This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A particle is moved along the x-axis by a force that measures 9/(6 x) pounds at a point x feet from the origin. Find the work W done in moving the particle from the origin to a distance of 12 ft.

Answer:

the probability that his mother would return on time to catch Gabe stealing is given by:[tex]\frac{15}{60}=0.25[/tex].

Step-by-step explanation:

probability is given by; the required outcome over the number of possible outcome.

Step1; we have to subtract the first initial 30 secs that Gabe spent wondering if he will be caught. i.e 90 sec-30 secs= 60 Sec.

Step2: we divide the number of seconds that he can finish a cookie without being caught over the number of seconds left before his mother returns. that is why we have [tex]\frac{15}{60} = 0.25[/tex]

The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.

Judy could return at any time in the next 90 seconds with equal probability. For the first 30 seconds, Gabe sheepishly wonders if he will get caught trying to grab a nearby cookie. And, it takes Gabe 15 seconds to grab a cookie from the jar.

Now the probability is

[tex]= 15 \div (90 - 30)\\\\= 15 \div 60[/tex]

= 0.25

Hence, we can conclude that The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.

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Answer:

2xy

Step-by-step explanation:

10x²y² - 8xy

2xy(5xy - 4)

The greatest common factor (GCF) of the expressions 10x^2y^2 and 8xy is 2xy. The GCF for the numbers is 2, and for the variables x and y it's x and y, respectively.

The subject is the greatest common factor (GCF), applied to algebraic expressions. The GCF is the largest expression that can be multiplied by another expression to get the original expression. For the expressions 10x^2y^2 and 8xy, the numbers are 10 and 8. Their GCF is 2. The term x appears in both expressions, and the lowest power of x is 1. The term y also appears in both expressions, with the lowest power being 1. So, 2x and y are the GCFs for the numbers and the variables respectively. Combining them, we get 2xy, which is the GCF of 10x^2y^2 and 8xy.

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Answer:

A Not a probability B.Probability C.not a probability d.Probability E propability F.not a propability

Step-by-step explanation:

A propability of any event is a number between zero and one

A)175%  converted 1.75 so lies outside zero and one not a propability

B)5.91 lies outside the zero and one rage not a propability

C)0.66 lies in the zero and one rage therefore a propability

D)0.002 Lies within the zero and one range therefore a propability

E) -110% converted equals -1.1 so does not lies in this range not a propability

The following numbers could not be probabilities:

A. The number 175% could not be a probability because it is larger than 100%.

B. The number 5.91 could not be a probability because it is larger than 1.

E. The number negative 110% could not be a probability because it is negative.

Probabilities must be between 0 and 1, inclusive. This is because a probability represents the likelihood of an event occurring, and there is no event that can be more likely than certain (1) or less likely than impossible (0).

The numbers 0.66 and 0.002 are valid probabilities.

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Answer:

Probability = 0.23572 .

Step-by-step explanation:

We are given that the length of time needed to complete a certain test is normally distributed with mean 35 minutes and standard deviation 15 minutes.

Let X = length of time needed to complete a certain test

Since, X ~ N([tex]\mu,\sigma^{2}[/tex])

The z probability is given by;

            Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)    where, [tex]\mu[/tex] = 35  and  [tex]\sigma[/tex] = 15

So, P(31 < X < 40) = P(X < 40) - P(X <= 31)

P(X < 40) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{40-35}{15}[/tex] ) = P(Z < 0.33) = 0.62930

P(X <= 31) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{31-35}{15}[/tex] ) = P(Z < -0.27) = 1 - P(Z <= 0.27)

                                               = 1 - 0.60642 = 0.39358

Therefore, P(31 < X < 40) = 0.62930 - 0.39358 = 0.23572 .  

To find the probability that it takes between 31 and 40 minutes to complete the test when the test completion times are normally distributed with a mean (μ) of 35 minutes and a standard deviation (σ) of 15 minutes, we can use the properties of the normal distribution.

Firstly, we need to standardize our times (31 and 40 minutes) to find the corresponding z-scores. The z-score is a measure of how many standard deviations an element is from the mean. We use the following formula to calculate the z-score:

\[ z = \fraction{(X - \mu)}{\sigma} \]

where:
- \( X \) is the value for which we are finding the z-score
- \( \mu \) is the mean
- \( σ \) is the standard deviation

We will calculate the z-scores for both 31 minutes and 40 minutes.

