4.36 Stats final scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) Is the distribution of scores on this final exam symmetric, right skewed, or left skewed

Answer:

(a) 35 heads correspond to -3 on the standard scale.

(b) z = 2.4 corresponds to 62 heads on the number of heads scale.

Step-by-step explanation:

(a) If we flip a fair coin once, probability of getting head = 0.5

If we flip a fair coin 100 times, mean number of heads = 100(0.5) = 50

If there are N draws with a P probability of success, the standard deviation (SD) is given as:

[tex]SD = \sqrt{(N)(P)(1 - P)}[/tex]

Here, the probability of getting a head (P) is 0.5 while the number of draws (N) is 100. So,

[tex]SD = \sqrt{(100)(0.5)(1-0.5)}[/tex]

SD = 5

The standard scale value is: (35 - 50) / 5 = -3

Hence, 35 heads correspond to -3 on the standard scale.

(b) The standard scale value is 2.4 and we need to find the number of heads.

(X - 50) / 5 = 2.4

X - 50 = 12

X = 62

Hence, z = 2.4 on the standard scale corresponds to 62 on the number of heads scale.

The 35 heads corresponds to -3 on the standard scale, and the number of heads is 62

The head on the standard scale

The given parameter is:

n = 100

In a flip of a coin, the probability of a head is:

p = 1/2

So, the mean of the distribution is:

[tex]\bar x = np[/tex]

[tex]\bar x = 100 * 1/2[/tex]

[tex]\bar x = 50[/tex]

The standard deviation is:

[tex]\sigma = \sqrt{np(1 - p)}[/tex]

So, we have:

[tex]\sigma = \sqrt{100 * 1/2 * (1 - 1/2)}[/tex]

[tex]\sigma = 5[/tex]

The corresponding value on the standard scale is then calculated as:

[tex]z= \frac{x - \mu}{\sigma}[/tex]

For 35 heads, we have:

[tex]z= \frac{35 - 50}{5}[/tex]

[tex]z = -3[/tex]

Hence, the 35 heads corresponds to -3 on the standard scale

(b) The number of heads

We have:

[tex]z= \frac{x - \mu}{\sigma}[/tex]

When z = 2.4, the equation becomes

[tex]2.4= \frac{x - 50}{5}[/tex]

Multiply both sides by 5

[tex]12= x - 50[/tex]

Add 50 to both sides

[tex]x = 62[/tex]

Hence, the number of heads is 62

Read more about z-scores at:

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Answer:

Therefore the required probability is 0.6472.

Step-by-step explanation:

Poisson distribution: A Poisson distribution is discrete distribution.

Let X be a discrete variable the number of event. Let λ be the mean of X.

Then the probability of k event is

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

Here mean of each page is =0.03

Mean of 100 pages = (0.03×100)= 3

λ = 3

The required probability

P(X≤3)  

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

[tex]=\frac{3^0e^{-3}}{0!}[/tex] [tex]+\frac{3^1e^{-3}}{1!}[/tex][tex]+\frac{3^2e^{-3}}{2!}[/tex][tex]+\frac{3^3e^{-3}}{3!}[/tex]

[tex]=e^{-3}(1+3+\frac{9}{2}+\frac{27}{6})[/tex]

[tex]=e^{-3}(13)[/tex]

≈ 0.6472

Therefore the required probability is 0.6472

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% 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:

So, solution of  the differential equation is

[tex]y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

[tex]y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\[/tex]

It is a homogeneous solution:

[tex]y_h(t)=c_1e^{-2i t}+c_2e^{2i t}[/tex]

Now, we finding a particular solution.

[tex]y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\[/tex]

we get

[tex]y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]

So, solution of  the differential equation is

[tex]y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]

To set up the particular solution for the differential equation using the Method of Undetermined Coefficients, we propose a function similar to the nonhomogeneous term with unknown coefficients. The form is a combination of polynomial and trigonometric terms, reflecting the equation's right-hand side.

To set up the particular solution yp for the differential equation y″+4y=5xcos(2x) using the Method of Undetermined Coefficients, first identify the form of the nonhomogeneous term 5xcos(2x). Since it's a product of a polynomial and a trigonometric function, we anticipate a particular solution that mirrors this form, but with unknown coefficients to solve for.

