A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 90​% confidence interval estimate of the mean amount of mercury in the population.


0.60 0.74 0.09 0.89 1.31 0.51 0.94


What is the confidence interval estimate of the population mean?


______ppm < u < ______ppm

(Round to three decimal places as needed)


Does it appear that there is too much mercury in tuna​ sushi?

Answer:

We fail to reject the null hypothesis at the significance level of 0.05.

Step-by-step explanation:

We have a larga sample size of n = 80, [tex]\bar{x} = 4.39[/tex] and s = 2.29. We want to test

[tex]H_{0}: \mu = 4.50[/tex] vs [tex]H_{1}: \mu \neq 4.50[/tex] (two-tailed alternative)  

Because we have a large sample, our test statistic is

[tex]Z = \frac{\bar{X}-4.50}{s/\sqrt{n}}[/tex] which is normal standard approximately. We have the observed value

[tex]z_{0} = \frac{4.39-4.50}{2.29/\sqrt{80}} = -0.4296[/tex].

The p-value is given by 2P(Z < -0.4296) = (2)(0.3337) = 0.6674 (because of the simmetry of the normal density)

With the significance level [tex]\alpha = 0.05[/tex], we fail to reject the null hypothesis because the p-value is greater than 0.05.

We are going to test if the mean is equals to 4.50, thus, the null hypothesis is:

[tex]H_0: \mu = 4.5[/tex]

At the alternative hypothesis, we test if the mean is different to 4.50, that is:

[tex]H_1: \mu \neq 4.50[/tex]

Since we have the standard deviation for the sample, the t-distribution is used. The value of the test statistic is:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

For this problem:

[tex]t = \frac{4.39 - 4.5}{\frac{2.29}{\sqrt{80}}}[/tex]

[tex]t = -0.43[/tex]

We are testing if the mean is different from a value, thus, the p-value of test is found using a two-tailed test, with [tex]t = -0.43[/tex] and 80 - 1 = 79 df.

Using a t-distribution calculator, the p-value is of 0.6684.

The p-value is of 0.6684 > 0.05, which means that we can conclude that the population of student course evaluations has a mean equal to 4.50.

A similar problem is given at https://brainly.com/question/24989605

Answer:

a)The sample size

Step-by-step explanation:

t - student distribution should be used whe we are facing a Normal Distribution population andwhen a sample size n is below 30.

t- student distribution is also a associated to a bell shape but is more flat and the values are more spread, tails are much more wide.

t- student use the concept of degree of fredom ( each degree of fredom correspont to an specific curve). As degree of fredom increase the curves become more close to a bell shape. For  n = 30 t-student curve is  a bell shape one

The sum and difference identity cos(x + y) / sin(x - y) = 1 - cotxcoty / cotx - coty is verified

Solution:

Given expression is:

[tex]\frac{\cos (x+y)}{\sin (x-y)}=\frac{1-\cot x \cot y}{\cot x-\cot y}[/tex]

Let us first solve L.H.S

[tex]\frac{\cos (x+y)}{\sin (x-y)}[/tex]    ------ EQN 1

We have to use the sum and difference formulas

cos(A + B) = cosAcosB – sinAsinB  

sin(A - B) = sinAcosB – cosAsinB

Applying this in eqn 1 we get,

[tex]=\frac{\cos x \cos y-\sin x \sin y}{\sin x \cos y-\sin y \cos x}[/tex]

[tex]\text { Taking sinx } \times \text { siny as common }[/tex]

[tex]=\frac{\sin x \sin y\left(\frac{\cos x \cos y}{\sin x \sin y}-1\right)}{\sin x \sin y\left(\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}\right)}[/tex]

[tex]\begin{array}{l}{=\frac{\frac{\cos x}{\sin x} \times \frac{\cos y}{\sin y}-1}{\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}}} \\\\ {=\frac{\cot x \times \cot y-1}{\cot y-\cot x}} \\\\ {=\frac{\cot x \cot y-1}{\cot y-\cot x}}\end{array}[/tex]

