What is slope of line between each pair of points?
(1, -19) and (-2, -7)

The rate of change of the distance between the two cars at that instant is 100344 meters per second.

Here, we have,

To find the rate of change of the distance between the two cars at a certain instant, we can use the concept of relative velocity.

The relative velocity between the two cars is the difference between their velocities.

Since the cars are moving towards each other, the relative velocity will be the sum of their individual velocities.

Let's denote the distance between the two cars at the certain instant as

d (in meters). The rate of change of this distance d with respect to time (t) is the derivative of d with respect to t (i.e., d/dt).

At the given instant, the first car's velocity (V₁) is 101010 meters per second, and the second car's velocity (V₂) is 666 meters per second.

The distance traveled by the first car (d₁) at the given instant is 444 meters, and the distance traveled by the second car (d₂) is 333 meters.

The distance between the two cars at the given instant (d) is the difference between the distances traveled by the first and second cars:

d=d₁−d₂

Now, let's find the rate of change of d with respect to time (t):

d/dt = d₁/dt - d₂/dt

Since the velocities of the cars are constants (not changing with time), the derivatives of d₁ and d₂ with respect to time are simply their velocities:

d/dt =V₁−V₂

Now, substitute the given values for V₁ and V₂:

d/dt=101010meters per second−666meters per second

d/dt =100344meters per second

Therefore, the rate of change of the distance between the two cars at that instant is 100344 meters per second.

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The rate of change of the distance between the cars at the given instant is 1210 m/s. We used the Pythagorean theorem to express the distance between the two cars at any time 't' and then differentiated this distance with respect to 't' to get the relative speed of the cars.

To calculate the rate of change of the distance between the two cars at a certain instance, we can use the Pythagorean theorem. Consider the two cars as points moving along perpendicular paths. At any time 't', car 1 is '1010t' m away from the intersection and car 2 is '666t' m away from the intersection. The distance 'd' between them is √[(1010t)² + (666t)²] which simplifies to d = 1210t.

The rate of change of the distance with respect to time, i.e., the derivative of 'd' with respect to 't' (dd/dt) would be the relative speed of the two cars at any instant. This could be found by differentiating 'd' with respect to 't'. Thus, at the moment when car 1 is 444 meters away from the crossing and car 2 is 333 meters away from the crossing, the rate of change of the distance between the cars (the relative velocity) at that instant would be d/dt = 1210 m/s.

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When calculating A×B, the number of columns of A = number of rows of B.

Thus, only last case is possible.

Answer:

a) [tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X = 27.2[/tex]

b) For this case we have n =48 observations and we can calculate the median with the average between the 24th and 25th values on the dataset ordered.

20.4 20.4 20.5 21.7 22.3 23.3 23.6 23.7 23.7 23.7  23.9 23.9 24.1 24.3 24.7 24.9 25.1 25.1 25.1 25.2  25.3 25.6 26.1 26.1 26.4 26.5 26.7 27.1 27.1 27.4  27.6 28.4 28.6 28.6 28.7 28.8 28.8 29.6 31.0 32.0  32.0 32.4 32.5 32.9 33.1 34.5 38.4 44.1

For this case the median would be:

[tex] Median = \frac{26.1+26.4}{2}=26.25 \approx 26.3[/tex]

c) [tex] Mode= 23.2, 25.1[/tex]

And both with a frequency of 3 so then we have a bimodal distribution for this case

Step-by-step explanation:

For this case we have the following dataset:

23.6, 26.5, 28.6, 28.6, 23.7, 25.3, 24.9, 28.7, 26.7, 32.4, 20.4, 23.9, 32.0, 32.5, 23.7, 26.1, 21.7, 26.1, 38.4, 24.1, 20.5, 25.2, 31, 26.4, 27.1 ,25.1, 25.1, 23.7, 23.9, 32.9, 28.8, 25.6, 28.8, 28.4, 32, 27.6, 29.6, 44.1, 27.1, 24.7, 22.3, 24.3, 23.3, 27.4, 20.4, 25.1, 34.5, 33.1

Part a

We can calculate the mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X = 27.2[/tex]

Part b

For this case we have n =48 observations and we can calculate the median with the average between the 24th and 25th values on the dataset ordered.

20.4 20.4 20.5 21.7 22.3 23.3 23.6 23.7 23.7 23.7  23.9 23.9 24.1 24.3 24.7 24.9 25.1 25.1 25.1 25.2  25.3 25.6 26.1 26.1 26.4 26.5 26.7 27.1 27.1 27.4  27.6 28.4 28.6 28.6 28.7 28.8 28.8 29.6 31.0 32.0  32.0 32.4 32.5 32.9 33.1 34.5 38.4 44.1

For this case the median would be:

[tex] Median = \frac{26.1+26.4}{2}=26.25 \approx 26.3[/tex]

Part c

For this case the mode would be:

[tex] Mode= 23.2, 25.1[/tex]

And both with a frequency of 3 so then we have a bimodal distribution for this case

The mean commute time for the 48 cities is 25.6 minutes, the median commute time is 26.5 minutes and the mode of the commute times is 28.6 minutes.

To address the student's question about commute times, let's go through the calculations step by step:

(a) Mean Commute Time

The mean (average) is calculated by summing all the commute times and dividing by the number of cities (48).

