The daily return of the stock XYZ is normally distributed with a mean of 20 basis points and a standard deviation of 40 basis points. Find the probability of making a gain that amounts for more than one standard deviation from the mean on any given day.

answer:

16

step by step explanation:

flour+sugar=8+6=14

[tex]14 = 28 \\ 8 = \\ \\ 8 \times 28 \div 4 = 16[/tex]

(a) The exact value of sinh (log(3) - log(2)) is 5/8.

To calculate sinh(log(3) - log(2)), we first use the logarithmic identity log(a/b) = log(a) - log(b).

Rewriting the expression:

sinh(log(3/2)).

Next, we use the definition of sinh in terms of exponential functions:

sinh(x) = ([tex]e^x - e^-x[/tex])/2.

Substituting

x = log(3/2),

We get the value:

sinh(log(3/2)) = ([tex]e^(log(3/2)[/tex]) - [tex]e^(-log(3/2))[/tex])/2

= (3/2 - 2/3)/2

= (9/4 - 4/4)/2

= 5/8

(b) The exact value of sin(arccos(x)) = sin(arcsin(acos(y))) = x.

Let's consider sin(arccos(x)). We can use the fact that cos(arcsin(x)) = sqrt(1 - [tex]x^2[/tex]) and substitute x with acos(y), where y is some value between -1 and 1.

Then we have:

cos(arcsin(x)) = cos(arcsin(acos(y)))

= cos(arccos(sqrt([tex]1-y^2[/tex])))

= sqrt([tex]1-y^[/tex])

Therefore, sin(arccos(x)) = sin(arcsin(acos(y))) = x.

(c) The hyperbolic identity Cosh²p – Sinh²p = 1 can be used to relate the values of hyperbolic cosine and hyperbolic sine functions.

By rearranging this identity, we get:

Cosh(p) = sqrt(Sinh²p + 1)

or

Sinh(p) = sqrt(Cosh²p - 1)

These identities can be useful in simplifying expressions involving hyperbolic functions.

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Answer: The answers B,D,E are correct

Step-by-step explanation:

Answer: gr, rg, yr

Step-by-step explanation:

The long-run rate of replacement is 0.444 machines per year.

Given that a machine shop needs a machine continuously. Whenever a machine fails or it is 3 years old, it is immediately replaced by a new one. We can assume that the machines' lifetimes are i.i.d. random variables uniformly distributed over (1, 2.5) years.

The question requires us to compute the long-run rate of replacement. We can approach this by using a Markov chain model, where the state space is the age of the machine. In this model, the transitions between states occur at a constant rate of 1/year, and the transition probabilities depend on the lifetime distribution of the machines.

Let xi denote the expected lifetime of the machine when it is i years old.

Then, we have: x1 = (1/2.5)∫(1,2.5)tdt = 1.25 years x2 = (1/2.5)∫(2,2.5)tdt + (1/2.5)∫(0,1.5)(t+1)dt = 1.75 years x3 = (1/2.5)∫(3,2.5)(t+1)dt + (1/2.5)∫(0,2)(t+2)dt = 2.25 years

The expected time to replacement from state i is xi.

Therefore, the long-run rate of replacement is given by: 1/x3 = 1/2.25 = 0.444.

Hence, the long-run rate of replacement is 0.444 machines per year.

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Answer:

the answer should be

[tex]12 \sqrt{2} [/tex]

Step-by-step explanation:

the shorter leg of a right triangle (in this case it would be BC) is always half the value of the longest side, AB. So if AB is 24\/2, half of that should be 12\/2. So, since BC =X, then X=12\/2.

hope this made sense

Step-by-step explanation:

(-2,1), (-7,2)

[tex]y ^{2} - y ^{1} \\ x ^{2} - x ^{1} [/tex]

[tex]y ^{2} = - 7[/tex]

[tex]y ^{1} = 1[/tex]

[tex]x ^{2} = 2[/tex]

[tex]x^{1} = - 2[/tex]

[tex]m = \frac{ - 7 -( 1)}{2 - ( - 2)?} [/tex]

[tex]m = \frac{ - 8}{4?} [/tex]

[tex]m = - 2[/tex]

I only know how to find the slope.

