An agricultural field trial compares the yield of two varieties of tomatoes for commercial use – one is the common variety used by farmers and the other is a new hybrid. The researchers plant each type of tomato on one half of each of 10 plots. At harvest time the difference in the yield of the hybrid tomato minus the common variety in pounds per plant (ppp) is calculated for each plot. The average difference is 1.05 ppp with a standard deviation of 1.23 ppp. In testing whether the hybrid tomato has a higher yield with a significance level of α = 0.05, what is the p-value? You may assume the conditions to carry out the relevant hypothesis test are satisfied.

Answer:

z=1.461, fail to reject the null hypothesis since [tex]p_v>\alpha[/tex] at 5% of singificance.

Step-by-step explanation:

1) Data given and notation

n=53 represent the random sample taken

X=41 represent the adults with damage liability claims by single people under the age of 25.

[tex]\hat p=\frac{41}{53}=0.774[/tex] estimated proportion of adults with damage liability claims by single people under the age of 25.

[tex]p_o=0.68[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test that the insurance claims of single people under the age of 25 is higher than the national percent reported by the large insurance company (68%).:  

Null hypothesis:[tex]p\leq 0.68[/tex]  

Alternative hypothesis:[tex]p > 0.68[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.774 -0.68}{\sqrt{\frac{0.68(1-0.68)}{53}}}=1.461[/tex]

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a unilateral right tailed test the p value would be:  

[tex]p_v =P(z>1.461)=0.072[/tex]  

So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with damage liability claims by single people under the age of 25 is not significantly higher from 0.68 or 68% .  

The Laplace transform of the function f(t) = eat sinh bt is F(s) = (e^a)/((s - a)^2 + b^2) after manipulating and simplifying using the standard Laplace transform rules.

The Laplace transform of the function f(t) = eat sinh bt is given by the formula:

F(s) = L{f(t)}(s) = Int0->infinity[ e^(at - s)*sinh(bt) dt]

By splitting the hyperbolic sine function, sinh(bt), into exponentials, we get:

F(s) = 0.5*Int0->infinity [ (e^(-s + a + b)*t - e^(-s + a - b)*t) dt]

As a result, after applying Laplace transform rules and simplifying, we get:

F(s) = (e^a)/((s - a)^2 + b^2).

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Answer:

See attached image for the graph of the function

Step-by-step explanation:

Notice that this is the product of a power function ([tex]4x^2[/tex]) times the trigonometric and periodic function cos(x). So the zeros (crossings of the x axis will be driven by the values at which they independently give zero. That is the roots of the power function (only x=0) and the many roots of the cos function: [tex]x= \frac{\pi}{2} , \frac{3\pi}{2} ,...[/tex], and their nagetiva values.

Notice that the blue curve in the graph represents the original function f(x), with its appropriate zeros (crossings of the x-axis), while the orange trace is that of "-f(x)". Of course for both the zeroes will be the same, while the rest of the curves will be the reflection over the x-axis since one is the negative of the other.

Answer:

1/4

Step-by-step explanation:

Parents            WwDd                                        wwdd

Gametes  WD   Wd   wD  wd                       wd wd wd wd

Offspring possibilities: WwDd WwDd WwDd WwDd

                                     Wwdd Wwdd Wwdd Wwdd

                                      wwDd wwDd wwDd wwDd

                                      wwdd wwdd wwdd wwdd

P (wavy red hair) =4/16 = 1/4

Answer:

The probability is one half (1/2)

But it seems to be a mistake in the statement .."married a wavy red haired woman(homozygous recessive for both traits)". If the woman has wavy hair, the trait can't be homozygous recessive, although she certainly is homozygous recessive for the second trait because is red hair.

Step-by-step explanation:

As wavy hair is dominant over straight hair, it means only one allele is necessary to be visible: WW (homozygous), and Ww (heterozygous) produce the wavy hair trait. An homozygous recessive (ww) doesn't show the wavy trait character; he or she would have straight hair

The woman, if she's homozygous and wavy hair, would be (WW). As she is homozygous for red hair, she has red hair.

If the statement "A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits)" is correct, meaning that she is not wavy hair but "STRAIGHT", the probability of having a wavy red hair changes, being only 1/4

Calculation can be performed with Punnett squares

For example for the wavy dark hair male (heterozygous for both traits), the genotype is WwDd and the possible allele combinations would be: WD, Wd, wD, and wd

For the woman, if she's wavy red hair (homozygous), the genotype would be WWdd and the possible allele combinations would be only Wd.

Then you need cross the male allele combinations against Wd

If the woman is straight red hair (homozygous for both traits), the genotype would be wwdd and the possible allele combinations would be only wd.

Then you need cross the male allele combinations against wd, and obtain the proportions produced from the cross

An example of Punnett square below

Answer: 19/12 pounds or 1.583 pounds

Step-by-step explanation:

Let a pound be represented by x. So x = 1 pound.

