Which behavior was observed when one and of the earthworm was placed on a wet paper towel while the other end was placed on a dry paper towel

Answer:

93% confidence interval for the mean number of days of sick leave is [10.39 days , 14.01 days].

Step-by-step explanation:

We are given that the firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days and the sample standard deviation is 10 days.

Assume the population standard deviation is 10.

Firstly, the pivotal quantity for 93% confidence interval for the population mean is given by;

                          P.Q. = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\bar X[/tex] = sample mean = 12.2 days

            [tex]\sigma[/tex] = population standard deviation = 10 days

            n = sample of files = 100

            [tex]\mu[/tex] = population mean

Here for constructing 93% confidence interval we have used One-sample z  test statistics because we know about population standard deviation.

So, 93% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-1.8121 < N(0,1) < 1.8121) = 0.93  {As the critical value of z at 3.5%

                                           level of significance are -1.8121 & 1.8121}  

P(-1.8121 < [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.8121) = 0.93

P( [tex]-1.8121 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X -\mu}[/tex] < [tex]1.8121 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.93

P( [tex]\bar X-1.8121 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.8121 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.93

93% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.8121 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X +1.8121 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]

                                       = [ [tex]12.2-1.8121 \times {\frac{10}{\sqrt{100} } }[/tex] , [tex]12.2+1.8121 \times {\frac{10}{\sqrt{100} } }[/tex] ]

                                       = [10.39 days , 14.01 days]

Therefore, 93% confidence interval for the mean number of days of sick leave is [10.39 days , 14.01 days].

Answer:

2/3 of a pound.

Step-by-step explanation:

10 pounds per 15 bowls = 2 pounds per 3 bowls, this is equal to 2/3 of ice cream a pound in a single bowl.

Answer:

0.4801

Step-by-step explanation:

This is a binomial distribution question.

It can be approximated using normal distribution if the following conditions are met:

np > 10

n(1-p) > 10

Here,

n = 29

p = 0.487

So,

np = 14.12

n(1-p) = 14.88

So, we can use normal approximation here:

Binomial:  X ~ B(n,p)  becomes

Normal Approx:  X~ N([tex]np,\sqrt{np(1-p)}[/tex])

Mean is:

[tex]\mu=np=14.123[/tex]

Standard Deviation is:

[tex]\sigma=\sqrt{np(1-p)} =2.69[/tex]

We need probability of less than or equal to 14, so we can say:

P(x ≤ 14)

Using  [tex]z=\frac{x-\mu}{\sigma}[/tex], we have:

P(x ≤ 14) = [tex]P(\frac{x-\mu}{\sigma} \leq \frac{14-14.123}{2.69})\\=P(z \leq -0.05)\\=0.4801[/tex]

Note: We used z table in the last line

So the probability is 0.4801

Answer:

72

Step-by-step explanation:

multiply by 16

Answer:

4 1/2 lb = 72 oz

brainliest?

Answer :

Step-by-step explanation:

Given that a multiple-choice test contains 10 questions and there are 4 possible answers for each question.

∴ Answers=4 options for each question.

Solving this by product rule

Product rule :

If one event can occur in m ways and a second event occur in n ways, the number of ways of two events can occur in sequence is then m.n

From the given the event of choosing the answer of each question having 4 options is given by

The 1st event of picking the answer of the 1st question=4 ,

2nd event of picking the answer of the 2nd question=4 ,

3rd event of picking the answer of the 3rd question=4

,....,

10th event of picking the answer of the 10th question=4.

It can be written as  by using the product rule

[tex]=4.4.4.4.4.4.4.4.4.4[/tex]

[tex]=4^{10}[/tex]

[tex]=1048576[/tex]

The distance between the two ships is changing at approximately 26.87 km/h at 4:00 PM, calculated using the Pythagorean theorem and the differentiation principles of related rates in calculus.

To determine how fast the distance between the two ships is changing at 4:00 PM, we need to apply the concept of related rates in calculus. Since at noon ship A is 130 km west of ship B and moving east, and ship B is moving north, we can imagine their paths as legs of a right triangle, where the hypotenuse represents the distance between the two ships.

Let's denote the distance that ship A has traveled east as x (in kilometers), the distance that ship B has traveled north as y (in kilometers), and the distance between the two ships as z (in kilometers). At 4:00 PM, ship A has been sailing east for 4 hours at 25 km/h, so x = 25 km/h * 4 h = 100 km. Ship B has sailed north for the same amount of time at 20 km/h, so y = 20 km/h * 4 h = 80 km.

