"The times for the mile run of a large group of male college students are approximately Normal with mean 7. 06 minutes and standard deviation 0. 75 minutes. Use the 68-95-99. 7 rule to answer the following questions. (Start by making a sketch of the density curve you can use to mark areas on. ) (a) What range of times covers the middle 95% of this distribution

The answer is <1,-1,1>.The given surface is In(xyz) + x² + y² + z² = 3.The gradient of the function f(x,y,z) = In(xyz) + x² + y² + z² = 3 is given by:grad f(x,y,z) = At the point P = (1,-1,-1), we have grad f(P) = <-1,-3,-2>.

Hence the equation of the tangent plane to the given surface at P is given by:-1(x - 1) - 3(y + 1) - 2(z + 1) = 0Simplifying we get x - 3y - 2z = -4Taking dot product of this normal vector <1,-3,-2> with each of the given vectors we get the following results:<1,1,0>.<1,-3,-2> = -5 ≠ 0<1,1,-1>.<1,-3,-2> = 0  [Answer]<-1,1,1>.<1,-3,-2> = 0<1,0,-1>.<1,-3,-2> = -5 ≠ 0<1,-1,1>.<1,-3,-2> = 0

Therefore the vector 0 <1,1,-1> is orthogonal to the tangent plane of the given surface at the point (1,-1,-1).Hence the correct option is 0 <1,1,-1>.

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Answer:

x^2

Step-by-step explanation:

The answer is x square because the evalutation at the top directly prooves its inexctriacble.

A. X and Y might be independent. Knowing the value of one independent variable tells us nothing about the value of the other independent variable. The mean and variance of X and Y being different does not necessarily mean that knowing the value of one tells us nothing about the other.

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The factorization of the polynomial x² - 4x + 3 are (x - 1)(x - 3)

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

x² - 4x + 3

Also, we have the graph

From the graph, we can see that the graph intersects the x-axis ar

x = 1 and 3

This means that the factors are

(x - 1) * (x - 3)

So, we have

(x - 1)(x - 3)

Hence, the factorization of x² - 4x + 3 are (x - 1)(x - 3)

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Answer:

$275/mont 2.

Step-by-step explanation:

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For n ≥ 459, Algorithm A is better than Algorithm B when B uses n√n operations.

To determine the value of n₀ for which Algorithm A is better than Algorithm B when B uses n√n operations, we need to find the point at which the number of operations for Algorithm A is less than the number of operations for Algorithm B.

Algorithm A: 10n log n operations

Algorithm B: n√n operations

Let's set up the inequality and solve for n₀:

10n log n < n√n

Dividing both sides by n gives:

10 log n < √n

Squaring both sides to eliminate the square root gives:

100 (log n)² < n

To solve this inequality, we can use trial and error or graph the functions to find the intersection point. After calculating, we find that n₀ is approximately 459. Therefore, For n ≥ 459, Algorithm A is better than Algorithm B when B uses n√n operations.

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R-1.3: For \($n \geq 14$\), Algorithm A is better than Algorithm B when B uses \($n^2$\) operations.

R-1.4: Algorithm A is always better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

R-1.3:

Algorithm A: \($10n \log n$\) operations

Algorithm B: \($n^2$\) operations

We want to determine the value of \($n_0$\) such that Algorithm A is better than Algorithm B for \($n \geq n_0$\).

We need to compare the growth rates:

\($10n \log n < n^2$\)

\($10 \log n < n$\)

\($\log n < \frac{n}{10}$\)

To solve this inequality, we can plot the graphs of \($y = \log n$\) and \($y = \frac{n}{10}$\) and find the point of intersection.

By observing the graphs, we can see that the two functions intersect at \($n \approx 14$\). Therefore, for \($n \geq 14$\), Algorithm A is better than Algorithm B.

R-1.4:

Algorithm A: \($10n \log n$\) operations

Algorithm B: \($n\sqrt{n}$\) operations

We want to determine the value of \($n_0$\) such that Algorithm A is better than Algorithm B for \($n \geq n_0$\).

We need to compare the growth rates:

\($10n \log n < n\sqrt{n}$\)

\($10 \log n < \sqrt{n}$\)

\($(10 \log n)^2 < n$\)

\($100 \log^2 n < n$\)

To solve this inequality, we can use numerical methods or make an approximation. By observing the inequality, we can see that the left-hand side \($(100 \log^2 n)$\) grows much slower than the right-hand side \($(n)$\) for large values of \($n$\).

Therefore, we can approximate that:

\($100 \log^2 n < n$\)

For large values of \($n$\), the left-hand side is negligible compared to the right-hand side. Hence, for \($n \geq 1$\), Algorithm A is better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

So, for R-1.4, the value of \($n_0$\) is 1, meaning Algorithm A is always better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

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Answer:

The volume of the soup can is 602. 88 cubic centimeters

Step-by-step explanation:

Volume of a cylinder = π {r} ^{2} h

Where π = 3.14

r = diameter ÷ 2

r= 8 ÷ 2

r= 4 cm

h= 12 cm

Volume of the soup can= 3.14 × 4 × 4 × 12

= 602. 88 cubic centimeters

Answer:the until prince is 4/1

Step-by-step explanation:

The graph is proportional since it’s a straight line through origin and the rate is 4/1 on the bottom

Sorry if I’m wrong

Let x = 6+y. We can replace 6+y with x getting the equation |x| = 2.