5/6 of a pound is the same as multiplying 5/6 × x. It becomes 5/6 × x = 5x/6

3/4 of a pound is the same as multiplying 3/4 × x. It becomes 3/4 × x = 3x/4

We want to look for 5/6 of a pound + 3/4 of a pound.

It becomes

5x/6 + 3x/4

Taking lowest common multiple(LCM) of 12, it becomes

= (2×5x + 3×3x) / 12

= (10x +9x)/12

= 19x /12

Substituting x = 1 pound,

It becomes 19/12 pounds

Expressing in decimal, 19/12 = 15.83

We can also solve from the beginning in terms of decimals

5/6 pounds = 0.833 pounds

3/4pounds = 0.75 pounds

5/6 pounds + 3/4 pounds = 0.833 + 0.75 = 1.583 pounds

The correct answer is: A,

On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).

Step-by-step explanation:

Figure represents graph of [tex]f(x) = x^{2}[/tex] and [tex]g(x) = (x-2)^{2} -3[/tex]

Here, [tex]f(x) = x^{2}[/tex] is " Translated " or " Transformation " to [tex]g(x) = (x-2)^{2} -3[/tex]

In process of transformation,

You should remember that shape of curve remain same and only changes we get in vertex shift

Now, Vertex can be shift in two direction, we are going to discuss both the cases

(A). Shifting of Vertex in X-Axis:

A new function g(x) = f(x - c) represents to X-axis shift and In graph of f(x), Curve is shifted c units along right side of the X-axis

(B). Shifting of Vertex in Y-axis:

A new function g(x) = f(x) + b represents to Y-axis shift and In graph of f(x), Curve is shifted b units along the upward direction of Y-axis

Looking at the figure, You can see that vertex of f(x) is shifted 2 Units in X-axis and Negative 3 units in Y-axis and result into g(x)

Now,  [tex]g(x) = (x-2)^{2} -3[/tex]  = f(x-2) + (-3)

Therefore, new vertex we get is (2,-3)

Also, [tex]g(x) = (x-2)^{2} -3[/tex]

[tex]g(0) = (0-2)^{2} -3[/tex]

[tex]g(0) = (4} -3[/tex]

[tex]g(0) = 1 [/tex]

So. g(x) passes through (0,1)

The correct answer is: A On a coordinate plane, a parabola opens up. It goes through (0, 1), has a vertex at (2, negative 3), and goes through (4, 1).

Answer:

Graph A is the correct answer

Step-by-step explanation:

I juust took the test and got it right :D hope this helps

Answer:

  a) 1.30656∠(3π/8), 0.541196∠(15π/8)

  b) 1.30656∠(π/8), 0.541196∠(5π/8)

Step-by-step explanation:

The critical points can be found in polar coordinates by considering ...

  [tex]\displaystyle{dy\over dx}={dy/d\theta\over dx/d\theta}={r\cos\theta + r'\sin\theta\over -r\sin\theta + r'\cos\theta} \quad\text{where $r'=dr/d\theta$}[/tex]

We can simplify the effort a little bit by rewriting r as …

  [tex]r=\sqrt{2}\sin{(\theta+\pi /4)}[/tex]

Then, filling in function and derivative values, we have …

  [tex]\dfrac{dy}{dx}=\dfrac{\sqrt{2}(\sin{(\theta+\pi /4)}\cos{(\theta)}+\cos{(\theta+\pi /4)}\sin{(\theta)})}{\sqrt{2}(-\sin{(\theta+\pi /4)}\sin{(\theta)}+\cos{(\theta+\pi /4)}\cos{(\theta)})}\\\\=\dfrac{\sin{(2\theta+\pi /4)}}{\cos{(2\theta+\pi /4)}}\\\\\dfrac{dy}{dx}=\tan{(2\theta +\pi /4)}[/tex]

__

(a) For horizontal tangents, dy/dx = 0, so we have …

 [tex]\tan{(2\theta+\pi /4)}=0\\\\2\theta+\dfrac{\pi}{4}=k\pi \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}-\dfrac{\pi}{8}[/tex]

We can use reference angles for the “r” expressions and write the two horizontal tangent point (r, θ) values of interest as …

  [tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{3\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{15\pi}{8})[/tex]

__

(b) For vertical tangents, dy/dx = undefined, so we have …

 [tex]2\theta+\dfrac{\pi}{4}=k\pi +\dfrac{\pi}{2} \quad\text{for some integer k}\\\\\theta=k\dfrac{\pi}{2}+\dfrac{\pi}{8}[/tex]

Again using reference angles for “r”, the two vertical tangent point values of interest are …

  [tex](\sqrt{2}\sin{\dfrac{3\pi}{8}},\dfrac{\pi}{8})\ \text{and}\ (\sqrt{2}\sin{\dfrac{\pi}{8}},\dfrac{5\pi}{8})[/tex]

__

The attached graph shows the angle values in degrees and the radius values as numbers. The points of tangency are mirror images of each other across the line y=x.