Sum of all commute times = 23.6 + 26.5 + 28.6 + 28.6 + 23.7 + 25.3 + 24.9 + 28.7 + 26.7 + 32.4 + 20.4 + 23.9 + 32.0 + 32.5 + 23.7 + 26.1 + 21.7 + 26.1 + 38.4 + 24.1 + 20.5 + 25.2 + 31.0 + 26.4 + 27.1 + 25.1 + 25.1 + 23.7 + 23.9 + 32.9 + 28.8 + 25.6 + 28.8 + 28.4 + 32.0 + 27.6 + 29.6 + 44.1 + 27.1 + 24.7 + 22.3 + 24.3 + 23.3 + 27.4 + 20.4 + 25.1 + 34.5 + 33.1 = 1229.3 minutes

Mean = Sum / Number of cities = 1229.3 / 48 ≈ 25.6 minutes

(b) Median Commute Time

To find the median, list the commute times in numerical order and find the middle value. Since we have an even number of cities (48), the median will be the average of the 24th and 25th values.

Ordered list: 20.4, 20.4, 20.5, 21.7, 22.3, 23.3, 23.6, 23.7, 23.7, 23.9, 23.9, 24.1, 24.3, 24.7, 24.9, 25.1, 25.1, 25.1, 25.2, 25.3, 25.6, 26.1, 26.1, 26.4, 26.5, 26.7, 27.1, 27.1, 27.4, 27.6, 28.4, 28.6, 28.6, 28.7, 28.8, 28.8, 29.6, 31.0, 32.0, 32.0, 32.4, 32.5, 32.9, 33.1, 34.5, 38.4, 44.1

The 24th and 25th values are 26.4 and 26.5.

Median = (26.4 + 26.5) / 2 = 26.45 ≈ 26.5 minutes

(c) Mode(s) Commute Time

The mode is the value that appears most frequently in the list.

From the list, we see that the most frequently occurring value is 28.6, which occurs 2 times.

Therefore, the mode is 28.6 minutes.

So, (a) the mean commute time is 25.6 minutes, (b) the median commute is 26.5 minutes and (c) the mode of the commute times is 28.6 minutes.

Answer:

Step-by-step explanation:

Hello!

You have two events.

A: The employee is bilingual.

The probability of the employee being bilingual is P(A)= 30/85= 0.35

And

B: The employee has a graduate degree.

Additionally, you know that the probability of an employee having a graduate degree given that he is bilingual is:

P(B/A)= 0.37

You need to calculate the probability of the employee being bilingual and having a graduate degree. This is the intersection between the two events, symbolically:

P(A∩B)

The events A and B are not independent, which means that the occurrence of A modifies the probability of occurrence of B.

Applying the definition of conditional probability you have that:

P(B/A)= [P(A∩B)]/P(A)

From this definition, you can clear the probability of the intersection between A and B

P(A∩B)= P(B/A)* P(A)= 0.37*0.35= 0.1295≅ 0.13

I hope it helps!

Final answer:

The probability that a randomly selected employee from the division is bilingual and has a graduate degree is approximately 12.9%.

Explanation:

The question asks for the probability that a randomly selected employee from a division with 85 employees is bilingual and has a graduate degree. We know that 30 employees are bilingual, and 37% of these bilingual employees have a graduate degree. To find this probability, we will perform a two-step calculation:

First, calculate the number of bilingual employees with a graduate degree by taking 37% of 30: number of bilingual employees with graduate degrees = 0.37 × 30.

Then, divide this number by the total number of employees to get the probability: probability that an employee is bilingual with a graduate degree = (number of bilingual employees with graduate degrees) / 85.

Now let's calculate the values:

0.37 × 30 = 11.1. Since we cannot have a fraction of a person, we'll round down to 11 bilingual employees with a graduate degree.

(11 / 85) = approximately 0.129, or 12.9%.

So, the probability that a randomly selected employee is bilingual and has a graduate degree is roughly 12.9%.

Answer:

Probability of having two hits in the same region = 0.178

mu: average number of hits per region

x: number of hits

e: mathematical constant approximately equal to 2.71828.

Step-by-step explanation:

We can describe the probability of k events with the Poisson distribution, expressed as:

[tex]P(x=k)=\frac{\mu^ke^{-\mu}}{k!}[/tex]

Being μ the expected rate of events.

If 535 bombs hit 553 regions, the expected rate of bombs per region (the events for this question) is:

[tex]\mu=\frac{\#bombs}{\#regions} =\frac{535}{553}= 0.9674[/tex]

For a region to being hit by two bombs, it has a probability of:

[tex]P(x=2)=\frac{\mu^2e^{-\mu}}{2!}=\frac{0.9674^2e^{-0.9674}}{2!}=\frac{0.9359*0.38}{2}=0.178[/tex]

Answer:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:  

[tex]n=(\frac{1.96(0.25)}{0.1})^2 =24.01 [/tex]  

So the answer for this case would be n=25 rounded up to the nearest integer  

Step-by-step explanation:

Previous 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 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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma=0.25[/tex] represent the population standard deviation

n represent the sample size (variable of interest)  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]  

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 ME =0.1 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)  

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got [tex]z_{\alpha/2}=1.96[/tex], replacing into formula (2) we got:  

[tex]n=(\frac{1.96(0.25)}{0.1})^2 =24.01 [/tex]  

So the answer for this case would be n=25 rounded up to the nearest integer  

Final answer:

To estimate the mean circumference of soccer balls within 0.1 in at a 95% confidence level, with a population standard deviation of 0.25 in, the minimum sample size required is 24.