The equation |x| = 2 has two solutions. The expression |x| represents the distance from x to 0 on the number line.

The two solutions to |x| = 2 are x = 2 and x = -2.

Going from x = 2 to 0 is a distance of 2 units, so is the distance from 0 to x = -2.

The solutions in terms of x are then used to find the solutions in terms of y

For example, if x = 2, then

x = 6+y

2 = 6+y

2-6 = y

y = -4

The other y solution is handled in a similar fashion.

----------

Side note: The 2 at the end of the original equation does not determine how many solutions there are. We could easily have |6+y| = 3 and still have two solutions.

Let x = 6+y. We can replace 6+y with x getting the equation |x| = 2.

The equation |x| = 2 has two solutions. The expression |x| represents the distance from x to 0 on the number line.

The two solutions to |x| = 2 are x = 2 and x = -2.

Going from x = 2 to 0 is a distance of 2 units, so is the distance from 0 to x = -2.

The solutions in terms of x are then used to find the solutions in terms of y

For example, if x = 2, then

x = 6+y

2 = 6+y

2-6 = y

y = -4

The other y solution is handled in a similar fashion.

If something melts in your mouth and not in your hand, then it must be something that has a low melting point.

Converse,Inverse and Contrapositve

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Answer:

What is the equation of the graphed line written in standard form?" has four options: 2x + 3y = -6, 2x + 3y = 6, y=-2/3x-2, and y= 2/3x-2. The standard form of a linear equation is Ax + By = C, where A, B, and C are constants. In this case, the first two options, 2x + 3y = -6 and 2x + 3y = 6 are already in standard form. The last two options, y=-2/3x-2 and y= 2/3x-2 are not in standard form. However, without additional information such as a graph or coordinates of points on the line, it is not possible to determine which of these options is the correct equation for the graphed line.

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Answer:

b,d

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

Answer:

15.946

Step-by-step explanation:

Just divide 48/3.01 using long division

The slope and the intercepts of the graph are determined.

Given:

the equation is :

2x+3y=6

we first find the slope of the equation:

m = -a/b

m = -2/3

now solve the intercepts:

to find x intercept, substitute in 0 for y and solve for x.

⇒ 2x+3y=6

⇒ 2x+3(0)=6

⇒ 2x+0=6

⇒ 2x=6

⇒ x=6/2

⇒ x=3

hence we get (3,0)

now, to find y intercept, substitute in 0 for x and solve for y.

⇒ 2x+3y=6

⇒ 2(0)+3y=6

⇒ 0+3y=6

⇒ 3y=6

⇒ y=6/3

⇒ y=2

hence we get (0,2) as the intercepts.

Therefore we plot the required graph.

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Answer: Vertex of 6 (x-9)^2 + 11 minimum (9,11)

Step-by-step explanation:

The addition equation to represent Jackson's net change in money is x + (-y) = 4.63

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

Jackson receives 4.63 as his change at the grocery store. He places it into a charity donation jar at the register.

WE need to Write an addition equation to represent Jackson's net change in money.

Given that :

The change received = 4.63

The Net change in money :

Let initial amount before purchase is represented by  x

The Cost of item purchased = y (negative as it is incurred)

The Net change in money:

Initial amount + cost of item purchased = change received

x + (-y) = 4.63

Therefore, an addition equation to represent Jackson's net change in money is x + (-y) = 4.63

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Answer:

\(A=\) $\(966.306\)

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

\(A=P(1+\frac{r}{n})^n^{t}\)

where

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

\(t= 5\) \(years\)

\(P=\) $\(600.00\)  

\(r=10\)% \(=10/100=0.10\)

\(n=1\)

substitute in the formula above

\(A=600(1+\frac{0.10}{1})^1^*^5\)

\(A=600(1.10)^5\)  

\(A=\) $\(966.306\)

Answer:

n/3 - 4 = 10

Step-by-step explanation:

hope this helps

After 200 years, 9.84 grams of Cesium would remain.

Let N represent the amount of substance after t years

The half-life of a cesium is 30 years., hence, this can be represented by exponential function:

\(N(t)=ab^t\)

where a is the initial value and b is the multiplier.

There were initially 1000g of the substance

a) The exponential model for this situation is:

\(N(t) =1000(\frac{1}{2} )^{\frac{1}{30} t}\\\\N(t) =1000(\frac{1}{2} )^{\frac{t}{30}\)

b) After 200 years (t = 200), the amount of Cesium remaining is:

\(N(200)=1000(\frac{1}{2} )^\frac{200}{30} \\\\N(200) = 9.84g\)

Hence after 200 years, 9.84 grams of Cesium would remain

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The probability that at least 75 rolls are necessary to exceed a total sum of 275 is approximately 0.293.