To find points with horizontal tangents for the polar curve r = cos(θ) + sin(θ), one must set the derivative with respect to θ to zero; for vertical tangents, find where the derivative is undefined. The angles found are then substituted back into the equation to determine r values, and points are listed within the specified domain of θ and r.

The question deals with finding points where the tangent lines to the polar curve r = cos(θ) + sin(θ) are horizontal or vertical. To find where the tangent lines are horizontal, we need to look for points where the derivative of the function with respect to θ is zero, since a horizontal tangent line would have a slope of zero. Conversely, for vertical tangent lines, we want to find where the derivative is undefined or infinite.

A general method for finding this for a polar curve involves taking the derivative of r with respect to θ and setting it equal to zero for horizontal tangents, and finding where it is undefined for vertical tangents:

Horizontal tangent: [tex]\frac{dr}{dθ}[/tex]= 0

Vertical tangent:  [tex]\frac{dr}{dθ}[/tex] is undefined

Once the relevant angles are determined, we substitute them back into the original equation to find the corresponding r value, ensuring that we only consider points within the given domain of θ from 0 to 2π and r ≥ 0.

For the given polar curve r = cos(θ) + sin(θ), we take the derivative of r with respect to θ, which gives us -sin(θ) + cos(θ). Setting this equal to zero yields angles where horizontal tangents occur. Points where this derivative is undefined will give us the angles for vertical tangents. The corresponding points (r, θ) are then listed according to the instructions.

Answer:

E. The area to the left of 5.5 and D. The area to the right of 5.5

Step-by-step explanation:

Please see attachment .

Answer:

a=55

b=5

T=17

Step-by-step explanation:

The general form of the equation is:

[tex]N(t)=ab^{\frac{t}{T}}[/tex]

For t = 0:

[tex]N(0)=ab^{\frac{0}{T}}\\N(0) = a = 55\\a=55[/tex]

Since there has been a fivefold increase after 17 years, at t = 17, N(17) = 55*5

[tex]N(17)=55b^{\frac{17}{T}}\\55*5 = 55b^{\frac{17}{T}}\\b^{\frac{17}{T}} = 5[/tex]

If at every 17*n years there in an increase of 5^n, one can deduct that the values for T and b are respectively 17 and 5:

[tex]b^{\frac{t}{T}}= 5^{\frac{17n}{17}}[/tex]

Therefore, the function that represents N(t) is:

[tex]N(t)=55*5^{\frac{t}{17}}[/tex]

The constants for the function representing the number of households with cable TV service are: a = 55 (initial number of households), b = 5 (fivefold increase), and T = 17 (the number of years for each fivefold increase). The function is N = 55 x 5^(t/17).

The student is asking to define the constants a, b, and T for the function representing the number of households N with cable TV service. The function is of the form N = abt/T, where t is the time in years since the households were first measured, N initially is 55, and the number of households increases fivefold every 17 years.

To find constants a, b, and T, first note that a is the initial value of households with cable TV service, so a = 55. Since the number increases fivefold every 17 years, b is the factor of increase which is 5. The time period for this increase, T, is every 17 years, so T = 17.

Therefore, the function representing the number of households with cable TV service is N = 55 x 5t/17.

Answer:

63

Step-by-step explanation:

The lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)

Suppose we're given that:

Then, the sample proportion of favorable cases is:

[tex]\hat{p} = \dfrac{X}{N}[/tex]

The critical value at the level of significance [tex]\alpha[/tex] is [tex]Z_{1- \alpha/2}[/tex]

The corresponding confidence interval is:

[tex]CI = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right)[/tex]

We need to construct the 99% confidence interval for the population proportion.

We have been provided with the following information about the number of favorable cases(the number of tears located between 3 and 6 mm from the meniscus rim were healed):

The sample proportion is computed as follows, based on the sample size N = 37 and the number of favorable cases X = 25

[tex]\hat p = \displaystyle \frac{X}{N} = \displaystyle \frac{25}{37} = 0.676[/tex]

The critical value for [tex]\alpha = 0.01[/tex] is [tex]z_c = z_{1-\alpha/2} = 2.576[/tex]

The corresponding confidence interval is computed as shown below:

[tex]\begin{array}{ccl} CI & = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right) \\\\ \\\\ & = & \displaystyle \left( 0.676 - 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}}, 0.676 + 2.576 \times \sqrt{\frac{0.676 (1- 0.676)}{37}} \right) \\\\ \\\\ & = & (0.477, 0.874) \end{array}[/tex]

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.477<p<0.874, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.477, 0.874).

