you randomly select 100 drivers ages 16 to 19 from example 4. what is the probability that the mean distance traveled each day is between 19.4 and 22.5 miles?

Answer:

$86.94

Step-by-step explanation:

All we need to do is subtract his weekly play from his pay after taxes and his other taxes.

   $625

   $499.61

_ $38.45

____________

   $86.94

Have a nice day, fam. Spread The Love.

Answer:

12x^2 + 30x+12    

Step-by-step explanation:

3x+6) (4x+2)

3x          6

12x^2       24x              4x

6x           12                   2

use a punnet square type deal here.

12x^2 + 24x + 6x + 12

12x^2 + 30x+12    

And so your answer should be: 12x^2 + 30x+12  

Answer:

7.07106781187

Step-by-step explanation:

Let us use the  distance formula:

[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}[/tex]

x2=2

x1=7

y2=4

y1=9

[tex]d = \sqrt{(2 - 7)^{2} + (4-9)^2}[/tex]

[tex]d = \sqrt{(-5)^2 + (-5)^2}[/tex]

[tex]d = \sqrt{25 + 25}[/tex]

[tex]d = \sqrt{50}[/tex]

d=7.07106781187 (round to whatever digit neccesary)

Hope this helps!!

Knowing the roots first write the equation in factored form:

(X - 3/4)(x +5) = 0

Now use the FOIL method ( multiply each term in one set of parentheses by each term the other set:

X•x + x•5 -3/4•x -3/4•5

Simplify:

X^2 + 5x -3/4x -3 3/4

Combine like terms:

X^2+ 4 1/4x - 3 3/4

Answer:

Fr. It’s annoying. My mom yells at me for not passing lolz.

Step-by-step explanation:

Answer:

Ignore them and re-upload the questions. The links are very bad.

Step-by-step explanation:

Answer:

first add the numbers that have the same variable s x7+8x=15x

then the numbers first add and then you subtract

1+7=8

3-8= -5

15x-5

Answer:

8x+x7+9

Step-by-step explanation:

subtracted one term from another

subtracted 1 from 3 to get 2

x7=+8x+2+7

add two terms together

add 2 and 7 toget 9

x7+8x+9

Answer:

What I think it might be is 62.3

The value of K is given as follows:

K = 1/12.

A probability is calculated as the division of the number of desired outcomes by the number of total outcomes.

For a distribution, we have that the sum of all the probabilities in the distribution must be of 1.

This means that the cumulative distribution at the least value is of 1, hence the equation is given as follows:

17/12 - 5K = 1

Then the value of K is obtained as follows:

5K = 17/12 - 1

5K = 17/12 - 12/12

5K = 5/12

K = 1/12.

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Answer:

I'm not too sure what you mean but since there was no image I tried to look up how to tell if angles are corresponding based on angle degree and the most I got was the angles will be equal but only if there is a transversal cur but none of the answers equal each other

Step-by-step explanation:

so what advice I can give you is corresponding angles are usually the same degree but on a different level. like on the first parallel line there is an angle and on the second (or whatever) there is another angle (on the same side of the first) -meaning they will both be on the right or the left- and yeah there you go

if you have any questions comment and I'll try my best to help

The congruent 10 - labeled color spinner isn't attached and could not be found.

However, we could solve the problem hypothetically.

Answer:

Kindly check explanation

Step-by-step explanation:

The events here are independent events :

Probability = required outcome / Total possible outcomes

Total possible outcomes = number of divisions on spinner = 10

Required outcome = number of specified color

Therefore, Probability that first spin lands on yellow :

Number of yellow divisions on spinner / total number of divisions

Let number of yellow divisions = 3

Therefore. Probability that first spin lands in yellow = 3 / 10

Probability that second spin lands on green :

Number of green divisions on spinner / total number of divisions

Let number of green divisions = 2

Therefore. Probability that first spin lands on green = 2 / 10

P(first) * P(second)

3/10 * 2/10 = 3/10 * 1/5 = 3/50

Recall ; this is an hypothetical situation and not the original problem.