To find the rate at which the distance z is changing, we use the Pythagorean theorem z =  x^2 +  y^2.

Differentiating both sides with respect to time t, we get 2zdz/dt = 2xdx/dt + 2ydy/dt. We can cancel the 2's and plug in the values for x, y, dx/dt (25 km/h) and dy/dt (20 km/h) to find dz/dt, which represents the rate at which the distance between the ships is changing.

Solving for dz/dt, we have:

z dz/dt = x dx/dt + y dy/dt
z dz/dt = 100 km * 25 km/h + 80 km * 20 km/h

First, we must find z, which is the distance between the ships at 4:00 PM:

z =
sqrt{130^2 + 80^2} =
sqrt{16900 + 6400} =
sqrt{23300} ≈ 152.65 km

Now, we solve for dz/dt:
152.65 km * dz/dt = 100 km * 25 km/h + 80 km * 20 km/h
dz/dt ≈ (2500 km²/h + 1600 km²/h) / 152.65 km
dz/dt ≈ 4100 km²/h / 152.65 km
dz/dt ≈ 26.87 km/h

Therefore, the distance between the ships is changing at approximately 26.87 km/h at 4:00 PM.

at 4:00 PM, the rate of change of the distance between the ships is[tex]\( \frac{7350}{10\sqrt{593}} \)[/tex] km/h.

To find the rate of change of the distance between the ships at 4:00 PM, we'll first find expressions for the positions of each ship at that time. Then, we'll differentiate the distance formula with respect to time and evaluate it at 4:00 PM.

Let ( t ) be the time in hours since noon.

Ship A's position [tex]\( x_A \)[/tex] at time \( t \) is given by:

[tex]\[ x_A = 130 + 25t \][/tex]

And ship B's position [tex]\( y_B \)[/tex] at time \( t \) is given by:

[tex]\[ y_B = 20t \][/tex]

Using these positions, the distance between the ships ( D ) at time ( t ) is given by the distance formula:

[tex]\[ D(t) = \sqrt{(130 + 25t)^2 + (20t)^2} \][/tex]

Now, let's differentiate [tex]\( D(t) \)[/tex] with respect to time [tex]\( t \)[/tex] using the chain rule:

[tex]\[ \frac{dD}{dt} = \frac{1}{2\sqrt{(130 + 25t)^2 + (20t)^2}} \times \left(2(130 + 25t)(25) + 2(20t)(20)\right) \][/tex]

Now, plug in ( t = 4 ) to find the rate of change of the distance between the ships at 4:00 PM:

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{(130 + 25(4))^2 + (20(4))^2}} \times \left(2(130 + 25(4))(25) + 2(20(4))(20)\right) \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{(130 + 100)^2 + (80)^2}} \times \left(2(130 + 100)(25) + 2(80)(20)\right) \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{230^2 + 80^2}} \times \left(2(230)(25) + 2(80)(20)\right) \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{52900 + 6400}} \times \left(2(5750) + 2(1600)\right) \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{59300}} \times (11500 + 3200) \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{1}{2\sqrt{59300}} \times 14700 \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{14700}{2\sqrt{59300}} \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{7350}{\sqrt{59300}} \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{7350}{\sqrt{100 \times 593}} \][/tex]

[tex]\[ \frac{dD}{dt} \Bigg|_{t=4} = \frac{7350}{10\sqrt{593}} \][/tex]

So, at 4:00 PM, the rate of change of the distance between the ships is[tex]\( \frac{7350}{10\sqrt{593}} \)[/tex] km/h. This can be simplified further, but it's best to leave it in this form to retain the exact value.

Answer:

The answer is D

Step-by-step explanation:

An equilateral triangle can not be a right triangle. Here is why: In an equilateral triangle, all angles are congruent, since all 3 of the sides are congruent.This is because in a right triangle, one angle equals 90 degrees. Therefore, you can't have 3 angles equal 90 degrees.

Answer:

B.The matrix is invertible because the determinant is defined.

Step-by-step explanation:

Definition: A matrix is invertible if the determinant is not equal to zero.

