobacco - 100acres

X2≤100

Also, since outcomes can be only positive, X1X2 ≥0

So, what we have:

MAX = 300X1 + 520X2

X1+X2≤410

105X1 +210X2≤52,500

X2≤100

X1X2 ≥0

To maximize profit, the Bradleys can formulate a linear programming model based on the costs and profits of each crop, subject to budget and government constraints.

Linear programming model formulation:

Answer: a) margin of error = 61.25, b) sample size when margin of error is 45 = 30

Step-by-step explanation:

The formulae to get the margin of error of a confidence interval is given as

Margin of error = critical value * (σ/√n)

Where σ = population standard deviation = 125

n = sample size = 16

Critical value =Zα/2 = 1.96 ( this is so because we are performing a 95% confidence level test then level of significance (α) will be 5% and since our test is of two values, it will be 2 tailed).

Margin of error = 1.96 * (125/√16)

Margin of error = 1.96 * 125/4

Margin of error = 1.96 * 31.25

Margin of error = 61.25

Question b)

Margin of error = 45

Critical value =Zα/2 = 1.96

Population standard deviation = σ = 125

Sample size =n =??

By recalling the formulae

Margin of error = critical value * (σ/√n)

45 = 1.96 * (125/√n)

45 = (1.96 * 125)/√n

45 = 245/√n

45 * √n = 245

√n = 245/ 45

√n = 5.444

n = (5.444)²

n= 29.64 which is approximately 30.

Final answer:

The margin of error calculation for estimating the mean weight of male baluga whales, determining the necessary sample size, and confirming the requirements for a confidence interval.

Explanation:

The margin of error in estimating the true mean weight of male baluga whales in the Arctic Ocean can be calculated using the formula:

Margin of Error = Z * (Standard Deviation / √sample size)

Given Z value for 95% confidence interval is 1.96,

Margin of Error = 1.96 * (125 / √16) = 61.25 kg.

To find the required sample size for a margin of error of 45 kg:

The requirements for the use of a confidence interval are considered met when the distribution of sample means is normal.

In this case, the distribution of sample means is normal because the sample size is large, ensuring the validity of utilizing a confidence interval for estimation.

Answer:

the line equation is 2*x - y = 3

Step-by-step explanation:

for the line equation in implicit form

A*x + B*y = C

then if the point (x=4,y=5) belong to the line

4*A + 5*B = C

and  if the point (x=7,y=11) belong to the line

7*A + 11*B = C

then since we can choose C freely , we set C=1 for simplicity , then

4*A + 5*B = 1 → B= (1-4*A)/5

7*A + 11*B = 1

7*A + 11*(1-4*A)/5  = 1

7*A - 44/5*A + 11/5 = 1

-9/5*A = -6/5

A= 2/3

B= (1-4*A)/5 = (1-4*2/3)/5 = -1/3

therefore

2/3*x - 1/3*y = 1

2*x - y = 3

Final answer:

Events B and D are disjoint; events B and C, A and C, B and D are independent; events A and B, A and C, and B and D are neither disjoint nor independent.

Explanation:

The pairs of events that are disjoint (have no intersection) are:

Events B and D are disjoint because if the white die is odd (event B), then it cannot match with the red die (event D).

Events A and D are disjoint because if the sum of the dice is 7 (event A), then the dice cannot match (event D).

The pairs of events that are independent are:

Events B and C are independent because the outcome of one die does not affect the outcome of the other die.

Events A and C are independent for the same reason.

The pairs of events that are neither disjoint nor independent are:

Events A and B are neither disjoint nor independent. If the white die is odd (event B), it is still possible for the sum of the dice to be 7 (event A).

Events A and C are also neither disjoint nor independent. If the sum of the dice is 7 (event A), it is still possible for the red die to have a larger number showing than the white die (event C).

Events B and D are neither disjoint nor independent. If the white die is odd (event B), it is still possible for the dice to match (event D).

Answer:

The probability that more than 10 parts will be defective is 0.99989.

Step-by-step explanation:

Let X = a part in the shipment is defective.

