The lengths of a certain species of fish are approximately normally distributed with a given mean Mu and standard deviation Sigma. According to the Empirical Rule, what percentage of the fish will have lengths within 1 standard deviation of the mean? 34% 47. 5% 68% 99. 7%.

Answer: 2

Step-by-step explanation:

Answer:

x = -2

Step-by-step explanation:

Step-by-step on how to solve it is in the image below.

Hope this helps! :)

The value of $×$ is 3.14 units

Perimeter is the distance around the edge of a shape. The algebraic sum of the length of each side is the perimeter of that shape. We have formulas available for the various shapes in geometry.

The perimeter of a square is expressed as 4l

The circumference of a circle = 2πr

This means that;

4l = 2πr

4l = 2× 3.142 × 2

4l = 12.57

divide both sides by 4

l = 12.57/4

l = 3.14 units(nearest hundredth)

therefore the value of the side length is 3.14 units

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Using Tap working problem solving,

when faucet and drain working together, the time to fill up the tub is 60 minutes.

We have the following information about question,

The time taken by faucet to fill up the tub

= 15 minutes

The time taken by drain to empty the tub

= 20 minutes

we have to calculate the time in which the tub is fully fill the tube with drain open.

Rate of fill up the tub by faucet = 1/15 i.e 1/15 th of tub in one minute .

Rate of empty the tub by drain = 1/20 i.e 1/20 th of tub in one minute.

In one minute both are working together and will fill the tub = 1/15 - 1/20 = 1/60 i.e 1/60 th of tub

Since, it takes one minute to fill 1/60th of the tub and it will take 60 minutes to fill the tub.

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Answer:

-8/5×-7/4 is your answerrrrrrrrrrr

Step-by-step explanation:

(-8/5)=-(8/5)

=8/5. ×4/7

The initial condition is $3.

The constant rate of change is 0.75 for every mile.

Based on the given information, we can express the following

Where x is miles, and y is the cost.

Answer:

1/8 or 0.125

Step-by-step explanation:

So, I started solving this problem by writing all the possible options, then pairing them each up with a fruit and drink..

...But when I read the question again, I noticed I forgot to leave out the fruit, as this question states that you're trying to retrieve "a turkey sandwich and a bottle of water.."

Well, oof.

Despite that, I'm still giving this answer as if it never left out the fruit. So, let's see what we need to do.

To find all the possible lunch options, I started out by writing down one of our meats, Turkey. As per requested, I made every possible turkey combination that included a fruit and drink, which gave me this:

Same for the ham:

Putting these together, this gives up 8 different lunchbox combinations. If we're trying to get one and we randomly select it, then we have a 1/8 chance of grabbing the turkey sandwich and a bottle of water....

..and a fruit.

Hope this helped!

If you need me to show my work, just comment me and I will attach a screenshot!

Source: N/A

Answer:

20ft.

Step-by-step explanation:

The perimeter is 40ft. You are starting at the perimeter of the pool swimming inward, you will have to swim half of that distance to reach the center.

Answer:

A , B , D

Step-by-step explanation:

Answer:

See explanation.

Step-by-step explanation:

Usually you have to add or subtract the number to get the variable by itself, and then divide the coefficient of the variable on both sides to solve.

Answer:

Take v/11 +3=9 for example

Step-by-step explanation:

First you need to get rid of the +3 so you do the opposite and do -3 to even that out. You do 9-3 and get 6. Then we are left with v/11 =6. See how I got that? To undo the division we will do multiplication. So do 6 times 11 and v/11 times 11. the 11 on the fraction will cancel out with the 11 it was multiplied by. leaving the v. with the 6 on the other side just multiply by 11 to get 66. The answer to this equation is v = 66     Always remember, what you do to one side of the equation do to the other.

Answer:

\(10.4\)

Step-by-step explanation:

The ladder leaning against the wall forms a right triangle with the wall and the ground. Therefore, we can use basic trig for a right triangle to solve this problem.