For \( X = 31 \):
\[ z_{31} = \fraction{(31 - 35)}{15} = \fraction{-4}{15} \approx -0.267 \]

For \( X = 40 \):
\[ z_{40} = \fraction{(40 - 35)}{15} = \fraction{5}{15} \approx 0.333 \]

Next, we look up these z-scores in the standard normal distribution (z-distribution) table, which will give us the area to the left of each z-score.

If we don't have the z-distribution table available, we might use statistical software or a calculator with a normal distribution function. But let's proceed as if we are using the z-table.

Let's say our z-table gives us the following areas:

Area to the left of \( z_{31} \approx -0.267 \): Approximately 0.3944 (please note that actual values will depend on the specific z-table or calculator you are using).

Area to the left of \( z_{40} \approx 0.333 \): Approximately 0.6304.

Now we want the area between these two z-scores, which represents the probability that the test completion time is between 31 and 40 minutes. To find this, we can subtract the area for \( z_{31} \) from the area for \( z_{40} \):

Probability \( P(31 < X < 40) = P(z_{40}) - P(z_{31}) \)

Substituting the values,

\[ P(31 < X < 40) = 0.6304 - 0.3944 = 0.2360 \]

So the probability that it will take between 31 and 40 minutes to complete the test is approximately 0.2360, or 23.60%.

Answer:

Probability a team from national football conference wins the same year will be equal to 0.53

Step-by-step explanation:

Probability that a National League team wins baseball world series = 0.44

Probability that a National Football conference team wins the           Superbowl = 0.53

The two events are independent, since the sports played are different (baseball and football). This is a logical assumption to make.

Since these events are independent, the occurrence of one event will not change the probability of the other event. This means that it does not matter whether the national league team won or lost the world series. The probability of the football team winning the Superbowl will remain the same, which is 0.53.

Answer:

(a) The probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.

(b) The probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.

Step-by-step explanation:

Let X = number of dandelions per square meter.

The average number of dandelions per square meter is, λ = 2.

The random variable X follows a Poisson distribution with parameter λ = 2.

The probability mass function of X is:

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

(a)

Compute the value of P (X = 0) as follows:

[tex]P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.13534\times 1}{1}=0.1353[/tex]

Thus, the probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.

(b)

Compute the value of P (X ≥ 1) as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             [tex]=1-0.1353\\=0.8647[/tex]

Thus, the probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.

To find the probability of no dandelions, use the Poisson distribution formula with the mean value of 2. To find the probability of at least one dandelion, use the complement rule.

To find the probability of no dandelions in a randomly selected area of 1 square meter in this region, we need to use the Poisson distribution. The mean number of dandelions per square meter is 2. The probability of no dandelions is given by the formula: P(X=0) = e^(-2) * (2^0 / 0!). Calculating this gives us approximately 0.1353.

To find the probability of at least one dandelion, we can use the complement rule. The probability of at least one dandelion is equal to 1 minus the probability of no dandelions. So, P(X >= 1) = 1 - P(X=0). Calculating this gives us approximately 0.8647.

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Answer:

20 liters of the 40% solution

60-20=40 liters of the 70% solution

Step-by-step explanation:

40%x+70%(60-x)=60%(60)

0.4x+0.7(60-x)=0.6(60)

0.4x+42-0.7x=36

-0.3x+42=36

-0.3x=-42+36

-0.3x=-6

0.3x=6

3x=60

x=60/3

x=20 liters of the 40% solution

60-20=40 liters of the 70% solution

check: 0.4*20=8

0.7*40=28

8+28=36 and 0.6*60=36 also

Answer:

f'(x) is increasing in the open interval (-1/2, +∞), and it is decreasing in the interval (-∞, -1/2).

Step-by-step explanation:

f'(x) is a quadratic function with positive main coefficient, then it will be decreasing until the x-coordinate of its vertex and it will be increasing from there onwards. The x-coordinate of the vertex is given by the equation -b/2a = -1/2. Hence

- f'(x) is incresing in the interval (-1/2, +∞)

- f'(x) is decreasing in the interval (-∞, -1/2)

Final answer:

To find intervals where f'(x) is increasing or decreasing, locate the critical points by solving f'(x) = 0 and test the signs of f'(x) around these points. This process determines where the original function f(x) is increasing or decreasing.