Given the matrix:

[tex]A=\left(\begin{array}{ccc}20&5&3\\4&-15&6\\0&25&-9\end{array}\right)[/tex]

The determinant of the matrix:

[tex]|A|=20\left|\begin{array}{ccc}-15&6\\25&-9\end{array}\right|-5\left|\begin{array}{ccc}4&6\\0&-9\end{array}\right|+3\left|\begin{array}{ccc}4&-15\\0&25\end{array}\right|[/tex]

[tex]|A|=20(-15*-9-25*6)-5(-9*4-6*0)+3(4*25-0*-15)\\=20(135-150)-5(-36-0)+3(100+0)\\=20(-15)-5(-36)+3(100)\\=-300+180+300\\|A|=180[/tex]

Since the determinant, [tex]|A|\neq 0[/tex], the matrix is invertible because the determinant is defined.

The matrix in question is invertible because its determinant is not zero. This is an essential rule in linear algebra, where a non-zero determinant implies an invertible matrix.

In linear algebra, a square matrix is invertible if and only if its determinant is not zero. The determinant being different from zero implies that the matrix has full rank and hence is invertible. In your case, the determinant of the matrix is 180180, which is not zero. Thus, the matrix is invertible.

The option: B. The matrix is invertible because the determinant is defined is not entirely correct because a matrix could have a defined determinant that equals zero which would make it non-invertible. The correct answer should be: C. The matrix is invertible because the determinant is not zero.

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Answer:

y = -xe^x + xe^x ln(x) +C

See attachment for step by step guide please

Answer:

Combining the method of undetermined coefficients with the method of variation of parameters, the solution to the differential equation

y'' - 2y' + y = 4x² - 6 + x^(-1)e^x

is

y = (C1 + C2x)e^x + 4x² + 16x + 18 -xe^x + xe^x lnx

Step-by-step explanation:

Given the differential equation:

y'' - 2y' + y = 4x² - 6 + x^(-1)e^x........(1)

Firstly, we solve the homogeneous part of (1)

y'' - 2y' + y = 0

Let the characteristic equation be

m² - 2m + 1 = 0

(m - 1)(m - 1) = 0

m = 1 twice.

The complementary function

y_c = (C1 + C2x)e^x...........................(2)

Now, consider the differential equation:

y'' - 2y' + y = 4x² - 6 ..........................(3)

Solve (3) using the method of UNDETERMINED COEFFICIENTS.

The nonhomogeneous part is 4x² - 6, so we assume a particular solution of the form

y_p = Ax² + Bx + C

y'_p = 2Ax + B

y''_p = 2A

Using these in (3), we have

y''_p - 2y'_p + y_p

= 2A - 2(2Ax + B) + Ax² + Bx + C = 4x² - 6

Ax² + Bx - 4Ax - 2B + 2A + C = 4x² - 6

Comparing the coefficients of various powers of x, we have that

A = 4

B - 4A = 0 => B = 16

2A - 2B + C = -6

=> C = -6 - 8 + 32 = 18

Therefore,

y_p = 4x² + 16x + 18 .....................(4)

Next, consider the differential equation:

y'' - 2y' + y = x^(-1)e^x ....................(3)

We solve using the method of VARIATION OF PARAMETERS.

With g(x) = x^(-1)e^x

Using the complementary solution, we have

y1 = e^x, and y2 = xe^x

Find the Wronskian of y1 and y2.

Let their Wronskian be W, then

W = |y1............y2|

........|y1'...........y2'|

= |e^x.....................xe^x|

...|e^x...........xe^x + e^x|

= xe^(2x) + e^(2x) - xe^(2x)

W = e^(2x)

y_q = Py1 + Qy2

Where P = integral of y2g(t)/W dx

= integral of xe^x.x^(-1)e^x/e^(2x) dx

= -x

Where Q = integral if y1g(t)/W dx

= integral of e^x.x^(-1)e^x/e^(2x) dx

= lnx

y_q = -xe^x + xe^x lnx

Finally, the general solution is

y = y_c + y_p + y_q

= (C1 + C2x)e^x + 4x² + 16x + 18 -xe^x + xe^x lnx

Answer: 3, 7 and 14

Step-by-step explanation:

We want 3 numbers with a mean of 8, and a MAD of 4.

if we have the numbers a, b and c.