Thus, the lower limit and upper limit of 99% confidence interval on the proportion of such tears that will heal is (0.477, 0.874)

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Answer:

a) [tex]\nabla F=0, \nabla\times F=(0,0,-2)[/tex]

b)  [tex]\nabla F=0, \nabla\times F=(0,0,0)[/tex], [tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].

c)  [tex]\nabla F=4x, \nabla\times F=(0,-4z,0)[/tex]

Step-by-step explanation:

Remember, if F= <f,g,h> is a vector field and [tex]\nabla[/tex] is the operator [tex]\nabla=<\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}>[/tex]

a) [tex]F=(x+y,-x+y,-2z)[/tex]

The divergence of F is

[tex]\nabla F=\frac{\partial}{\partial x}(x+y)+\frac{\partial}{\partial y}(-x+y)+\frac{\partial}{\partial z} (-2z)=1+1-2=0[/tex]

The curl of F is

[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x+y&-x+y&-2z\end{array}\right])\\=i(\frac{\partial}{\partial y}(-2z)-\frac{\partial}{\partial z}(-x+y))-j(\frac{\partial}{\partial x}(-2z)-\frac{\partial}{\partial z}(x+y))+k(\frac{\partial}{\partial x}(-x+y)-\frac{\partial}{\partial y}(x+y))=0i-0j-2k=(0,0,-2)[/tex]

b) [tex]F=(2y,2x+3z,3y)[/tex]

The divergence of F is

[tex]\nabla F=\frac{\partial}{\partial x}(2y)+\frac{\partial}{\partial y}(2x+3z)+\frac{\partial}{\partial z} (3y)=0+0+0=0[/tex]

The curl of F is

[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\2y&2x+3z&3y\end{array}\right])\\=i(\frac{\partial}{\partial y}(3y)-\frac{\partial}{\partial z}(2x+3z))-j(\frac{\partial}{\partial x}(3y)-\frac{\partial}{\partial z}(2y))+k(\frac{\partial}{\partial x}(2x+-3z)-\frac{\partial}{\partial y}(2y))=0i-0j+0k=(0,0,0)[/tex]

Since the curl of F is 0 the we will try find f such that the gradient of f be F.

Since [tex]f_x=2y[/tex], [tex]f=2xy+g(y,z)[/tex].

Since [tex]f_y=2x+3z[/tex], [tex]2x+3z=f_y=2x+g_y(y,z)\\3z=g_y[/tex].

Since [tex]f_z=3y[/tex] and [tex]f_z=3y+h'(z)[/tex], then [tex]h'(z)=0[/tex]. Thins means that [tex]h(z)=C, C\in\mathbb{R}[/tex]

Therefore,

[tex]f=2xy+3yz+C, C\in\mathbb{R}[/tex].

c) [tex]H=(x^2-z^2,2,2xz)[/tex]

The divergence of F is

[tex]\nabla F=\frac{\partial}{\partial x}(x^2-z^2)+\frac{\partial}{\partial y}(2)+\frac{\partial}{\partial z} (2xz)=2x+0+2x=4x[/tex]

The curl of F is

[tex]\nabla\times F=det(\left[\begin{array}{ccc}i&j&k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2-z^2&2&2xz\end{array}\right])\\=i(\frac{\partial}{\partial y}(2xz)-\frac{\partial}{\partial z}(2))-j(\frac{\partial}{\partial x}(2xz)-\frac{\partial}{\partial z}(x^2-z^2))+k(\frac{\partial}{\partial x}(2)-\frac{\partial}{\partial y}(x^2-z^2))=0i-(2z-(-2z))j-0k=(0,-4z,0)[/tex]

Answer:

The solution has been given in the following attachment .

Step-by-step explanation:

Answer:

(a) According to the  Albert Michelson's measurements, two-sided 95% confidence interval for velocity of light in the air would be 299,852.4±15.5

(b) Since true velocity of light in the air is not included in the 95% confidence interval, we can say that Michelson’s measurements were not accurate.

Step-by-step explanation:

Confidence Interval can be calculated using M±ME where

And margin of error (ME) from the mean can be calculated using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

Then ME=[tex]\frac{1.96*79.01}{\sqrt{100} }[/tex] ≈ 15.5

According to the  Albert Michelson's measurements, 95% confidence interval for velocity of light in the air would be 299,852.4±15.5

Answer:

At least 1537 samples needed to estimate the true average number of eggs a queen bee lays with 95 percent confidence

Step-by-step explanation:

Minimum sample size required can be found using the formula

N≥[tex](\frac{z*s}{ME} )^2[/tex] where

then N≥[tex](\frac{1.96*10}{0.5} )^2[/tex] =1536.64

Then, at least 1537 samples needed to estimate the true average number of eggs a queen bee lays with 95 percent confidence

The minimum sample size needed is 1537 for a margin of error of 0.5.

