Natalie is skiing along a circular ski trail that has a radius of 2.6 km. She starts at the 3-o'clock position and travels in the CCW direction. Natalie stops skiing when she is 0.942 km to the right and 2.423 km above the center of the ski trail. Imagine an angle with its vertex at the center of the circular ski trail that subtends Natalie's path.
a. How many radians has the angle swept out since Natalie started skiing? ______radians
b. How many km has Natalie skied since she started skiing? _________km

Answer:

  A. ∠A

Step-by-step explanation:

The shortest side is BC. The smallest angle is opposite the shortest side, so is angle A.

∠A is the smallest in measure.

The expected number of children who get his or her own coat among 10 students using expected deviation will be 6.

Probability is defined as the possibility of the occurrence of an event.

Probability lies between 0 and 1.

A low standard deviation suggests that data are grouped around the mean, whereas a large standard deviation shows that data are more dispersed.

Given that:

Number of students = 10

The probability for 10 students is 0.1

The pay-off table is as follows:

The expected deviation can be calculated as:

[tex]E(x) = 1\times0.1 + 2\times0.1+ 3\times0.1+ 4\times0.1 + 5\times0.1 + 6\times0.1 + 7\times0.1 + 8\times0.1 + 9\times0.1 + 10\times0.1[/tex]

E(x) = 5.5

The expected number of children who get his or her own coat is 6 when rounded off.

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The table showing the relation between the number of students to its probability is shown below.

Final answer:

The expected number of children who will get their own coat when the coats are handed out randomly to 10 students is 6, based on the concept of expected value in probability.

Explanation:

To calculate the expected number of children who get their own coat among 10 students using expected deviation:

First, assign a probability to each possible outcome (number of children getting their own coat). Since there are 10 students and each has an equal probability of getting their own coat, the probability for each outcome is 0.1.

Next, multiply each outcome by its respective probability and sum them up. This gives the expected value (E(x)).

E(x) = (1 * 0.1) + (2 * 0.1) + (3 * 0.1) + (4 * 0.1) + (5 * 0.1) + (6 * 0.1) + (7 * 0.1) + (8 * 0.1) + (9 * 0.1) + (10 * 0.1)

E(x) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 + 0.8 + 0.9 + 1.0

E(x) = 5.5

When rounded off, the expected number of children who get their own coat is 6.

Answer:

[693,38:725,62]hs

Step-by-step explanation:

Hello!

Your study variable is X: the lifespan of a light bulb. This variable is said to have an approximately normal distribution.

X≈N(μ;σ²)

Were

μ= populatiom mean

σ= 45 hs standard deviation

A sample of n= 35

To estimate the population mean with a confidence interval you have to use the Z statistic:

X{bar} ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)

[tex]Z_{1-\alpha /2} = Z_{0.98} = 2.054[/tex]

710 ± 2.054*(45/√35

[693,38:725,62]hs

I hope you have a SUPER day!

Answer:

the probability is 0.28

Step-by-step explanation:

using Bayes's theorem

P(A|B)=P(A∩B)/P(B)

where

P(A∩B) = probability that events A and B happen

P(A|B) = probability that event A happen if B already happened

P(B)= probability of event B

therefore

P(A∩B)=P(A|B)*P(B)

if event A= selection of a student that lives in a dormitory  and event B = selection of a freshmen student

P(A|B) = 0.8 (live in a dormitory knowing that is a freshmen student )

P(B) = 0.35 (freshmen student)

P(A∩B)=P(A|B)*P(B) = 0.8* 0.35 =0.28

The probability that a randomly selected student is a freshman who lives in a dormitory is 0.28. So the correct option is  A.

To find the probability that a randomly selected student is a freshman who lives in a dormitory, we need to multiply the probability that a student is a freshman by the probability that a freshman lives in the dormitory. According to the given information, 35% of the students are freshmen, and among them, 80% live in the dormitory.

We calculate this probability using the formula:

Probability of (Freshman who lives in dorm) = Probability of (Freshman) × Probability of (Dorm | Freshman)

Probability of (Freshman who lives in dorm) = 0.35 × 0.80

Probability of (Freshman who lives in a dorm) = 0.28

Thus, the correct answer is A) 0.28.