The probability of a defective part is, P (Defect) = p = 0.03.

The size of the sample is: n = 1000.

Thus, the random variable [tex]X\sim Bin(1000, 0.03)[/tex].

But the sample size is very large.

The binomial distribution can be approximated by the Normal distribution if the following conditions are satisfied:

Check the conditions:

[tex]np=1000\times0.03=30>10\\n(1-p)=1000\times(1-0.03)=970>10[/tex]

Thus, the binomial distribution can be approximated by the Normal distribution.

The sample proportion (p) follows a normal distribution.

Mean: [tex]\mu_{p}=0.03[/tex]

Standard deviation: [tex]\sigma_{p}=\sqrt{\frac{p(1-p)}{n} } =\sqrt{\frac{0.03(1-0.03)}{1000} } =0.0054[/tex]

Compute the probability that there will be more than 10 defective parts in this shipment as follows:

The proportion of 10 defectives in 1000 parts is: [tex]p=\frac{10}{1000}=0.01[/tex]

The probability is:

[tex]P(p>0.01)=P(\frac{p-\mu_{p}}{\sigma_{p}}> \frac{0.01-0.03}{0.0054}) =P(Z>-3.704)=P(Z<3.704)[/tex]

Use the standard normal table for the probability.

[tex]P(p>0.01)=P(Z<3.704)=0.99989[/tex]

Thus, the probability that more than 10 parts will be defective is 0.99989.

Answer:

The systems of equation is CONSISTENT

Step-by-step explanation:

The detailed steps using crammers rule to ascertain the CONSISTENCY is as shown in the attached file

Answer:

After 167 miles the Budget rental deal would be better than the U-haul deal

Step-by-step explanation:

Let x be the number of miles.

Considering the U-haul plan the cost can be expressed as:

[tex]u=29.95+0.28x[/tex]

Consider the Budget rental plan the cost can be expressed as:

[tex]b=34.95+0.25x[/tex]

For the Budget rental plan to be better than the U-haul plan the relation between the two cost equation will be:

[tex]b<u[/tex]

Substitute the equations of u and b and solve for x:

[tex]b<u\\34.95+0.25x<29.95+0.28x\\34.95-29.95<0.28x-0.25x\\5<0.03x\\0.03x>5\\x>166.67\approx167[/tex]

So after 167 miles the Budget rental deal would be better than the U-haul deal.

After approximately 167 miles, Budget Truck rental becomes a cheaper option than U-Haul for Bryce and Lauren's move.

To find the point at which Budget Truck rentals becomes cheaper than U-Haul, we must set up the cost equation for each company and solve for the number of miles (m) that makes them equal. For U-Haul, the cost (C) is given by C = 29.95 + 0.28m, and for Budget it’s C = 34.95 + 0.25m.

Setting these two equations equal, we get: 29.95 + 0.28m = 34.95 + 0.25m.

We then solve for m: 0.03m = 5. Subtracting 29.95 from both sides, then dividing by 0.03 gives: m ≈ 167 miles.

So, after 167 miles, Budget Truck becomes the cheaper option.

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The water level in the bowl is rising at a rate of approximately 0.00269 meters per second when the water is 4 meters deep.

Let's address each part of the problem:

a. To find the volume of water in a hemispherical bowl of radius "a" to a depth "h," we can use the formula for the volume of a spherical cap. The volume of a spherical cap is given by:

V = (1/3)πh^2(3a - h)

In this case, "a" is the radius of the hemisphere, and "h" is the depth of the water.

So, the volume of water in the bowl is:

V = (1/3)πh^2(3a - h)

b. To find how fast the water level in the bowl is rising, we can use related rates. Let's denote the radius of the water-filled hemisphere as "R" (which is equal to "a" since it's the same hemisphere), and the depth of the water as "h."

Given that water is running into the bowl at a rate of 0.2 cubic meters per second, we can express the change in volume with respect to time:

dV/dt = 0.2 m^3/sec

We want to find dh/dt, the rate at which the water level is rising when the water is 4 meters deep.