Margin of error is used to determine by what value there is deviation from the real value. Margin of error (E) is given by:

[tex]E=Z_\frac{\alpha}{2} *\frac{standard\ deviation}{\sqrt{sample\ size} }[/tex]

The 95% confidence level have a z score of 1.96. Hence for E = 0.5, standard deviation = 10. hence:

[tex]0.5=1.96*\frac{10}{\sqrt{sample\ size}}\\ \\sample\ size=1537[/tex]

The minimum sample size needed is 1537 for a margin of error of 0.5.

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Answer:

Greatest area: One square enclosure with side 50 m

Least area: Two square enclosures with side 25 m each

Step-by-step explanation:

We know we have 200 m of fencing to make the required enclosures. Since the fence will run surrounding any enclosure, it is called the perimeter.

The perimeter for any square area of side z is computed as

P=4z

And its area is

[tex]A=z^2[/tex]

Now, let's analyze both options (shown in the figure below)

Option 1: One square enclosure

Knowing P=200 m, we can determine the length of the side

z=200 m /4 = 50 m

The area is easily computed

[tex]A=50^2=2500 m^2[/tex]

Option 2: Two separate square enclosures of any size

Let's say the side of one of them is x and the side of the other one is y

Assuming both enclosures have no sides in common, the total perimeter is

P=4x+4y

We have 200 m to make the job, so

4x+4y=200

Or equivalently

x+y=50 => y=50-x

The total area of both enclosures is

[tex]A=x^2+y^2[/tex]

Replacing the expression of y

[tex]A=x^2+(50-x)^2[/tex]

To know what the best value is for x to maximize or minimize the area, we use derivatives with respect to x

[tex]A'=2x+2(50-x)(-1)[/tex]

[tex]A'=2x-100+2x=4x-100[/tex]

We equate A'=0 to find the critical point

4x-100=0

x=25 m

Since y=50-x

y=25 m

And the total area is

[tex]A=25^2+25^2=1250\ m^2[/tex]

Note: if we set any other combination for x and y, say x=20 m and y=30m we would get greater areas

[tex]A=20^2+30^2=1300\ m^2[/tex]

The first option gives us the greatest area of 2500 m^2 and the second option has the least area of 1250 m^2

To get the greatest area with 200 meters of fencing, construct one single square enclosure, which will provide an area of 2500 m^2. For the least area, construct two separate square enclosures, which will give a total area of 312.5 m^2.

The question is asking how to use 200 meters of fencing to create square enclosures for the least and greatest area. For a given perimeter, a square has the largest area of any rectangle, so if we want to maximize the area enclosed, we would create one large square.

To find the size of the square, we would divide the total length of the fence by the number of sides of our square. So, 200 m / 4 sides = 50 m per side. Hence, the area of this square would be 50 m * 50 m = 2500 m^2.

For the least area, we would make 2 separate squares, each with half the fencing. So, each square would have 50 m of fence, or 12.5 m per side. The area for each square would then be 12.5 m * 12.5 m = 156.25 m^2, for a total area of 312.5 m^2 when you add both squares together.

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Final answer:

The independent variable is time and the dependent variable is elevation in the context of a linear function modeling climbers ascending a mountain at a rate of 1500 feet per hour.

Explanation:

The situation described by the student can be modeled by a linear function in which the independent variable is time and the dependent variable is elevation.

As the group of climbers begins at an elevation of 5000 feet and climbs at a steady rate of 1500 feet per hour, the relationship between the time spent climbing and the elevation gained is direct.

The function that models this situation is f(t) = 1500t + 5000, where 't' represents the time in hours, and 'f(t)' represents the elevation in feet.

A) the z score for 95% is 1.96

Multiply by the deviation:

1.96 x 0.005 = 0.0098

Now add and then subtract that from the mean:

4.035 - 0.0098 = 4.0252

4.035 + 0.0098 = 4.0448

The interval is (4.0252, 4.0448)

B) 4.035 +/- 1.96 x sqrt( 0.005/sqrt(25))

= 4.035 +/-0.00196

Answer: ( 4.033, 4.037)

C) the conclusion is that both 4.020 and 4.055 are out of the range.

4.020 is below the lowest range and 5.055 is higher than highest range.

The 95% confidence interval for the sample means is; CI = (4.033, 4.037)

A) We are given;

Mean; x' = 4.035 mm

standard deviation; σ = 0.005 mm

sample size; n = 25

Formula for confidence interval is;

CI = x' ± z(σ/√n)

The z score for 95% confidence level is 1.96. Thus;

CI = 4.035 ± 1.96(0.005)

CI = 4.035 ± 0.0098

CI = (4.025, 4.045)

B) From the formula for confidence interval earlier stated, we have;

CI = 4.035 ± 1.96(0.005/√(25))

CI = 4.035 ± 0.00196

CI = (4.033, 4.037)

C)i) The conclusion is that both 4.020 is out of the range of the confidence interval.

ii) The conclusion is that 4.055 is higher than the confidence interval

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Answer:

See below

Step-by-step explanation:

(a) There is a number whose cube is equal to 2.