Answer:

x = 34°

Step-by-step explanation:

Given AC and BD are perpendicular bisectors, we can say that at point E, there are 4 right angles [perpendicular bisectors intersect to create 4 90 degree angles].

Now, if we look at the triangle AED, we know that it is a right triangle, meaning that angle E is a right angle.

Also,

We know sum of 3 angles in a triangle is 180 degrees. Thus, we can write:

∠A + ∠E + ∠D = 180

Note: Angle A and Angle D are just the half part of the diagram. More exactly we can write:

∠EAD + ∠ADE + ∠DEA = 180

Given,

∠EAD = 56

∠DEA = 90

We now solve:

∠EAD + ∠ADE + ∠DEA = 180

56 + ∠ADE + 90 = 180

146 + ∠ADE = 180

146 + x = 180

x = 180 - 146

x = 34°

Answer: 500 miles

Step-by-step explanation:

Given : Sergio and Lizeth have planned to rent a car from a company that charges $75 a week plus $0.25 a mile.

i.e. Fixed charge= $75

Rate per mile = $0.25

Let x denotes the number of miles.

Then, Total charges = Fixed charge+ Rate per mile x No. of miles traveled

=  $75+ $0.25x

To keep budget within $200, we have following equation.

[tex]75+0.25x=200\\\\\Rightarrow\ 0.25=200-75\\\\\Rightarrow\ 0.25=125\\\\\Rightarrow\ x=\dfrac{125}{0.25}=\dfrac{12500}{25}=500[/tex]

Hence, they can travel 500 miles and still keep within their $200 budget.

Answer:

98% Confidence interval:  (31.74,38.4)

Step-by-step explanation:

We are given the following data set:

29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{526}{15} = 35.07[/tex]

Sum of squares of differences = 506.93

[tex]S.D = \sqrt{\displaystyle\frac{506.93}{14}} = 6.02[/tex]

98% Confidence interval:  

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.02} = \pm 2.145[/tex]  

[tex]35.07 \pm 2.145(\frac{6.02}{\sqrt{15}} ) = 35.07 \pm 3.33 = (31.74,38.4)[/tex]  

Thus, there is 98% confidence that the population mean number of minutes spent traveling by workers is between 31.74 mins and 38.40

To determine the number of cards each class parent helper will receive, divide the total number of cards (24) by the number of helpers (3), resulting in each helper receiving 8 cards.

Ms. Taylor's students are giving out cards to each of the 3 class parent helpers equally, and there are a total of 24 cards. To find out how many cards each helper gets, we can divide the total number of cards by the number of helpers.

Step 1: Determine the total number of cards: 24.

Step 2: Determine the number of class parent helpers: 3.

Step 3: Divide the total number of cards by the number of helpers: 24 ÷ 3.

Step 4: Calculate: 24 ÷ 3 = 8.

So, each class parent helper will receive 8 cards.

This solution calculates the probability of different outcomes when rolling a fair die three times. The results are obtained by defining the successful outcomes versus the total possible outcomes for each specific event.

This question is about probability. A fair die has 6 equally likely outcomes. Let's address each part:

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The probability of rolling all 4's is 1/64. The probability of rolling all even numbers is 7/64. The probability of not rolling all 2's is 63/64.

a) The probability of rolling all 4's is ⅛3, or 1/64. There is only one way to roll a 4 on a fair die and a total of 6 possible outcomes on each roll, so the probability is 1/6. Since the rolls are independent, the probability of getting a 4 on all three rolls is (1/6)(1/6)(1/6) = 1/64.

b) The probability of rolling all even numbers is also 1/64. There are 3 even numbers on a die (2, 4, and 6), so the probability of rolling an even number on any single roll is 3/6 or 1/2. Since the rolls are independent, the probability of getting an even number on all three rolls is (1/2)(1/2)(1/2) = 1/8. However, we need to subtract the probability of rolling all 4's from this, which is also 1/64. So the final probability is 1/8 - 1/64 = 7/64.

c) The probability of none of the rolls getting a number divisible by 2 is 1 - (1/2)(1/2)(1/2) = 1 - 1/8 = 7/8. This is the complement of rolling all even numbers.

d) The probability of rolling at least one 2 is 1 - the probability of rolling no 2's. The probability of not rolling a 2 on any single roll is 5/6, so the probability of not rolling a 2 on all three rolls is (5/6)(5/6)(5/6) = 125/216. Therefore, the probability of rolling at least one 2 is 1 - 125/216 = 91/216.

e) The probability of rolling all 2's is 1/64. Therefore, the probability of not rolling all 2's is 1 - 1/64 = 63/64.

