2+2x+x2 but x=5 need help

a. Probability of losing the second thumb war is also 0.38.

b. Probability of losing the third thumb war is also 0.38.

Given that Nate loses 38% of all thumb wars.

We can use the probability of losing the thumb war as

p=0.38.

The probability of winning is

1-0.38=0.62.

a) Probability of losing two thumb wars in a row is:

P(loses two in a row) = P(loses first) × P(loses second)

The probability of losing the first thumb war is 0.38.

The probability of losing the second thumb war is also 0.38

So,

P(loses two in a row) = (0.38) × (0.38)

= 0.1444

b) Probability of losing three thumb wars in a row is:

P(loses three in a row) = P(loses first) × P(loses second) × P(loses third)

The probability of losing the first thumb war is 0.38.

The probability of losing the second thumb war is also 0.38

The probability of losing the third thumb war is also 0.38.

So,

P(loses three in a row) = (0.38) × (0.38) × (0.38)

= 0.054872

c) When events are independent, their complements are independent as well.

If the probability of winning the thumb war is p, then the probability of losing the thumb war is 1-p.

The complement of "losing three in a row" is "not losing three in a row".

P(not loses three in a row) = 1 - P(loses three in a row)P(not loses three in a row)

= 1 - 0.054872

= 0.945128

The probability that Nate loses three thumb wars in a row, but does not lose four in a row is the probability of losing three minus the probability of losing four in a row.

P(loses three but not four in a row) = P(loses three in a row) - P(loses four in a row)

P(loses four in a row) = P(loses three in a row) × P(does not lose the next)

P(loses four in a row) = (0.38) × (0.38) × (0.38) × (0.62)

= 0.02081416

P(loses three but not four in a row) = 0.054872 - 0.02081416

= 0.03405784

The probability that Nate loses two thumb wars in a row is 0.1444

The probability that Nate loses three thumb wars in a row is 0.054872.

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Answer:

Triangle JKLM ~ Triangle STUV

Answer:

Point form -

[tex](2-\frac{5}{2})[/tex]

Equation form -

[tex]x=2, y=-\frac{5}{2}[/tex]

Step-by-step explanation:

if you have any questions feel free to ask

Answer:

i know right? It's so difficult till I'm brainless when I try to solving it

Answer:

same tbh

Step-by-step explanation:

Answer: 60

Step-by-step explanation:

The equation is separable. The solution to the initial value problem is y(t) = y(0) = -3.

A. The equation is separable. The solution to the initial value problem is y(t) = ___

To determine whether the given differential equation is separable, we need to check if it can be written in the form dy/dt = g(t) * h(y), where g(t) is a function of t only and h(y) is a function of y only.

In this case, the equation is dy/dt = 2ty + 1. We can rearrange it as:

dy = (2ty + 1) dt

Now, we can see that we have both y and t terms on the right-hand side, indicating that the equation is not yet separable.

Therefore, the correct choice is B. The equation is not separable.

Unfortunately, no solution can be provided for the initial value problem y(0) = -3 since the equation is not separable.

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A 98% confidence interval for the variance of all student's grades is [9.41, 31.41].

The degrees of freedom (df) are (n - 1) = (20 - 1) = 19.

A confidence interval is given by the formula: $[{\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f},\ { \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}]$ where ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}$ is the lower percentile of the chi-square distribution with (n - 1) degrees of freedom, ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}$ is the upper percentile of the chi-square distribution with (n - 1) degrees of freedom, and s² is the sample variance.

Lower percentile (L.P) = ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}$

Upper percentile (U.P) = ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}$

Now, α = 0.02, n = 20, s² = 16, and df = 19.

Lower percentile:L.P = ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f} = { \chi }_{\frac{0.02}{2},19}^{2}\frac{16}{19} = 9.410$

Upper percentile:U.P = ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f} = { \chi }_{1-\frac{0.02}{2},19}^{2}\frac{16}{19} = 31.410$

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The 98% confidence interval for the variance of all student's grades is (10.12, 30.29). Hence, option A is the correct answer.