[tex]\large \exists x \in \mathbb{R}\;|\;x^3=2[/tex]

(b) The square of every number is at least 0.

[tex]\large \forall x\in \mathbb{R},\;x^2\geq 0[/tex]

(c) There is a number that is equal to its square.

[tex]\large \exists x \in \mathbb{R}\;|\;x^2=x[/tex]

(d) Every number is less than or equal to its square.

[tex]\large \forall x\in \mathbb{R},\;x\leq x^2[/tex]

(By the way, this last statement is not true when 0 < x < 1)

Answer: Consumption of,

Japan = 4.6 million barrel,

China = 7.8 million barrel,

The US = 19.8 million barrel.

Step-by-step explanation:

Let x be the quantity of oil per day consumed by Japan,

∵ China consumes 3.2 million barrels a day than Japan.

So, the consumption of China = ( x + 3.2) million Barrel,

Also, The United States consumes 12 million more barrels a day than China.

So, the consumption of the US = (x + 3.2 + 12) = (x + 15.2) million barrel,

Thus, the total consumption of these three countries

= x + x + 3.2 + x + 15.2

= (3x + 18.4) million barrel,

According to the question,

3x + 18.4 = 32.2

3x = 32.2 - 18.4

3x = 13.8

⇒ x = 4.6

Hence, Japan, China and the US consume 4.6 million barrel, 7.8 million barrel and 19.8 million barrel respectively.

Answer:

Confidence interval for the proportion mean is between 0.0113 and 0.0587. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test.

Step-by-step explanation:

We have a large sample size of n = 400 random tests conducted. Let p be the true proportion of employees who failed the test. A point estimate of p is [tex]\hat{p} = 14/400 = 0.035[/tex], we can estimate the standard deviation of [tex]\hat{p}[/tex] as [tex]\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.035(1-0.035)/400}=0.0092[/tex]. A [tex]100(1-\alpha)%[/tex] confidence interval is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n[/tex], then, a 99% confidence interval is [tex]0.035\pm z_{0.005}0.0092[/tex], i.e., [tex]0.035\pm (2.5758)(0.0092)[/tex], i.e., (0.0113, 0.0587). [tex]z_{0.005} = 2.5758[/tex] is the value that satisfies that there is an area of 0.005 above this and under the standard normal curve. B. Yes, it is reasonable to conclude that 5% of the employees cannot pass the employee test, because this inverval contain 0.05.

The 99% confidence interval for the proportion of employees that fail the test is between 0.016 and 0.054. Since 5% is within this range, it is reasonable to conclude that 5% of the employees cannot pass the test.

To compute a 99% confidence interval for the proportion of employees that fail the test, we first need to calculate the sample proportion (p). Here, 14 employees failed the tests out of 400, so p = 14/400 = 0.035. The 99% confidence interval requires Z-score of 2.576 (as 99% of the data lies within 2.576 standard deviations of the mean in a normal distribution).

The estimation error (E) can be calculated using the formula E = Z * √( (p*(1-p)) / n), where n is the total number of tests. Substituting the values, E = 2.576 * √(0.035 * (1 - 0.035) / 400) = 0.019

So, the 99% confidence interval is (p - E, p + E) = (0.035 - 0.019, 0.035 + 0.019) = (0.016, 0.054). Thus, we are 99% confident that the true proportion of employees that fail the test is between 0.016 and 0.054.

Given that 5% (or 0.05) is within the 99% confidence interval we calculated, it would be reasonable to conclude that 5% of the employees cannot pass the employee test.

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Final answer:

To assess if the sample of fish with high mercury levels supports the hypothesis of a 24% proportion in the population, we set up null and alternative hypotheses, calculate the z-test statistic, compare it to a critical value, compute the p-value, and then make a conclusion based on whether the p-value is less than the chosen significance level of 0.05.

Explanation:

To test the hypothesis about the percentage of freshwater fish with high levels of mercury, we must first establish our null hypothesis (H0) and alternative hypothesis (Ha). The null hypothesis is that the true proportion of fish with dangerous mercury levels is 24% (H0: p = 0.24), while the alternative hypothesis is that the proportion is not 24% (Ha: p ≠ 0.24).

We then calculate the z-test statistic using the sample proportion (p' = 80/350 = 0.2286) and the assumed population proportion under H0 (p = 0.24), along with the standard deviation of the sampling distribution (σp = sqrt[p(1-p)/n]). This result is compared to the critical z-score of 1.96, which corresponds to our significance level (α) of 0.05 in a two-tailed test.

The p-value is calculated to determine the probability of observing our sample statistic, or one more extreme if the null hypothesis is true. If this p-value is less than 0.05, we reject H0 and conclude that the sample provides enough evidence against the null hypothesis.

Answer:

There is exactly one real root.

Step-by-step explanation:

There are two parts to arrive at the solution.

(i) The polynomial has at least one real root.