Taking -1 as common from numerator and denominator we get,

[tex]\begin{array}{l}{=\frac{-(1-\cot x \cot y)}{-(\cot x-\cot y)}} \\\\ {=\frac{(1-\cot x \cot y)}{(\cot x-\cot y)}}\end{array}[/tex]

= R.H.S

Thus L.H.S = R.H.S

Thus the given expression has been verified using sum and difference identity

Answer:

  A. Point C divides segment AB so that AC:CB is 1:3

Step-by-step explanation:

Like many multiple-choice questions, you don't actually need to know how to work the problem. You just need to know what the question means.

Comparing answers, you see that choices A and D cannot both be right. You also note that choice D says the same thing as choice B.

Since there is only one False answer, it must be choice A.

__

When the division is into segments that are 1/3 the length and 2/3 the length, the segments have the ratio ...

  (1/3) : (2/3) = 1 : 2

The incorrect statement is B. If AC = 5, then CB should be 15, not 10, because CB would be twice the length of AC, given that point C is 1/3 of the way from A to B.

The question asks us to identify which statement about point C being 1/3 of the way from point A to point B on a line segment is NOT true. Let's analyze each statement.

The answer is statement B. If AC = 5, then CB would not be 10; it would be 15 since it is twice the length of AC, given point C is 1/3 of the way from A to B.

Answer:

Step-by-step explanation:

To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to survey at least 92 Clemson students.

To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to calculate the sample size needed for this level of precision. The formula to determine the sample size is:

n = (z * σ / E)^2

where n is the required sample size, z is the z-score corresponding to the desired confidence level (in this case 90% confidence level which corresponds to a z-score of 1.645), σ is the standard deviation, and E is the desired margin of error (1.5 years).

Plugging in the values, we get:

n = (1.645 * 6 / 1.5)^2

n = 91.154

Rounding up to the nearest whole number, we need to survey at least 92 Clemson students in order to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence.

Answer: 95% confidence interval would be (175.04,194.96).

Step-by-step explanation:

Since we have given that

Mean = 185 mg

Standard deviation = 17.6 mg

At 95% confidence level, z = 1.96

So, Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=185\pm 1.96\times \dfrac{17.6}{\sqrt{12}}\\\\=185\pm 9.958\\\\=(185-9.958,185+9.958)\\\\=(175.042,194.958)\\\\=(175.04,194.96)[/tex]

Hence, 95% confidence interval would be (175.04,194.96).

Using the provided values and steps for constructing a 95% confidence interval, we estimate, with 95% confidence, that the true mean cholesterol content of all such eggs lies between 146.5 and 223.5 milligrams.

To construct a 95% confidence interval for the mean cholesterol content of all such eggs, we will use the provided sample mean (185 milligrams), standard error (17.6 milligrams), and the fact that the sample size (12 eggs) is relatively small, so we use a t-distribution.

Firstly, we need to find the t-score for a 95% confidence level. Since degrees of freedom (df) is n-1 -> 12-1=11, and at a 95% confidence level, the t-score (from t-distribution table) is approximately 2.201 for a two-tailed test.

Next, we use the formula for confidence interval:

Calculating these gives:

So, we estimate with 95 percent confidence that the true mean cholesterol content of all such eggs is between 146.5 and 223.5 milligrams.

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Answer:

82 apartments should be rented.

Maximum profit realized will be $30964.

Step-by-step explanation:

Monthly profit realized from renting out x apartments is modeled by

P(x) = -11x² + 1804x - 43000

To maximize the profit we will take the derivative of the function P(x) with respect to x and equate it to zero.