Explanation:

To determine the open intervals in which the derivative of a function f(x), denoted as f'(x), is increasing or decreasing, one must:

In the scenario one would solve the equation for f'(x) = 0 to find the critical points. Once these are found, the signs of f'(x) to the left and right of these points can be tested to see where the function is increasing or always decreasing.

Answer:

Probability that the sample contains at least five Roman Catholics = 0.995 .

Step-by-step explanation:

We are given that In Quebec, 90 percent of the population subscribes to the Roman Catholic religion.

The Binomial distribution probability is given by;

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

Here, n = number of trials which is 8 in our case

         r = no. of success which is at least 5 in our case

         p = probability of success which is probability of Roman Catholic of

                 0.90 in our case

So, P(X >= 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= [tex]\binom{8}{5}0.9^{5}(1-0.9)^{8-5} + \binom{8}{6}0.9^{6}(1-0.9)^{8-6} + \binom{8}{7}0.9^{7}(1-0.9)^{8-7} + \binom{8}{8}0.9^{8}(1-0.9)^{8-8}[/tex]

= 56 * [tex]0.9^{5} * (0.1)^{3}[/tex] + 28 * [tex]0.9^{6} * (0.1)^{2}[/tex] + 8 * [tex]0.9^{7} * (0.1)^{1}[/tex] + 1 * [tex]0.9^{8}[/tex]

= 0.995

Therefore, probability that the sample contains at least five Roman Catholics is 0.995.

The probability that a random sample of eight Quebecois contains at least five Roman Catholics is approximately 0.9950 or 99.50%.

Given:

- The probability of success (subscribing to the Roman Catholic religion) p = 0.9 .

- The number of trials (sample size)  n = 8 .

- We want to find the probability of getting at least 5 successes.

The binomial probability formula is:

[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]

Where:

- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient [tex]\(\frac{n!}{k!(n-k)!} \)[/tex],

- k is the number of successes,

- p is the probability of success,

- (1-p)  is the probability of failure.

We need to find [tex]\( P(X \geq 5) \)[/tex], which is the sum of the probabilities of getting exactly 5, 6, 7, and 8 successes.

[tex]\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \][/tex]

Calculating each term individually:

1.

[tex]\( P(X = 5) \):\[ P(X = 5) = \binom{8}{5} (0.9)^5 (0.1)^3 \]\[ \binom{8}{5} = \frac{8!}{5!3!} = 56 \]\[ P(X = 5) = 56 \times (0.9)^5 \times (0.1)^3 \]\[ P(X = 5) = 56 \times 0.59049 \times 0.001 \]\[ P(X = 5) = 56 \times 0.00059049 \]\[ P(X = 5) \approx 0.0331 \][/tex]

2.  P(X = 6) :

[tex]\[ P(X = 6) = \binom{8}{6} (0.9)^6 (0.1)^2 \]\[ \binom{8}{6} = \frac{8!}{6!2!} = 28 \]\[ P(X = 6) = 28 \times (0.9)^6 \times (0.1)^2 \]\[ P(X = 6) = 28 \times 0.531441 \times 0.01 \]\[ P(X = 6) = 28 \times 0.00531441 \]\[ P(X = 6) \approx 0.1488 \][/tex]

3.  P(X = 7) :

[tex]\[ P(X = 7) = \binom{8}{7} (0.9)^7 (0.1)^1 \]\[ \binom{8}{7} = \frac{8!}{7!1!} = 8 \]\[ P(X = 7) = 8 \times (0.9)^7 \times 0.1 \]\[ P(X = 7) = 8 \times 0.4782969 \times 0.1 \]\[ P(X = 7) = 8 \times 0.04782969 \]\[ P(X = 7) \approx 0.3826 \][/tex]

4. P(X = 8) :

[tex]\[ P(X = 8) = \binom{8}{8} (0.9)^8 (0.1)^0 \]\[ \binom{8}{8} = 1 \]\[ P(X = 8) = 1 \times (0.9)^8 \times 1 \]\[ P(X = 8) = (0.9)^8 \]\[ P(X = 8) = 0.43046721 \]\[ P(X = 8) \approx 0.4305 \][/tex]

Now sum these probabilities:

[tex]\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \]\[ P(X \geq 5) \approx 0.0331 + 0.1488 + 0.3826 + 0.4305 \]\[ P(X \geq 5) \approx 0.9950 \][/tex]

So, the probability that the sample contains at least five Roman Catholics is approximately 0.9950 or 99.50%.