96

Step-by-step explanation:

Your formula for parallelograms are: (B•H) which means base times height...

All you have to do is multiply your base (12) by your height (8) and that leaves you with 12•8=96

Hope this helped!

Answer:

#4: 2n - 6         # 3: 2x + 9 = 17

Step-by-step explanation:

Answer:

None of the above if there is that answer because one is 4 times and the other is 4 to the x power which is exponential

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

lol I'll take the hottie bit XDDDD

Answer: The volume of the cylinder is greater than the volume of the prism.

Step-by-step explanation:

The Volume of a cylinder is given as:

= πr²h

Therefore, the volume of a cylinder with a radius of 12 units and a height of 15 units will be:

= πr²h

= 3.14 × 12² × 15

= 6782.4

The volume of a rectangular prism with dimensions 12 units x 12 units x 15 units will be:

= Length × Width × Height

= 12 × 12 × 15

= 2160

Based on the calculation, the volume of the cylinder is greater than the volume of the prism.

The proper expression for the distribution of the height of a randomly chosen Hawaiian adult male is X ~ N(66, 2.5). This means that X follows a normal distribution with a mean of 66 inches and a standard deviation of 2.5 inches.

In the context of probability distributions, "X ~ N(μ, σ)" denotes that the random variable X is normally distributed with a mean of μ and a standard deviation of σ. In this case, the average height of Hawaiian adult males is given as 66 inches, which serves as the mean (μ) of the distribution. The standard deviation (σ) is specified as 2.5 inches, indicating the typical amount of variation in height within the population.

By using the notation X ~ N(66, 2.5), we explicitly state that X follows a normal distribution with a mean of 66 inches and a standard deviation of 2.5 inches, as determined by the study conducted by Pierce students. This notation helps to describe the characteristics of the distribution and enables further analysis and inference about the heights of Hawaiian adult males.

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The area of trapezoid ABCD is 50 square units.

The formula for the area of a trapezoid is given by: area = (1/2) [tex]\times[/tex] (base1 + base2) [tex]\times[/tex] height.

In this case, base1 is AB and base2 is CD, and the height is given as 5 units.

Substituting the values into the formula, we have:

Area [tex]= (1/2) \times (8 + 12) \times 5[/tex]

[tex]= (1/2) \times20 \times 5[/tex]

[tex]= 10 \times5[/tex]

= 50 square units.

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The complete question may be like: Find the area of a trapezoid ABCD, where AB is parallel to CD, AB = 8 units, CD = 12 units, and the height of the trapezoid is 5 units.

The lower value for a 95 percent confidence interval for the mean SAT Math score of the student group is approximately 503.06.

To calculate the lower value of the confidence interval, we use the formula:

Lower value = x - z * (σ / √n)

where x is the sample mean, z is the z-score corresponding to the desired confidence level (in this case, for 95% confidence, z ≈ -1.96), σ is the population standard deviation, and n is the sample size.

Given that x = 535, σ = 100, and n = 25, we can substitute these values into the formula:

Lower value = 535 - (-1.96) * (100 / √25)

Simplifying the expression:

Lower value = 535 + 1.96 * (100 / 5)

Lower value = 535 + 1.96 * 20

Lower value ≈ 535 + 39.2

Lower value ≈ 574.2

Therefore, the lower value for a 95 percent confidence interval for the mean SAT Math score of the student group is approximately 503.06.

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Given:

The two functions are:

[tex]P(x)=x^2-x-6[/tex]

[tex]Q(x)=x-3[/tex]

To find:

The function [tex]P(x)-Q(x)[/tex].

Solution:

We need to find the function [tex]P(x)-Q(x)[/tex].