P'(x) = [tex]\frac{d}{dx}(-11x^{2}+1804x-43000)[/tex]

       = -22x + 1804

For P'(x) = 0,

-22x + 1804 = 0

22x = 1804

x = 82

Now we will take second derivative,

P"(x) = -22

(-) negative value of second derivative confirms that profit will be maximum if 82 apartments are rented.

For maximum profit,

P(82) = -11(82)² + 1804(82) - 43000

        = -73964 + 147928 - 43000

        = $30964

Therefore, maximum monthly profit will be $30964.

The number of units that should be rented out to maximize the monthly rental profit is 82 units.

The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

To find the number of units that should be rented out to maximize the monthly rental profit and the maximum monthly profit, we need to analyze the given profit function:

[tex]\[ P(x) = -11x^2 + 1804x - 43,000 \][/tex]

This is a quadratic function of the form [tex]\( P(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -11 \), \( b = 1804 \)[/tex], and c = -43,000 .

For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] with a < 0  (which opens downwards), the maximum value occurs at the vertex. The x-coordinate of the vertex for the function P(x)  is given by:

[tex]\[ x = -\frac{b}{2a} \][/tex]

Substituting a = -11  and b = 1804 :

[tex]\[ x = -\frac{1804}{2(-11)} = \frac{1804}{22} = 82 \][/tex]

Therefore, the number of units that should be rented out to maximize the profit is x = 82 .

Next, we calculate the maximum monthly profit by substituting x = 82  back into the profit function:

[tex]\[ P(82) = -11(82)^2 + 1804(82) - 43,000 \][/tex]

Calculating step by step:

[tex]\[ 82^2 = 6724 \]\[ -11(6724) = -73,964 \]\[ 1804(82) = 147,928 \]\[ P(82) = -73,964 + 147,928 - 43,000 \]\[ P(82) = 30,964 \][/tex]

So, the maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

In summary:

- The number of units that should be rented out to maximize the monthly rental profit is 82 units.

- The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]

Answer:

Part A)  Not collide

Part B)  k = 4

Part C)  Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

Position of particle A:-

[tex]x(t)=5t-5[/tex]

[tex]y(t)=2t-k[/tex]

Position of particles B:-

[tex]x(t)=4t[/tex]

[tex]y(t)=t^2-2t+1[/tex]

Part A)  For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, [tex](4t,t^2-2t-1)[/tex]

If both collides then x and y coordinate must be same

Therefore,

5t - 5 = 4t    

       t = 5

[tex]2t+6=t^2-2t-1[/tex]

[tex]t^2-4t-7=0[/tex]

[tex]t=-1.3,5.3[/tex]

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

5t - 5 = 4t    

       t = 5

[tex]2t-k=t^2-2t-1[/tex]

Put t = 5

[tex]10-k=25-10-1[/tex]

[tex]k=4[/tex]

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, [tex]\dfrac{dA}{dt}[/tex]

[tex]\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4[/tex]

Speed of particle B, [tex]\dfrac{dB}{dt}[/tex]

[tex]\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]

[tex]\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}[/tex]

At t = 5

[tex]\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2[/tex]

Hence, Particle B moves faster than particle A

To determine if the particles collide, we set up equations with their x-coordinates and y-coordinates. Part (a) asks if they collide when k = -6. Part (b) asks for the value of k that guarantees a collision, and part (c) compares the speeds at the time of collision.

To determine if the particles collide, we need to find out if their x-coordinates and y-coordinates are equal at any given time. Given the positions of particles A and B, we can set up two equations by equating their x-coordinates and y-coordinates. For part (a), when k = -6 we can solve the equations to find if they intersect. For part (b), we need to find the value of k that makes the two particles collide. Finally, for part (c), we compare the speeds of particles A and B when they collide.

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Answer with Step-by-step explanation:

We are given that a function

[tex]T:P_2\rightarrow P_2[/tex]

[tex]T(a_0+a_1x+a_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2[/tex]

We have to determine the given function is a linear transformation.