Explanation:

The question involves estimating the minimum sample size required for a 95% confidence interval with a given precision and known population standard deviation. In this scenario, the soccer ball manufacturer wants to ensure that the mean circumference of soccer balls is estimated within 0.1 in, given a population standard deviation of 0.25 in. To find the minimum sample size, we use the formula: n = (Z*σ/E)², where Z is the Z-score corresponding to a 95% confidence level (~1.96), σ is the population standard deviation (0.25 in), and E is the margin of error (0.1 in). Plugging in the values gives us: n = (1.96*0.25/0.1)² = 23.04. Since we can't have a fraction of a sample, we round up to get a minimum sample size of 24.

Answer:for FRR method the difference equation is given by:

u(k+1)=0.8125u(k)+(-24.68)e(k)+25e (k+1)

Step-by-step explanation:see the pictures attached

Answer:

0% probability cab tires depths would be shallower than 0.25cm.

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 2.2, \sigma = 0.33[/tex]

What is the probability cab tires depths would be shallower than 0.25cm.

This probability is the pvalue of Z when X = 0.25. So

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

[tex]Z = \frac{0.25 - 2.2}{0.33}[/tex]

[tex]Z = -5.9[/tex]

[tex]Z = -5.9[/tex] has a pvalue of 0.

0% probability cab tires depths would be shallower than 0.25cm.

Final answer:

The Z-score for a tire tread depth of 0.25 cm when compared to the city's average tread depth is -5.91, which suggests an extremely low probability, nearly zero, of finding a tire tread depth shallower than 0.25 cm.

Explanation:

To determine the probability that the tire tread depth for cab tires is shallower than 0.25 cm, we will use the standard normal distribution and the process of finding a Z-score. We are given the population mean (μ) to be 2.2 cm and the population standard deviation (σ) to be 0.33 cm.

The Z-score is calculated as:

Z = (X - μ) / σ

Where:

So, the Z-score for 0.25 cm is:

Z = (0.25 - 2.2) / 0.33 = -5.91

A Z-score of -5.91 is significantly beyond the typical range for a standard normal distribution table, which indicates that the probability of having a tire tread depth of less than 0.25 cm is extremely low, essentially approaching zero.

Answer:

i) [tex] P(9.5 < X<11)[/tex]

And we can solve this problem using 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(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]

And we can find this probability with this difference:

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]

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

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]

ii) The z scores for this case are:

[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]

[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]

And we can check the answer with the following excel code:

=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)

iii) [tex]P(X>a)=0.975[/tex]   (a)

[tex]P(X<a)=0.025[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.96<\frac{a-10}{1.414}[/tex]

And if we solve for a we got

[tex]a=10 -1.96*1.414=7.228[/tex]

So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.

iv) [tex] z = \frac{7.228-10}{1.414}= -1.96[/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 X the random variable that represent the blade length of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(10,\sqrt{2})[/tex]  

Where [tex]\mu=10[/tex] and [tex]\sigma=1.414[/tex]

Part i

For this case we want this probability:

[tex] P(9.5 < X<11)[/tex]

And we can solve this problem using 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(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]

And we can find this probability with this difference:

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]

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

[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]

Part ii

The z scores for this case are:

[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]

[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]

And we can check the answer with the following excel code:

=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)

Part iii

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.975[/tex]   (a)

[tex]P(X<a)=0.025[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.96<\frac{a-10}{1.414}[/tex]

And if we solve for a we got

[tex]a=10 -1.96*1.414=7.228[/tex]

So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.

Part iv

The z score for this value is given by:

[tex] z = \frac{7.228-10}{1.414}= -1.96[/tex]

Final answer:

The question involves calculating probabilities and finding values within a normal distribution regarding the length of grass blades. It covers finding specific probabilities, standardizing values, and locating a value below which a certain percentage of data lies, all based on a given mean and variance.

Explanation:

A meticulous gardener is interested in the length of blades of grass on his lawn. He believes that blade length X follows a normal distribution centered on 10 mm with a variance of 2 mm.

Find the probability that a blade of grass is between 9.5 and 11 mm long.

What are the standardized values of 9.5 and 11 in the context of this distribution? Using the standardized values, confirm that you can obtain the same probability you found in (i) with the standard normal density.

Below which value is the shortest 2.5 percent of blade lengths found?

Standardize your answer from (iii).

The standard deviation (sqrt(variance)) is sqrt(2) mm. The standardized value, or z-score, is computed as Z = (X - μ)/σ, where X is the value, μ is the mean (10 mm), and σ is the standard deviation.

For X = 9.5, Z = (9.5 - 10) / sqrt(2) = -0.3536.

For X = 11, Z = (11 - 10) / sqrt(2) = 0.7071.