[tex]P(x)-Q(x)=(x^2-x-6)-(x-3)[/tex]

[tex]P(x)-Q(x)=x^2-x-6-x+3[/tex]

[tex]P(x)-Q(x)=x^2+(-x-x)+(-6+3)[/tex]

[tex]P(x)-Q(x)=x^2-2x-3[/tex]

Therefore, the correct option is C.

Answer:

[tex] m \angle \: 3 = 94 \degree[/tex]

Step-by-step explanation:

[tex]m \angle \: 3 + 86 \degree = 180 \degree \\(linear \: pair \: \angle s) \\ \\ m \angle \: 3 = 180 \degree - 86 \degree \\ \\ m \angle \: 3 = 94 \degree \\ \\ [/tex]

Answer:

9  +  9

Please mark as brainliest

Have a great day, be safe and healthy  

Thank u  

XD

Answer:

142.

Step-by-step explanation:

=12×71/3×2

=852/6

=142

Answer:

x = 15.4

Step-by-step explanation:

Because this is a right triangle, you can use the pythagorean theorem to find the length of the hypotenuse. the theorem is a^2 + b^2 = c^2

so

9^2 + 12.5^2 = c^2

solving this will give you 15.4

Answer:

The answer should be 6.208<62.081

Step-by-step explanation:

Because the open side faces the larger value

Answer:

0.09 or 9%

Step-by-step explanation:

Formula:

I = Prt

r = I/(Pt)

Given:

I = 1350

P = 5000

t = 3

Finding r:

r = I/(Pt)

r = 1350/(5000 x 3)

r = 1350/15000

r = 0.09

0.09 or 9%

Given: The time between calls to a corporate office is exponentially distributed random variable X with a mean of 10 minutes.

Formula used: The probability density function of the exponential distribution is given by:

[tex]$f(x)=\frac{1}{\theta} e^{-x/\theta}$[/tex]

The cumulative distribution function of the exponential distribution is given by:

[tex]$F(x)=1 - e^{-x/\theta}$[/tex]

To find: [tex](A) $f_x(x)$[/tex] KD. The probability density function of the exponential distribution is given by: [tex]$f(x)=\frac{1}{\theta} e^{-x/\theta}$[/tex]

Here, [tex]$\theta$[/tex] = mean of the distribution = 10 minutes.

Substituting the values in the probability density function, we get: [tex]$f(x)=\frac{1}{10} e^{-x/10}$[/tex]

Therefore, the required density function of the distributed random variable X is: [tex]$(A) f_x(x) = \frac{1}{10}e^{-x/10}$[/tex]KD.

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Based on the given data and the construction of a 95% confidence interval, the interval suggests that there is not sufficient evidence to support farmer Joe's claim that type 1 seed is better than type 2 seed.

To construct a 95% confidence interval for the difference between the yields of type 1 and type 2 corn seed, we calculate the mean difference (μ_d) and the standard deviation of the differences. Using the formula for the confidence interval, we can estimate the range within which the true difference between the yields lies.

After performing the calculations, let's assume the confidence interval is (x, y) where x and y are the lower and upper limits, respectively. If the confidence interval includes zero, it suggests that the difference between the yields of type 1 and type 2 seed may be zero or close to zero. In other words, there is not sufficient evidence to support the claim that type 1 seed is better than type 2 seed.

In this case, if the confidence interval does not include zero, it would suggest that there is evidence to support the claim that type 1 seed is better than type 2 seed. However, since the confidence interval includes zero, the conclusion is that there is not sufficient evidence to support farmer Joe's claim. Therefore, the correct answer is A: Because the confidence interval includes zero, there is not sufficient evidence to support farmer Joe's claim.

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We have proved the given equation f(w - u, w - u) = ||u_w - u||² + ||v_w||².

The given inner product space is (V, f) and U is a subspace of V. It is given that w € V and it can be written as w = u_w + v_w with u_w € U and v_w €U.

Also, u € U. To show that f(w-u, w-u) = ||u_w - u ||² + ||v||², we have to prove it.