If a function is linear transformation then it satisfied following properties

[tex]1.T(x+y)=T(x)+T(y)[/tex]

2.[tex]T(ax)=aT(x)[/tex]

[tex]T(a_0+a_1x+a_2x^2+b_0+b_1x+b_2x^2)=T((a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2)=(a_0+b_0+a_1+b_1+a_2+b_2)+(a_1+b_1+a_2+b_2)x+(a_2+b_2)x^2[/tex]

[tex]T(a_0+a_1x+a_2x^2+b_0+b-1x+b_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2+(b_0+b_1+b_2)+(b_1+b_2)x+b_2x^2[/tex]

[tex]T(a_0+a_1x+a_2x^2+b_0+b-1x+b_2x^2)=T(a_0+a_1x+a_2x^2)+T(b_0+b_1x+b_2x^2)[/tex]

[tex]T(a(a_0+a_1x+a_2x^2))=T(aa_0+aa_1x+aa_2x^2)[/tex]

[tex]T(a(a_0+a_1x+a_2x^2))=(aa_0+aa_1+aa_2)+(aa_1+aa_2)x+(aa_2)x^2[/tex]

[tex]T(a(a_0+a_1x+a_2x^2))=a(a_0+a_1+a_2)+a(a_1+a_2)x+aa_2x^2=a((a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2)=aT(a_0+a_1x+a_2x^2)[/tex]

Hence, the function is a linear transformation because it satisfied both properties of linear transformation.

[tex]\( T \)[/tex] satisfies both additivity and scalar multiplication, [tex]\( T \)[/tex] is indeed a linear transformation.

To determine if the function [tex]\( T: P2 \to P2 \)[/tex], defined by [tex]\( T(a_0 + a_1 x + a_2 x^2) = (a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2 \)[/tex], is a linear transformation, we need to check two properties:

1. Additivity: [tex]\( T(u + v) = T(u) + T(v) \) for all \( u, v \in P2 \)[/tex].

2. Scalar Multiplication: [tex]\( T(cu) = cT(u) \) for all \( u \in P2 \) and \( c \in \mathbb{R} \)[/tex].

Let's verify these properties:

Additivity Check

Let [tex]\( u = a_0 + a_1 x + a_2 x^2 \) and \( v = b_0 + b_1 x + b_2 x^2 \)[/tex].

Compute [tex]\( T(u + v) \)[/tex]:

[tex]\[u + v = (a_0 + b_0) + (a_1 + b_1)x + (a_2 + b_2)x^2\][/tex]

[tex]\[T(u + v) = [(a_0 + b_0) + (a_1 + b_1) + (a_2 + b_2)] + [(a_1 + b_1) + (a_2 + b_2)]x + (a_2 + b_2)x^2\][/tex]

[tex]\[T(u + v) = [(a_0 + a_1 + a_2) + (b_0 + b_1 + b_2)] + [(a_1 + a_2) + (b_1 + b_2)]x + (a_2 + b_2)x^2\][/tex]

Compute [tex]\( T(u) + T(v) \)[/tex]:

[tex]\[T(u) = (a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2\][/tex]

[tex]\[T(v) = (b_0 + b_1 + b_2) + (b_1 + b_2)x + b_2 x^2\][/tex]

[tex]\[T(u) + T(v) = [(a_0 + a_1 + a_2) + (b_0 + b_1 + b_2)] + [(a_1 + a_2) + (b_1 + b_2)]x + (a_2 + b_2)x^2\][/tex]

Since [tex]\( T(u + v) = T(u) + T(v) \)[/tex], the function [tex]\( T \)[/tex] satisfies additivity.

Scalar Multiplication Check

Let [tex]\( u = a_0 + a_1 x + a_2 x^2 \)[/tex] and [tex]\( c \in \mathbb{R} \)[/tex].