The mean is:

M = (a + b + c)/3

MAD = ( Ia - MI + Ib - MI + Ic - MI)/3

now, we know that M = 8 and MAD = 4, so we can replace those values:

8 = (a + b + c)*/3

3*8 = 24 = a + b + c

and

4 = ( Ia - 8I + Ib - 8I + Ic - 8I)/3

4*3 = 12 = Ia - 8I + Ib - 8I + Ic - 8I

So now we have the two equations:

24 = a + b + c

12 = Ia - 8I + Ib - 8I + Ic - 8I

Let's took the numbers 3, 7 and 14

I selected those because are the numbers where the distance between them is the biggest, and they add up to 24, and you can see that the MAD is a big number, which implies that our 3 numbers are not close between them.

3 + 7 + 14 = 10 + 14 = 24

and for the MAD

I3 -8I + I7 - 8I + I14 - 8I = 5 + 1 + 6 = 12

So the correct options are 3, 7 and 14

Answer:

Systems of linear equations are used in the real world by economists and entrepreneurs to find out when supply equals demand.It's all about the mullah, and if you don't know the numbers when you have a business, it might fail.

Step-by-step explanation:

Answer:

The index of refraction for Medium 2 would be 1.25

Step-by-step explanation:

We have the following data:

Speed of light in Medium 1 = V₁

Speed of light in Medium 2 = V₂

The speed of Light in Medium 2 is related to speed of light in Medium 1 as:

V₂ = 0.8 V₁

Moreover, it is given that Medium 1 is air with index of refraction equal to 1. i.e.

n₁ = 1

We need to find the index of refraction for medium 2. Lets name it is as n₂.

Snell's law related the above mentioned 4 quantities by following relation:

The ratio of indices of refraction of 2 mediums is equivalent to the reciprocal of ratio of  phase velocities in those 2 mediums. Mathematically, this can be written as:

[tex]\frac{n_{1}}{n_{2}} = \frac{V_{2}}{V_{1}}[/tex]

Substituting the values we have gives us:

[tex]\frac{1}{n_{2}}=\frac{0.8V_{1}}{V_{1}} \\\\\frac{1}{n_{2}}=0.8\\\\ 1=0.8n_{2}\\\\\frac{1}{0.8}=n_{2}\\\\ n_{2}=1.25[/tex]

This means, the index of refraction for Medium 2 would be 1.25

To find the index of refraction for medium 2 (n2), one must use the relationship n = c/v. Since V2 = 0.8 V1 where V1 approximates c, the index of refraction n2 is calculated as n2 = 1/0.8, resulting in an index of 1.25.

The question asks about determining the index of refraction for a medium when a light beam travels from air into this medium. According to the given information, the speed of light in medium 2 (V2) is 0.8 times the speed of light in medium 1 (V1), with air being medium 1 where the speed of light is approximately the speed of light in a vacuum and the index of refraction (n1) is 1.

The law of refraction, or Snell's Law, relates the index of refraction of a medium to the speed of light in that medium through the formula
n = c/v, where
c is the speed of light in vacuum and
v is the speed of light in the medium. To find the index of refraction for medium 2 (n2), you would use the relation that n = c/v. Since the speed of light in medium 2 is given as 0.8 V1, and V1 is the speed of light in air (which is approximately equal to c since n1 = 1), you can deduce that
n2 = c/(0.8c). This simplifies to n2 = 1/0.8, which means that the index of refraction for medium 2 is 1.25.

Answer:

57

Step-by-step explanation:

Given:

In Ms. Perron’s class, 75% of the students are boys. There are 18 boys in the class.

We need to determine the total number of students in Ms. Perron's class.

Total number of students:

Let n denote the total number of students in Ms. Perron's class.

Thus, we have;

[tex]\frac{75}{100}n=18[/tex]

Multiplying both sides of the equation by [tex]\frac{100}{75}[/tex], we get;

[tex]n=18\times \frac{100}{75}[/tex]

Simplifying, we get;

[tex]n=\frac{1800}{75}[/tex]

[tex]n=24[/tex]

Therefore, the total number of students in Ms. Perron's class are 24.