We have the formula for the volume of water in the hemisphere from part (a):

V = (1/3)πh^2(3a - h)

Differentiate both sides of this equation with respect to time (t):

dV/dt = (1/3)π(2h)(dh/dt)(3a - h) - (1/3)πh^2(d(3a - h)/dt)

Now, plug in the values we know:

dV/dt = 0.2 m^3/sec

h = 4 m

a = 5 m

Now, solve for dh/dt:

0.2 = (1/3)π(24)(dh/dt)(35 - 4) - (1/3)π(4^2)(d(3*5 - 4)/dt)

0.2 = (8/3)π(dh/dt)(15 - 4) - (16/3)π(d(15 - 4)/dt)

0.2 = (8/3)π(11)(dh/dt) - (16/3)π(d(11)/dt)

0.2 = (88/3)π(dh/dt) - (16/3)π(0)

Now, solve for dh/dt:

(88/3)π(dh/dt) = 0.2

dh/dt = 0.2 / [(88/3)π]

Now, calculate dh/dt:

dh/dt ≈ 0.00269 meters per second

for such more question on hemispherical bowl

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Final answer:

To find the volume of water in a hemispherical bowl, subtract the volume of the upper portion of the hemisphere from the volume of the entire hemisphere. Therefore, the volume of the water in the hemispherical bowl is (2/3)πa³ - (1/3)π(h²)(3a - h).

Explanation:

To find the volume of water in a hemispherical bowl, we need to consider the shape of the bowl. The volume of a hemisphere is given by the formula V = (2/3)πr³, where r is the radius of the hemisphere. However, we only need to find the volume of the water in the bowl, not the entire bowl. To do this, we can subtract the volume of the upper portion of the hemisphere that is not filled with water.

Let's break down the steps:

Therefore, the volume of the water in the hemispherical bowl is (2/3)πa³ - (1/3)π(h²)(3a - h).

Answer:the equation representing his daily salary is

s = 100 + c

Step-by-step explanation:

Let s represent the Salary that Martins earns per day.

Let c represent the commission that Martins earns on his sales on that day.

He earns $100 per day, plus

any commission on his sales. Since

his daily salary in dollars depends on the amount of commission, then an equation to represent his daily salary would be

s = 100 + c

Answer: you need to add each side .45+45+90=180

Step-by-step explanation:

Answer:

wait time in order = 0,1,4,3,0,1,3,3,3,2

Step-by-step explanation:

We need to make a table for easy computation of wait time for each customer.

Customer    Arrival   Time taken      Checkout      Wait

Serial no.       time     for checkout      time             time

1                        1              4                    1+4=    5        0

2                       4             4                    4+5=    9       5-4=   1

3                       5             4                    9+4=   13       9-5=  4

4                       10            4                    13+4=  17       13-10=3

5                       20           3                    20+3= 23      0

6                       22            3                   23+3= 26      23-22=1

7                       23            5                   26+5= 31       26-23=3

8                       28            1                    31+1=   32      31-28=3

9                       29            5                   32+5= 37      32-29=3

10                     35                                                        37-35=2

Answer:

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, p = 0.40.

(1)

Let X = number of computers to be tested before the 1st defect is found.

Then the random variable [tex]X\sim Geo(p)[/tex].

The probability function of a Geometric distribution for k failures before the 1st success is:

[tex]P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...[/tex]

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

              = 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

              [tex]=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078[/tex]

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let Y = number of computers infected.

The number of computers in the company is, n = 20.

Then the random variable [tex]Y\sim Bin(20,0.40)[/tex].

The probability function of a binomial distribution is:

[tex]P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...[/tex]

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

             = 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)]               [tex]=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904[/tex]

Thus, the probability at least 5 computers are infected is 0.949.

Answer:

[tex]n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689[/tex]

So the answer for this case would be n=689 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: "Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.

Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year."