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800  and  [tex]\sigma[/tex] = 750.

Let X = randomly selected element of the population

The z probability is given by;

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

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75

The z probability is given by;

           Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

Using the normal distribution and the central limit theorem, it is found that:

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

In this problem:

The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:

X = 58000:

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

[tex]Z = \frac{58000 - 57800}{750}[/tex]

[tex]Z = 0.27[/tex]

[tex]Z = 0.27[/tex] has a p-value of 0.6064.

X = 57000:

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

[tex]Z = \frac{57000 - 57800}{750}[/tex]

[tex]Z = -1.07[/tex]

[tex]Z = -1.07[/tex] has a p-value of 0.1423.

0.6064 - 0.1423 = 0.4641.

0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.

For samples of size 100, [tex]n = 100[/tex], and then:

[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]

For samples of size 100, the mean is of 57800 and the standard deviation is 75.

Then, the probability is:

X = 58000:

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

By the Central Limit Theorem

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

[tex]Z = \frac{58000 - 57800}{75}[/tex]

[tex]Z = 2.7[/tex]

[tex]Z = 2.7[/tex] has a p-value of 0.9965.

X = 57000:

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

[tex]Z = \frac{57000 - 57800}{75}[/tex]

[tex]Z = -10.7[/tex]

[tex]Z = -10.7[/tex] has a p-value of 0.

0.9965 - 0 = 0.9965.

0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.

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Using the combination formula, the student can make 165 different possible schedules for next semester if she is planning to take three classes from the eleven available that she has met the prerequisites for.

This problem is a combination problem where students need to select 3 classes from 11 available classes to plan her schedule. The combination formula is used when the order of election does not matter. Here, the formula to find a combination is written as C(n, r) = n! / [r!(n - r)!].

In this particular case, the student has 11 classes (n=11) and plans to take 3 classes (r=3). So, the combination becomes C(11,3) = 11! / [3!(11-3)!] = (11*10*9) / (3*2*1) = 165.

This means, the student can make 165 different possible schedules for next semester assuming she is planning to take three classes from the eleven available classes.

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The student could make 165 different schedules.

The formula for combinations is given by:

[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]

where n! (n factorial) is the product of all positive integers up to n, and r is the number of items to choose.

 Given that the student has 11 classes left and she plans to take 3 classes next semester, we substitute n = 11 and r = 3 into the combination formula:

[tex]\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} \][/tex]

 Now, we simplify the factorial expressions:

[tex]\[ \binom{11}{3} = \frac{11!}{3! \times 8!} \][/tex]

We can cancel out the common terms in the numerator and the denominator:

[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \][/tex]

Now, we perform the arithmetic:

[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \][/tex]

Thus, the student has 165 different possible schedules for next semester if she is to take 3 classes out of the remaining 11.

By first reducing the following function f(x, y) to g(x)=x3 with y=0 and then determining its derivative at x=1, we can obtain fx(1, 0) equals 3.

You are required to determine the value of fx(1, 0) in the given function f(x, y) = x(x(x2 + y2)3/2 e)(sin(x2y)). The tip suggests that you can address this problem by computing g'(1) where g(x) = f(x, 0). When y = 0, the function changes to f(x,0) = x * x2 * e0, which is then expressed as x3. G(x)=x3 has a derivative, g(x)=3x2. Consequently, fx(1, 0) equals g'(1), which equals 3.

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Answer: 9.16666666667

Step-by-step explanation:

Answer:

Step-by-step explanation:

Hello!

You have 4 events A, B, C and D

With probabilities:

P(A∪B)= 5/8

P(A)= 3/8

P(C∩D)= 1/3

P(C)= 1/2

A and B are disjoint events, this means that there are no shared elements between then and their intersection is void, symbolically A∩B= ∅, in consequence, these events are mutually exclusive.