The lowest degree polynomial in the nonhomogeneous term is x, so we begin with Ax and will also need to account for the derivative of cosine, which is sine. Therefore, our yp will be of the form:

yp = x(Acos(2x) + Bsin(2x)) + Cx²cos(2x) + Dx²sin(2x).

Here, A, B, C, and D are the undetermined coefficients that will be found by differentiating this assumed form and substituting back into the original differential equation.

Answer:

16.38%

Step-by-step explanation:

Given

Frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000

Let p = Probability of homozygous recessive individuals having character extra-long eyelashes = 0.09

p + q = 1 where q = Probability of homozygous recessive individuals not having character extra-long eyelashes

0.09 + p = 1

p = 1 - 0.09

p = 0.91

The frequency of the carrier individual is calculated by npq

Where n = mating partners = 2

Frequency = 2 * 0.09 * 0.91

Frequency = 0.1638 or 16.38%

Answer:

The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals which is

0.42 or 42 %

Step-by-step explanation:

To solve the question, we note that

p²+2pq+q²=1  and p + q =1

where

p = frequency of the dominant allele

q= frequency of the recessive allele

p² = frequency of homozygous dominant allele

q² = frequency of homozygous recessive allele

2·p·q = frequency of hetrozygous individuals

The frequency of the homozygous recessive individuals q² is 0.09

Therefore the frequency of q = √(0.09) = 0.3, therefore p = 1 - 0.3 =0.7

The  percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals or 2×p×q = Ae = 2*0.3*0.7 = 0.42

→ 0.42× 100 = 42 %

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:

Step-by-step explanation:

Given that an ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken from the four populations.

Hence total observations are 30*4 =120

No of groups = 3

Hence numerator df = 3-1 =2

Now total degrees of freedom = 120-1 =119

So denominator degrees of freedom = 119-2 = 117

Thus F statistic will have numerator as 2 degrees of freedom and denominator as 117 degrees of freedom.

Answer:

[tex]z=-0.674<\frac{a-12}{1.25}[/tex]

And if we solve for a we got

[tex]a=12 -0.674*1.25=11.157[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 11.157.  

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 mean life of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(12,1.25)[/tex]  

Where [tex]\mu=12[/tex] and [tex]\sigma=1.25[/tex]

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

[tex]P(X>a)=0.75[/tex]   (a)

[tex]P(X<a)=0.25[/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 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

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

[tex]z=-0.674<\frac{a-12}{1.25}[/tex]

And if we solve for a we got

[tex]a=12 -0.674*1.25=11.157[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 11.157.  

The lifespan of the dishwashers in the bottom 25% is approximately 11.16 years, found by calculating the corresponding value to the 25th percentile z-score from the normal distribution of the appliance's lifespan.

The problem is asking for the life of the dishwasher that falls in the bottom 25% of the normally distributed lifespan range. In other words, we need to find the dishwasher lifespan that is the cutoff for the bottom quartile. For a normal distribution, the 25th percentile (the cutoff for the bottom quartile) is commonly found by using a z-score, which is a measure of how many standard deviations a value is from the mean.

The z-score for the 25th percentile is typically -0.675. We can find the lifespan corresponding to this z-score using the formula: lifespan = mean + z*sd. Here, 'mean' is the average lifespan, 'z' is the z-score, and 'sd' is the standard deviation.

Therefore, substituting the known values into the formula, we get: lifespan = 12 + (-0.675)*1.25 = 11.15625 years.

Therefore, the number of years the bottom 25% of dishwashers would last is approximately 11.16 years.

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Answer:

The argument is valid.

Step-by-step explanation:

We are told "Loretta's hobby is stamp collecting."

"If her husband likes to fish, then Loretta's hobby is not stamp collecting."

Her hobby is stamp collecting, so her husband does not like to fish.

"If her husband does not like to fish, then Nathan likes to read."

Her husband does not like to fish; therefore, Nathan likes to read.

The argument is valid.

The argument is invalid as it fails the informal test of validity; the premises about Loretta and her husband's hobbies do not logically lead to the conclusion about Nathan's reading habits.