In a right triangle, \(\tan\theta=\frac{\text{opp}}{\text{adj}}\). Therefore, we have the following equation, where \(x\) is distance between the bottom of the ladder to the base of the wall:

\(\tan 30^{\circ}=\frac{6}{x},\\x=\frac{6}{\tan 30^{\circ}},\\x\approx \boxed{10.4}\).

Answer:

I think it might be 3 and 8 because I used the dots and used the number that connects to it with out an un even line I'm not sure if this will make sense or if it is correct but I'm just trying to help. dont listen to me unless you think it's the right answer.

Answer:

graph 19 because since it is that means also like the thingy equals that number si then like when the thing is at the highest point it cab maybe actually it will equal that number

f(x) = 13e^(2x) – e^(x^2)

f'(x) = 26e^(2x) – 2xe^(x^2)

f'(3) = 26e^(2*3) – 2(3)e^(3^2) = 26e^6 – 18e^9

a) f(x) = 4 ln x

b) g(x) = ln(x^4)

We use the Power rule and the chain rule

g'(x) = 4x^(4-1)(1/x) = 4x^3(1/x) = 4x^2

To find f'(x), we need to use the power rule and the chain rule.

f(x) = 13e^(2x) – e^(x^2)

f'(x) = 26e^(2x) – 2xe^(x^2)

To find f'(3), we substitute x=3 into the derivative expression:

f'(3) = 26e^(2*3) – 2(3)e^(3^2) = 26e^6 – 18e^9

To compute the derivatives of the given functions:

a) f(x) = 4 ln x

We use the logarithmic differentiation rule:

f'(x) = 4(1/x) = 4/x

b) g(x) = ln(x^4)


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Answer:

Step-by-step explanation:

Step-by-step explanation:

2x + 8y = 12 ......... Equation 1

x - 2y = 0 .............. Equation 2

Make x the subject in equation 2 and substitute it into equation 1

That's

x = 2y

We have

2(2y) + 8y = 12

4y + 8y = 12

12y = 12

Divide both sides by 12

y = 1

Substitute y = 1 into x = 2y

That's

x = 2(1)

x = 2

The answers are

x + y = 7......... Equation 1

2x + y = 5 ......... Equation 2

Make y the subject in equation 1 and substitute it into equation 2

We have

y = x - 7

2x + x - 7 = 5

3x = 7 + 5

3x = 12

Divide both sides by 3

x = 4

Substitute x = 4 into y = x - 7

y = 4 - 7

y = - 3

We have the answers as

y = -1/2 + 1 ......... Equation 1

2x + 3y = 6 ........ Equation 2

From equation 1

y = 1/2

Substitute y = 1/2 into Equation 2

So we have

2x + 3(1/2) = 6

2x + 3/2 = 6

2x = 6 - 3/2

2x = 9/2

Divide both sides by 2

x = 9/4

The answers are

2x - 1/3y = - 9.......... Equation 1

-3x + y = 15 ........... Equation 2

Make y the subject in equation 2 and substitute it into equation 1

That's

y = 15 + 3x

2x - 1/3( 15 + 3x) = - 9

2x - 5 - x = - 9

x = - 9 + 5

x = - 4

Substitute x = - 4 into y = 15 + 3x

That's

y = 15 + 3(-4)

y = 15 - 12

y = 3

The answers are

Hope this helps you

The balance sheet that correctly represents the engineer's income, expenses and balance is illustrated below.

We should know that a balance sheet is a summary of assets and liabilities and income and expenses of a  legal person. We are told that the balance sheet shows a bi-weekly net pay.

This implies that a month has two bi-weeks, therefore every income and expense is multiplied by 2 to get the following figures.