The probability that the first spin will land on yellow and the second spin will land on green is; 2/5

The spinner is missing but from online search, it is discovered that from the 10 congruent sections labeled, the colors are as follows;

Yellow - 2 sections

Green - 2 sections

Red - 3 sections

Blue - 3 sections

Thus, probability of getting yellow on first spin is;

P(yellow on first spin) = 2/10

Probability of getting green on second spin is;

P(green on second spin) = 2/10

Thus, probability that the first spin will land on yellow and the second spin will land on green is;

P(1st on yellow and 2nd on green) = 2/10 + 2/10

P(1st on yellow and 2nd on green) = 2/5

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The probability of Sunshine Delight winning given that it is raining is approximately 0.79. The probability that a document is spam if it contains the word "llwws" is approximately 0.50.

In the case of Sunshine Delight winning, given that it is raining, the probability is directly provided as 0.79.

For the probability of a document being spam if it contains the word "llwws," the information shows that the probability of the word "llwws" appearing in a spam document is 0.50. However, without further information about the overall prevalence of spam and non-spam documents, we cannot directly determine the probability that a document containing "llwws" is spam. Additional context about the overall occurrence of spam and non-spam documents is required to calculate the requested probability.

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Answer:

.........

it is -7

Before the price change, the optimal consumption choice (A) is determined by equalizing the marginal utilities of x and y.

Using the utility function U(x, y) = xy, and considering prices of Px = 1 and Py = 2, along with an income of 40, we find that the utility equalization condition leads to y = x/2.

With the budget constraint of x + 2y = 40, we solve for x = 20 and y = 10, giving the optimal consumption choice (20, 10) labeled as point A on the graph.

After the price change, with the price of y falling to 1, the consumer aims to maintain the same utility level. Setting MUX/Px = MUY/Py, we find x = y, and using the budget constraint x + y = 40, we get x = y = 20, resulting in the optimal consumption choice (20, 20) labeled as point B.

The substitution effect is illustrated by a movement from A to B along the budget constraint. The new optimal consumption choice (C) after the price change is determined by the new budget constraint x + y = 40, where x = 30 and y = 10, labeled as point C on the graph.

The income effect is depicted by the parallel shift of the budget constraint line outward from the origin.

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Answer:

From greatest to least it would be 3.66666,[tex]\sqrt{11}[/tex],2(1/4),-2.5,-3.97621

Step-by-step explanation:

Just type in a calc

Answer:

[tex]\sqrt{11}=3.31[/tex], -2.5, [tex]2\frac{1}{2}= 2.25[/tex], 3.6, -3.97621...

Step-by-step explanation:

Greatest to least would be:

3.6,  [tex]\sqrt{11}[/tex], [tex]2\frac{1}{4}[/tex], -2.5, -3.97621...

Least to greatest would be:

-3.97621, -2.5, [tex]2\frac{1}{4}[/tex], [tex]\sqrt{11}[/tex], 3.6

Hopefully, that helps.

Answer:

C and D.

Step-by-step explanation:

For C, 4 * 2x = 8x and 4*1 = 4.

For D, 5x+3x = 8x and 2+2 = 4.

Answer:

A)  8( x + 2 ) is the only equal expression.

The calculated t-value is greater than 1.734, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

An engine additive is being tested to see whether it can effectively increase gas mileage for a number of vehicles. Twenty assorted vehicles had their gas mileage, in miles per gallon, measured. Then, the engine additive was placed into each of the engines, and the gas mileage was measured again. Let µd be the true mean difference in the gas mileage before and after the engine additive is placed into the engines. We want to test the hypothesis that the engine additive has no effect. The null hypothesis is: H0: µd = 0The alternative hypothesis is: Ha: µd > 0 (one-tailed test)Assuming that the difference in gas mileage before and after the engine additive is approximately normally distributed, we can use a one-sample t-test. The test statistic is given by: $$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$$Where, $\bar{x}$ is the sample mean difference in gas mileage, μ is the hypothesized population mean difference, s is the sample standard deviation of the differences, and n is the sample size. We can use a significance level of α = 0.05.To determine the critical value of the t-distribution, we need to find the degrees of freedom (df). Since we have a sample size of n = 20, we have n - 1 = 19 degrees of freedom. Using a t-distribution table with 19 degrees of freedom and a significance level of 0.05 for a one-tailed test, we get a critical value of t = 1.734. If the calculated t-value is greater than 1.734, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