The task involves assessing the validity of an argument by applying the informal test of validity. Let's deconstruct the argument provided. It states that Loretta's hobby is stamp collecting, and presents two conditional statements involving her husband's hobbies and Nathan's interest. Finally, it concludes that Nathan likes to read.

To apply the informal test of validity, we must determine if it's possible for all premises to be true while the conclusion could still be false. In this argument, even if all the stated conditions about Loretta and her husband's hobbies are true, they don't logically necessitate the conclusion about Nathan's reading habits. Thus, the argument is invalid because it's possible to imagine a scenario where all premises are true, but Nathan does not like to read. The premises about Loretta and her husband have no logical connection to Nathan's interests, making it invalid.

A valid argument ensures that if the premises are true, the conclusion must also be true. However, this argument fails to meet that criterion, indicating its invalidity based on the informal test.

Answer:

Step-by-step explanation:

Given that of all airline flight requests received by a certain discount ticket broker, 90% are for domestic travel (D) and 10% are for international flights (I).

Let x be the number of requests among the next three requests received that are for domestic flights.

X can take values as 0,1, 2 or 3.

Since independence of successive requests is assumed we find that X having two outcomes and independence

X is binomial with n =3 and p= 0.90 (constant for each trial)

PDF of X would be

[tex]P(X=x) = 3Cx (0.9)^r (0.1)^{3-r} , r=0,1,2,3[/tex]

Thus pdf of X is

X           0           1             2            3

p         0.001    0.027   0.243   0.729

Final answer:

The probability distribution of x, where x is the number of domestic flight requests among the next three requests, follows a binomial distribution with n=3 and p=0.9. It can be calculated using the formula for binomial probabilities, considering the value of x can be 0, 1, 2, or 3. We get P = 0.081.

Explanation:

The question involves calculating the probability distribution of the random variable x, which represents the number of domestic flight requests among the next three requests received by a discount ticket broker. Since each request is independent and only two outcomes are possible (domestic or international), a binomial distribution can be used to determine the probabilities for x. For instance, the probability for one particular sequence of requests such as DID (domestic, international, domestic) is calculated as (0.9)(0.1)(0.9), which is the product of the probabilities for each request in that sequence. To find the probability distribution of x, we consider all possible sequences of three requests and their associated probabilities, which are determined by multiplying the individual probabilities for each request in the sequence following the rules of binomial distribution.

The random variable x can take on the following values: 0, 1, 2, or 3, representing the number of domestic flight requests out of three. We calculate probabilities for each of these values using the binomial formula:

[tex]P(x=0) = (0.1)^3\\P(x=1) = 3 * (0.9)(0.1)^2\\P(x=2) = 3 * (0.9)^2(0.1)\\P(x=3) = (0.9)^3[/tex]

P = 0.081.

Answer: x = - 16

y = - 11

Step-by-step explanation:

The given system of linear equations is expressed as

2x - 3y = 1- - - - - - - - - - - - -1

x - 2y = 6 - - - - - - - - - - - -2

We would eliminate x by multiplying equation 1 by 1 and equation 2 by 2. It becomes

2x - 3y = 1- - - - - - - - - - -3

2x - 4y = 12- - - - - - - - - - - -4

Subtracting equation 4 from equation 3, it becomes

y = - 11

Substituting y = - 11 into equation 2, it becomes

x - 2 × - 11 = 6

x + 22 = 6

Subtracting 22 from the left hand side and the right hand side of the equation, it becomes

x + 22 - 22 = 6 - 22

x = - 16

Final answer:

To obtain the desired mixture, you will need 24 lbs of the first type of chocolate and 6 lbs of the second type of chocolate.

Explanation:

To find the amount of each chocolate needed to obtain the desired mixture, we can set up a system of equations. Let's say x represents the pounds of the first type of chocolate and y represents the pounds of the second type of chocolate. We know that the total weight of the mixture is 30 lbs, so we have the equation:

x + y = 30

We also know that the cost per pound of the mixture is $4.20, so the total cost is:

2.40x + 9.90y = 4.20 * 30

Simplifying the second equation, we get:

2.40x + 9.90y = 126

We can solve this system of equations using substitution, elimination, or a calculator. The solution is x = 24 lbs and y = 6 lbs. Therefore, you will need 24 lbs of the first type of chocolate and 6 lbs of the second type of chocolate to obtain the desired mixture.