To find the probability between 9.5 and 11 mm, we look up these z-scores in the standard normal distribution table or use a calculator.

To find the value below which the shortest 2.5 percent of blade lengths are found, we look up the z-score that corresponds to the cumulative area of 0.025 in the standard normal distribution table. Then, we use the z-score formula in reverse to find the original value in mm.

[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-14}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-2)}}}\implies \cfrac{-24}{10+2}\implies \cfrac{-24}{12}\implies -2[/tex]

[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-2}[x-\stackrel{x_1}{(-2)}]\implies y-10=-2(x+2) \\\\\\ y-10=-2x-4\implies y=-2x+6[/tex]

Answer:

Step-by-step explanation:

a.

The initial length of the candle is 16 inch. It also given that, it burns with a constant rate of 0.8 inch per hour.

After one hour since the candle was lit, the length of the candle will be (16 - 0.8) = 15.2 inch.

After two hour since the candle was lit, the length of the candle will be (15.2 - 0.8) = 14.4 inch. The length of the candle after two hours can also be represented by {16 - 2(0.8)}.

Hence, the length of the candle after t hours when it was lit can be represented by the function, [tex]f(t) = 16 - 0.8t[/tex]. [tex]f(t) = 0[/tex] at t = 20.

b.

The domain of the function is 0 to 20.

c.

The range is 0 to 16.

The function formula for the given context is f(t) = 16 - 0.8*t. The domain of this function is from 0 to 20 hours, and the range is from 0 to 16 inches.

The function you want to define for the scenario presented represents the remaining length, in inches, of the candle after being lit for a certain number of hours (t). Given that the initial length of the candle is 16 inches and burns at the constant rate of 0.8 inches per hour, the function will follow a linear format: f(t) = 16 - 0.8*t.

The domain of this function, referring to the permissible values for t in this context, would be any value greater than or equal to 0 and less than or equal to 20 (since it would take 20 hours for the candle to completely burn down). Therefore, the domain of this function is [0, 20].

The range of this function will be the possible lengths the candle could have, depending on the time it has burned. As the candle was 16 inches long initially and reduces to 0 after burning for 20 hours, the range lies between 0 and 16 (inclusive). Thus, the range of this function is [0, 16].

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Answer:

P(T∩E) = 0.017

Step-by-step explanation:

Since every fourth CD is tested. Thus if T is the event that represents 4 disks being tested,

P(T) = 1/4 = 0.25

Let Fi represent event of failure rate. So from the question,

P(F1) = 0.01 ; P(F2) = 0.03 ; P(F3) =0.02 ; P(F4) = 0.01

Also Let F'i represent event of success rate. And we have;

P(F'1) = 1 - 0.01 = 0.99 ; P(F'2) = 1 - 0.03 = 0.97; P(F'3) = 1 - 0.02 = 0.98; P(F'4) = 1 - 0.01 = 0.99

Since all programs run independently, the probability that all programs will run successfully is;

P(All programs to run successfully) =

P(F'1) x P(F'2) x P(F'3) x P(F'4) =

0.99 x 0.97 x 0.98 x 0.97 = 0.932

Now, that all 4 programs failed will be = 1 - 0.932 = 0.068

Let E be denote that the CD fails the test. Thus P(E) = 0.068

Now, since testing and CD's defection are independent events, the probability that one CD was tested and failed will be =P(T∩E) = P(T) x P(E)= 0.25 x 0.068 = 0.017

Final answer:

The probability that a CD fails any of the four independent tests, given individual failure rates of 0.01, 0.03, 0.02, and 0.01, is approximately 6.88%.

Explanation:

To calculate the probability that a CD fails any test, we should first understand that the probability of failing any particular test is the same as 1 minus the probability of passing that test. Given the failure rates of 0.01, 0.03, 0.02, and 0.01 for the four independent tests, the probabilities of a CD passing each test are 0.99, 0.97, 0.98, and 0.99, respectively.

The probability that the CD passes all four tests is the product of the individual probabilities of passing each test (since the tests are independent):

Subtracting this from 1 gives the probability that a CD fails at least one test:

Let's perform the calculation:

P(pass all tests) = 0.99 * 0.97 * 0.98 * 0.99 ≈ 0.9312

P(fail any test) = 1 - 0.9312 ≈ 0.0688

Therefore, the probability that a CD fails any test is approximately 0.0688 or 6.88%.

Answer:

(a) The PMF of X is: [tex]P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....[/tex]

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let X = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable X follows a Geometric distribution.

The probability mass function of X is:

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

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of X is:

[tex]P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....[/tex]

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

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

              [tex]=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64[/tex]

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

[tex]E(X)=\frac{1}{p}[/tex]

Compute the expected number of opponents contested in a game as follows:

[tex]E(X)=\frac{1}{p}=\frac{1}{0.20}=5[/tex]

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

              = 1 - P (X = 1) - P (X = 2) - P (X = 3)

              [tex]=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512[/tex]

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

[tex]E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2[/tex]

Thus, the expected number of game plays until a player contests four or more opponents is 2.

Answer:

yes its C

stop scrolling !!

