okay last one . help

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:

Answer:

pi, and square root of 2

I don't think you're ACTUALLY going to give me brainliest

Step-by-step explanation:

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

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.

The particular solution that satisfies the given condition is s = (16/3)t³ + (9/2)t² - 6t + 120

To find the particular solution of the differential equation ds/dt = 16t² + 9t - 6 with the condition s = 120 when t = 0, we need to integrate the right-hand side of the equation with respect to t and then solve for the constant of integration using the given condition.

First, let's integrate the right-hand side of the equation:

∫(ds/dt) dt = ∫(16t² + 9t - 6) dt

Integrating term by term, we get:

s = (16/3)t³ + (9/2)t² - 6t + C

Now, we can use the given condition s = 120 when t = 0 to determine the value of the constant of integration C:

120 = (16/3)(0)³ + (9/2)(0)² - 6(0) + C

120 = C

Therefore, the particular solution that satisfies the given condition is:

s = (16/3)t³ + (9/2)t² - 6t + 120

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Answer:

x = 138

Step-by-step explanation:

Exterior angle thm:

68 + 70 = x

x = 138

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

The concentration of sugar in the solution in the tank after 33 minutes is approximately 0.00734 kg/L or 7.34 g/L.

(a) Initially, the tank contains pure water, so there is no sugar in the tank. Therefore, the amount of sugar in the tank initially is 0 kg.

(b) The amount of sugar in the tank after t minutes can be calculated using the formula:

amount = 166.6(1 - [tex]e^{-t/340}[/tex])

Here, t is the time in minutes. This formula is derived from the exponential decay model, where the rate of sugar leaving the tank is proportional to the amount of sugar present. The constant 166.6 represents the equilibrium amount of sugar in the tank.

As time passes, the term [tex]e^{-t/340}[/tex] approaches 1, and the amount of sugar in the tank approaches the equilibrium amount of 166.6 kg.

(c) To find the concentration of sugar in the solution in the tank after 33 minutes, we need to divide the amount of sugar by the volume of the tank. The volume of the tank is given as 2380 L.

First, let's calculate the amount of sugar in the tank after 33 minutes using the formula from part (b):

amount = 166.6(1 - [tex]e^{-33/340}[/tex])

amount ≈ 166.6(1 - 0.895)

amount ≈ 166.6(0.105)

amount ≈ 17.493 kg

Now, we can calculate the concentration:

concentration = amount / volume

concentration = 17.493 kg / 2380 L

concentration ≈ 0.00734 kg/L

Therefore, the concentration of sugar in the solution in the tank after 33 minutes is approximately 0.00734 kg/L or 7.34 g/L.

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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

Answer:

Step-by-step explanation:

se gana 10000 y 2

Answer:

1047.12in³

Step-by-step explanation:

The volume of the sphere is:

V= 4/3 πr³

Where r is radius

Therefore the volume of each pinata is:

V= 4/3 x π x 5³

V= 523.6in³

The total:

V = 523.6 + 523.6 = 1047.12in³

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. 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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The 90% confidence interval is approximately 2,400 and 3,400.

The 90% confidence interval of the true population mean can be calculated using the formula: sample mean ± (critical value * standard error). Since the sample size is small (n=27), we need to use the t-distribution and its corresponding critical value. For a 90% confidence level with 26 degrees of freedom (n-1), the critical value is approximately 1.706.

Using the given values, the standard error is calculated as the sample standard deviation divided by the square root of the sample size: 900 / sqrt(27) ≈ 171.97.

Substituting the values into the formula, the 90% confidence interval is approximately 3,000 ± (1.706 * 171.97), which yields a range of approximately 2,424.5 to 3,575.5.

Therefore, the closest option to the 90% confidence interval is 2,400 and 3,400.

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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 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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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

Answer:

x = 1/2

y = 3

Step-by-step explanation:

Remark

The diagonals of a parallelogram cut each other in half. That means that x lengths are equal and the y lengths are equal

Equations

3y = y + 6

10x = 6x + 2

Solution for x

10x = 6x + 2                     Subtract 6x from both sides

10x - 6x = 6x - 6x + 2

4x = 2                              Divide by 4

4x/4 = 2/4

x = 1/2

Solution for y

3y = y + 6

3y - y = y- y + 6

2y = 6

y = 3

Answer:

Step-by-step explanation:

Answer:

5.86

Step-by-step explanation: its right on delta math

Answer:

Angle EHG

Step-by-step explanation:

If you want explanation tell me in comments

Answer:

the correct answer is angle EHB

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:

40

Step-by-step explanation:

2 cups = 1 pt

therefore

1/2 is the equation

...

80/2 = 40

T-value is not always greater than Z-value. Therefore, the statement is false

False. The relationship between a T critical value and a Z critical value is not that a T value is always greater than a Z value. The choice between using a T critical value or a Z critical value depends on the specific context and assumptions of the statistical analysis.

In general, when the sample size is small (typically below 30) and the population standard deviation is unknown, a T critical value is used. The T distribution accounts for the additional uncertainty introduced by the smaller sample size, resulting in wider confidence intervals and more conservative hypothesis tests compared to the Z distribution.

On the other hand, when the sample size is large (typically above 30) or the population standard deviation is known, a Z critical value is used. The Z distribution assumes a large sample size, and it is based on the known population standard deviation or the approximation of the sample standard deviation to the population standard deviation.

Therefore, it is incorrect to state that a T-value is always greater than a Z-value. The choice between T and Z critical values depends on the specific conditions and assumptions of the statistical analysis being performed.

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The absolute maximum and minimum of the function z = f(x, y) = 3x² - 12x + 3y² - 12y on the domain x² + y² < 4 are not provided.

To find the absolute maximum and minimum of the function z = f(x, y) = 3x² - 12x + 3y² - 12y on the domain D: x² + y² < 4, we need to locate the critical points and examine the boundaries of the domain.

Since the absolute maximum and minimum values are not provided, further calculations are needed to determine these values for the given function on the given domain.

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