O There were 9 bags of
candy donated for the
neighborhood party.
Each bag contained
245 pieces. How much
candy did they have
for the party?

Answer:

angle 3=angle (vertical opposite angle)

angle 3 +angle 2=180°(by linear pair)

136°+angle 2=180°

angle 2=180°-136°

angle 2=44°

angle 2=angle 4(by vertical opposite angle)

Answer:

1256 square inches

Step-by-step explanation:

Area of a circle:

A = πr²

Given:

r = 20

Work:

A = πr²

A=(3.14)20²

A = 3.14(400)

A = 1256

Answer:

1256 square inches.

Step-by-step explanation:

20 squared * π = 1256 square inches

Answer:

Step-by-step explanation:

x+3 - x = 3 = width

2(x+3) = 2x + 6 = length

2x + 6 +2x +6 +3 +3 = 4x + 18 = Perimeter of T

The given arguments are proved using logical inference rules.

To show that each of the following arguments is valid, we need to construct a proof using logical inference rules. Here is a proof for the given arguments:

Argument 1:

1. C ⊃ D (Premise)

2. ~(A ∨ B) ∨ C (Premise)

3. ~B ∨ D (Premise)

4. ~(A ∨ B) (Assumption)

5. ~A ∧ ~B (De Morgan's Law, 4)

6. ~B (Simplification, 5)

7. D (Disjunctive Syllogism, 3, 6)

8. ~(A ∨ B) ∨ D (Disjunction Introduction, 7)

9. C (Disjunction Elimination, 2, 8)

10. ~(A ∨ B) ∨ C (Disjunction Introduction, 9)

Therefore, the argument is valid.

Argument 2:

1. C ⊃ D (Premise)

2. ~(A ∨ B) ∨ C (Premise)

3. ~B ∨ D (Premise)

4. ~A ∨ ~B (Assumption)

5. ~(A ∨ B) (De Morgan's Law, 4)

6. C (Disjunction Elimination, 2, 5)

7. C ⊃ D (Premise)

8. D (Modus Ponens, 6, 7)

9. ~B ∨ D (Disjunction Introduction, 8)

10. ~(A ∨ B) ∨ D (Disjunction Introduction, 9)

Therefore, the argument is valid.

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Answer: -9

2 to the power of 0 = 1 then all you have to do is plug in the numbers and simplify

Answer:

-9

Step-by-step explanation:

Well you just subsitute it all and solve from there

s^0= 2^0

p^-2= -3^-2

Anything squared to the power of 0 is 1

so its already 1/smth

the second part is just 3^-2 first which is 1/9 then the negative sign which is -1/9

The number of students that are in college is 7200 if ​Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in​ dormitories.

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

It is given that:

​Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in​ dormitories

Let x be the number of students that are in college.

Then from the question:

The value of x can be found as follows:

x = (6000/5)×6

x = (1200)×6

x = 7200

Thus, the number of students that are in college is 7200 if ​Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in​ dormitories.

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Answer:

1695579

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

The GCF of 42 and 12 is 6

It is advised that lower pressure test should be administered.

Probability of survival of pressure vessel the pressure vessel is tested under the lower pressure test at 58 bar, the probability of survival is given by the sum of the probability of survival if the vessel is one of the 60% that will survive 70 bar and the probability of survival if the vessel is one of the 40% that will fail at 70 bar but will survive 58 bar or less. If P1 represents the probability of survival of the vessel if it is one of the 60% that will survive 70 bar, and P2 represents the probability of survival if it is one of the 40% that will fail at 70 bar but will survive 58 bar or less, then the probability of survival of the vessel, if it is tested under the lower pressure test at 58 bar, is given by:

P = 0.60 x 1 + 0.40 x (1 - 0.60) = 0.76

If the pressure vessel is tested under the higher pressure test at 65 bar, the probability of survival is given by the sum of the probability of survival if the vessel is one of the 60% that will survive 70 bar, and the probability of survival if the vessel is one of the 40% that will fail at 70 bar but will survive 65 bar or less. If P3 represents the probability of survival of the vessel if it is one of the 40% that will fail at 70 bar but will survive 65 bar or less, then the probability of survival of the vessel, if it is tested under the higher pressure test at 65 bar, is given by:

P = 0.60 x 1 + 0.40 x (1 - 0.80 x P3)

The condition that the probability of survival of the vessel, if it is tested under the higher pressure test at 65 bar, is at least 0.95 is therefore:

0.60 + 0.40 x (1 - 0.80 x P3) ≥ 0.95

This simplifies to: P3 ≤ 0.625

Using the above values for P1, P2, and P3, it is clear that the probability of the vessel surviving if tested at the lower pressure of 58 bar is greater than 0.95. Therefore, the lower pressure test should be carried out.

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The AIC, or the Akaike Information Criterion, strikes a balance between model complexity and goodness of fit.

In statistical modeling, it is crucial to find a balance between the complexity of a model and its ability to accurately capture the underlying patterns in the data. On one hand, a complex model with numerous parameters may be able to fit the data very closely, resulting in a low error or residual.

However, such a model runs the risk of overfitting, meaning it may become too specific to the training data and perform poorly when applied to new, unseen data.

On the other hand, a simpler model with fewer parameters may not capture all the nuances of the data and may have a higher error or residual. This is known as underfitting, as the model fails to capture the underlying complexity of the data.

The AIC addresses this trade-off by considering both the goodness of fit and the complexity of the model. It penalizes models with a higher number of parameters, encouraging a balance between model complexity and goodness of fit.

The AIC takes into account the residual sum of squares (RSS) or the likelihood of the model, and adjusts it based on the number of parameters used. The goal is to select the model with the lowest AIC value, indicating a good compromise between complexity and fit.

By striking this balance, the AIC provides a reliable criterion for model selection, allowing researchers and statisticians to choose the most appropriate model for their data while avoiding both overfitting and underfitting.

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Answer:

Hello, Brainly users, hi hows your day going. Great. Yeah thanks for asking. Anyway, the answer is 5.2.

Step-by-step explanation:

Will provide step-by-step explanation as to how I figured it out if I can get brainliest *HINT HINT*. Have a good day. And keep good vibes amid the pandemic

:D

Answer:

5.2

Plz mark me as brainliest.

The division of 32x^3 - 48x^2 - 40x by 8x results in the quotient 4x^2 - 6x - 5 on solving the given equation.

To divide 32x^3 - 48x^2 - 40x by 8x, we divide each term of the dividend by the divisor, 8x.

Dividing 32x^3 by 8x gives us 4x^2, as x^3/x = x^2 and 32/8 = 4.

Dividing -48x^2 by 8x gives us -6x, as -48x^2/8x = -6x.

Dividing -40x by 8x gives us -5, as -40x/8x = -5.

Combining these results, the quotient is 4x^2 - 6x - 5.

The quotient represents the result of dividing the dividend by the divisor, resulting in a polynomial expression without any remainder. Therefore, when dividing 32x^3 - 48x^2 - 40x by 8x, the quotient is 4x^2 - 6x - 5.

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Answer:

400

Step-by-step explanation:

The area of sides and add then up

Based on the illustration above, the value of margin of error M.E is 6.961

Margin of error (M.E) is calculated as the product of critical value (CV) and standard error (SE) of sample mean.

The formula for standard error of sample mean is:

SE = σ/√n

where σ is the population standard deviation and n is the sample size. The formula for margin of error is:

M.E. = CV x SE

where CV is the critical value.

The critical value for a 95% confidence level with 22 degrees of freedom (sample size 23 - 1) is 2.074 (rounded to 3 decimal places).

The sample mean is 37.6 and the population standard deviation is 16.1.

Sample size, n = 23.

Using the formula,

SE = σ/√n

SE = 16.1/√23

SE = 3.365 (rounded to 3 decimal places)

Now, using the calculated value of SE and CV,

ME = CV x SE

ME = 2.074 × 3.365

ME = 6.961 (rounded to 1 decimal place)

Therefore, the margin of error (M.E.) is 6.961 (rounded to 1 decimal place).