Answer:

There is a 1/2 chance the coin will land on heads and there is a 1/6 chance that the number cube will land on 6. hope this helps

Answer:

[tex]t=52.39[/tex]

Step-by-step explanation:

Answer:

a) [tex] f(x) = \frac{1}{40-30}, 30 \leq x \leq 40[/tex]

b) [tex] E(X) = \frac{a+b}{2}= \frac{30+40}{2}=35[/tex]

c) [tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(40-30)^2}{12}= 8.33[/tex]

And the deviation would be:

[tex] Sd(X) = \sqrt{8.33}= 2.887[/tex]

d) [tex]P(34< X <38) = F(38) -F(34)= \frac{38-30}{10} -\frac{34-30}{10}= 0.8-0.4=0.4[/tex]

Step-by-step explanation:

For this case we define the random variable X with this distribution:

[tex] X \sim Unif (a=30, b=40)[/tex]

Part a

The density function since is an uniform distribution is given by:

[tex] f(x) = \frac{1}{40-30}, 30 \leq x \leq 40[/tex]

Part b

The expected value is given by:

[tex] E(X) = \frac{a+b}{2}= \frac{30+40}{2}=35[/tex]

Part c

The variance is given by:

[tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(40-30)^2}{12}= 8.33[/tex]

And the deviation would be:

[tex] Sd(X) = \sqrt{8.33}= 2.887[/tex]

Part d

For this case we want this probability:

[tex] P(34< X <38)[/tex]

And we can use the cumulative distribution function given by:

[tex] F(x)= \frac{x-30}{40-30}, 30 \leq X \leq 40[/tex]

And using this we got:

[tex]P(34< X <38) = F(38) -F(34)= \frac{38-30}{10} -\frac{34-30}{10}= 0.8-0.4=0.4[/tex]

Final answer:

The probability density function, expected price, standard deviation, and the probability of the stock price being between $34 and $38 are calculated.

Explanation:

To find the probability density function, f(x), for the price of the stock, we need to determine the range of the variable x, and then divide the range by the total probability. In this case, the range of x is from $30 to $40, and the total probability is equal to 1. So, the probability density function is:

f(x) = 1 / (40 - 30) = 1/10 = 0.1

To determine the expected price of the stock, we can calculate the average of the range of x:

Expected price = (30 + 40) / 2 = $35

To determine the standard deviation for the stock, we can use the formula:

Standard deviation = (40 - 30) / sqrt(12) = 10 / sqrt(12) ≈ 2.8879

To find the probability that the stock price will be between $34 and $38, we need to find the area under the probability density function curve between these two prices. Since the probability density function is a uniform distribution, the probability is equal to the width of the range divided by the total width of the distribution:

Probability = (38 - 34) / (40 - 30) = 4 / 10 = 0.4

In a right triangle, the tangent of an angle is defined as the ratio of the opposite side to the adjacent side. Therefore, for triangle ΔWXY, the tangent of ∠X is the ratio of side WY to YX, which is 8/15.

To understand this question, we need to know that in the context of a right triangle, the tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side. In ΔWXY, where the measure of ∠Y=90°, ∠X is the angle we are considering. The side opposite to ∠X is WY and the side adjacent to ∠X is YX. Therefore, the tangent of ∠X can be calculated using the formula: tan(X) = WY / YX.

In this scenario, we know that WY = 8 and YX = 15. So, the tangent of ∠X is given by: tan(X) = WY / YX = 8 / 15. Hence, the ratio that represents the tangent of ∠X is 8 / 15.

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Answer:

32.40

Step-by-step explanation:

Answer:

1176.5 square feet

Step-by-step explanation:

Let us recall from the following question,

Because  fuel tank is a cylinder, the first step to take is  get the surface area of this cylinder.

The Surface area of  a cylinder = 2. π. r. 2. +. 2. π.

Then,

The given values (h = 50, r = 7/2 = 3.5) as applied from the formula given,

The area Surface a of fuel tank  = 1176.5 square feet

Answer:

Surface area A = 76.979 + 21.994h

Where h will be the value of the lenght of the fuel tanker.

Step-by-step explanation:

The fuel tank is cylindrical so we solve as a cylinder.

Total surface area of a cylinder = the areas of the circles on the top and the bottom and the area of the body.

Area = (2¶d^2)/4 + ¶dh

= (¶d^2)/2 + ¶dh

= ¶d( d/2 + h)

If it has a diameter of 7 ft, then,

A = 3.142 x 7 ( 7/2 + h)

A = 21.994 ( 3.5 + h)

A = 76.979 + 21.994h

Answer:

The pole is 15 feet tall

Step-by-step explanation:

Pythagora's Theorem

Let's call x the distance from the base of the pole to the spot where the guy wire is anchored. The height of the pole is 7 feet more, i.e. x+7.