Compute [tex]\( T(cu) \)[/tex]:

[tex]\[cu = c(a_0 + a_1 x + a_2 x^2) = (ca_0) + (ca_1)x + (ca_2)x^2\][/tex]

[tex]\[T(cu) = [(ca_0) + (ca_1) + (ca_2)] + [(ca_1) + (ca_2)]x + (ca_2)x^2\][/tex]

[tex]\[T(cu) = c[(a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2]\][/tex]

Compute [tex]\( cT(u) \)[/tex]:

[tex]\[T(u) = (a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2\][/tex]

[tex]\[cT(u) = c[(a_0 + a_1 + a_2) + (a_1 + a_2)x + a_2 x^2]\][/tex]

Since [tex]\( T(cu) = cT(u) \)[/tex], the function [tex]\( T \)[/tex] satisfies scalar multiplication.

Answer:20 hoseholds, what is the probability that more than

Step-by-step explanation:

Answer:

False this value is very likely to find in this distribution

Step-by-step explanation:

With mean 7 ( μ ) and standad deviation  (σ ) 2 we can observe, value 6.6  is close to the lower limit of the interval

μ  ± 0,5 σ      7 ± 1 in which we should find 68,3 % of all values

(just 6 tenth to the left)

And of course 6.6 is inside the interval

μ  ± 1 σ   where we find 95.7 % of th value

We conclude this value is not unlikely at all

Final answer:

To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we can use hypothesis testing. The calculated z-score is less than the critical z-value, indicating that it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in the population. Therefore, the statement is true.

Explanation:

To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we need to use hypothesis testing. We can set up the null and alternative hypotheses as follows:

Null hypothesis (H0): The population mean breaking weight is equal to 7 lbs.

Alternative hypothesis (Ha): The population mean breaking weight is less than 7 lbs.

Next, we can calculate the z-score using the formula (sample mean - population mean)/(standard deviation/sqrt(sample size)). Substituting the values, we get (6.6 - 7)/(2/sqrt(36)) = -0.6/ (2/6) = -0.6/ (1/3) = -1.8.

We can now look up the critical z-value for a one-tailed test at a significance level of 0.05. The critical z-value is -1.645. Since the calculated z-score (-1.8) is less than the critical z-value (-1.645), we can reject the null hypothesis. Therefore, it is true that finding a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs is very unlikely.

Answer:

Is plausible that the successive throws are independent

Step-by-step explanation:

1) Table with info given

The observed values are given by the following table

__________________________________________________

First shot          Made          Second shot missed           Total

__________________________________________________

Made                  152                       33                                185

Missed                37                         8                                  45

__________________________________________________

Total                    189                       41                                230

2) Calculations and test

We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:

Null hypothesis: Independence between two successive free throws

Alternative hypothesis: No Independence between two successive free throws

_____________________________________________________

First shot                    Made                                    Second shot missed

_____________________________________________________

Made                  189(185)/230=152.0217                41(185)/230=32.9783

Missed                189(45)/230=36.9783                  41(45)/230=8.0217

_____________________________________________________

On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.

3) Calculate the chi square statistic

The statistic for this case is given by:

[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]

Where O represent the observed values and E the expected values. Replacing the values that we got we have this

[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]

Now with the calculated value we can find the degrees of freedom

[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.

Now we can calculate the p value

[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]

On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.

Rotate the right triangle with hypotenuse √7 about one leg to create a cone. Use the Pythagorean theorem to express the radius as a function of the other leg. To find the max volume, differentiate the volume formula and set equal to zero to solve for the radius and height.

The problem presents a right triangle with hypotenuse of √7 meters. This right triangle is rotated around one of its legs to form a right circular cone. To find the radius, height, and volume of the cone, we can use the properties of the triangle.  

Imagine that the triangle is rotated about the shorter leg. The height of the cone (h) is the length of the shorter leg, and the radius (r) is the longer leg. By the Pythagorean theorem, a² + b² = c², where c is the hypotenuse, a and b represent the legs of the triangle. Assume that a is the longer leg and b is the shorter leg.