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Answer:

I don't know how to do it the subject

Answer:

The experimental probability should get closer to the theoretical probability of 1/6 with more trials.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Experimental probability:

The number of desired outcomes is taken from the results of an experiment.

Theoretical probability:

Found before the experiment happens.

For a large number of trials, the experimental probability will be closer to the theoretical probability.

In this question:

A standard die has 6 sides, one which is 3. So the theoretical probability of rolling a 3 is 1/6.

After six rolls of a standard die, the experimental probability of rolling a 3 is 2/6.

The experimental probability, after six rolls, is 2/6 = 1/3.

What do you expect will happen to the experimental probability if the die is rolled 90 more times?

As the number of trials increase, the experimental probability is expected to get closer to the theoretical probability, which in this case is 1/6.

Answer:

Georgio has to earn $1,800 and borrow $2,700.

Answer:

$18.5

Step-by-step explanation:

You just add up all the costs and divide them by two and then you would get your answer

The given second-order linear differential equation, 1y'' + 4y' + y = x^2 - 4x, can be solved using the method of undetermined coefficients. The particular solution is obtained by assuming a form for the solution and determining the coefficients based on the right-hand side of the equation.

To solve the given differential equation by undetermined coefficients, we first consider the homogeneous equation, which is obtained by setting the right-hand side equal to zero: 1y'' + 4y' + y = 0. The characteristic equation associated with this homogeneous equation is [tex]r^2[/tex]+ 4r + 1 = 0, where r represents the roots of the equation. Solving this quadratic equation, we find two complex conjugate roots: r = -2 ± i.

Since the right-hand side of the original equation is a polynomial of degree 2, we assume a particular solution of the form y_p = A[tex]x^{2}[/tex] + Bx + C. Substituting this assumed form into the original equation, we differentiate it twice to obtain the expressions for y''_p and y'_p, and substitute them back into the original equation. This allows us to equate the coefficients of like powers of x on both sides of the equation.

By comparing coefficients, we find that A = 1 and B = -2. However, the term C is a constant and does not contribute to the differential equation. Hence, the particular solution is y_p = [tex]x^{2}[/tex] - 2x.

Finally, the general solution of the differential equation is given by the sum of the homogeneous solution and the particular solution: y = y_h + y_p. Since the homogeneous solution contains complex roots, it can be expressed as y_h =[tex]e^{-2x}[/tex](C_1cos(x) + C_2sin(x)), where C_1 and C_2 are arbitrary constants. Thus, the complete solution is y = [tex]e^{-2x}[/tex]C_1cos(x) + C_2sin(x)) + [tex]x^2[/tex] - 2x.

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Answer:

324

...................

The null hypothesis (H0) assumes that the proportion of adults annoyed by television violence is equal to or greater than 0.5, while the alternative hypothesis (Ha) assumes that the proportion is less than 0.5.

In this case, the sample proportion is calculated as the number of adults indicating annoyance divided by the total sample size: 1,152/2,400 = 0.48.

Next, we can calculate the test statistic, which follows a standard normal distribution under the null hypothesis. The test statistic formula is z = (p - P) / sqrt(P(1-P)/n), where p is the sample proportion, P is the hypothesized proportion under the null hypothesis (0.5 in this case), and n is the sample size.

Using the given values, we can calculate the test statistic:

z = (0.48 - 0.5) / sqrt(0.5(1-0.5)/2400) ≈ -1.67.

Finally, we compare the test statistic to the critical value. At a significance level of 0.10, the critical value for a one-tailed test is approximately -1.28. Since the test statistic (-1.67) is smaller than the critical value, we reject the null hypothesis. This suggests that there is evidence to support the claim that less than half of all adults are annoyed by violence on television.

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 The answer is not provided in the given options.

To simplify the expression (4x^3 - 3x - 7)(3x^3 + 5x + 3), we can use the distributive property of multiplication.