Step-by-step explanation:

2020 edg

In between the given options expression 6⁻ᵃ  in the exponential form, So Option A is correct

An exponential function is a mathematical function which we write as a

f(x) = aˣ, where a is constant and x is variable term. The most commonly used exponential function is eˣ , where e is constant having value 2.7182

Given that,

The expressions,

6⁻ᵃ

6a

a-6

6a-1

Exponential form is a mathematical representation where a number (the base) is raised to a power (the exponent).

The exponent tells us how many times the base is multiplied by itself. For example, in the expression "6a", 6 is the base and "a" is the exponent.

This means that the expression can be written as 6 * 6 * 6 * ... * 6,

where there are "a" number of 6's being multiplied together.

The expression "6⁻ᵃ" is in exponential form.

To know more about exponential function check:

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Answer:

$0.09 tax per every dollar spent.

Step-by-step explanation:

45.78-42 which gives you the amount of money added on by the tax. which = $3.78 then divide that buy how much the game cost before tax was added to get the amount of tax per each $$$

Answer:

$0.09 tax per every dollar spent.

Step-by-step explanation:

You take the $45.78, and subtract that by $42 (the original cost) and then get $3.78. Then divide by every dollar spent, and you get the final rate, $0.09 tax rate.

Answer:

Step-by-step explanation:

Let us assume that the test scores of the students were normally distributed. we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = test scores of students.

µ = mean test scores

σ = standard deviation

From the information given,

µ = 510

σ = 100

We want to find the probability that a student scored below 300. It is expressed as

P(x ≤ 300)

For x = 300

z = (300 - 510)/100 = - 2.1

Looking at the normal distribution table, the probability corresponding to the z score is 0.018

Therefore, the percentage of students that scored below 300 is

0.018 × 100 = 1.8%

Answer:

(a) The probability that the inspection procedure will pass the shipment is 0.9185.

(b) The expected number of defectives in this process of inspecting 5 items is 0.50.

(c) The probability that there will be 4 defectives in a sample of 5 is 0.0016.

Step-by-step explanation:

Let X = number of defective items.

The probability of selecting a defective item is, p = 0.10.

A sample of n = 5 items is selected at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.10.

The probability mass function of X is:

[tex]P(X=x)={5\choose x}0.10^{x}(1-0.10)^{5-x};\ x=0,1,2,3....[/tex]

It is provided that the shipment will pass if there are no more than 1 defective items in the selected 5 units.

(a)

Compute the probability that the inspection procedure will pass the shipment as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             [tex]={5\choose 0}0.10^{0}(1-0.10)^{5-0}+{5\choose 1}0.10^{1}(1-0.10)^{5-1}\\=(1\times 1\times 0.59049) + (5\times 0.10\times 0.6561)\\=0.59049+0.32805\\=0.91854\\\approx0.9185[/tex]

Thus, the probability that the inspection procedure will pass the shipment is 0.9185.

(b)

The expected value of a Binomial distribution is:

[tex]E(X)=np[/tex]

Compute the expected number of defectives in this process of inspecting 5 items as follows:

[tex]E(X)=5\times0.10=0.50[/tex]

Thus, the expected number of defectives in this process of inspecting 5 items is 0.50.

(c)

The probability of finding a defective is now changed to 0.20.

Compute the probability that there will be 4 defectives in a sample of 5 as follows;

[tex]P(X=4)={5\choose 4}0.20^{4}(1-0.20)^{5-4}\\=5\times0.0016\times0.20\\=0.0016[/tex]

Thus, the probability that there will be 4 defectives in a sample of 5 is 0.0016.

Answer:

(a) Probability that the inspection procedure will pass the shipment = 0.91854

(b) E(X) = [tex]5 \times 0.1[/tex] = 0.5

(c) Probability of 4 defectives in a sample of 5 is 0.0064      

Step-by-step explanation:

We are given that a company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall.

The above situation can be represented through Binomial distribution;

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

where, n = number of trials (samples) taken = 5 items

            r = number of success

           p = probability of success which in our question is % of items that are

                  defective, i.e., 10%

LET X = Number of defective items

Also, it is given that a sample of 5 items is taken,

So, it means X ~ [tex]Binom(n=5,p=0.10)[/tex]

(a) Probability that the inspection procedure will pass the shipment is given by the fact that if no more than 1 item is found to be defective, then only shipment is passed, i.e.;

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

             = [tex]\binom{5}{0}0.1^{0} (1-0.1)^{5-0} + \binom{5}{1}0.1^{1} (1-0.1)^{5-1}[/tex]

             = [tex]1 \times 1 \times 0.9^{5} +5 \times 0.1 \times 0.9^{4}[/tex]

             = 0.59049 + 0.32805 = 0.91854

(b) The expected number of defectives in this process of inspecting 5 items is given by = E(X) or Mean of X

So, mean of binomial distribution is, E(X) = [tex]n \times p[/tex]

      So,  E(X) = [tex]5 \times 0.1[/tex] = 0.5

Therefore, the expected number of defectives in this process of inspecting 5 items is 1 (after rounding to nearest integer).

(c) Now, the probability of success which is % of items that are defective is changed to 20% rate overall, i.e., p = 0.20 now.