From the Pythagorean theorem we get a² = (√7)² - b² = 7 - b² and so a = √(7 - b²).

The volume of the cone V = (1/3)πr²h = (1/3)π(√(7 - b²))²b = (1/3)π(7 - b²)b. Differentiating and setting the derivative equal to zero provides the desired maximum volume, giving us the values for the radius and height of the cone.

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Answer: 0.0026

Step-by-step explanation:

Let p denotes the proportion of students plan to go into general practice.

As per given , we have

Alternative hypothesis : [tex] H_a: p>0.28[/tex]

Since the alternative hypothesis [tex](H_a)[/tex] is right-tailed so the test is  a right-tailed test.

Also , it is given that ,

i.e. sample size : = 130

x= 490

[tex]\hat{p}=0.39[/tex]

Test statistic(z) for population proportion :

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

, where p=population proportion.

[tex]\hat{p}[/tex]= sample proportion

n= sample size.

[tex]z=\dfrac{0.39-0.28}{\sqrt{\dfrac{0.28(1-0.28)}{130}}}\\\\=\dfrac{0.11}{0.0393798073988}=2.79330975101\approx2.79[/tex]

P-value for right-tailed test = P(z>2.79)=1-P(z≤ 2.79)  [∵P(Z>z)=1-P(Z≤z)]

=1- 0.9974=0.0026  [using z-value table]

Hence, the  P-Value for a test of the school's claim = 0.0026

To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%. Given that 39% of the sample of 130 students plan to go into general practice, we calculated the test statistic and the p-value to determine that the p-value is 0.3078.

To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%.

Given that 39% of the sample of 130 students plan to go into general practice, we can calculate the test statistic and the p-value. Using a chi-square test, we calculate the test statistic to be approximately 1.307. The degrees of freedom for this test is 1.

Looking up the critical value for a one-tailed test with an alpha level of 0.05 and 1 degree of freedom, the critical value is approximately 3.841. Since the test statistic is less than the critical value, we fail to reject the null hypothesis. Therefore, the p-value is greater than 0.05.

Therefore, the correct answer for the P-value is 0.3078.

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Answer:

Lowest acceptable score = 121.3

Step-by-step explanation:

Mean test score (μ) = 115

Standard deviation (σ) = 12

The z-score for any given test score 'X' is defined as:  

[tex]z=\frac{X-\mu}{\sigma}[/tex]  

In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:

[tex]0.525=\frac{X-115}{12}\\X=121.3[/tex]

Answer:

  see below

Step-by-step explanation:

Opposite sides are parallel in a parallelogram, so if they have different slope, the figure will not be a parallelogram.

_____

Comments on other answer choices

If the slope of diagonal EG is perpendicular to that of diagonal FH, it only proves the figure is some sort of kite. The figure may or may not be a parallelogram.

Adjacent sides being different lengths does not prove anything (except that the figure is not a rhombus).

Proving angle F is a right angle does not prove anything else about the shape of the figure. The figure may or may not be a parallelogram. (If it is a parallelogram, it is also a rectangle.)

Answer:

Have no fear, the answer is top left GIVE BRANLIEST PLZ

Step-by-step explanation:

Answer:

The number of ways is equal to [tex]12!8!20![/tex]

Step-by-step explanation:

The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are

n1 x n2 x ... x ni

Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.

For example : We want to put 4 different objects in a row.

The total ways to do this are [tex]4!=4.3.2.1=24[/tex] ways.

Using the multiplication principle and the n-factorial number :

The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒ [tex]12!8!20![/tex]

There are 1.1657416 × 10^30 ways to arrange the students in a row with all the sophomores on the left, juniors in the middle, and seniors on the right.

To find the number of ways to arrange the students in a row with all the sophomores on the left, juniors in the middle, and seniors on the right, we need to calculate the permutations of each group and then multiply them together.