The guy wire is 17 feet long. These dimensions form the sides of a right triangle where the guy wire is the hypotenuse.

Applying Pythagora's Theorem

[tex]x^2+(x+7)^2=17^2[/tex]

Operating

[tex]x^2+x^2+14x+49=289[/tex]

Rearranging and simplifying by 2

[tex]x^2+7x-120=0[/tex]

Factoring

[tex](x-8)(x+15)=0[/tex]

Solving

[tex]x=8,\ x=-15[/tex]

Only the positive solution is valid, thus x=8

The height of the pole is x+7=15 feet

The pole is 15 feet tall

Final answer:

Using the Pythagorean theorem, we find that the height of the pole is approximately 15 feet when solving the derived quadratic equation.

Explanation:

The question involves applying the Pythagorean theorem to solve for the height of the pole. Let's denote the distance from the base of the pole to the spot where the guy wire is anchored as x. Then, the height of the pole is x + 7 feet. Since the guy wire creates a right triangle with the pole and the ground, the Pythagorean theorem states that the square of the hypotenuse (guy wire) is equal to the sum of the squares of the other two sides (height of the pole and the distance from the base). Therefore, we can write the equation 17^2 = (x + 7)^2 + x^2. By solving this equation, we will find the value of x and then determine the height of the pole.

Step by step, we solve the equation: 289 = (x + 7)^2 + x^2, which simplifies to 289 = 2x^2 + 14x + 49. Subtracting 289 from both sides results in 0 = 2x^2 + 14x - 240. By factoring or using the quadratic formula, we find that x is approximately 8 feet. Therefore, the height of the pole is x + 7 feet, which is 15 feet tall.

Answer:

The frequency of the purple and pink alleles in this field population are 84.7% and 15.3% respectively.

Step-by-step explanation:

According to the law of large numbers, in probability concept, as we increase the sample size (n), the sample statistic value approaches the value of the population parameter.

For example, as the sample size increases the sample mean ([tex]\bar x[/tex]) approaches the true population mean (μ).

Similarly, as the sample size is increased the the population proportion can be estimated by the sample proportion.

In this case, a random sample of n = 1000 tulips are selected.

Of these 1000 tulips, the number of purple flowers was 847 and the number of pink flower was 153.

Compute the proportion of purple flower as follows:

[tex]P(purple)=\frac{847}{1000}=0.847[/tex]

The frequency of purple flower is 0.847.

Since, the sample size is quite large, this proportion can be used to estimate the true proportion of purple alleles.

Compute the proportion of pink flower as follows:

[tex]P(pink)=\frac{153}{1000}=0.153[/tex]

The frequency of pink flower is 0.153.

Again since, the sample size is quite large, this proportion can be used to estimate the true proportion of pink alleles.

Thus, the frequency of the purple and pink alleles in this field population are 84.7% and 15.3% respectively.

Answer:

Hello

The answer is  30% increase in texts since last month.

IF you feel any problem in understanding , do comment pls.

Step-by-step explanation:

Let

X = last month sent texts

y = this month sent texts

First of all find the no. of increased texts,

by subtracting x from y

=> y-x= 7085- 5450

        = 1635

We want to find these 15 texts % with respect to 5450 texts

i.e.    1635/X

=0.30

for answer in % multiply with 100

i.e. 30%

Complete question:

Researchers are studying two populations of sea turtles. In population D, 30 percent of the turtles have a shell length greater than 2 feet. In population E, 20 percent of the turtles have a shell length greater than 2 feet. From a random sample of 40 turtles selected from D, 15 had a shell length greater than 2 feet. From a random sample of 60 turtles selected from E, 11 had a shell length greater than 2 feet. Let pˆD represent the sample proportion for D, and let pˆE represent the sample proportion for E.

(a) What is the value of the difference pˆD−pˆE? Show your work.

(b) What are the mean and standard deviation of the sampling distribution of the difference in sample proportions pˆD−pˆE? Show your work and label each value.

(c) Can it be assumed that the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal? Justify your answer.

(d) Consider your answer in part (a). What is the probability that pˆD−pˆE is greater than the value found in part (a)?