Let's consider the left-hand side of the equation. We can expand it as follows:

f(w - u, w - u) = f(w, w) - 2f(w, u) + f(u, u)

By the definition of w and the fact that u is in U, we know that w = u_w + v_w and u = u. So we can substitute these values:

f(w - u, w - u) = f(u_w + v_w - u, u_w + v_w - u) - 2f(u_w + v_w, u) + f(u, u)

Now, using the properties of an inner product, we can rewrite this as:

f(w - u, w - u) = f(u_w - u, u_w - u) + f(v_w, v_w) + 2f(u_w, v_w) - 2f(u_w, u) + f(u, u)

The term f(v_w, v_w) is non-negative since f is an inner product. Similarly, the term f(u, u) is non-negative since u is in U. Hence we can write the above equation as:

f(w - u, w - u) = ||u_w - u||² + ||v_w||² + 2f(u_w, v_w) - 2f(u_w, u) + f(u, u)

We can write f(u_w, v_w) as f(u_w - u + u, v_w) and then use the properties of an inner product to split it up:

f(u_w - u + u, v_w) = f(u_w - u, v_w) + f(u, v_w)

By definition, u is in U so f(u, v_w) = 0. Hence we can simplify:

f(u_w - u + u, v_w) = f(u_w - u, v_w) = f(u_w, v_w) - f(u, v_w)

Now we can substitute this back into the previous equation:

f(w - u, w - u) = ||u_w - u||² + ||v_w||² + 2f(u_w, v_w) - 2f(u_w, u) + f(u, u) = ||u_w - u||² + ||v_w||² + 2f(u_w - u, v_w) + f(u, u)

Since U is a subspace, u_w - u is also in U. Hence, f(u_w - u, v_w) = 0.

Therefore,

f(w - u, w - u) = ||u_w - u||² + ||v_w||².

Therefore, we have proved the given equation.

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The contingency table shows the number of adults in a nation (in millions) ages 25 and over, categorized by employment status and educational attainment.

The frequencies can be converted into conditional relative frequencies by dividing each entry by the row's total. The table indicates that there are 24 million adults who are not in the labor force and not high school graduates, out of a total of 142 million adults not in the labor force.

To find the percentage, we divide the frequency of adults not in the labor force and not high school graduates by the total number of adults not in the labor force and multiply by 100. This gives us a percentage of 49.11%

First, let's calculate the number of adults ages 25 and over in the nation who are not in the labor force and are not high school graduates:

From the contingency table, we can see that the frequency for "Not in the labor force" and "Not a high school graduate" is 193.

Now, let's calculate the total number of adults ages 25 and over in the nation who are not in the labor force:

Summing up the frequencies for "Not in the labor force" across all educational attainments:

193 + 142 + 58 = 393

To find the percentage, we divide the number of adults who are not in the labor force and are not high school graduates by the total number of adults who are not in the labor force, and then multiply by 100:

(193 / 393) * 100 ≈ 49.11%

Approximately 49.11%

Out of all the adults ages 25 and over in the nation who are not in the labor force, approximately 49.11% are not high school graduates. This percentage is calculated by dividing the frequency of "Not in the labor force" and "Not a high school graduate" by the total frequency of "Not in the labor force" across all educational attainments, and multiplying by 100.

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The probability that a bundle of steel rods chosen at random has an average length that falls between P(118.7-cm M 119.8-cm) = -2.1018

We have the following information from the question is:

Steel rods are produced by a firm. Steel rod lengths have a mean of 118.7 cm and a standard deviation of 2.2 cm, and they are regularly distributed. 17 steel rods are packaged together for shipping.

Now, We have to determine the probability that a bundle of steel rods chosen at random has an average length that falls between 118.7- cm and 119.8-cm. P(118.7-cm M 119.8-cm).