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Answer:

a₁, a₂, a₃, . . .

Step-by-step explanation:

When we name something in Mathematics it is always advisable to number them as [tex]$ a_1, a_2, a_3, ...$[/tex].

Because the alphabets are only 26 in number we might run out of notations.

The simple logic behind the notion of [tex]$ a_1, a_2, a_3, ... $[/tex] is that the numbers have no end and we can number as many variables we want using this logic.

In this problem, you can continue your notation from [tex]$ a_{27}, a_{28}, a_{29} $[/tex] so that you don't have to make a change in your figure.

Answer:  The required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]

Step-by-step explanation:  Given that a particle moves in a straight line and has acceleration given by

[tex]a(t)=6t+2.[/tex]

The initial velocity of the particle is v(0) = −5 cm/s and its initial displacement is s(0) = 7 cm.

We are to find the position function s(t).

We know that the acceleration function a(t) is the derivative of the velocity function v(t). So,

[tex]v^\prime(t)=a(t)\\\\\Rightarrow v^\prime(t)=6t+2\\\\\Rightarrow v(t)=\int (6t+2) dt\\\\ \Rightarrow v(t)=3t^2+2t+A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

Also, the velocity function v(t) is the derivative of the position function s(t). So,

[tex]s^\prime(t)=v(t)\\\\\Rightarrow s^\prime(t)=3t^2+2t+A\\\\\Rightarrow s(t)=\int(3t^2+2t+A) dt \\\\\Rightarrow s(t)=t^3+t^2+At+B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

From equation (i), we get

[tex]v(0)=0+0+A\\\\\Rightarrow A=-5,~\textup{where A is a constant}[/tex]

and from equation (ii), we get

[tex]s(0)=0+0+0+B\\\\\Rightarrow B=7,~\textup{where B is a constant}.[/tex]

Substituting the values of A and B in equation (ii), we get

[tex]s(t)=t^3+t^2-5t+7.[/tex]

Thus, the required position function is [tex]s(t)=t^3+t^2-5t+7.[/tex]

Velocity: [tex]\( v(t) = 3t^2 + 2t - 5 \)[/tex] cm/s. Position: [tex]\( s(t) = t^3 + t^2 - 5t + 7 \)[/tex] cm, starting at 7 cm with initial velocity of -5 cm/s.

let's solve this step by step.

Given that ( a(t) = 6t + 2 ), we need to find the velocity function ( v(t) ) by integrating the acceleration function with respect to time.

1. **Find velocity function ( v(t) )**:

[tex]\[ a(t) = \frac{dv}{dt} \][/tex]

  So, integrating ( a(t) ) with respect to ( t ) will give us ( v(t) ):

[tex]\[ \int a(t) \, dt = \int (6t + 2) \, dt \][/tex]

[tex]\[ v(t) = \int (6t + 2) \, dt = 3t^2 + 2t + C_1 \][/tex]

  Here, ( C_1 ) is the constant of integration.

  Given that ( v(0) = -5 ) cm/s, we can find the value of ( C_1 ):

[tex]\[ v(0) = 3(0)^2 + 2(0) + C_1 = C_1 = -5 \][/tex]

  So, [tex]\( v(t) = 3t^2 + 2t - 5 \).[/tex]

2. **Find position function ( s(t) )**:

  We know that velocity is the rate of change of displacement. So, we need to integrate the velocity function with respect to time to find the position function.

[tex]\[ v(t) = \frac{ds}{dt} \][/tex]

  Integrating ( v(t) ) with respect to ( t ) will give us ( s(t) ):

[tex]\[ \int v(t) \, dt = \int (3t^2 + 2t - 5) \, dt \][/tex]

[tex]\[ s(t) = \int (3t^2 + 2t - 5) \, dt = t^3 + t^2 - 5t + C_2 \][/tex]

  Here, [tex]\( C_2 \)[/tex] is the constant of integration.