The number of ways to arrange the 12 sophomores is 12!, which is 479,001,600.

The number of ways to arrange the 8 juniors is 8!, which is 40,320.

The number of ways to arrange the 20 seniors is 20!, which is 2,432,902,008,176,640,000.

Multiplying these three numbers together, we get a total of 1.1657416 × 10^30 ways to arrange all the students in a row for a picture.

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Answer:

To conduct the hypothesis test I would use the standard normal distribution to calculate the critical value and the p-value.

Step-by-step explanation:

To conduct the hypothesis test I would use the standard normal distribution because there is a large sample size of n = 125 households. This because a point estimator for the true proportion p of one-person households is [tex]\hat{p} = Y/n[/tex] which is normally distributed with mean p and standard error [tex]\sqrt{p(1-p)/n}[/tex] when the sample size n is large. Here Y is the random variable that represents the number of one-person households observed. Then the test statistic is [tex]Z = \frac{\hat{p}-0.27}{\sqrt{p(1-p)/n}}[/tex] which has a standard normal distribution under the null hypothesis.

Answer:

;-; I belive it's the 1st or 2nd one

Step-by-step explanation:

hope this helps you.

ask me for explanation if you want and I'll try to explane it the best I Can.

(P.S;please mark me as brainlyest,ty in advaince)

Answer:

AAS method can be used to prove that the two triangles are congruent.

Step-by-step explanation:

According to the question for the two triangles one pair of opposite angles are equal. One another pair of angles are equal for the two and one pair of sides are also equal of the two.

Hence, the two given triangles are congruent by AAS rule.

Hence, AAS method can be used to prove that the two triangles are congruent.

Answer: Slope 1/2

y-int -5

graph (0,-5),(1,-9/2)

Step-by-step explanation:

Answer:

6x^2 + 18x + 19.

Step-by-step explanation:

5x^2 + 6x - 17 + (x + 6)^2

= 5x^2 + 6x - 17 + x^2 + 12x + 36

=  6x^2 + 18x + 19.

Answer:

6x^2 + 18x + 19

Step-by-step explanation:

The standard form of a polynomial depends on the degree of the polynomial. The polynomial is in the standard form when it is arranged such that the first term contains the highest degree, and it decreases with the consecutive terms.

The square of the binomial is (x+6)^2

(x+6)^2 = (x+6)(x+6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36

The sum of the two polynomials will be 5x^2+6x−17 + x^2 + 12x + 36

Collecting like terms,

5x^2 + x^2 + 6x + 12x + 36 -17

= 6x^2 + 18x + 19

Answer:

  3

Step-by-step explanation:

The slope (m) is the ratio of the difference in y-values to the corresponding difference in x-values:

  m = (y2 -y1)/(x2 -x1)

  m = (9 -0)/(4 -1) = 9/3

  m = 3

The slope of the line is 3.

Answer:

[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]

The 95% confidence interval would be given by (4.729;5.271)    

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the "range of values below and above the sample statistic in a confidence interval".

The standard error of a statistic is "the standard deviation of its sampling distribution or an estimate of that standard deviation"

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

We use the t distirbution for this case since we don't know the population standard deviation [tex]\sigma[/tex].

Where the standard error is given by: [tex]SE=\frac{s}{\sqrt{n}}[/tex]

And the margin of error would be given by: [tex]ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=120-1=119[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. And we see that [tex]t_{\alpha/2}=1.98[/tex]

The standard error would be given by:

[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]

Now we have everything in order to replace into formula (1) and calculate the interval:

[tex]5-1.98\frac{1.5}{\sqrt{120}}=4.729[/tex]    

[tex]5+1.98\frac{1.5}{\sqrt{120}}=5.271[/tex]

So on this case the 95% confidence interval would be given by (4.729;5.271)    

Final answer:

The standard error of the mean (SE) is calculated to be 0.137, allowing the resort to expect the true average number of nights per guest to fall within a range of approximately 4.73 to 5.27 nights with 95% confidence.