Given data are: The sample size, n is 20, the mean of the sample is 72 and the variance of the sample, s² is 16.

Option A is the correct answer.

Step-by-step explanation: Given, Sample size, n = 20

Mean of the sample, [tex]\bar x = 72[/tex]

Variance of the sample, [tex]s^{2} =16[/tex].

We have to find a 98% confidence interval for the variance of all student's grades.

Assumption: Grades are distributed normally.

So, the formula for the confidence interval for the variance is: [tex]((n-1)s^{2} / \chi^{2} \alpha /2, (n-1)s^{2} / \chi^{2} 1-\alpha /2)[/tex]

Where, [tex]\alpha = 1-0.98[/tex]

= 0.02

n-1 = 19

Degrees of freedom [tex]\chi^{2}\alpha /2[/tex] and [tex]\chi^{2}²1-\alpha /2[/tex] are the critical values of chi-square distribution with (n-1) degrees of freedom.

From the chi-square distribution table, we can find the values of [tex]\chi^{2}\alpha /2[/tex] and [tex]\chi^{2}²1-\alpha /2[/tex] such that the area between these values is 0.98.

Here, [tex]\alpha = 0.02[/tex] (as 98% confidence interval)

[tex]\alpha /2 = 0.02/2[/tex]

= 0.01

Using the above values, we get,

[tex]\chi^{2}0.01/2,19 = 8.907[/tex] and

[tex]\chi^{2}1-0.01/2,19 = 32.852[/tex]

Now, using the formula of confidence interval for the variance, we get,

[tex]((n-1)s^{2} / \chi^{2} \alpha /2, (n-1)s^{2} / \chi^{2} 1-\alpha /2)=((20-1) \times 16 / 32.852, (20-1) \times 16 / 8.907)[/tex]

We get, (10.12, 30.29).

Therefore, the 98% confidence interval for the variance of all student's grades is (10.12, 30.29). Hence, option A is the correct answer.

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Answer:

1 1/16

Step-by-step explanation:

you you have to add 7/8 + 3/16 to get the answer of one whole and 1/16

(a) The possible remainders when 12 + 16 + 20 is divided by 11 are 8 and 19.

(b) For every integer n, the expression 121 + n^2 + 16n + 20 is always divisible by 11.

In the first statement, we are asked to prove that a - c divides ab + cd if and only if a - c divides ad + bc. To prove this, we can use the property of divisibility. If a - c divides ab + cd, then there exists an integer k such that ab + cd = (a - c)k. Similarly, if a - c divides ad + bc, there exists an integer m such that ad + bc = (a - c)m. By rearranging the terms, we can express k and m in terms of a, b, c, and d. By substituting these expressions into the equation ab + cd = (a - c)k, we can show that ab + cd = (a - c)(ad + bc)m. Thus, proving the equivalence.

In the second statement, we are asked about the possible remainders when 12 + 16 + 20 is divided by 11. To find the remainder, we can calculate the sum as 48 and then divide it by 11. The quotient is 4 with a remainder of 4. Hence, 12 + 16 + 20 leaves a remainder of 4 when divided by 11.

To prove that 121 + n^2 + 16n + 20 is divisible by 11 for every integer n, we can use algebraic manipulation. By factoring the expression as (n + 4)(n + 4) + 16(n + 4), we can rewrite it as (n + 4)^2 + 16(n + 4). Now, we notice that both terms have a common factor of (n + 4). Factoring it out, we get (n + 4)(n + 4 + 16). This simplifies to (n + 4)(n + 20). Since both factors contain n + 4, we can conclude that the expression is divisible by 11 for any integer n.

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The combination of pies she can make are:

one cherry pie, one peach pie and one apple pie

A bar graph is defined as a diagram in which the numerical values of variables are represented by the height or length of lines or rectangles of equal width.