C and D are independent events, this means that the occurrence of one of them does not affect the probability of occurrence of the other one in two consecutive repetitions.

a.

i. P(A∩B)= 0

⇒ Since A and B are disjoint events, the probability of their intersection is zero.

ii. A and B are mutually exclusive events, this means that P(A∪B)= P(A)+P(B)

⇒ From this expression, you can clear the probability of b as P(B)= P(A∪B)-P(A)= 5/8-3/8= 1/4

iii. If Bc is the complementary event of B, its probability would be P(Bc)= 1 - P(B)= 1 - 1/4= 3/4. If the events A and B are mutually exclusive and disjoint, it is logical to believe that so will be the events A and Bc, so their intersection will also be void:

P(A∩Bc)= 0

vi.P(A∪Bc)= P(A) + P(Bc)= 3/8+3/4= 9/8

b.

If A and B are independent then the probability of A is equal to the probability of A given B, symbolically:

P(A)= P(A/B)

[tex]P(A/B)= \frac{P(AnB)}{P(B)}= \frac{0}{1/4}= 0[/tex]

P(A)= 3/8

P(A) ≠ P(A/B) ⇒ A and B are not independent.

c.

i. P(D) ⇒ Considering C and D are two independent events, then we know that P(C∩D)= P(C)*P(D)

Then you can clear the probability of D as:

P(D)= P(C∩D)/P(C)= (1/3)/(1/2)= 2/3

ii. If Dc is the complementary event of D, then its probability is P(Dc)= 1 - P(D) = 1 - 2/3= 1/3

P(C∩Dc)= P(C)*P(Dc)= (1/2)*(1/3)= 1/6

iii. Now Cc is the complementary event of C, its probability is P(Cc)= 1 - P(C)= 1 - 1/2= 1/2

P(Cc∩Dc)= P(Cc)*P(Dc)= (1/2)*(1/3)= 1/6

vi. and e.

[tex]P(C/D)= \frac{P(CnD)}{P(D)} = \frac{1/3}{2/3} = 1/2[/tex]

P(C)=1/2

As you can see the P(C)=P(C/D) ⇒ This fact proves that the events C and D are independent.

d.

i. P(C∪D)= P(C) + P(D) - P(C∩D)= 1/2 + 2/3 - 1/3= 5/6

ii. P(C∪Dc)= P(C) + P(Dc) - P(C∩Dc)= 1/2 + 1/3 - 1/6= 2/3

I hope it helps!

a. Expectation is linear, so that

[tex]E[27X_1+125X_2+512X_3]=27E[X_1]+125E[X_2]+512E[X_3]=98,630[/tex]

The variance of a sum of independent random variables is equal to a weighted sum of the variances, with weights equal to the squares of the coefficients of the [tex]X_i[/tex]:

[tex]V[27X_1+125X_2+512X_3]=27^2V[X_1]+125^2V[X_2]+512^2V[X_3]=2,306,838[/tex]

b. If the [tex]X_i[/tex] were dependent on one another, we would have the same expectation, but now the variance of a sum of random variables becomes the sum of their covariances:

[tex]\displaystyle\sum_{i=1}^3V[\alpha_iX_i]=\sum_{i=1}^3\sum_{j=1}^3\mathrm{Cov}[X_i,X_j]=\sum_{i=1}^3{\alpha_i}^2V[X_i]+2\sum_{i\neq j}\alpha_i\alpha_j\mathrm{Cov}[X_i,X_j][/tex]

where

[tex]\mathrm{Cov}[X_i,X_j]=E[(X_i-E[X_i])(X_j-E[X_j])]=E[X_iX_j]-E[X_i]E[X_j][/tex]

When we assumed independence, we were granted [tex]E[X_iX_j]=E[X_i]E[X_j][/tex], but this may not be the case if the variables are dependent.

The expected value of the total volume shipped is 98630 ft3. The variance of the total volume shipped is 17423640 ft6. If the variables were not independent, only the expected value would remain correct.