  Given that [tex]\( s(0) = 7 \)[/tex] cm, we can find the value of [tex]\( C_2 \):[/tex]

[tex]\[ s(0) = (0)^3 + (0)^2 - 5(0) + C_2 = C_2 = 7 \][/tex]

  So, [tex]\( s(t) = t^3 + t^2 - 5t + 7 \).[/tex]

Therefore, the position function of the particle is [tex]\( s(t) = t^3 + t^2 - 5t + 7 \)[/tex]cm.

Answer:

The correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.

Step-by-step explanation:

Consider the provided information.

The t-distribution of the Student is a distribution of probability that is used when when the sample size is small and/or when the population variance is unknown to estimate population parameters.

The number of independent observations is equal to the sample size minus one when calculating a mean score or a proportion from a single sample.

Since µ and σ determine the shape of the distribution so we use standard normal distribution if σ is known.

Hence, the correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.

Final answer:

The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function would be all positive real numbers.

Explanation:

The function y = 1.29x + 3 represents the price y a website store charges for shipping x items. The reasonable range for this function depends on the context. Since the function represents the price of shipping, the range should be positive, as shipping cannot have a negative price.

Therefore, the reasonable range for this function would be J- all positive real numbers.

Final answer:

The reasonable range for the shipping cost function starts from 3 and includes values that increase by 1.29 for each additional item, corresponding to whole numbers of items shipped. Therefore, the answer is G - {4.29, 5.58, 6.87, ...}.

Explanation:

The function given is y = 1.29x + 3, which represents the price y that a website store charges for shipping x items. Since shipping cannot have a negative cost and the minimum number of items shipped is either zero or a positive integer, the reasonable range for this function would begin at the point where x is zero. Therefore, we start our range by calculating the shipping cost for zero items: 1.29(0) + 3 = 3. As x increases, the cost will also increase linearly according to the function. Hence, all subsequent shipping prices will be greater than 3.

Now, let's consider what the reasonable range for shipping items would be. It would be abnormal to have a fractional number of items shipped because items are discrete entities. Thus, the list of shipping prices should only include charges for whole numbers of items. So, the range should only include prices that correspond to whole numbers of items shipped.

As a result, the reasonable range would include values starting from y = 3 onwards at intervals of 1.29 times an integer value. Option G, which starts from 4.29 and increases at a constant rate of 1.29, represents these intervals since 4.29 is the price for shipping one item (1.29*1 + 3). Any positive number of items shipped will result in a corresponding shipping price that is greater than 3, and since it is discreetly incremented, the prices will form a sequence of specific numbers, not all positive real numbers. Therefore, option G - {4.29, 5.58, 6.87, ...} - is the most appropriate answer.

Answer:

[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=\sqrt{10} (\frac{m-7.5}{s})[/tex]

Step-by-step explanation:

1) Notation

n=10 represent the sample size

[tex]\bar X=m[/tex] represent the sample mean  

[tex]s[/tex] represent the sample standard deviation

m represent the margin of error

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".  

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean for the population is 7.5 or no, the system of hypothesis would be:    

Null hypothesis:[tex]\mu =7.5[/tex]    

Alternative hypothesis:[tex]\mu \neq 7.5[/tex]    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=(\sqrt{10})\frac{m-7.5}{s}[/tex]

and we have our statistic in terms of m (mean) and the sample standard deviation s.

Number of subscriber the magazine will have after 3 years from now approximately be 8767

Solution:

Given that magazine currently has 8700 subscribers for its online web version

[tex]\begin{array}{l}{\mathrm{R}(\mathrm{t})=190 \mathrm{e}^{0.03 \mathrm{t}} \text { subscribers/month }} \\\\ {\mathrm{S}(\mathrm{t})=\mathrm{e}^{-0.06 \mathrm{t}}}\end{array}[/tex]

After 3 years, time(t) = 36 month

Total number of subscribers after 3 years from now :

Substitute "t" = 36

[tex]\begin{array}{l}{\mathrm{R}(36)=190 \mathrm{e}^{0.03 \times(36)}=190 \times(2.944)} \\\\ {\mathrm{R}(36) \approx 560} \\\\ {\mathrm{S}(36)=\mathrm{e}^{-0.06 \times(36)}=0.12}\end{array}[/tex]

Subscribers remaining = 0.12 x 560 = 67.2

The magazine currently has 8700 subscribers

Added Subscriber = 8700 + 560 = 9260

Remaining Subscriber = 8700 + 67.2 = 8767.2

Therefore number of subscriber the magazine will have after 3 years from now approximately be 8767

Answer:

(0.6728, 0.7672)

(0.1330, 0.1870)

Step-by-step explanation:

Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.

Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183[/tex]

Margin of error for 99% we would use the value [tex]z = 2.576[/tex]

Margin of error = [tex]2.576*SE\\=2.576*0.0183\\=0.0472[/tex]

Confidence interval lower bound = [tex]0.72-0.0472=0.6728[/tex]

Upper bound = [tex]0.72+0.0472=0.7672[/tex]

99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)

b) n =500

Sample proportion p=[tex]\frac{80}{500} =0.16[/tex]

Margin of error for 90% = [tex]1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270[/tex]

90% confidence interval for the true population proportion of people with kids. =[tex](0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)[/tex]

Answer:  [tex]1.0848<\mu<1.3152[/tex]

Step-by-step explanation:

Confidence interval for population mean is given by :-

[tex]\overline{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}< mu< \overline{x}+ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]z_{\alpha/2}[/tex] = two -tailed z-value for [tex]{\alpha[/tex] (significance level)

n= sample size .

[tex]\sigma[/tex] = Population standard deviation.

[tex]\overline{x}[/tex] = Sample mean

By considering the given information , we have

[tex]\sigma=0.2\text{ mg}[/tex]

[tex]\overline{x}=1.2\text{ mg}[/tex]

n= 20

[tex]\alpha=1-0.99=0.01[/tex]

Using z-value table ,

Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]

The 99 percent two-sided confidence interval for the mean nicotine content of a cigarette will be :-

[tex]1.2- (2.576)\dfrac{0.2}{\sqrt{20}}<\mu<1.2+ (2.576)\dfrac{0.2}{\sqrt{20}}\\\\=1.2- 0.1152<\mu<1.2+ 0.1152\\\\=1.0848<\mu<1.3152 [/tex]

Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette:  [tex]1.0848<\mu<1.3152[/tex]

Answer: 6

Step-by-step explanation:

Let  S= Total colleges

A = colleges are expensive.

B= colleges are  far from home( more than 200 miles away).

Given : n(S)= 20

n(A)=8

n(B)=8

n(A∩B) =2

Then, the number of  college that are not expensive and within 200 miles from home :-

[tex]n(A'\cap B')=n(S)-n(A\cup B)\\\\=20-(n(A)+n(B)-N(A\cap B))\ \ [\because\ n(A\cup B)=n(A)+n(B)-N(A\cap B)]\\\\=20-(8+8-2)\\\\=20-14=6[/tex]

i.e.  the number of  college that are not expensive and within 200 miles from home=6

Hence, the number of possible selections are 6 .

Answer:

so (x,y) = (2,3)

Step-by-step explanation:

we have

x +y= 5  ------ equation 1

2x+y= 7-------equation 2

Multiplying equation--1 by 2, we will have equation---3

2x +2y= 10-----equation 3

now subtracting equation 2 from equation 3

2x + 2y= 10

2x + y = 7

     y = 3

adding value of y in equation 1

x + y = 5

x= 5 -y

x= 5 - 3

x =2

so (x,y) = (2,3)

Answer: the correct answer is

5 - x = 7 - 2x

Step-by-step explanation:

Answer:

Step-by-step explanation:

Since 6 cakes serve 120 people, each cake serves 120/6 = 20 people.  Each cake weighs (5 lb)(16 oz/lb) = 80 oz. Then each person gets ...

  (80 oz)/(20 persons) = 4 oz/person

__

150 servings will require 150/20 = 7.5 cakes. Peter already has 6 cakes, so needs to make 2 more.

5lb per cake *6 cakes= 30lb

30lb / 120 people = 0.25lb per person

150 people - 120 people = 30 people

0.25lb per person * 30 people = 7.5lb

5lb___1 cake

7.5lb___X=1.5 cakes

(7.5lb * 1 cake)/5lb = 1.5 cakes

The probability that the ant will be at 1 after 3 seconds is 49/64.