We are given the following parameters:

Number of apples to make one apple pie = 6

Number of peaches to make one peach pie = 9

Number of cherries to make one cherry pie = 32

From the bar graph, she has:

26 apples

12 peaches

34 cherries

Since she wants to make 3 pies, then she can make:

one cherry pie, one peach pie and one apple pie

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The calculated value of the standardized score for this sample average is 2

From the question, we have the following parameters that can be used in our computation:

Population mean = 2

Population standard deviation = 0.6

Sample size = 36

Sample mean = 2.2

The standardized score for this sample average can be calculated using

z = (Score - Sample mean)/(Sample standard deviation)

Where

Standard deviation = Population standard deviation/√n

So, we have

Standard deviation = 0.6/√36

Evaluate

Standard deviation = 0.1

So, we have

z = (2.2 - 2)/0.1

Evaluate

z = 2

Hence, the standardized score for this sample average is 2

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Answer:

Below.

Step-by-step explanation:

The formula for a GS is:

a1, a1r,  a1 r^2, a1r^3 .....

Examples are:

1. First  term a1 =2 and common ratio r = 3, the first 4 terms are:

2,  2*3,  2*3^2,  2*3^3

= 2, 6, 18, 54

2. a1 = 1 , r = 0.5:

1, 0.5,  0.25,  0.125  .....

3. a1 = -4,  r = -2:-

-4,  8, -16,  32  ...

4.  a1 = 0.1,  r = -3:

0.1,  -0.3,  0.9, -2.7 .....

5. a1 = 2, r = 1/3:

2,  2/3,  2/9,  2/27 .....

Answer:

Step-by-step explanation:

se gana 10000 y 2

Answer:

1.4

Step-by-step explanation:

Probability of selecting a red counter :

Number of red counter / total number of counters

5 / (5 + 3) = 5 /8 = 0.625

Probability of selecting a blue counter = 0.375

Let number of red counters :

0, 1 or 2

0 red counters :

Expected value = x * p(x)

0 + (0.625*1) + (0.390625 * 2)

0 + 0.625 + 0.78125 = 1.40625

Expected number of red counters = 1.4

Answer:

yes and no....It depends if u got the answer correct then u got ur mark but if the teacher expects your answer a certain way then u might lose maybe just 1 mark but dont stress it looks good and u should get full marks

Since the mean and variance of y are the same as those of a normal random variable, y is a normal random variable.

The given moment generating function is Mx(t) = e^(t(4 + 5t + 17t^2/2)), where t is the moment. The moment generating function of a normal random variable is given by Mx(t) = e^(μt + σ^2t^2/2).Comparing the two, we get:μ = 4σ^2 = 17/2We can now compute the first and second moments of y, using the moment generating function: My(t) = e^(t(34 + 5t)) × e^(t(4 + 5t + 17t^2/2))= e^(34t + 9t^2 + 17t^3/2 + 4t + 5t^2) = e^(34t + 14t^2 + 17t^3/2)So,μy = My(0) = 1 and σy^2 = My''(0) - My'(0)^2= (17/2) - 4 = 9/2

Since the mean and variance of y are the same as those of a normal random variable, y is a normal random variable.

This completes the proof using the moment generating function uniqueness theorem.

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The moment generating function (MGF) of a random variable is a mathematical function that uniquely defines the probability distribution of that variable. Therefore, if the MGF of a random variable is the same as that of a known distribution, the random variable follows that distribution.

This is known as the uniqueness theorem of moment generating functions.

Using the given moment generating function of a random variable X (Max+b)(+).etMy.lat), we need to prove that X follows a normal distribution. Here, Max+b = a, My = m, and  = s (standard deviation).

Using the MGF, we can find the moments of the distribution. Differentiating the MGF 'n' times gives the nth moment about zero. We can use this to find the mean and variance of the distribution.

Therefore, the mean of the distribution is the first derivative of the MGF at t=0,

and the variance is the second derivative of the MGF at t=0.