Solution for the problem

First we need ti find the estimation for the standard deviation using the Rule of thumb, with the following formula:

[tex] s \approx \frac{R}{4}[/tex]

Where R is the range defined as :

[tex] R = Max - Min = 8-0 = 8[/tex]

So then the deviation would be approximately:

[tex] s \approx 4[/tex]

Important concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error (ME) is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (1)

And on this case we have that [tex]ME =\pm 0.25[/tex] and we are interested in order to find the value of n, if we solve n from equation (1) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (2)

We can assume that the estimator for the population deviation from the rule of thumb is [tex]\hat \sigma = s= 4[/tex]

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.64[/tex], replacing into formula (2) we got:

[tex]n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689[/tex]

So the answer for this case would be n=689 rounded up to the nearest integer

To estimate textbook ages, an assumed range of 1 to 20 years might give an estimated standard deviation of 4.75 years. Using the formula for sample size, about 275 textbooks would need to be sampled to estimate the mean age to within 0.25 years with 90% confidence.

To estimate the minimum and maximum ages for typical textbooks currently used in college courses, we need more specific information. However, if we assume a range of 1 to 20 years, this would give us a range of 19 years. We can use the range rule of thumb for estimating the standard deviation, which is the range divided by 4. So the estimated standard deviation of the textbook ages is 19 / 4 = 4.75 years.

To find the required sample size to estimate the mean age of textbooks, we use the equation n = (Z*σ/E)^2 where Z is the z-score corresponding to the desired confidence level, σ is the standard deviation, and E is the maximum tolerable error. For a 90% confidence level, the z-score is 1.645. Thus with σ = 4.75 and E = 0.25. Filling these values into our equation we get: n = (1.645*4.75/0.25)^2 = 274.68. Rounded up, we need a sample of size 275 textbooks to estimate the mean age to within 0.25 years with 90% confidence.

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Answer:The statement is true

Step-by-step explanation:

This is an implication, then if we by starting from the premise we can get to the conclusion, then the implication is true. First let us remember that a given set of n vectors, say [tex]\{u_1,\ldots, u_n\}[/tex] is l.i (linearly independent), by definition if the only solution to [tex]\alpha_1\cdot u_1 + \cdots + \alpha_n\cdot u_n = \textbf{0}, is (\alpha_1, \ldots, \alpha_n)= \textbf{0}\in \mathcal {R}^n, \mathcal {R}[/tex] the set of real numbers, that is the only linear combination equal to the null vector is the null combination, all the scalars must be zero. And since it is a definition it is  if and only if.

Then if we say that the premise is true, then u4 is not a linear combination of {u1,u2,u3}, this means there do not exist [tex]\alpha_1, \alpha_2, \alpha_3 \in \mathcal{R}[/tex] not all zero, such that, [tex]\alpha_1\cdot u_1+ \alpha_2\cdot u_2+ \alpha_3\cdot u_3=u_4[/tex] which is equivalent to say that for every [tex]\alpha_1, \alpha_2, \alpha_3,\alpha_4 \in \mathcal{R}[/tex] with

[tex]\alpha_1\cdot u_1 + \alpha_2\cdot u_2+ \alpha_3\cdot u_3+ \alpha_4\cdot u_4 = \textbf{0}[/tex] implies [tex]\alpha_1= \alpha_2= \alpha_3= \alpha_4=0[/tex], then the given set is linearly independent as by definition and just as it is stated in the conclusion, therefore the affirmation is true.

Answer:

Step-by-step explanation:

The formula for finding the sum of the measure of the interior angles in a regular polygon is expressed as (n - 2) × 180.

Where

n represents the number of sides of the polygon.

5a) The polygon has 11 sides. Therefore, sum of angles is

(11 - 2) × 180 = 1620

The measure of each angle is

1620/11 = 148.3°

b) n = 24

Therefore, sum of angles is

(24 - 2) × 180 = 3960

The measure of each angle is

3960/24 = 165°

5a) n = 7

Therefore, sum of angles is

(7 - 2) × 180 = 900

The measure of each angle is

900/7 = 128.6°

5b) n = 10

Therefore, sum of angles is

(10 - 2) × 180 = 1440

The measure of each angle is

1440/10 = 144°