Income                                             Amount

Bi-weekly income                          $2,175.25

Total Monthly Income                   $4,350.50

Expenses                 Percent                              Amount

Housing                        35%                                $1,522.68

Food/Household          10%                               $435.05

Savings                         10%                                $435.05

Transportation              15%                                $652.58

Debt                              5%                                   $217.53

Entertainment               6%                                   $261.03

Medical/Personal Care 5%                                  $217.53

Giving                             5%                                 $217.53

Clothing                         4%                                  $174.02

Miscellaneous               5%                                 $217.53

                                                                          $4,350.50

Balance

Income $4,350.50

Expenses $4,350.50

Balance $0

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2. You see them in your face:

\(\displaystyle m∠F ≅ m∠C;\)\($\overline{FD}$\) ≅ \($\overline{CA}$\)

\(m∠E ≅ m∠B;\)\($\overline{FE}$\) ≅ \($\overline{CB}$\)

\(m∠D ≅ m∠A;\)\($\overline{ED}$\) ≅ \($\overline{BA}$\)

To sum up, corresponding parts of congruent triangles are always congruent [CPCTC].

1. The edge and angle markings of both triangles represent all corresponding parts. Each angle and edge corresponds with another angle and edge.

I am joyous to assist you at any time.

Answer:

37 degrees

Step-by-step explanation:

Answer:

A=37

Step-by-step explanation:

Answer:

198 in²

Step-by-step explanation:

triangle (A=1/2)bh [A=1/2(22×8)] b=22 H=8

rectangle (A=bh) [A= 22×5] B=22 H=5

triangle = 88

rectangle = 110

So, 88+110 = 198

Answer:

13 y^2 - 13 y + 11

Step-by-step explanation:

Simplify the following:

15 y^2 - 2 y^2 + 2 y - 15 y + 4 + 7

Grouping like terms, 15 y^2 - 2 y^2 + 2 y - 15 y + 4 + 7 = (15 y^2 - 2 y^2) + (-15 y + 2 y) + (7 + 4):

(15 y^2 - 2 y^2) + (-15 y + 2 y) + (7 + 4)

15 y^2 - 2 y^2 = 13 y^2:

13 y^2 + (-15 y + 2 y) + (7 + 4)

2 y - 15 y = -13 y:

13 y^2 + -13 y + (7 + 4)

7 + 4 = 11:

Answer: 13 y^2 - 13 y + 11

Answer:

9

Step-by-step explanation:

9/9=1

27/9=3

The maximum capacity of the cup depends on the radius R and is given by\((1/3) \times \pi \times R^3\).

To find the maximum capacity of the cup, we need to maximize its volume. The volume of a cone is given by the formula:

\(V = (1/3) \times \pi \times r^2 \times h\)

where r is the radius of the base, h is the height, and π is a constant equal to approximately 3.14159.

In our case, the radius of the base is R, and the height of the cone is h.

To find the height h, we need to use the Pythagorean theorem. Let's call the angle CAB θ, and let's call the length of the segment AB x. Then we have:

sin θ = h / R

cos θ = x / R

Using the Pythagorean theorem, we have:

\(x^2 + h^2 = R^2\)

Substituting h in terms of θ and x, we get:

\(sin^2 \theta \times R^2 + x^2 = R^2\\x^2 = R^2 - R^2 \times sin^2 \theta\\x = R \times cos \theta\)

Now we can express the volume of the cone in terms of θ and x:

\(V = (1/3) \times \pi \times R^2 \times h\\V = (1/3) \times \pi \timesR^2 \times sin \theta \times R\\V = (1/3) \times \pi \times R^3 \times sin \theta\)

To find the maximum volume, we need to find the value of θ that maximizes V. We can do this by taking the derivative of V with respect to θ and setting it equal to zero:

\(dV/d\theta = (1/3) \times \pi \times R^3 \times cos \theta = 0\)

This gives us cos θ = 0, which implies θ = π/2.