Answer:

(D) 9a

Step-by-step explanation:

Given

4a^2 + 9

Add 9a to the expression

4a^2 + 9 + 9a

This can be written as

(2a)^2 + (3)^2 + (3)^2 a

Answer:

E. 12a

Step-by-step explanation:

TBH no idea why thats correct.

Answer:

No

Step-by-step explanation:

Because the model was developed using only pre-statistics course grades between 70% and 95% so it is risky to assume that linear trend will continue far beyond that span of values.

Answer / Step-by-step explanation:

(1) Given t = 0.0588s ¹.¹²⁵

where s is the distance and t is the time to run that distance.

The second portion asks us to find the derivative of the equation when our s value is equal to 20 and interpret.

(2) First, we try to convert the unit from miles to meters

Therefore, 1 mile = 1609 meters

Then,

         t = 0.0588 ( 1609 ) ¹.¹²⁵

             =238 . 09

This gives us the instantaneous rate of change of seconds between every 20 meters ran.

The last portion asks us to compare this estimate to current world records. And have they been surpassed?

As of today, the fastest official record for a standard mile is held by a man from Morocco named Hichan El Guerrouj. The time was recorded at 3.43 minutes in Rome, Italy on July 7th, 1999.

Now, keep in mind that this is almost a full minute slower than the estimated time. However, how do these projections hold up against Usain Bolt, the man that is considered the fastest man in the world ?

Although, Usain Bolt does run long distances, he holds records in nearly every sprinting event that he has ever competed in.

Hence, Kennelly's estimate for the fastest mile is 238.09

(3) Now, noting that since dt / ds = 0.0588 ( 1.25 ) s  ⁰.¹²⁵

Then,

           dt / ds I 100 = 0.0588 (1.25) (20)  ⁰.¹²⁵

                              = 0.1176

Answer:

(a) 0.4242

(b) 0.0707

Step-by-step explanation:

The total number of ways of selecting 8 herbs from 12 is

[tex]\binom{12}{8}=495[/tex]

(a) If 2 herbs are selected, then there are 8 - 2 = 6 herbs to be selected from 12 - 8 = 10. The number of ways of the selection is then

[tex]\binom{10}{6}= 210[/tex]

Note that this is the number of ways that both are included. We would have multiplied by 2! if any of them were to be included.

The probability = [tex]\dfrac{210}{495}=0.4242[/tex]

(b) If 5 herbs are selected, then there are 8 - 5 = 3 herbs to be selected from 12 - 5 = 7. The number of ways of the selection is then

[tex]\binom{7}{3}= 35[/tex]

This is the number of ways that both are included. We would have multiplied by 5! if any of them were to be included. In that case, our probability will exceed 1; this implies that certainly, at least, one of them is included.

The probability = [tex]\dfrac{35}{495}=0.0707[/tex]

The probability that the 2 tiles with a blue tint will be included in the splash-guard is 0.0455. The probability that all 5 of the herbs grown in the backyard garden will be included on the splash-guard is 0.0058.

The probability that the 2 tiles with a blue tint will be included in the splash-guard is 0.0455.

The probability that all 5 of the herbs grown in the backyard garden will be included on the splash-guard is 0.0058.

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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:

41 degrees

Step-by-step explanation:

step 1

Find the measure of arc DGF

we know that

The inscribed angle is half that of the arc comprising

so

[tex]m\angle E=\frac{1}{2}[arc\ DGF][/tex]

we have

[tex]m\angle E=123^o[/tex]

substitute

[tex]123^o=\frac{1}{2}[arc\ DGF][/tex]

[tex]arc\ DGF=246^o[/tex]

step 2

Find the measure of arc EF

we know that

[tex]arc\ DGF+arc\ DE+arc\ EF=360^o[/tex] ----> by complete circle

substitute the given values

[tex]246^o+73^o+arc\ EF=360^o[/tex]