(ii) The polynomial has exactly one real root.

We prove (i) using Intermediary Value Theorem.

f(x) = x³ + x - 1 = 0 is a polynomial. So, it is continuous.

At x = 1, f(1)   = 1

At x = o, f(0) = -1

Since, there is a change of sign it should have crossed through zero.

Now, to prove there is exactly one real root we use Rolle's theorem.

Let us assume there are two real roots to the polynomial, say 'a' and 'b'.

Then f(a) = 0 and f(b) = 0.

⇒ f(a) = f(b)

To use Rolle's theorem we need the function to be continuous, differentiable and for any two points a,b f(a) = f(b) there should exist a 'c' such that f'(c) = 0.

Now, f'(x) = 3x² + 1

Note that f'(x) is always greater than equal to 1.

It can never be zero for any c. This contradicts Rolle's Theorem. o, our assumption that two real roots exist must be wrong.

Hence, we conclude that there is exactly one real root to the polynomial.

Answer:

s=29.93m/s

h=22.88m

Step-by-step explanation:

we must find the initial speed,  we will determine its position (x-y).

x component [tex]s=0+v_{0}cos\alpha.t+0=v_{0}cos\alpha.t[/tex]

y component [tex]h=0+v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}\\[/tex]  since the ball is caught at the same height then h=0

[tex]h=v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0\\v_{0}sin\alpha.t-\frac{1}{2}gt^{2}=0;v_{0}sin\alpha.t=\frac{1}{2}gt^{2}\\t=\frac{2v_{0}sin\alpha}{g}\\[/tex]

where t= flight time;[tex]s=v_{0}cos\alpha.t[/tex], replacing t:

[tex]v_{0}=\sqrt{\frac{sg}{sin2\alpha}}[/tex]

[tex]s=v_{0}cos\alpha(\frac{2sin\alpha }{g})=\frac{v_{0} ^{2}2sin\alpha.cos\alpha}{g}=\frac{v_{0} ^{2}sin(2\alpha)}{g}]

: the values ​​must be taken to the same units

[tex]300ft*0.3048m/ft=91.44m[/tex]

[tex]v_{0}=\sqrt{\frac{91.44m*9.8\frac{m}{s^{2}}}{sin2(45)}}=\sqrt{895.112(\frac{m}{s} )^{2} }=29.93\frac{m}{s}[/tex]

To calculate the height you should know that this is achieved when its component at y = 0

[tex]v_{y}=v_{0}sin\alpha-gt=0;gt=v_{0}sin\alpha\\\\ t=\frac{v_{0}sin\alpha  }{g}\\h=v_{0}sin\alpha  .t-\frac{1}{2}gt^{2}[/tex]

replacing t;[tex]h=v_{0}sin\alpha(\frac{v_{0}sin\alpha}{g})-\frac{1}{2}g(\frac{v_{0sin\alpha}}{g}) ^{2}\\[/tex]

finally

[tex]h=\frac{(v_{0}sin\alpha)^{2}}{2g}=\frac{(29.95*sin45)^{2}}{2*9.8}=22.88m[/tex]

The alternative hypothesis whereby mud = the population mean difference is expressed as; Ha: mud > 0

An alternative hypothesis is defined as one in which the observers or researchers anticipate a difference (or an effect) between two or more variables.

Now, in this case, the null hypothesis is that eating milk chocolate lowered someone's cholesterol levels. This means the alternate hypothesis is that eating milk chocolate increases someone's cholesterol levels.

Thus, alternative hypothesis in this case is expressed as;

Ha: mud > 0

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Answer:

c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2

Step-by-step explanation:

Given that a  statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2.

Group   Group One     Group Two  

Mean 75.8600 75.4100

SD 16.9100 19.7300

SEM 2.8583 3.2436

N 35       37      

*SEM is std error/sqrt n

Mean difference = 0.4500

[tex]H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y[/tex]

(two tailed test)

Std error for difference = 4.342

Test statistic t = [tex]\frac{0.45}{4.342} \\=0.1036[/tex]

df =70

p value = 0.9178

Since p >0.05 we accept H0

c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2

Final answer:

The concluding statement is that there is insufficient evidence to conclude a significant difference in the means of statistics day students and statistics night students on Exam 2.

Explanation:

The hypothesis test being conducted in this scenario is a two-sample t-test. The null hypothesis, denoted as H0, states that there is no significant difference between the mean scores of statistics day students and statistics night students on Exam 2. The alternative hypothesis, denoted as Ha, states that there is a significant difference between the means of the two groups. To determine whether there is sufficient evidence to support the alternative hypothesis, we can compare the t-statistic with the critical t-value from a t-distribution table.