Multiplying each term in the first expression by each term in the second expression, we get:

(4x^3)(3x^3) + (4x^3)(5x) + (4x^3)(3) + (-3x)(3x^3) + (-3x)(5x) + (-3x)(3) + (-7)(3x^3) + (-7)(5x) + (-7)(3)

Simplifying each term, we have:

12x^6 + 20x^4 + 12x^3 - 9x^4 - 15x^2 - 9x - 21x^3 - 35x - 21

Combining like terms, we get:

12x^6 + (20x^4 - 9x^4) + (12x^3 - 21x^3) + (-15x^2) + (-9x - 35x) + (-21)

Simplifying further, we have:

12x^6 + 11x^4 - 9x^3 - 15x^2 - 44x - 21

Therefore, the correct simplification of (4x^3 - 3x - 7)(3x^3 + 5x + 3) is:

12x^6 + 11x^4 - 9x^3 - 15x^2 - 44x - 21.

Therefore, the answer is not provided in the given options.

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Given that we randomly select 100 drivers ages 16 to 19 from example 4. We are to determine the probability that the mean distance traveled each day is between 19.4 and 22.5 miles. The probability that the mean distance traveled each day is between 19.4 and 22.5 miles is approximately 1.00.

Probability distribution is a function which represents the probabilities of all possible values of a random variable.

When the probability distribution of a random variable is unknown, we can use the Central Limit Theorem (CLT) to estimate the mean of the population.

Let X be the mean distance traveled each day by the 100 drivers ages 16 to 19.

Then, the distribution of X is approximately normal with the mean μ = 20.4 miles and the standard deviation σ = 3.8 miles.

Therefore, we can calculate the z-score as follows; z = (X - μ) / (σ / √n), where X = 19.4 and n = 100.

z₁ = (19.4 - 20.4) / (3.8 / √100)

z₁ = -2.63 and

z₂ = (22.5 - 20.4) / (3.8 / √100)

z₂ = 5.53

Hence, the probability that the mean distance traveled each day is between 19.4 and 22.5 miles is;

P(19.4 < X < 22.5) = P(z₁ < z < z₂).

Using the z-table, the probability is found to be; P(-2.63 < z < 5.53) ≈ 1.00.

Therefore, the probability that the mean distance traveled each day is between 19.4 and 22.5 miles is approximately 1.00.

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Given the heights of the female adults in a country is represented by a random variable that follows the normal distribution N(170,30)1. To enter the tallest 20% of the female adults, a man must be at least 184.87 cm tall.

Solution:It is given that, the heights of the female adults in a country can be represented by a random variable that follows the normal distribution N(170,30)Let X be the height of female adults, then X ~ N(170, 30)

Let P be the probability of the tallest 20% female adults.To find the value of x we need to use the standard normal distribution formula which is given byz = (x - μ) / σWhere,z = standard score or z-scorex = the raw scoreμ = the meanσ = the standard deviation

Now, the probability of the tallest 20% female adults is P = 0.20 or 20%We know that the total area under the normal curve is 1 which means P(X < μ) = 0.5So, P( X > μ) = 1 - P(X < μ) = 1 - 0.5 = 0.5Therefore, 0.5 = P(Z < z) at z = 0.84 from standard normal distribution table,0.84 = (x - μ) / σOn substituting the values,0.84 = (x - 170) / 30x - 170 = 0.84 x 30x - 170 = 25.2x = 195.2So, to enter the tallest 20% of the female adults, a man must be at least 184.87 cm tall.2.

To enter the tallest 1% of the female adults, a man must be at least 201.17 cm tall.

Solution: It is given that, the heights of the female adults in a country can be represented by a random variable that follows the normal distribution N(170,30)Let X be the height of female adults, then X ~ N(170, 30)

Let P be the probability of the tallest 1% female adults.

Now, the probability of the tallest 1% female adults is P = 0.01 or 1%We know that the total area under the normal curve is 1 which means P(X < μ) = 0.5So, P( X > μ) = 1 - P(X < μ) = 1 - 0.5 = 0.5Therefore, 0.5 = P(Z < z) at z = 2.33 from standard normal distribution table,2.33 = (x - μ) / σOn substituting the values,2.33 = (x - 170) / 30x - 170 = 2.33 x 30x - 170 = 69.9x = 239.9 cm

So, to enter the tallest 1% of the female adults, a man must be at least 201.17 cm tall.

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To enter the tallest 1% of the female adults, a man must be at least 104.1 cm tall.