[tex]319^o+arc\ EF=360^o[/tex]

[tex]arc\ EF=360^o-319^o=41^o[/tex]

Answer: A)41 Degrees

Step-by-step explanation:

It was right on edu 2023

Answer with Step-by-step explanation:

S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),..,(2,6),(3,1),(3,2),...,(3,6),(4,1),..,(4,6),(5,1),(5,2),(5,3),...,(5,6),(6,1),(6,2),..,(6,6)}

Total number of cases=36

M=Maximum of the two tosses

M=1,

(1,1)

Therefore, [tex]P(1)=\frac{1}{36}[/tex]

Using the formula, P(E)=[tex]\frac{favorable\;cases}{Total\;number\;of\;cases}[/tex]

M=2

(1,2),(2,1),(2,2)

Favorable cases=3

[tex]P(2)=\frac{3}{36}=\frac{1}{12}[/tex]

M=3

(1,3),(3,1),(3,2),(2,3),(3,3)

[tex]P(3)=\frac{5}{36}[/tex]

M=4

(1,4),(4,1),(2,4),(3,4),(4,2)(4,3),(4,4)

[tex]P(4)=\frac{7}{36}[/tex]

M=5

(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)

[tex]P(5)=\frac{9}{36}=\frac{1}{4}[/tex]

M=6

(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

[tex]P(6)=\frac{11}{36}[/tex]

Answer:

The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

Step-by-step explanation:

The (1 - α) % confidence interval for population mean (μ) is:

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

Here the population standard deviation (σ) is not provided. So the confidence interval would be computed using the t-distribution.

The (1 - α) % confidence interval for population mean (μ) using the t-distribution is:

[tex]CI=\bar x\pm t_{\alpha /2,(n-1)}\frac{s}{\sqrt{n}}[/tex]

Given:

[tex]\bar x=27.97\\s=10.04\\n=25\\t_{\alpha /2, (n-1)}=t_{0.01/2, (25-1)}=t_{0.005, 24}=2.797[/tex]

*Use the t-table for the critical value.

Compute the 99% confidence interval as follows:

[tex]CI=27.97\pm 2.797\times\frac{10.04}{\sqrt{25}}\\=27.97\pm5.616\\=(22.354, 33.586)\\\approx(22.35, 33.59)[/tex]

Thus, the 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).

Answer:

mL

Step-by-step explanation:

Ml

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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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:

(C) Systematic Sampling

Step-by-step explanation:

The reporter decides to interview every fifth student of Ohio State University that passes by. This is referred to as Systematic Sampling.

In Systematic Sampling, the members of the population are place in an ordered list and the members of the sample are taken at a stated interval. This interval is called the Sampling Interval.

Although the members of the sample are random, they are chosen at stated intervals.

The slope of the line is [tex]m=-4[/tex]

Explanation:

Given that the pair of points are [tex](1,-19)[/tex] and [tex](-2,-7)[/tex]

We need to determine the slope of the line.

The slope of the line can be determined using the formula,

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Substituting the points [tex](1,-19)[/tex] and [tex](-2,-7)[/tex] for the coordinates [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] in the above formula, we have,

[tex]m=\frac{-7-(-19)}{-2-1}[/tex]

Simplifying the terms, we get,

[tex]m=\frac{-7+19}{-2-1}[/tex]

Adding the terms, we have,

[tex]m=\frac{12}{-3}[/tex]

Dividing, we get,

[tex]m=-4[/tex]

Thus, the slope of the line between the pair of points is [tex]m=-4[/tex]

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.