Answer:

a) 0.1917

b) The mean is 0.1917 and the standard deviation is 0.0914.

c)Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal

d) 0.1580

Step-by-step explanation:

a) for p`D we have:

[tex] \frac{15}{40} [/tex]

= 0.375

For p`E we have:

[tex] \frac{11}{60} [/tex]

= 0.1833

Therefore, p`D - p`E, we have:

0.375 - 0.1833

=0.1917

b) The Mean can be calculated as p`D - p`E =

0.375 - 0.1833

=0.1917

For standard deviation:

[tex] s.d = \sqrt{\frac{p`D (1-p`D)}{N_D} + \frac{p`E(1-p`E)}{N_E}}[/tex]

[tex] s.d = \sqrt{\frac{0.375(1 - 0.375)}{40} + \frac{0.1833(1 - 0.1833)}{60}}= 0.0914[/tex]

The mean is 0.1917 and the standard deviation is 0.0914.

c) Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal, because for normal condition, we have:

i) np ≥ 10

ii) n(1-p) ≥ 10.

From the expressions, we can see the samples satisfy the condition for normality.

d) To get the probability, wen need to find the Z score.

The Z score can be calculated using the formula:

[tex] Z = \frac{(p`D - p`E) -(pD - pE)}{s.d}[/tex]

[tex] = \frac{(0.1917) -(0.1)}{0.0914}[/tex]

= 1.0029

Therefore,

P(Z > 1.0029) = 1 - P(Z ≤ 1.0029)

From the z distribution table, we have:

P (Z > 1.0029) = 1 - 0.8420 = 0.1580

The probability is 0.1580

The value of the difference pˆD−pˆE is 0.1917. The mean and standard deviation is 0.1917 and 0.0914 respectively and the probability that pˆD−pˆE is greater than the 0.1917 is 0.1580.

Given :

a) The value of pˆD is:

[tex]=\dfrac{15}{40}[/tex]

The value of pˆE is:

[tex]=\dfrac{11}{60}[/tex]

So, the value of (pˆD - pˆE) is:

[tex]=\dfrac{15}{40}-\dfrac{11}{60}[/tex]

= 0.375 - 0.1833

= 0.1917

b) Mean is given by the formula:

pˆD - pˆE = 0.1917

For standard deviation using the formula:

[tex]\rm SD =\sqrt{ \dfrac{p\hat{}D(1-p\hat{}D)}{N_D}+\dfrac{p\hat{}E(1-p\hat{}E)}{N_E}}[/tex]

[tex]\rm SD = \sqrt{\dfrac{0.375(1-0.375)}{40}+\dfrac{0.1833(1-0.1833)}{60}}[/tex]

SD = 0.0914

c). Yes, the sampling distribution of the difference of the sample proportions pˆD−pˆE is approximately normal.

d). To determine the probability, first evaluate the z-score.

[tex]\rm Z=\dfrac{(p\hat{}D-p\hat{}E)-(pD-pE)}{SD}[/tex]

[tex]\rm Z = \dfrac{0.1917-0.1}{0.0914}[/tex]

Z = 1.0029

Now, P(Z>1.0029) = 1 - P(Z [tex]\leq[/tex] 1.0029)

                              = 1 - 0.8420

                              = 0.1580

For more information, refer to the link given below:

https://brainly.com/question/2561151

Answer:

The work done is 2.5(b-a)* ln(d/c).

Step-by-step explanation:

Steps are in the following attachments                    

The work done by an ideal Carnot engine is equal to the area enclosed by the region in the pV diagram.

The work done by an ideal Carnot engine is equal to the area enclosed by the region in the pV diagram. This region is bounded by two isothermal curves and two adiabatic curves. The work done by the engine can be calculated by finding the area under the isothermal curves and subtracting the area under the adiabatic curves.

To find the work done, you can divide the region into smaller shapes, such as rectangles or triangles, and calculate the area of each shape. Then, sum up the areas of all the shapes to get the total work done by the engine.

Remember to use the equations for the isothermal and adiabatic processes to relate the pressure and volume of the gas at different points in the cycle.

https://brainly.com/question/34820223

#SPJ11

Final answer:

To calculate the number of days needed to travel at least 4,325 miles at a rate of about 75 miles per day, divide 4,325 by 75, which gives approximately 57.67. Therefore, participants must ride for at least 58 days to cover the distance.