Therefore, the maximum volume of the cone-shaped cup is:

\(V = (1/3) \times \pi \times R^3 \times sin(\pi /2) = (1/3) \times\pi \times R^3\)

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The cubic equation is used to describe the glide path of a plane. The given information states that the plane is in level flight at an altitude of 1 mile and starts its descent 6 miles east of the airport.

The question provides information about the altitude of the plane is in and starts its descent 6 miles east of the airport. However, since the equation itself is not provided, we cannot determine the exact shape or characteristics of the glide path.
To better understand the glide path described by the cubic equation, we need to know the equation itself. Unfortunately, the equation is not provided in the question. Without the equation, we cannot determine the exact shape or characteristics of the glide path.
However, a cubic equation typically has the form:
y = ax^3 + bx^2 + cx + d
Where x represents the horizontal distance and y represents the vertical distance (altitude) from the ground. The coefficients a, b, c, and d determine the specific shape of the curve.
Without the specific equation, we cannot provide explanation or examples based on the given information. We would need the equation to calculate and visualize the glide path accurately.
In summary, the question provides information that the plane is in level flight at an altitude of 1 mile and starts its descent 6 miles east of the airport, about a plane's glide path being described by a cubic equation. However, since the equation itself is not provided, we cannot determine the exact shape or characteristics of the glide path.

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Probability of at least one six is 0.6651. Probability of at least two sixes is 0.6187. Probability of at least three sixes is 0.5973.

To estimate the probabilities of rolling at least a certain number of sixes, we can use the concept of complementary probabilities. The complementary probability of an event A is equal to 1 minus the probability of the event not occurring (denoted as A').

For a fair six-sided die, the probability of rolling a six in a single roll is 1/6. Therefore, the probability of not rolling a six in a single roll is 1 - 1/6 = 5/6.

Let's calculate the probabilities for each case:

At least one six when rolling 6 six-sided dice:

To calculate the probability of rolling at least one six, we need to calculate the complementary probability of not rolling any sixes in all six rolls.

Probability of not rolling a six in a single roll: 5/6

Probability of not rolling a six in all six rolls: \((5/6)^6\)

Now, the probability of rolling at least one six is the complementary probability:

Probability of at least one six = 1 - \((5/6)^6\)

At least two sixes when rolling 12 six-sided dice:

Similar to the previous case, we need to calculate the complementary probability of not rolling at least two sixes in all twelve rolls.

Probability of not rolling a six in a single roll: 5/6

Probability of not rolling at least two sixes in all twelve rolls:\((5/6)^{12\)

Probability of at least two sixes = 1 - \((5/6)^{12\)

At least three sixes when rolling 18 six-sided dice:

Again, we calculate the complementary probability of not rolling at least three sixes in all eighteen rolls.

Probability of not rolling a six in a single roll: 5/6

Probability of not rolling at least three sixes in all eighteen rolls: \((5/6)^{18\)

Probability of at least three sixes = 1 - \((5/6)^{18\)

The estimated probabilities are approximately:

Probability of at least one six when rolling 6 six-sided dice: 0.6651 or 66.51%

Probability of at least two sixes when rolling 12 six-sided dice: 0.6187 or 61.87%

Probability of at least three sixes when rolling 18 six-sided dice: 0.5973 or 59.73%

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The time it will it take him to grade 150 papers is 32 minutes, 4 seconds

It is important to note that proportion is a method of comparison in which two expressions or equations are made equal to each other.

From the information given, we have that;

Mr. B grades a total of 12.5 papers in 2.7 minutes.

Then, for 150 papers, we would have;

If 12. 5 papers = 2.7 minutes

Then 150 papers = x

Cross multiply the values

12.5x = 2.7(150)

multiply the values

12.5x = 405

Divide the values by the coefficient of x, we get;

x = 405/12.5

x = 32 minutes, 4 seconds

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In all cases, we can find a pumped string \(xy^kz\) that does not belong to L₂, which contradicts the assumption that L₂ is regular. Therefore, L₂ is not regular.