To calculate the expected value of the total volume shipped, we use the linearity of the expectation:

E(Total Volume) = E(27X1 + 125X2 + 512X3)

E(Total Volume) = 27E(X1) + 125E(X2) + 512E(X3)

E(Total Volume) = 27(μ1) + 125(μ2) + 512(μ3)

E(Total Volume) = 27(220) + 125(250) + 512(120)

E(Total Volume) = 5940 + 31250 + 61440

E(Total Volume) = 98630 ft3

To calculate the variance of the total volume shipped, given that X1, X2, X3 are independent:

V(Total Volume) = V(27X1 + 125X2 + 512X3)

V(Total Volume) = 272V(X1) + 1252V(X2) + 5122V(X3)

V(Total Volume) = 729(σ12) + 15625(σ22) + 262144(σ32)

V(Total Volume) = 729(92) + 15625(132) + 262144(82)

V(Total Volume) = 729(81) + 15625(169) + 262144(64)

V(Total Volume) = 59049 + 2639375 + 16777216

V(Total Volume) = 17423640 ft6

If the Xi's were not independent, the expected value would still be correct but the variance would not be correct because the calculation of variance for non-independent variables includes covariance terms that are not present for independent variables.

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Answer:

The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

Step-by-step explanation:

Given:

Weight of a given sample (x) = 2.33 oz

Mean weight (μ) = 1.75 oz

Standard deviation (σ) = 0.22 oz

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

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

Plug in the values and solve for 'z'. This gives,

[tex]z=\frac{2.33-1.75}{0.22}=2.64[/tex]

So, the z-score of the distribution is 2.64.

Now, we need the probability [tex]P(x\geq 2.33 )=P(z\geq 2.64)[/tex].

From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.

But, we need area more than the z-score value. So, the area is:

[tex]P(z\geq 2.64)=1-0.9959=0.0041=0.41\%[/tex]

Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].

Solution:

Given data:

Number between 1 and 15 is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14.

Total numbers between 1 and 15 = 13

N(S) = 13

Multiple of 5 between 1 and 15 = 5, 10

Number of multiples of 5 between 1 and 15 = 2

N(A) = 2

Probability of multiple of 5 between 1 and 15:

[tex]$P(A)=\frac{N(A)}{N(S)}[/tex]

[tex]$P(A)=\frac{2}{13}[/tex]

The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].

Final answer:

To determine the probability of selecting a multiple of 5 between 1 and 15, count the multiples (5 and 10) and divide by the total number (15), resulting in a probability of 2/15.

Explanation:

The question is related to the concept of probability in mathematics. To find the probability that a randomly chosen number between 1 and 15 is a multiple of 5, you simply count how many multiples of 5 are in that range and divide by the total number of possibilities (15). The multiples of 5 between 1 and 15 are 5 and 10, so there are 2 favorable outcomes. Hence, the probability is 2 divided by 15, which equals 2/15.

Answer:

68% of the sample can be expected to fall between 28 and 32 cm

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 = 30

Standard deviation = 2

What proportion of the sample can be expected to fall between 28 and 32 cm

28 = 30 - 2

28 is one standard deviation below the mean

32 = 30 + 2

32 is one standard deviation above the mean.

By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm

Answer:

A sampling method is  independent qualitative when an individual selected for one sample does not dictate which individual is to be in the second sample.

Step-by-step explanation:

individuals selected for one sample do not dictate which individuals are to be in the second sample, while dependent quantitative involves individuals selected to be in one sample are used to determine the individuals in the second sample.

Final answer:

An independent sampling method allows for selection of individuals for one sample without influencing the selection for another, maintaining the integrity of quantitative research.

Explanation:

A sampling method is considered independent when an individual selected for one sample does not dictate which individual is to be in the second sample. This type of sampling strategy ensures that the selection of participants for one sample does not influence the selection for another, which is crucial for maintaining the validity and reliability of research findings, particularly in quantitative research.

In research, particularly in studies involving statistical analysis, it is essential to use appropriate sampling methods to ensure that the results are representative of the wider population. Probability sampling methods are often employed in quantitative research as they provide a means to make generalizations from the sample to the population. Types of probability samples include simple random samples, stratified samples, and cluster samples.

Answer:

This statement can be made with a level of confidence of 97.72%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.1 mm

Standard Deviation, σ = 0.5 mm

Sample size, n = 100

We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

Standard error due to sampling:

[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05[/tex]

P(mean thickness is less than 8.2 mm)

P(x < 8.2)

[tex]P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 8.2) =0.9772 = 97.72\%[/tex]

This statement can be made with a level of confidence of 97.72%.