So, probability that you will find 4 defectives in a sample of 5 = P(X = 4)

 P(X = 4) = [tex]\binom{5}{4}0.2^{4} (1-0.2)^{5-4}[/tex]

               =  [tex]5 \times 0.2^{4} \times 0.8^{1}[/tex] = 0.0064        

Answer:

The code is as given below to be copied in a new matlab script m file. The screenshots are attached.

Step-by-step explanation:

As the question is not complete, the complete question is attached herewith.

The code for the problem is as follows:

%Defining the given matrices:

%P is the matrix showing the percentage of changes in voterbase

P = [ 0.8100 0.0800 0.1600 0.1000;

0.0900 0.8400 0.0500 0.0800;

0.0600 0.0400 0.7400 0.0400;

0.0400 0.0400 0.0500 0.7800];

%x0 is the vector representing the current voterbase

x0 = [0.5106; 0.4720; 0.0075; 0.0099];

%In MATLAB, the power(exponent) operator is defined by ^

%After 3 elections..

x3 = P^3 * x0;

disp("The voterbase after 3 elections is:");

disp(x3);

%After 6 elections..

x3 = P^6 * x0;

disp("The voterbase after 6 elections is:");

disp(x3);

%After 10 elections..

x10 = P^10 * x0;

disp("The voterbase after 10 elections is:");

disp(x10);

%After 30 elections..

x30 = P^30 * x0;

disp("The voterbase after 30 elections is:");

disp(x30);

%After 60 elections..

x60 = P^60 * x0;

disp("The voterbase after 60 elections is:");

disp(x60);

%After 100 elections..

x100 = P^100 * x0;

disp("The voterbase after 100 elections is:");

disp(x100);

The output is as well as the code in the matlab is as attached.

Answer:

The voter-base after 3 elections is:

0.392565, 0.400734, 0.109855, 0.096846

The voter-base after 6 elections is:

0.36168, 0.36294, 0.14176, 0.13362

The voter-base after 10 elections is:

0.35405, 0.34074, 0.15342, 0.15178

Step-by-step explanation:

This question is incomplete. I will proceed to give the complete question. Then I will add a screenshot of my code solution to this question. After which I will give the expected outputs.

Let's use the results of the 2012 presidential election as our x0. Looking up the popular vote totals, we find that our initial distribution vector should be (0.5106, 0.4720, 0.0075, 0.0099)T. Enter the matrix P and this vector x0 in MATLAB:

P = [ 0.8100 0.0800 0.1600 0.1000;

0.0900 0.8400 0.0500 0.0800;

0.0600 0.0400 0.7400 0.0400;

0.0400 0.0400 0.0500 0.7800];

x0 = [0.5106; 0.4720; 0.0075; 0.0099];

According to our model, what should the party distribution vector be after three, six and ten elections?

Please find the code solution in the images attached to this question.

The voter-base after 3 elections is therefore:

0.392565, 0.400734, 0.109855, 0.096846

The voter-base after 6 elections is therefore:

0.36168, 0.36294, 0.14176, 0.13362

The voter-base after 10 elections is therefore:

0.35405, 0.34074, 0.15342, 0.15178

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: number of bars of service on your cell phone.

The posible values of this variable are {0, 1, 2, 3, 4, 5}

a. F(X) is the cummulative distribution function P(X≤x₀) you can calculate it by adding all the point probabilities for each value:

X: 0; 1; 2; 3; 4; 5

P(X): 0.15; 0.25; 0.25; 0.15; 0.1; 0.1

F(X): 0.15; 0.4; 0.65; 0.8; 0.9; 1

The maximum cumulative probability of any F(X) is 1, knowing this, you can calculate the probability of X=5 as: 1 - F(4)= 1 - 0.9= 0.1

b.

The mean of the sample is:

E(X)= ∑Xi*Pi= (0*0.15)+(1*0.25)+(2*0.25)+(3*0.15)+(4*0.1)+(5*0.1)

E(X)= 2.1

V(X)= E(X²)-(E(X))²

V(X)= ∑Xi²*Pi- (∑Xi*Pi)²= 6.7- (2.1)²= 2.29

∑Xi²*Pi= (0²*0.15)+(1²*0.25)+(2²*0.25)+(3²*0.15)+(4²*0.1)+(5²*0.1)= 6.7

c.

P(X<2)

You can read this expression as the probability of having less than 2 bars of service. This means you can have either one or zero service bars, you can rewrite it as:

P(X<2)= P(X≤1)= F(1)= 0.4

d.

P(X≤3.5)

This expression means "the probability of having at most (or less or equal to) 3.5 service bars", this variable doesn't have the value 3.5 in its definition range so you have to look for the accumulated probability until the lesser whole number. This expression includes the probabilities of X=0, X=1, X=2, and X=3, you can express it as the accumulated probability until 3, F(3):

P(X≤3.5)= P(X≤3)= F(3)= 0.80

I hope it helps!

Answer:

Step-by-step explanation:

Given that a study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 P.M. on weekdays there is an average of four customers waiting in line.

Let X be the number of customers waiting in line

X is Poisson with parameter = 4

the probability that you visit Safeway today during this period and find:

a. No customers are waiting

[tex]P(X=0) = 0.018316[/tex]

b. Four customers are waiting?

[tex]=P(x=4) =0.193567[/tex]

c. Four or fewer are waiting?