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The intersection of N(x) and N'(x), denoted by N(x)∩N'(x), is an open neighborhood of x. Since N(x)∩N'(x) ⊆ N(x) ⊆ U and N(x)∩N'(x) ⊆ N'(x) ⊆ X\U, we can conclude that N(x)∩N'(x) ⊆ UA.

Since N(x)∩A is a non-empty set contained in A\U, we have shown that every point in (A\U)' has a neighborhood contained in A\U. Therefore, (A\U)' is open in X, which implies that A\U is closed in X.

To show that if U is open in X and A is closed in X, then UA is open in X and A\U is closed in X, we need to prove two statements:

Let's prove these statements one by one:

Let x be an arbitrary point in UA. Since x is in UA, it must belong to U as well as A. Since U is open in X, there exists an open neighborhood N(x) of x that is completely contained within U. Now, since x is in A, it is also in X\U (complement of U in X). As A is closed in X, X\U is closed in X, which means its complement, U, is open in X. Therefore, there exists an open neighborhood N'(x) of x that is completely contained within X\U.

Now, consider the intersection of N(x) and N'(x), denoted by N(x)∩N'(x). This intersection is an open neighborhood of x. Since N(x)∩N'(x) ⊆ N(x) ⊆ U and N(x)∩N'(x) ⊆ N'(x) ⊆ X\U, we can conclude that N(x)∩N'(x) ⊆ UA.

Since N(x)∩N'(x) is an open neighborhood of x completely contained within UA, we have shown that UA is open in X.

Let x be an arbitrary point in (A\U)'. Since x is not in A\U, it means that x must either be in A or in U (or both). If x is in A, then x is not in A\U. Therefore, x is in U.

Since x is in U and U is open in X, there exists an open neighborhood N(x) of x that is completely contained within U. Now, consider the intersection of N(x) and A. Since x is in A, N(x)∩A is a non-empty set. Let y be any point in N(x)∩A.

We know that N(x)∩A ⊆ U∩A ⊆ A\U, because if y was in U, it would contradict the assumption that y is in A. Therefore, N(x)∩A is a subset of A\U.

Since N(x)∩A is a non-empty set contained in A\U, we have shown that every point in (A\U)' has a neighborhood contained in A\U. Therefore, (A\U)' is open in X, which implies that A\U is closed in X.

Hence, we have shown that if U is open in X and A is closed in X, then UA is open in X, and A\U is closed in X.

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Answer: 1,953

Step-by-step explanation:

23,440 divided by 12

Required correct option is Since the two confidence intervals overlap, neither gender appears to be substantially more successful in their challenges.

a) In order to find the 99% confidence interval for the percentage of successful challenges, the formula is given below: Lower limit of CI: upper limit of CI: The confidence interval for the percentage of successful challenges is (19.68%, 34.97%).b) In part (a), we found the confidence interval for the percentage of successful challenges among women playing singles matches. 99% confidence interval for the percentage of successful challenges made by the men playing singles matches is 21.7%± 4.93%.Here, the two confidence intervals overlap, therefore, neither gender appears to be substantially more successful in their challenges.Therefore, option (C) is the correct answer.

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Answer:

trapazoid

Step-by-step explanation:

.....

Answer:

x = 10[tex]\sqrt{3}[/tex]

Step-by-step explanation:

Using the cosine ratio in the right triangle and the exact value

cos30° = [tex]\frac{\sqrt{3} }{2}[/tex] , then

cos30° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{x}{20}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )

2x = 20[tex]\sqrt{3}[/tex] ( divide both sides by 2 )

x = 10[tex]\sqrt{3}[/tex]

Answer:

If I'm correct I think its answer B

Step-by-step explanation:

I'm not sure but i hope this help

Answer:

second option

-∞ < x < 1

Step-by-step explanation:

The solution to the given initial value problem [tex]cos2x sin x dy + (cos?x)y = 5, 7(7/3) = 4 dx[/tex]  is [tex]y(x) = (5/6) * (cos^2(x) - 1) + (28/9) * sin(x) + (40/9) * cos(x)[/tex].