In this case, since the standard deviations are assumed to be equal, we can calculate the pooled standard deviation and use it to calculate the t-statistic. With the given sample means, standard deviations, and sample sizes, the calculated t-statistic is -0.246. The critical t-value for a two-tailed test with a significance level of 0.05 and 70 degrees of freedom is approximately 1.994. Since the calculated t-statistic (-0.246) falls within the range between -1.994 and 1.994, we fail to reject the null hypothesis.

Therefore, the concluding statement is:

c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2.

Answer:

The amount of ounce used for paint is [tex]\frac{9}{15} ounce[/tex]  

Step-by-step explanation:

used amount for water is [tex]\frac{2}{3}[/tex] ounce.

used amount for sky is [tex]\frac{3}{5}[/tex] ounce.

thus, the total amount used is [tex]\frac{2}{3}[/tex] +  [tex]\frac{3}{5}[/tex]

= [tex]\frac{19}{15}[/tex]

Thus the total unused amount is 2 - [tex]\frac{19}{15}[/tex] = [tex]\frac{11}{15}[/tex] .

But the remaining amount is [tex]\frac{2}{15}[/tex].

Thus used amount for flag is [tex]\frac{11}{15}[/tex] - [tex]\frac{2}{15}[/tex]

= [tex]\frac{9}{15}[/tex]

9514 1404 393

Answer:

  No; 495

Step-by-step explanation:

For many folks, the order of application of toppings will not matter. For those folks, the number of possible 4-topping pizzas is

 C(12,4) = 495 = 12!/(8!×4!)

There are 495 possible 4-topping pizzas.

The picture is given below. The perimeter, total time, and area of the circuit will be 700 feet, 70 seconds, and 100 square meters, respectively.

Let r is the radius of the sector and θ be the angle subtended by the sector at the center.

Then the arc length of the sector of the circle will be

Arc = (θ/2π) 2πr

Then the area of the sector of the circle will be

Arc = (θ/2π) 2πr

We know that 100 yards = 300 feet.

The perimeter covered in 10 seconds will be given as,

10 x 10 = (θ/2π) 2π(300)

θ = 19.1°

The perimeter of the circuit will be given as,

P = 100 + 300 + 300

P = 700 feet

The total time is given as,

Time = 700 / 10

Time = 70 seconds

The area of the shape will be given as,

A = (19.1/360) 2π(300)

A = 100 square feet

The picture is given below. The perimeter, total time, and area of the circuit will be 700 feet, 70 seconds, and 100 square meters, respectively.

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Answer:

Mass

[tex]\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)[/tex]

Center of mass

Coordinate x

[tex]\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Coordinate y

[tex]\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Coordinate z

[tex]\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)  

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass  m would be the curve integral along the path W

[tex]m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt[/tex]

The density D(x,y,z) is given by

[tex]D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16[/tex]

on the other hand

[tex]||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}[/tex]

and we have

[tex]m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)[/tex]

The center of mass is the point [tex](\bar x,\bar y,\bar z)[/tex]

where

[tex]\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)[/tex]

We have

[tex]\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)[/tex]

so

[tex]\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

[tex]\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi[/tex]

[tex]\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

[tex]\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi[/tex]

[tex]\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}[/tex]

Final answer:

To find the mass and center of mass of the wire loop in the shape of a helix, we need to integrate the density function along the length of the wire. The formula for the mass of a wire is m = ∫(ρ dl), where ρ is the density function and dl is an infinitesimally small segment of the wire. To find the center of mass, we can use the formulas Cx = (1/m) ∫(x ρ dl), Cy = (1/m) ∫(y ρ dl), and Cz = (1/m) ∫(z ρ dl), where x, y, and z are the parametric equations and ρ is the density function.

Explanation:

To find the mass and center of mass of the wire loop in the shape of a helix, we need to integrate the density function along the length of the wire. Since the density is equal to the square of the distance from the origin to the point, we can use the formula for the mass of a wire:

m = ∫(ρ dl)

Where ρ is the density function and dl is an infinitesimally small segment of the wire. In this case, we have:

ρ = r² = (x² + y² + z²)

Using the given parametric equations for x, y, and z, we can substitute them into the formula and integrate along the range of t:

m = ∫(t² + (4cos(t))² + (4sin(t))²) dt

Next, to find the center of mass, we can use the formula:

Cx = (1/m) ∫(x ρ dl)

Cy = (1/m) ∫(y ρ dl)

Cz = (1/m) ∫(z ρ dl)

We can substitute the parametric equations for x, y, and z, and the density function into these formulas and integrate along the range of t to find the center of mass.

Answer:

The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 50

We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.06.

P(X > x)  = 6% = 0.06

[tex]P( X > x) = P( z > \displaystyle\frac{x - 500}{50})=0.06[/tex]  

[tex]= 1 - P( z \leq \displaystyle\frac{x - 500}{50})=0.06 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 500}{50})=1-0.06=0.94 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 1.555) = 0.94[/tex]

[tex]\displaystyle\frac{x - 500}{50} = 1.555\\x = 577.75[/tex]  

Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.