=[tex]P(X\leq 4) = 0.628837[/tex]

d. Four or more are waiting

=[tex]P(X\geq 4)=0.56653[/tex]

Answer:

For the exponential distribution:

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

[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]

[tex] \mu_{\bar X} = \bar X[/tex]

[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

For this case we have a large sample size n =60 >30

The exponential distribution is the probability distribution that describes the time between events in a Poisson process.

For the exponential distribution:

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

[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]

We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.

[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]

[tex] \mu_{\bar X} = \bar X[/tex]

[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]

Answer:

0.8894 is the probability that the test result comes back negative if the disease is present .

Step-by-step explanation:

We are given the following in the question:

P(Disco Fever) = P( Disease) =

[tex]P(D) = \dfrac{1}{2}\times 1\% = 0.5 \times 0.01 = 0.005[/tex]

Thus, we can write:

P(No  Disease) =

[tex]P(ND) =1 - P(D)= 0.995[/tex]

P(Test Positive given the presence of the disease) = 0.99

[tex]P(TP | D ) = 0.99[/tex]

P( false-positive) = 4%

[tex]P( TP | ND) = 0.04[/tex]

We have to evaluate the probability that the test result comes back negative if the disease is present, that is

P(test result comes back negative if the disease is present)

By Bayes's theorem, we can write:

[tex]P(ND|TP) = \dfrac{P(ND)P(TP|ND)}{P(ND)P(TP|ND) + P(D)P(TP|D)}\\\\P(ND|TP) = \dfrac{0.995(0.04)}{0.995(0.04) + 0.005(0.99)} = 0.8894[/tex]

0.8894 is the probability that the test result comes back negative if the disease is present .

Answer:

A) Total frequency = 30

B) Yes, the distribution is consistent with the label because the frequencies are greatest when the lengths are closest to the labeled size of 1/2 inches which is 0.5 inches.

Step-by-step explanation:

Since the class limit width is 0.010 from the question, we arrive at;

Class Limits Frequency

Length(Inches)

0.470 - 0.479 3

0.480 - 0. 489 5

0.490 - 0.499 9

0.500 - 0.509 12

0.510 - 0.519 1

Adding the frequency, total = 30

The frequency distribution for the screws shows that most lengths are around 0.5 inches, consistent with the label. The classes and frequencies support this observation. The analysis is based on screw measurements provided.

To construct a frequency distribution of the screw lengths, we start with a lower class limit of 0.470 inches and use a class width of 0.010 inches. Now, let's count the number of screws in each class:

The frequency distribution is consistent with the given label of 0.5 inches as most of the data are centered around the 0.5 inch length.

Answer:

Step-by-step explanation:

You can find your answer in attached document.

The number of passwords is the sum from i=12 to 16 of 66^i minus the sum from i=12 to 16 of 52^i, representing all combinations of characters from length 12 to 16, excluding all-letter combinations.

The subject of your question pertains to combinatorics, specifically about counting password possibilities. Given that we have 66 different characters (10 digits, 52 letters lower and upper-case and 10 special characters), we have 66^12 to 66^16 possibilities for the length. However, we need to exclude the passwords composed strictly of letters. There are 52 letter characters, so we have 52^12 to 52^16 such passwords. Hence, the number of permitted passwords, in un-evaluated/un-simplified form, is the sum from i=12 to 16 of 66^i minus the sum from i=12 to 16 of 52^i

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Answer:

Step-by-step explanation:

Lets take 1000 output in total.

From these 1000 outputs, 700 are from F1 and 300 are from F2.

Defective from F1 is [tex]\frac{700\times3}{100} = 21[/tex] and defective from F2 is [tex]\frac{300\times2}{100} = 6[/tex].

a.

Total received part is 1000.

Total defectives are (21+6) = 27.

The probability of a received part being defective is [tex]\frac{27}{1000}[/tex].

b.

The probability of the randomly chosen defective part from F1 is [tex]\frac{21}{27} = \frac{7}{9}[/tex].

The probability that a received part is defective is 2.7%, while the probability that a randomly chosen defective part came from factory F1 is 77.78%.

To find the probability that a received part is defective, we need to consider the probabilities of receiving a defective part from each factory. Let's assume that the VCR manufacturer receives 100 parts. From factory F1, 70% of the parts are received, which means 70 parts. The probability of a part from F1 being defective is 3%, so the number of defective parts from F1 is 70 * (3/100) = 2.1. From factory F2, which accounts for the remaining 30% of the parts (30 parts), the probability of a part being defective is 2%, resulting in 30 * (2/100) = 0.6 defective parts. Therefore, the total number of defective parts is 2.1 + 0.6 = 2.7. Since there are 100 parts in total, the probability that a received part is defective is 2.7/100 = 0.027, or 2.7%.

To find the probability that a randomly chosen defective part came from factory F1, we can use the concept of conditional probability. The probability of a part coming from F1 given that it is defective can be found using the formula P(F1|Defective) = P(F1 and Defective) / P(Defective). We already know that P(F1 and Defective) = 2.1/100 and P(Defective) = 2.7/100. Substituting these values, we get P(F1|Defective) = (2.1/100) / (2.7/100) = 2.1/2.7 = 0.7778, or 77.78%.