To find the probability that the ant will be at 1 after 3 seconds, we need to consider all possible paths it can take. After each second, the ant can either move left with a probability of 1/4 or move right with a probability of 3/4.

Let's analyze all the possible paths:

Adding up the probabilities from each path, the total probability that the ant will be at 1 after 3 seconds is (27/64) + (9/64) + (9/64) + (3/64) + (1/64) = 49/64.

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Answer:

The sample should be 1,068.

Step-by-step explanation:

Consider the provided information.

Confidence level is 95% and margin of error is 0.03.

Thus,

1-α=0.95

α=0.05, E=0.03 and planning value [tex]\hat p=0.5[/tex]

Formula to calculate sample size is: [tex]n=\frac{\hat p(1-\hat p)(z_{\alpha/2})^2}{E^2}[/tex]

From the table we can find:  

[tex]z_{\alpha/2}=z_{0.05/2}\\z_{0.025}=1.96[/tex]

Substitute the respective values in the above formula we get:

[tex]n=\frac{0.5(0.5)(1.96)^2}{(0.03)^2}[/tex]

[tex]n=\frac{0.25(1.96)^2}{(0.03)^2}\approx 1067.111[/tex]

Hence, the sample should be 1,068.

Final answer:

To find the added width of the rectangular plot, set up the equation (32 + x)(25 + x) = 2(32)(25) based on the fact that the area is doubled. Solving this equation will give the added width.

Explanation:

To find the added width of the rectangular plot, we need to determine the increase in length and width that will double the area. The original dimensions of the plot are 32 feet long and 25 feet wide. Since equal width is added to each side, the new length and width will be 32 + x and 25 + x, respectively. To find x, we can set up the equation (32 + x)(25 + x) = 2(32)(25) using the fact that the area is doubled. Solving this equation will give us the added width.

Expanding (32 + x)(25 + x), we get 800 + 57x + x^2 = 1600. Rearranging and simplifying, we get x^2 + 57x - 800 = 0. We can then solve this quadratic equation using factoring or the quadratic formula to find the value of x.

The correct equation to use in order to find the added width is (x + 32)(x + 25) = 800.

Answer:

If only one of these statements is true, #17 robbed the bank.

Step-by-step explanation:

1st Scenario: #26 tells the truth

If #26 is telling the truth, that means #14 is guilty and, consequently, lying. However, that would also mean that #17 is telling the truth and then more than one statement would be true.

2nd Scenario: #17 tells the truth

If #17 is telling the truth, #17 is innocent, but then again either #26 or #14 are lying and two statements would be true.

3rd Scenario: #14 tells the truth

If #14 is telling the truth, #14 is innocent and, consequently, #26 is lying. That leaves us with #17 claiming innocence, but since only #14 can be telling the truth, #17 is lying and robbed the bank.

Answer:

101, 3, 0.025, 0.7416

Step-by-step explanation:

Given that  a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.

As per central limit theorem we have

a) Mean of sample mean = [tex]E(\bar x) =\\ \mu =101[/tex]

Std deviation of sample mean = [tex]\frac{\sigma}{\sqrt{n} } =3[/tex]

Mean = 101

Std dev =3.00

b) [tex]P(\bar x >107) = P(Z>\frac{107-101}{3} )\\= P(Z>2) = 0.025[/tex]

c) the probability that the sample mean deviates from the population mean μ = 101 by no more than 4.

=[tex]P(|\bar x-101|) \leq 4\\= P(|z|\leq 1.13)\\= 2(0.3708)\\=0.7416[/tex]

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

Step-by-step explanation:

Answer:

There was no error

Step-by-step explanation:

There are two types of statistical errors, the type 1 error and the type 2 error. In this case we refute the null hypothesis when the hypothesis is, in fact, false, because the mean process is 9 days instead of 7. Therefore we made no errors.

If the null hypothesis were True, a type 1 error would have ocurred. If the null hypothesis were false and we didnt refute it, then a type 2 error would have ocurred.

To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, use the sample variance, sample size, and chi-square distribution. The coating layer thickness should be approximately normally distributed for the confidence interval to be valid.