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Answer:

b. 10.4

c.  26.9

Step-by-step explanation:

Let the universal set U = 100% which is the total no of people in the American community

Let A = 14.6% which is the total no of people living below poverty line

Let B = 20.7% which is the total no of people speaking foreign Language

C = 4.2% no of people who both speak foreign language and live below poverty line

X = no of people who neither live below poverty line nor speak foreign language

P (A) = 14.6%

P (B) = 20.7%

P (C) = P (A ∩ B) = 4.2%

P (A – C) = P (A ∩ U) = 14.6 – 4.2 = 10.4%

P (B – C) = P (B ∩ U) = 20.7 – 4.2 = 16.5%

P (X) = P (A ᴜ B) c =100 – (10.4 + 4.2 + 16.5) = 68.9%

a. The venn diagram is as shown above

b. Percent of Americans who live below poverty line and Speak English at home(minus foreign lang speakers living below poverty line) that is A only

= A – C

= 14.6 – 4.2

= 10.4%

c. Percentage of Americans Living below poverty line or Speaking foreign language

= A only + B only

A only = A – C ( People living below poverty line only)

= 14.6 -4.2

= 10.4%

B only = B – C ( people speaking foreign languages only)

= 20.7 – 4.2

= 16.5%

Hence

A only + B only = 10.4 + 16.5 = 26.9%

Final answer:

The question pertains to calculating probabilities from a given discrete Cumulative Distribution Function (CDF) for the number of ATMs in use. By utilizing the step function provided in the CDF, probabilities for specific intervals or exact values can be determined.

Explanation:

The student's question involves understanding the Cumulative Distribution Function (CDF) for a discrete random variable, specifically the number of ATMs in use at a particular time.

The CDF is used to find the probability that the random variable X, representing the number of ATMs used, is less than or equal to a certain value x.

In this scenario, we have a step function showing the probabilities associated with different intervals of x. The function increases as x increases, which is typical of a CDF, as it accumulates the probabilities for all values up to x.

To calculate probabilities for specific intervals from the given CDF, one would subtract the CDF value just before the interval from the CDF value at the end of the interval. For example, to find the probability that exactly four ATMs are in use (P(X=4)), one would calculate P(X<5) - P(X<4), which is 0.83 - 0.62 = 0.21.

Answer:

Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex]  °C/m.

Step-by-step explanation:

Given that, the temperature at a point (x,y,z) is

[tex]T(x,y,z)=200e^{-x^2-3y^2-9z^2}[/tex].

Rate change of temperature at the point P(2,-1,2) in the direction toward the point Q (3,-3,3) is [tex]D_uT(2,-1,2)[/tex]

[tex]T_x= 200(-2x)e^{-x^2-3y^2-9z^2}[/tex][tex]=-400x200.2xe^{-x^2-3y^2-9z^2}[/tex]

[tex]T_y = 200.(-6y)e^{-x^2-3y^2-9z^2}[/tex][tex]=-1200ye^{-x^2-3y^2-9z^2}[/tex]

[tex]T_z=200(-18z)e^{-x^2-3y^2-9z^2}[/tex][tex]=-3600ze^{-x^2-3y^2-9z^2}[/tex]

The gradient of the temperature

[tex]\bigtriangledown T(x,y,z)= (-400x200.2xe^{-x^2-3y^2-9z^2},-1200ye^{-x^2-3y^2-9z^2},-3600ze^{-x^2-3y^2-9z^2})[/tex]                   [tex]=-400e^{-x^2-3y^2-9z^2}(x,3y,9z)[/tex]

[tex]\bigtriangledown T(2,-1,2) = -400e^{-2^2-3(-1)^2-9.2^2}(2,3.(-1),9.2)[/tex]

                   [tex]=-400 e^{-43}(2,-3,18)[/tex]

V=[tex]\overrightarrow {PQ}= \vec{Q}-\vec{P}[/tex]=(3,-3,3)-(2,-1,2)=(1,-2,1)

The unit vector of V is [tex]\frac{(1,-2,1)}{\sqrt{1^2+(-2)^2+1^2}}[/tex]

                                   [tex]=\frac{1}{\sqrt6}(1,-2,1)[/tex]

Therefore,

[tex]D_uT(2,-1,2) = \bigtriangledown T(2,-1,2).\frac{1}{\sqrt6}(1,-2,1)[/tex]

                       [tex]=-400 e^{-43}(2,-3,18).\frac{1}{\sqrt6}(1,-2,1)[/tex]

                       [tex]=-\frac{400}{\sqrt6}e^{-43}(2.1+(-3).(-2)+18.1)[/tex]

                       [tex]=-\frac{10400}{\sqrt6}e^{-43}[/tex]  

                        [tex]=-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m

Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex]  °C/m.