The z-score calculation reveals that the mean thickness is less than 8.2 mm with approximately 97.50% confidence.

To determine the level of confidence that the mean thickness is less than 8.2 mm, we can calculate the z-score of the sample mean and then find the corresponding percentile in the standard normal distribution. The z-score is calculated by the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.

Using the given information, x = 8.1 mm, μ = 8.2 mm (the mean to test against), σ = 0.5 mm, and n = 100, we get:

z = (8.1 mm - 8.2 mm) / (0.5 mm / √100) = -0.1 / 0.05 = -2

The z-score corresponds to a percentile in the standard normal distribution, which can be found using a z-table or statistical software. A z-score of -2 corresponds to approximately the 2.5th percentile, which means that the sample mean is less than 8.2 mm with about 97.5% confidence.

Therefore, the statement that the mean thickness is less than 8.2 mm can be made with approximately 97.50% confidence.

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Answer:

the minimum is located in x = -5/3 , y= -5/3

Step-by-step explanation:

for the function

f(x,y)=2x + 2y

we define the function g(x)=9x² - 9xy + 9y² - 25  ( for g(x)=0 we get the constrain)

then using Lagrange multipliers f(x) is maximum when

fx-λgx(x)=0 → 2 - λ (9*2x - 9*y)=0 →

fy-λgy(x)=0 → 2 - λ (9*2y - 9*x)=0

g(x) =0 → 9x² - 9xy + 9y² - 25 = 0

subtracting the second equation to the first we get:

2 - λ (9*2y - 9*x) - (2 - λ (9*2x - 9*y))=0

- 18*y + 9*x + 18*x - 9*y = 0

27*y = 27 x  → x=y

thus

9x² - 9xy + 9y² - 25 = 0

9x² - 9x² + 9x² - 25 = 0

9x² = 25

x = ±5/3

thus

y = ±5/3

for x=5/3 and y=5/3 →  f(x)= 20/3 (maximum) , while for x = -5/3 , y= -5/3 →  f(x)= -20/3  (minimum)

finally evaluating the function in the boundary , we know because of the symmetry of f and g with respect to x and y that the maximum and minimum are located in x=y

thus the minimum is located in x = -5/3 , y= -5/3

The z-scores for the given data values are approximately:

560: 0.6

650: 1.5

500: 0

450: -0.5

300: -2

Given the mean (μ) of 500 and the standard deviation (σ) of 100, we can calculate the z-scores for the provided data values:

The z-score (also known as the standard score) measures how many standard deviations a data point is away from the mean. It is calculated using the formula:

z = (x - μ) / σ

Where:

x is the data value

μ is the mean of the sample

σ is the standard deviation of the sample

For x = 560:

z = (560 - 500) / 100 = 0.6

For x = 650:

z = (650 - 500) / 100 = 1.5

For x = 500:

z = (500 - 500) / 100 = 0

For x = 450:

z = (450 - 500) / 100 = -0.5

For x = 300:

z = (300 - 500) / 100 = -2

So, the z-scores for the given data values are approximately:

560: 0.6

650: 1.5

500: 0

450: -0.5

300: -2

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The z-scores for the data values 560, 650, 500, 450, and 300 in a sample with a mean of 500 and a standard deviation of 100 are 0.6, 1.5, 0, -0.5, and -2 respectively. They are computed using the formula z = (X - μ) / σ.

This question refers to the concept of z-scores in statistics, which is a part of Mathematics. A z-score indicates how many standard deviations a given data point is from the mean. The formula to calculate the z-score is: z = (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.

The given sample has a mean (μ) of 500 and a standard deviation (σ) of 100. Let's calculate the z-scores:

So the z-scores for the data values 560, 650, 500, 450, and 300 are 0.6, 1.5, 0, -0.5, and -2 respectively.

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bro I hope someone can help you like an expert or something...