Using facts (i) and (ii), we can write the MGF of X as:

(Mx(t)) = [(b + mt) + a(s^2 + m^2)/2] / [(s^2/2) + (t^2/2)]

Taking the first derivative of Mx(t) and substituting t=0, we get:

[tex]E(X) = [(b + m*0) + a(s^2 + m^2)/2] / [(s^2/2) + (0^2/2)]E(X) = (b + am) / (s^2/2) + 0E(X) = (2(b + am))/s^2[/tex]

This gives us the mean of the distribution as

(2(b + am))/s^2

Taking the second derivative of Mx(t) and substituting t=0, we get:

[tex]Var(X) = [(b + m*0) + a(s^2 + m^2)/2] / [(s^2/2) + (0^2/2)]Var(X) = [a(s^2 + m^2)/2] / (s^2/2) + 0Var(X) = a(s^2 + m^2)/s^2 - 1[/tex]

We know that the MGF of a normal distribution with mean m and variance s^2 is given by:

[tex](Mn(t)) = e^(mt + s^2t^2/2)[/tex]

Comparing this with the given MGF of X, we see that they are equal.

Therefore, X follows a normal distribution with mean (2(b + am))/s^2 and variance a(s^2 + m^2)/s^2 - 1.

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A

see attached for explanation

I hope it helps

a)The force on electron B doubles the force acting on electron A

The correct answer is (b)

b)The path radius for electron A is half that of electron B .

The correct option is (a)

Moving charges are sources of the magnetic field and also recipients of the magnetic interaction. Stationary charges interact with the electric field but are not influenced by the magnetic field. The magnetic force acting on a moving charge is proportional to the value of the charge, to its velocity, and the absolute value of the magnetic field. The absolute value of the force also depends on the relative direction of the magnetic field and the velocity.

The magnetic force acting on the electron equals:

[tex]F_A =ev_A(B).....(1)\\\\F_B =ev_B(B).....(2)[/tex]

proportional to the vector product between the velocity and the magnetic field. The magnetic field and the velocity vectors are perpendicular, consequently, the absolute value of the vector product reduces to the product of the absolute values,

[tex]|F_A| =e|v_A||B|\\\\|F_B| =e|v_B||B|[/tex]

The only difference between the forces strives in the value of the velocities. The velocity of electron B is twice that of electron A . Therefore, the force on electron B doubles the force acting on electron A

The correct answer is (b)

b) To analyze the orbit of the electrons due to the magnetic force let us use Newton's second law. The magnetic force is always perpendicular to the velocity not changing its absolute value but only its direction. The acceleration resulting from the force is centripetal. In scalar form

[tex]F_A =ev_A(B)=ma_c_A\\\\F_B =ev_B(B)=ma_c_B[/tex]

Substituting the centripetal acceleration in terms of the velocities and the radiuses,

[tex]ev_AB=m\frac{v^2_A}{R_A} \\\\ev_BB=m\frac{v^2_B}{R_B}[/tex]

Solving for the radius,

[tex]R_A=\frac{mv_A}{eB} \\\\R_B=\frac{mv_B}{eB}[/tex]

The orbital radius is directly proportional to the velocity of the electron. As a result the path radius for electron A is half that of electron B .

The correct option is (a)

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To find the confidence interval for p1 - p2 at a given level of confidence 95%, with x1 = 354, n1 = 545, x2 = 406, n2 = 596, we need to first calculate the point estimate for the difference in proportions:

$$\hat{p_1} = \frac{x_1} {n_1} = \frac {354}{545} \approx. 0.6495$$$$\hat{p_2} = \frac{x_2} {n_2} = \frac {406}{596} \approx. 0.6822$$

Therefore, the point estimate of the difference in proportions is: $$\hat{p_1} - \hat{p_2} = 0.6495 - 0.6822 \approx. -0.0327$$. Now, we can use the formula for the confidence interval for the difference in proportions:

$$\text{Confidence interval} = (\hat{p_1} - \hat{p_2}) \pm z_{\alpha/2} \sqrt{\frac{\hat{p_1}(1 - \hat{p_1})}{n_1} + \frac{\hat{p_2}(1 - \hat{p_2})}{n_2}}$$where z_{\alpha/2} is the z-score for the level of confidence 95% (or 0.95), which is approximately 1.96

Using this information, the confidence interval for p1 - p2 at a 95% level of confidence is: $$(0.6495 - 0.6822) \pm 1.96 \sqrt {\frac {0.6495(1 - 0.6495)}{545} + \frac {0.6822(1 - 0.6822)}{596}}$$$$\approx. -0.0327 \pm 0.0472$$

Therefore, we can conclude that the researchers are 95% confident the difference between the two population proportions, P1 - P2, is between -0.080 and -0.004 (using ascending order and rounding to three decimal places as needed).