To prove that the given languages L₁ and L₂ are not regular using the pumping lemma, we need to show that for any hypothetical regular language L.

There exists a pumping length p such that for any string s in L of length at least p, we can pump s in a way that the pumped string is not in L.

1. L₁ = {\(0^n1^n2^n\) | n ≥ 0, Σ = {0, 1, 2}}

Assume L₁ is regular and let p be the pumping length. Consider the string s = \(0^p1^p2^p\). This string is in L₁ because it has the form \(0^n1^n2^n\), where n = p.

By the pumping lemma, we can decompose s into three parts: s = xyz, such that:

1. |y| > 0

2. |xy| ≤ p

3. For all k ≥ 0, \(xy^kz\) is in L₁.

Let's consider different cases for the possible placement of y in s.

y contains only 0s (\(y = 0^m\), where 1 ≤ m ≤ p).

In this case, when we pump y (k > 1), the number of 0s will exceed the number of 1s and 2s, and hence the pumped string \(xy^kz\) will not be in L₁.

y contains both 0s and 1s (\(y = 0^m1^k\), where 1 ≤ m + k ≤ p).

In this case, when we pump y (k > 1), the number of 0s and 1s will not be balanced with the number of 2s, and hence the pumped string \(xy^kz\) will not be in L₁.

y contains only 1s (\(y = 1^k\), where 1 ≤ k ≤ p).

In this case, when we pump y (k > 1), the number of 1s will exceed the number of 0s and 2s, and hence the pumped string \(xy^kz\) will not be in L₁.

y contains both 1s and 2s (\(y = 1^m2^k\), where 1 ≤ m + k ≤ p).

In this case, when we pump y (k > 1), the number of 1s and 2s will not be balanced with the number of 0s, and hence the pumped string \(xy^kz\) will not be in L₁.

Thus, in all cases, we can find a pumped string \(xy^kz\) that does not belong to L₁, which contradicts the assumption that L₁ is regular. Therefore, L₁ is not regular.

2. L₂ = {ωωω | ω ∈ {a, b}*}

Assume L₂ is regular and let p be the pumping length. Consider the string \(s = a^pb^pa^pb^pa^pb\). This string is in L₂ because it has the form ωωω, where ω = \(a^pb^p\).

By the pumping lemma, we can decompose s into three parts: s = xyz, such that:

1. |y| > 0

2. |xy| ≤ p

3. For all k ≥ 0, \(xy^kz\) is in L₂.

Let's consider different cases for the possible placement of y in s.

y contains only a's (\(y = a^m\), where 1 ≤ m ≤ p).

In this case, when we pump

y (k > 1), the number of a's will exceed the number of b's in the first or second occurrence of ω, and hence the pumped string \(xy^kz\) will not be in L₂.

y contains only b's (\(y = b^m\), where 1 ≤ m ≤ p).

In this case, when we pump y (k > 1), the number of b's will exceed the number of a's in the second or third occurrence of ω, and hence the pumped string \(xy^kz\) will not be in L₂.

y contains both a's and b's (\(y = a^mb^n\), where 1 ≤ m + n ≤ p).

In this case, when we pump y (k > 1), the number of a's or b's will exceed the number of the corresponding symbol in the corresponding occurrence of ω, and hence the pumped string \(xy^kz\) will not be in L₂.

Thus, in all cases, we can find a pumped string \(xy^kz\) that does not belong to L₂, which contradicts the assumption that L₂ is regular. Therefore, L₂ is not regular.

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Complete Question

Use the pumping lemma to prove that the following languages are not regular:

L1 = {0^n 1^n 2^n | n ≥ 0, Σ = {0, 1, 2}}

L2 = {ωωω | ω ∈ {a, b}*}

For each language, apply the pumping lemma to show that there exists a pumping length (p) such that no matter how the string is divided into segments, it is not possible to pump the segments to generate all the strings in the language. This demonstrates that the languages are not regular.