To solve the initial value problem, we can use the method of integrating factors. The integrating factor is found by taking the exponential of the integral of the coefficient of y with respect to x. In this case, the coefficient is cos(x). So, the integrating factor is [tex]e^(^\int ^{ cos(x)} \, ^d^x^) = e^s^i^n^(^x^))[/tex].

Multiplying the given differential equation by the integrating factor, we obtain:

[tex]e^{sin(x)} * [cos^{2x} sin(x) dy + cos(x) y] = 5e^{sin(x)} dx[/tex]

By using the product rule and simplifying, we have:

[tex]d/dx [e^{sin(x)} * y * cos^{2x}] = 5e^{sin(x)}[/tex]

Integrating both sides with respect to x, we get:

[tex]e^{sin(x)} * y * cos^{2(x)} = \int {5e^{sin(x)}} \, dx = 5e^{sin(x)} + C,[/tex]

where C is the constant of integration.

Simplifying and solving for y, we obtain:

[tex]y(x) = [(5/6) * (cos^{2x} - 1) + (28/9) * sin(x) + (40/9) * cos(x)] / (e^{sin(x)} * cos^{2(x)} + Ce^{-sin(x)},[/tex]

where C is the constant of integration.

To find the value of C, we can use the initial condition y(7/3) = 4. Substituting this into the equation and solving for C, we can determine the specific solution.

In conclusion, the solution to the given initial value problem is [tex]y(x) = (5/6) * (cos^{2x} - 1) + (28/9) * sin(x) + (40/9) * cos(x)[/tex], subject to the specified initial condition.

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Answer:

D. 105°

Step-by-step explanation:

The events A and B are not independent. The occurrence of event B affects the probability of event A.

To determine whether events A and B are independent,

we need to check if the probability of event A occurring is affected by the occurrence of event B, and vice versa.

Probability of event A: Since we are flipping two coins,

the probability of getting heads on each flip is 1/2.

Therefore, the probability of getting two consecutive heads is

[tex](1/2) \times (1/2) = 1/4[/tex]

Probability of event B: The first or last flip yielding tails means there are two possibilities:

either the first flip is tails and the second flip is any outcome,

or the first flip is any outcome and the second flip is tails.

Each of these individual possibilities has a probability of

[tex](1/2) \times (1/2) = 1/4[/tex]

Hence, theprobability of event B is 1/4 + 1/4 = 1/2.

Since the probability of event A is 1/4 and the probability of event B is 1/2, and 1/4 ≠ 1/2,

we can conclude that events A and B are not independent.

The occurrence of event B (first or last flip yielding tails) affects the probability of event A (two consecutive flips yielding heads).

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(i)  When using three classifiers with a majority decision, the probability of correctly classifying the new object is 0.973.

(ii) The probability of correct classification increases as the number of classifiers increases

(i) All three classifiers make the correct classification: p * p * p = 0.7 * 0.7 * 0.7 = 0.343.

Two classifiers make the correct classification and one classifier makes an error:

(p * p * q) + (p * q * p) + (q * p * p) = 3 * (0.7 * 0.7 * 0.3) = 0.441.

One classifier makes the correct classification and two classifiers make errors:

(p * q * q) + (q * p * q) + (q * q * p) = 3 * (0.7 * 0.3 * 0.3)

= 0.189.

The probability of the new object being correctly classified is the sum of these probabilities:

0.343 + 0.441 + 0.189

= 0.973.

(ii)  The probability of correct classification increases as the number of classifiers increases. This is because the probability of a majority decision being correct is the probability that at least two of the classifiers make the correct decision. The more classifiers there are, the more likely it is that at least two of them will make the correct decision.

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