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To find the vector function that represents the curve of intersection between the paraboloid z = 2x^2y^2 and the parabolic cylinder y = 3x^2, we can parameterize the curve using a parameter t. The vector function r(t) will consist of x(t), y(t), and z(t), where each component is expressed in terms of t.

First, we need to find the relationship between x and y by setting the equation of the parabolic cylinder equal to the y-coordinate of the paraboloid. Substituting y = 3x^2 into z = 2x^2y^2, we get z = 2x^2(3x^2)^2 = 18x^6.

Now, we can express x and z in terms of t. Let's set x(t) = t, which allows us to write z(t) = 18t^6. The y-component can be obtained by substituting x(t) into the equation of the parabolic cylinder: y(t) = 3(t^2).

Finally, we can combine the x(t), y(t), and z(t) components to form the vector function r(t) = (x(t), y(t), z(t)). In this case, r(t) = (t, 3t^2, 18t^6) represents the curve of intersection between the two surfaces.

Note that this vector function parameterizes the curve and allows us to describe various points on the curve by plugging in different values of t.

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??? The answer is not able to be found for the fact that there are 3 numbers all coming from x.

A performance task is a learning activity that asks students to perform to demonstrate their knowledge, understanding and proficiency.

Your information is incomplete. Therefore, an overview will be given. A performance task measure whether a student can apply his or her knowledge to make sense of a new phenomenon.

It should be noted that performance tasks yield a tangible product and performance that serve as evidence of learning.

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Step-by-step explanation:

I think that is your answer because in the Himalayas it lies between latitudes 26.

The probability represents the combination consists only of even numbers as per given condition is equal to 0.020.

Total numbers in combination lock = 38

Numbers from 0 to 37.

To find the probability that the combination consists only of even numbers,

Determine the total number of combinations that can be formed using only even numbers

And divide it by the total number of possible combinations.

Total number of even numbers in the lock

= 19 (since there are 19 even numbers from 0 to 37)

Calculate the total number of combinations using only even numbers,

Use the concept of combinations (nCr).

Since there are 19 even numbers to choose from,

Choose 4 numbers without repetition, the number of combinations is,

Number of combinations

= ¹⁹C₄

= 19! / (4!(19-4)!)

= (19 × 18 × 17 × 16) / (4 × 3 × 2 × 1)

= 3876

Now, calculate the total number of possible combinations without any restrictions.

Since we have 38 numbers to choose from,

and  choose 4 numbers without repetition, the number of combinations is,

Number of total combinations

= ³⁸C₄

= 38! / (4!(38-4)!)

= (38 × 37 × 36 × 35) / (4 × 3 × 2 × 1)

= 194,580

Finally, find the probability by dividing the number of combinations using only even numbers by the total number of combinations,

Probability

= Number of combinations using only even numbers / Total number of combinations

= 3876 / 194580

≈ 0.0199

≈ 0.020 Rounded to three decimal places.

Therefore, the probability that the combination consists only of even numbers is approximately 0.020.

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Answer:

eight.

Step-by-step explanation:

2(6) ÷ 3

12÷3

4+4

8

Answer:

384

Step-by-step explanation:

v= L x W x H

v= 4 x 2 x 1.2

v= 9.6

9.6 x 40 = 384

lmk if you have any questions  :)

Answer:

A

Step-By-Step Explanation:

Step One: The first step is to figure out the volume of one. The volume formula is length times width times height. We have all the dimensions so: 4 times 2 times 1.2= 9.6.

Step Two: Now, we just mutiply the number by 40 since there are 40 slabs: 384.