Explanation:

To find out the number of days participants must ride to travel at least 4,325 miles on a bicycle, assuming they cover about 75 miles per day, we need to divide the total distance by the daily distance covered.

Total distance to travel: 4,325 miles
Average distance per day: 75 miles

Number of days (d) = Total distance / Average distance per day

d = 4,325 miles / 75 miles per day
d ≈ 57.67 days

Since participants cannot travel for a fraction of a day, they must ride for at least 58 days to cover 4,325 miles, assuming they ride about 75 miles each day.

Answer:

Time t = 2 seconds

It will reach the maximum height after 2 seconds

Completed question;

Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, after t seconds that Amir has thrown the ball is given by:

H (t) = -(t-2)^2+9

many seconds after being thrown will the ball reach its maximum height?

Step-by-step explanation:

The equation of the height!

h(t) = -(t-2)^2 + 9 = -(t^2 -4t +4) + 9

h(t) = -t^2 +4t -4+9

h(t) = -t^2 + 4t +5

The maximum height is at dh/dt = 0

dh/dt = -2t +4 = 0

2t = 4

t = 4/2 = 2

Time t = 2 seconds

It will reach the maximum height after 2 seconds

The ball hits the ground at [tex]\( t = 5 \)[/tex] seconds.

Let's analyze the given height function of the ball:

[tex]\[ H(t) = -(t-2)^2 + 9 \][/tex]

This function describes the height of the ball [tex]\( H \)[/tex] in meters at any time [tex]\( t \)[/tex] in seconds.

The function is in the vertex form of a quadratic equation, which is generally written as:

[tex]\[ H(t) = a(t - h)^2 + k \][/tex]

Here, [tex]\( a = -1 \), \( h = 2 \), and \( k = 9 \).[/tex]

Time When the Ball Hits the Ground

- The ball hits the ground when [tex]\( H(t) = 0 \):[/tex]

[tex]\[ 0 = -(t-2)^2 + 9 \] \\\ (t-2)^2 = 9 \][/tex]

 Taking the square root of both sides:

[tex]\[ t-2 = \pm 3 \\\ t = 2 + 3 \quad \text{or} \quad t = 2 - 3 \\\ t = 5 \quad \text{or} \quad t = -1 \][/tex]

 Since negative time doesn't make sense in this context, we discard [tex]\( t = -1 \):[/tex]

[tex]\[ t = 5 \, \text{seconds} \][/tex]

 Therefore, the ball hits the ground at [tex]\( t = 5 \)[/tex] seconds.

The complete question is:

Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x seconds after Amir threw it, is modeled by: h(x)=-(x-2)^2+16 How many seconds after being thrown will the ball hit the ground?

Answer:

x = [tex]14\frac{7}{9}[/tex]  or   14.78

Step-by-step explanation:

The lines are parallel, therefore, k(18) acts same as g(x)

That means that:

k(18) = g(x)

- 14 = [tex]-\frac{18}{7}x[/tex] + 24

- 14 - 24 =  [tex]-\frac{18}{7}x[/tex]

- 38 = [tex]-\frac{18}{7}x[/tex]

38(7) = 18x

266 = 18x

266 / 18 = x

133 / 9 = x

x = [tex]14\frac{7}{9}[/tex]  or   14.78

Answer:

The correct answer is wins and rebounds are correlated positively ,but we cannot decided that having more rebounds leads to more wins,on average.

Step-by-step explanation:

From the example given, the most appropriate conclusion is that, because  causation is not the same as correlation, If two variables are compared,this does not mean that one leads to the other.

An observed data is based on correlation,but for description of  causation ,we need to make experiments,as we update the  variable treatment regarding  to the changes in response variable.

Answer:

[tex]\dfrac{1}{3}[/tex]

Step-by-step explanation:

[tex]\text{P(Captain William Hits}|\text{Pirate Miss)=} \dfrac{1}{3}[/tex]

[tex]\text{P(Captain William Hits}|\text{Pirate Hits)=} 0[/tex]

[tex]\text{P(Pirate Emily Hits}|\text{Captain Misses)=} \dfrac{1}{7}[/tex]

[tex]\text{P(Pirate Emily Hits}|\text{Captain Hits)=} 0[/tex]

Therefore:

Now, if the Captain hits, the Pirate will always miss.

P(Captain Hits and the Pirate Miss)[tex]=\dfrac{1}{3}X1 =\dfrac{1}{3}[/tex]