Answer: Mathematically Bayes’ theorem is defined as

P(A\B)=P(B\A) ×P(A)

P(B)

Bayes theorem is defined as where A and B are events, P(A|B) is the conditional probability that event A occurs given that event B has already occurred (P(B|A) has the same meaning but with the roles of A and B reversed) and P(A) and P(B) are the marginal probabilities of event A and event B occurring respectively.

Step-by-step explanation: for example, picking a card from a pack of traditional playing cards. There are 52 cards in the pack, 26 of them are red and 26 are black. What is the probability of the card being a 4 given that we know the card is red?

To convert this into the math symbols that we see above we can say that event A is the event that the card picked is a 4 and event B is the card being red. Hence, P(A|B) in the equation above is P(4|red) in our example, and this is what we want to calculate. We previously worked out that this probability is equal to 1/13 (there 26 red cards and 2 of those are 4's) but let’s calculate this using Bayes’ theorem.

We need to find the probabilities for the terms on the right-hand side. They are:

P(B|A) = P(red|4) = 1/2

P(A) = P(4) = 4/52 = 1/13

P(B) = P(red) = 1/2

When we substitute these numbers into the equation for Bayes’ theorem above we get 1/13, which is the answer that we were expecting.

Bayes’ theorem, featuring components of likelihood and prior probability, is fundamental to probability theory and statistics. It's a tool that allows us to update our previous assumptions based on new data, making it essential in machine learning.

Bayes’ theorem is a fundamental theorem in probability theory and statistics, named after Thomas Bayes. It formalizes the process of updating probabilities based on new data. The theorem can be derived from the definition of conditional probability, P(A|B), which is the probability of event A given that event B has occurred.

Bayes' theorem is generally represented by the formula:

P(A|B) = (P(B|A) * P(A)) / P(B)

Bayes’ theorem has two main components:

In the realm of Machine Learning, Bayes' theorem allows us to update our previous assumptions (priors) with data to get updated, more accurate results (posterior). The prior and posterior are probability distributions over the same event space, but Bayesian inference allows us to adjust our initial assumptions in light of new evidence, making it a dynamic and flexible method.

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Answer:

The range of GPA of 95% students is from 1.9 to 3.18.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 3.04

Standard deviation = 0.57

Assuming that students' GAPs are normally distributed, what is the range of GPA of 95% students

Within 2 standard deviations of the mean

3.04 - 2*0.57 = 1.9

3.04 + 2*0.57 = 3.18

The range of GPA of 95% students is from 1.9 to 3.18.

Answer:

(a) Discrete

(b) Discrete

(c) Discrete

(d) Discrete

(e) Discrete

(f) Continuous

Step-by-step explanation:

A discrete random variable considers a finite set of values. Whereas a continuous random viable assumes infinite set of values.

One cannot count the values of a continuous random variable.

(a)

X = The number of visitors to the Museum of Science in Boston on a randomly selected day.

The number of visitors can be counted and on a selected day a finite number of people will visit the museum.

So this is a Discrete random variable.

(b)

X = The camber-angle adjustment necessary for a front-end alignment.

The angle measurements are countable values.

So this is a Discrete random variable.

(c)

X = The total number of pixels in a photograph produced by a digital camera.

The number of pixels in a photograph can be counted and fixed values.

So this is a Discrete random variable.

(d)

X = The number of days until a rose begins to wilt after purchase from a flower shop.

Roses usually wilt after 2 or 3 days from the date of purchase.

So this is a Discrete random variable.

(e)

X = The running time for the latest James Bond movie.

A movie is usually, on average, 96.5 minutes long. The number of minutes can be counted.

So this is a Discrete random variable.

(f)

X = The blood alcohol level

The blood alcohol level assumes values in a fixed interval. The values cannot be counted as there are infinite number of values in this interval.

So this is a Continuous random variable.

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

[tex]\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3[/tex]

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}[/tex]

Now, we calculate the sum

[tex]\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\[/tex]

Therefore, we get

[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98[/tex]

So, the sample standard deviation is 6.98.

Answer:

(35,0)

Step-by-step explanation:

Consider the diagram below, the starting point is given as A and the finish point given as C.

Using similar right-angle triangle, we have that:

[tex]\frac{|AM|}{|BM|}= \frac{|BM|}{|MC|}\\\frac{7}{14}= \frac{14}{x}\\7x=14 X 14\\x=196/7=28[/tex]

Therefore to find the point where Derek stops at C, we first determine the distance |AC|

|AC|=7+28=35

The Coordinates at C where Derek stops is (35,0)

Derek walks along a road described by the equation y = 2x. At the point (7,14), he turns right and walks perpendicularly to his original path until he reaches the x-axis. Upon reaching the x-axis, his stopping point is at (35,0).

When Derek reached the point (7,14), he turned right and started walking perpendicular to the original road. Given that this road is represented by the linear equation y = 2x, a perpendicular path would have a negative reciprocal slope. Therefore, the path he took after turning is represented by y = -1/2x + b. As he turned at the point (7, 14), substituting these coordinates into the equation provides b = 17.5. Since he stopped on the x-axis where y = 0, putting this into the equation gives x = 35. So, Derek stopped at (35,0).

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