To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution. First, we need to calculate the sample variance and sample size. Then, we can use the chi-square distribution table to find the critical values for a 99% confidence interval with (n-1) degrees of freedom. Finally, we can calculate the confidence interval using the formula:

CI = sqrt((n - 1) * s^2 / X^2)

where CI is the confidence interval, n is the sample size, s^2 is the sample variance, and X^2 is the critical value from the chi-square distribution.

In this case, the coating layer thickness should be approximately normally distributed for the confidence interval to be valid.

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In the given case, the answer No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.

To calculate the 99% confidence interval (CI) for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution.

Given that the sample size is small (n = 8), we'll use the chi-square distribution with  n - 1 = 7 degrees of freedom.

The formula for the confidence interval is:

[tex]\[ \left( \sqrt{\frac{(n-1)S^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)S^2}{\chi^2_{1-\alpha/2, n-1}}} \right) \][/tex]

Given the sample observations: 21, 16, 29, 36, 42, 25, 24, 25, we first need to calculate the sample standard deviation:

[tex]\[ S = \sqrt{\frac{\sum_{i=1}^{n}(X_i - \bar{X})^2}{n-1}} \][/tex]

Let's perform the calculations:

[tex]\[ S = \sqrt{\frac{(21-27.625)^2 + (16-27.625)^2 + \ldots + (25-27.625)^2}{7}} \][/tex]

[tex]\[ S \approx 9.38 \][/tex]

Now, plug the values into the formula:

[tex]\[ \left( \sqrt{\frac{7 \times 9.38^2}{18.48}}, \sqrt{\frac{7 \times 9.38^2}{2.17}} \right) \][/tex]

[tex]\[ \left( \sqrt{26.97}, \sqrt{229.78} \right) \][/tex]

[tex]\[ \left( 5.19, 15.15 \right) \][/tex]

Therefore, the 99% confidence interval for the standard deviation of the coating layer thickness distribution is approximately [tex]\( (5.19, 15.15) \)[/tex] micrometers.

Therefore, the answer is: No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.

Answer:  95% confidence interval would be (0.449, 0.511)

Step-by-step explanation:

Since we have given that

n = 1022

p = 48% = 0.48

We need to find the 95% confidence interval first.

z = 1.96

Margin of error would be

[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.48\times 0.52}{1022}}\\\\=0.031[/tex]

95% confidence interval would be

[tex]p\pm 0.031\\\\=(0.48-0.031,0.48+0.031)\\\\=(0.449,0.511)[/tex]

It means true proportion who felt that economic growth is more important than protecting the environment is within 0.449 and 0.511 using 95% confidence.

Answer:

p-value 0.0124

Step-by-step explanation:

Hello!

In this experiment, 10 plots were randomly sampled and each plot was divided in a half, one half was planted with the common variety of tomato and the other half was planted with the new hybrid variety. At harvest time it was measured the pounds per plant harvested and the variable difference was established. This study variable is defined as the difference between the pounds per plant of the hybrid variety minus the pound per plant of the common variety. Symbolized: Xd: Xnew - Xcommon.

The purpose of this experiment is to test whether there is a difference between the yield of both species. This is a classic example of a paired test, in which you want to put two dependent samples to test. The easiest way to recognize this type of test, if it is not specified in the problem, is that both variables are measured in the same sampling unit. In this case, the sampling unit is "one plot" that was divided and both verities of tomatoes were planted on it.

This test is also called "paired samples T-test" and as a statistic, you have to use the Student t.

The data given is:

sample: 10 plots

sample mean: Xd[bar]: 1.05ppp

sample standard deviation Sd: 1.23ppp

If the hybrid has a higher yield, that would mean that the difference between them would be positive if the difference is positive, its safe to assume that the population mean of the difference will be positive as well. Symbolically: μd>0

With this, you can state the hypothesis as:

H₀:μd≤0

H₁:μd>0

α: 0.05

t= Xd[bar] - μd ~ [tex]t_{n-1}[/tex]

       Sd/n

t[tex]t_{obs}[/tex]= (1.05 - 0) / (1.23/√10) = 2.69

This is a one-tailed test with critical value [tex]t_{9;0.95} = 1.83[/tex]

the p-value for this test is also one tailed.

p([tex]t_{9}[/tex]≥2.69) = 1-P([tex]t_{9}[/tex]<2.69) = 1 - 0.9876 = 0.0124

The decision is to reject the null hypothesis.

I hope it helps!