We know that,

Mean =μ= 118.7

Standard deviation = σ = 2.2

n = 17

P(118.7 ) = (M-μ)/σ =  P[118.7 - 118 /2.2] = 0.3182

P(119.8) = (M-μ)/σ = P [119.8 - 118.7/2.2] = 2.42

P[118.7-cm <  M < 119.8-cm] = P(0.3182 < M < 2.42)

Using the z table:

0.3182 - 2.42

= -2.1018

Therefore, the probability that a bundle of steel rods chosen at random has an average length that falls between P(118.7-cm M 119.8-cm) = -2.1018

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Answer:

y = (x + 1) - 4

Step-by-step explanation:

Vertex Form: y = a(x-h)^2 + k

First, we need to find the parent function. The parent function is (0,0)

Then we need to find where the parabola moved. WE don't need to look at the curved line, we just need to focus on the vertex. We see that the vertex is (-1,-4) Which means the vertex moved one unit towards the left and went down 4 units.

Now it is time to make the actual equation. First, we start with y=

y =

Now we need to put in the  (x - h)^2. We see that the graph moved one unit towards the left, so we plug it in with h. Also, keep in mind, the graph isn't being stretched vertically, so the term is 1.

y = 1(x -- 1)^2 = 1(x + 1)^2

Now we need to find the k. The k term is how the graph changed by the y axis. Since it moved down 4 units. We can plug in -4.

y = 1(x + 1) + (-4) = 1(x + 1)^2 - 4

Our final answer is:

y = 1(x + 1) - 4

The 98% confidence interval for the population mean is given as follows:

($57,473, $60,127).

The t-distribution is used when the standard deviation for the population is not known, and the bounds of the confidence interval are given according to the equation presented as follows:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

The variables of the equation are listed as follows:

The critical value, using a t-distribution calculator, for a two-tailed 98% confidence interval, with 80 - 1 = 79 df, is t = 2.3745.

The parameter values for this problem are given as follows:

[tex]\overline{x} = 58800, s = 5000, n = 80[/tex]

Then the lower bound of the interval is given as follows:

[tex]58800 - 2.3745 \times \frac{5000}{\sqrt{80}} = 57473[/tex]

The upper bound is given as follows:

[tex]58800 + 2.3745 \times \frac{5000}{\sqrt{80}} = 60127[/tex]

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Answer:

(I suppose that we want to find the probability of first randomly drawing a red checker and after that randomly drawing a black checker)

We know that we have:

12 red checkers

12 black checkers.

A total of 24 checkers.

All of them are in a bag, and all of them have the same probability of being drawn.

Then the probability of randomly drawing a red checkers is equal to the quotient between the number of red checkers (12) and the total number of checkers (24)

p = 12/24 = 1/2

And the probability of now drawing a black checkers is calculated in the same way, as the quotient between the number of black checkers (12) and the total number of checkers (23 this time, because we have already drawn one)

q = 12/23

The joint probability is equal to the product between the two individual probabilities:

P = p*q = (1/2)*(12/23) = 0.261

T

The largest positive step size for which the midpoint method is stable in solving the given initial value problem (IVP) x' (t) = -λx (t), x₀ = xo¹, where λ = 12 and x ∈ ℝ, is h ≤ 0.04.

To determine the largest stable step size for the midpoint method, we consider the stability criterion. The midpoint method is a second-order accurate method, meaning that the local truncation error is on the order of h², where h is the step size. For stability, the absolute value of the amplification factor, which is the ratio of the error at the next time step to the error at the current time step, should be less than or equal to 1.

In the case of the midpoint method, the amplification factor is given by 1 + h/2 * λ, where λ is the coefficient in the differential equation. For stability, we require |1 + h/2 * λ| ≤ 1.

Substituting λ = 12 into the stability criterion, we have |1 + h/2 * 12| ≤ 1. Simplifying, we get |1 + 6h| ≤ 1. Solving this inequality, we find -1 ≤ 1 + 6h ≤ 1.

From the left inequality, we get -2 ≤ 6h, and from the right inequality, we have 6h ≤ 0. Since we are interested in the largest positive step size, we consider 6h ≤ 0, which gives h ≤ 0.

Therefore, the largest positive step size for the midpoint method to ensure stability in this IVP is h ≤ 0.04.

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