The heights of female adults in a country can be represented by a random variable that follows the normal distribution N(170,30).

The questions and their solutions are:

1. To enter the tallest 20% of female adults, a man must be at least [ ] cm tall.

To solve this, we can use the standard normal distribution table.

Let Z be the standard normal distribution.

To find the corresponding Z-score to the 20th percentile, we use the standard normal distribution table.

P(Z < z) = 0.20, where P(Z < z) is the area under the standard normal distribution curve to the left of z.

z = -0.84 (rounded to 2 decimal places).

Using the formula z = (X - µ) / σ, we can solve for X, the height of the woman:

[tex]z = (X - µ) / σX = σz + µX = 30(-0.84) + 170X = 147.8[/tex] (rounded to the nearest tenth of a cm)

Therefore, to enter the tallest 20% of the female adults, a man must be at least 147.8 cm tall.

2. To enter the tallest 1% of female adults, a man must be at least [ ] cm tall.

P(Z < z) = 0.01

z = -2.33 (rounded to 2 decimal places).

Using the formula z = (X - µ) / σ, we can solve for X, the height of the woman:

[tex]z = (X - µ) / σX = σz + µX = 30(-2.33) + 170X = 104.1[/tex] (rounded to the nearest tenth of a cm)

Therefore, to enter the tallest 1% of the female adults, a man must be at least 104.1 cm tall.

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Given the random variable X with specific expected values and the equation Y = a + bx + cX, we are asked to find the correlation coefficient p(X,Y).

The correlation coefficient between two random variables X and Y is given by the formula:

p(X,Y) = Cov(X,Y) / sqrt(Var(X) * Var(Y))

To calculate the correlation coefficient, we need to find the covariance (Cov(X,Y)) and the variances (Var(X) and Var(Y)).

Given the expected values, we can calculate the required values as follows:

Cov(X,Y) = E[XY] - E[X]E[Y]

Var(X) = E[[tex]X^2[/tex]] - [tex](E[X])^2[/tex]

Var(Y) = E[tex][(Y - E[Y])^2][/tex]

Using the provided expected values, we can substitute them into the formulas:

Cov(X,Y) = E[XY] - E[X]E[Y] = E[(a + bx + cX)X] - (0)(E[a + bx + cX]) = E[aX + b[tex]X^2[/tex] + c[tex]X^2[/tex]] = a(E[X]) + b(E[[tex]X^2[/tex]]) + c(E[[tex]X^3[/tex]])

Var(X) = E[[tex]X^2[/tex]] - [tex](E[X])^2[/tex] = 1 - [tex](0)^2[/tex] = 1

Var(Y) = E[(Y - [tex]E[Y])^2[/tex]] = E[(a + bx + cX - [tex](E[a + bx + cX]))^2[/tex]] = E[[tex](a + bx + cX)^2[/tex]]

Using the provided values for E[[tex]X^3[/tex]] and E[[tex]X^4[/tex]], we can simplify the expressions further and calculate the values.

Once we have the values of Cov(X,Y), Var(X), and Var(Y), we can substitute them into the correlation coefficient formula to find p(X,Y).

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Answer:

Step-by-step explanation:

x = 6

arc RT = 49°

"It is a line that intersects the circle exactly at one point."

"It is a line that intersects circle at two points."

For given example,

arc QT = (27x + 3)°

arc RT = (9x – 5)°  

∠RST = (10x – 2)°

From figure we can observe that line ST is tangent and line SQ is secant.

∠RST is the angle subtended by tangent ST and secant SQ

We know, the angle subtended by the tangent and the secant is half the difference of the measures of the intercepted arcs.

⇒ ∠RST = (QT - RT)/2

⇒ 10x - 2 = [(27x + 3) - (9x - 5)] /2

⇒ 2(10x - 2) = 27x + 3 - 9x + 5

⇒ 20x - 4 = 18x + 8

⇒ 20x - 18x = 8 + 4

⇒ x = 6

So, arc RT would be,

⇒ 9x - 5 = 9(6) - 5

⇒ 9x - 5 = 49°

Therefore, arc RT = 49°

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