Answer:

B. x = -4/3 and x = 5

Step-by-step explanation:

2 / (x − 2) + 7 / (x² − 4) = 5 / x

2 / (x − 2) + 7 / ((x + 2) (x − 2)) = 5 / x

Multiply both sides by x − 2:

2 + 7 / (x + 2) = 5 (x − 2) / x

Multiply both sides by x + 2:

2 (x + 2) + 7 = 5 (x − 2) (x + 2) / x

Multiply both sides by x:

2x (x + 2) + 7x = 5 (x − 2) (x + 2)

Simplify:

2x² + 4x + 7x = 5 (x² − 4)

2x² + 4x + 7x = 5x² − 20

0 = 3x² − 11x − 20

Factor:

0 = (3x + 4) (x − 5)

x = -4/3 or x = 5

Answer:

a) 125 scoops b) 31 scoops

Step-by-step explanation:

The sink = 4000/3 pi

a) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...

(1/3)(pi)(8)(2^2)=32/3 pi, and...

(4000pi/3)/(32pi/3) =

4000/32 =

125

b) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...

(1/3)(pi)(8)(4^2) = 128pi/3, and...

(4000pi/3)/(128pi/3) =

4000/128 =

31.25, which is approx. 31

Answer:

a) 125

b) 31

Step-by-step explanation:

a) radius = 2

Volume of cup = ⅓×pi×r²×h

= 32pi/3

No. of scoops = 4000pi/3 ÷ 32pi/3

= 125

b) radius = 4

Volume of cup = ⅓×pi×r²×h

= 128pi/3

No. of scoops = 4000pi/3 ÷ 128pi/3

= 31.25

Nearest whole number: 31

Answer:

For part  ;a)

probability that the Atlanta Braves win the World Series=0.40

For part b ;b)

average number of games played regardless of winner =5.8

Step-by-step explanation:

using the  rand() function for the simulation of  the probabilities and  compare with given probabilites if less than above then Atlanta wins otherwise loses

a)

probability that the Atlanta Braves win the World Series=0.40

b)

average number of games played regardless of winner =5.8

Answer:

Step-by-step explanation:

Hello!

The definition of the Central Limi Theorem states that:

Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

If the variable of interest is X: the number of accidents per week at a hazardous intersection.

There is no information about the distribution of this variable, but a sample of n= 52 weeks was taken, and since the sample is large enough you can approximate the distribution of the sample mean to normal. With population mean μ= 2.2 and standard deviation σ/√n= 1.1/√52= 0.15

I hope it helps!

Answer:

Step-by-step explanation:

Hello!

Since the data on the text is missing, I've found a similar exercise with the same questions so I'm going to use the data of that one to explain how to calculate the probabilities.

We have a sample of 323000 high school students

154000 are girls, of those 47700 dropped out

16900 are boys, of those 10300 dropped out

a) To calculate this probability you have to divide the total number of girls by the total number of students:

P(G)= 154000/323000= 0.476 ≅ 0.48

b) In this item you have to calculate the probability of dropouts, the total of students that dropped out are the girls + boys that dropped out:

Dropouts: 47700+10300= 58000

Now you divide it by the total of students to reach the probability:

P(D)= 58000/323000= 0.179≅ 0.18

c) Now you have to calculate the probability of the intersection between the events "female" and "dropped out", symbolically:

P(G∩D)= P(G)*P(D)= 0.48*0.18= 0.0864

d) The probability of the student dropping out given that it is female is a conditional probability and you can calculate using the following formula:

P(D/G)= P(G∩D) = 0.0864 = 0.18

                 P(G)        0.48

e) Now you have to calculate the probability of the student being a femal, given that she dropped out of school, symbolically:

P(G/D)= P(G∩D) = 0.0864 = 0.48

                 P(D)        0.88

Note:

As you can see in d) P(D/G)=P(D)=0.18 and in e) P(G/D)=P(G)=0.48, this means that both events "the student is a girl" and "the student dropped out" are independent.

Remember two events are not independent when the occurrence of one modifies the probability of occurrence of the other.

I hope it helps!

Answer: It's C.5

Step-by-step explanation:

Answer:

the demand curve will shift upward by $20 and the price paid by buyers will decrease by $20.

Step-by-step explanation:

the reduction in the fixed amount of tax from $50 t0 $30 will bring about reduction  of $20 in the price of ticket. the reduction in the price of the ticket, other factors held constant, will brings about change in the demand curve.

The correct answer is D, as supply curve will shift downward by $20, and the effective price received by sellers will increase by $20.

This is so because by reducing the tax, the value that each airline will effectively increase. In this way, as they obtain greater income for each ticket, the airlines will in turn increase the offer of tickets.

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