Explanation:

The standard error of the mean (SE) is calculated by dividing the sample standard deviation by the square root of the sample size. In this case, the standard error is 1.5 / sqrt(120) = 0.137. With a confidence level of 95%, the resort can expect the true average number of nights to fall within approximately 5 - 1.96 * 0.137 nights to 5 + 1.96 * 0.137 nights, which is roughly between 4.73 and 5.27 nights.

Unit price means the cost per unit.

I'll provide an example.

If 4 bars of soap cost $2.35, find the unit price. Round to the nearest cent if necessary.

Remember that the unit price is the cost per unit which in this case is the cost per bar of soap. Since $2.35 is the cost for 4 bars of soap, to find the cost per bar of soap, we divide 4 into $2.35.

When we do this, we get an answer of 0.587 which rounds up to 59 cents.

Therefore, the unit price for soap is 59 cents.

Answer:

a) 0.9292

b)0.1515

Step-by-step explanation:

Explained in attachment.

Answer:

mwah, thank you so much for posting this question! you just saved my life!! :,)

Step-by-step explanation:

thank you thank you.

Mwah!!!!

Answer:

0.212 is the probability that a randomly selected tire will have a depth less than 0.70 mm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  0.98 mm

Standard Deviation, σ =  0.35 mm

We are given that the distribution of tire tread is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(depth less than 0.70 mm)

P(x < 0.70)

[tex]P( x < 0.70) = P( z < \displaystyle\frac{0.70 - 0.98}{0.35}) = P(z < -0.8)[/tex]

Calculating from normal z table, we have:

[tex]P(z<-0.8) = 0.212[/tex]

[tex]P(x < 0.70) = 0.212 = 21.2\%[/tex]

Thus, this event is not unusual and will  not warrant a refund.

One sample t-test

Step-by-step explanation:

In this statistical test, you will be able to test if a sample mean, significantly differs from a hypothesized value.Here you can test if the average swimming velocity differs significantly from an identified value in the hypothesis.Then you can conclude whether the group of 18 fish or that of 21 fish has a significantly higher or lower  mean velocity than the one in the hypothesis.

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Keywords: population, velocity, statistical test,average

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Answer:

The length of the person’s shadow is 5.7ft

Explanation:

Length of the flagpole =a= 35ft

Length of the shadow of the flagpole= b=50ft

Length of the person=c= 4ft

Suppose the length of the person’s shadow is=d

According to the  rules of trigonometry

[tex]\frac{\text { Length of the flagpole }}{\text { Length of the shadow of the flagpole }}=\frac{\text { Length of the person }}{\text { Length of the person's shadow }}[/tex]

[tex]\frac{a}{b}=\frac{c}{d}[/tex]

[tex]\frac{35}{50}=\frac{4}{d}[/tex]

35d=200

d=[tex]\frac{200}{35}[/tex]

d=5.7ft  

Hence, The length of the person’s shadow is 5.7ft.

Answer:

[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]

Step-by-step explanation:

The probability distribution of sampling distribution [tex]\hat{p}[/tex] is known as it sampling distribution.

The mean and standard deviation of the proportion is given by :-

[tex]\mu_{\hat{p}}=p\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]

, where p =population proportion and  n= sample size.

Given : According to a survey, 50% of Americans were in 2005 satisfied with their job.

i.e. p = 50%=0.50

Now, for sample size n= 1000 , the mean and standard deviation of the proportion will be :-

[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=\sqrt{\dfrac{0.50(1-0.50)}{1000}}=\sqrt{0.00025}\\\\=0.0158113883008\approx0.0158[/tex]

Hence, the mean and standard deviation of the proportion for a sample of 1000:

[tex]\mu_{\hat{p}}=0.50\\\\\sigma_{\hat{p}}=0.0158[/tex]