Let X be the random variable denoting the total number of rolls required to exceed 275. We want to calculate P(X >= 75). The probability of rolling a number greater than 5 is 1/3, and the probability of rolling a number less than or equal to 5 is 2/3. Therefore, the expected value of a single roll is

The variance of a single roll is

Using the properties of the geometric distribution, we can calculate the probability that it takes at least 75 rolls to achieve a sum greater than 275:

P(X >= 75) = (1 - P(X < 75)) = (1 - (1 - (275/3.67)/75\()^{(75)}\)) = 0.293.

Therefore, the probability that at least 75 rolls are required is approximately 0.293.

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Answer:

She earned $230.

Step-by-step explanation:

Each hour Nala earns $11.50.

If she works for 20 hours, we can do 11.50*20 to get the amount she earned.

11.50*20 = 230

Answer:

incenter

Step-by-step explanation:

The center of the circle inscribed in a triangle is the incenter of the triangle, the point where the angle bisectors of the triangle meet.

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

Nathaniel ate 25% of candies = 0.25

Total Candies = 160

Step 02:

Nathaniel ate:

160 * 0.25 = 40

The answer is:

Nathaniel ate 40 candies.

Answer:

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

The variance of the number of customers who will make a purchase is 2.4.

The variance of the number of customers who will make a purchase can be calculated using the formula:

Variance = n * p * (1 - p)

where n is the number of customers and p is the probability of a customer making a purchase.

In this case, n = 10 and p = 0.6. Substituting these values into the formula, we get:

Variance = 10 * 0.6 * (1 - 0.6)
Variance = 10 * 0.6 * 0.4
Variance = 2.4

Therefore, the variance of the number of customers who will make a purchase is 2.4.

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Answer:

\(3240^{0}\)

Step-by-step explanation

n = Number of sides

Sum of interior angles of a regular polygon = (n - 2) x \(180^{o}\)

                                                                        = (20 - 2) x  \(180^{o}\)

                                                                        = 18 x \(180^{o}\)

                                                                        = \(3240^{o}\)

The numerical solution of Simpson's 1/3 rule for a single segment, according to the given formula, is: ∫[x₁,x₂] f(x) dx ≈ (x₂ - x₁) / 6 * (f(x₁) + 4f((x₁ + x₂) / 2) + f(x₂))

Simpson's 1/3 rule is a numerical integration technique used to approximate the definite integral of a function over a given interval. It is based on approximating the function by a quadratic polynomial within each subinterval and then integrating that polynomial exactly. The formula provided represents the Simpson's 1/3 rule for a single segment.

In this formula, x₁ and x₂ represent the endpoints of the segment over which we want to approximate the integral. f(x₁) and f(x₂) are the function values at these endpoints. The term (x₂ - x₁) / 6 represents the width of the segment divided by 6, which is a constant factor used in the approximation.

The main approximation step in Simpson's 1/3 rule is to evaluate the function at the midpoint of the segment, which is given by (x₁ + x₂) / 2. This is denoted as f((x₁ + x₂) / 2) in the formula. By using this midpoint, we consider the behavior of the function in the middle of the segment as well.

The formula then combines these function values at the endpoints and the midpoint, weighted by specific coefficients (1, 4, 1), to compute an approximation of the integral over the segment. The coefficients are chosen such that they yield an accurate approximation for certain types of functions.

The Simpson's 1/3 rule for a single segment uses the function values at the endpoints and the midpoint, along with appropriate coefficients, to estimate the integral. This approximation provides a reasonable balance between accuracy and simplicity for many functions.

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The appropriate regression model depends on the stationarity properties of both the dependent and independent variables, as well as the error term. The researcher can use a standard OLS regression model with first-order differencing of both Y and X.

In the first scenario, both Y and X are I(0), which means they are stationary time series. In this case, the researcher can perform a standard linear regression analysis, as the stationary series would lead to a stable long-run relationship. The answer from this model will be reliable and less likely to suffer from spurious regressions. In the second scenario, Y is I(2) and X is I(0). This implies that Y is integrated of order 2 and X is stationary. In this case, the researcher should first difference Y twice to make it stationary before performing a regression analysis. However, this approach might not be ideal as the integration orders differ, which can lead to biased results.

In the third scenario, Y and X are both I(1) and the error term is I(0). This indicates that both Y and X are non-stationary time series, but their combination might be stationary. The researcher should employ a co-integration analysis, such as the Engle-Granger method or Johansen test, to identify if there is a stable long-run relationship between Y and X. If co-integration is found, then an error correction model can be used for more accurate predictions.

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The simple interest earned on the stated amount for 6 months is around $1675.5.

The earned simple interest can be calculated using the formula -

S.I. = P×R×T/100

Time period = 6 months

Time period = 6/12

Time period = 1/2 years

Using the formula to find the interest earned -

S.I. = (85000 × 3.9 × 1)/(100 × 2)

Performing multiplication in both numerator and denominator

S.I. = 3,31,500/200

Performing division on Right Hand Side of the equation

S.I. = $1657.5

The numbers are divisible and hence won't generate answer round to 2 decimal places.

Hence, the interest earned is $1675.5.

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