f(x) = 2x-1
What is f(13)?

Answer:

$24

Step-by-step explanation:

Since 25% is off the price of the shirt, she bought the shirt at 75% of it's cost price. This difference between the full cost and the 75% is what she saved (i.e. 25% was the savings)

So, if the shirt costs x dollars:

$6 = 25% of x

= ( 25/100)*X = 0.25X

X = $6/0.25 = $6x4

X = $24

Answer:

0.125

Step-by-step explanation:

Answer:

1/8

Step-by-step explanation:

-3/8 * -4/12

We can simplify the second fraction

Divide the top and bottom by 4

4/12 = 1/3

-3/8 * -1/3

The threes  in the top and bottom cancel

A negative times a negative cancel

1/8

Answer:

They Bisect

Step-by-step explanation:

They don't have the same slope.

They aren't instersecting at a right angle (they aren't perpendicular)

They aren't parallel because they touch.

The value of the car at the end of the fifth year is approximately $5695.62, determined by the initial value of $24000 and a 25% annual depreciation rate. The nth term formula reflects the cumulative effect of depreciation on the car's value over n years.

Finding the value of the car at the end of the fifth year and how the nth term of the sequence is related to the value of the car at the end of each year.

Finding the value at the end of the fifth year (a₅):

We are given the formula for the value of the car at the end of each year: a_n = 24000 (3/4)^n, where n is the year.

We want to find the value at the end of the fifth year, so we need to substitute n = 5 into the formula: a_5 = 24000 (3/4)^5.

Calculating this expression, we get: a_5 = 24000 * 0.2373 ≈ $5695.62.

Therefore, the value of the car at the end of the fifth year is approximately $5695.62.

Understanding the nth term of the sequence:

The nth term of the sequence, a_n, represents the value of the car at the end of the nth year.

The formula for the nth term shows that the value of the car is determined by two factors:

Initial value: The initial value of the car is represented by 24000 in the formula. This is the value of the car when it is brand new.

Depreciation rate: The depreciation rate is represented by the fraction 3/4. This fraction indicates that the value of the car decreases by 25% each year. The exponent n in the formula tells us how many times this depreciation rate is applied.

Therefore, the nth term of the sequence tells us how much the car's value has depreciated after n years, starting from its initial value of $24000. The higher the value of n, the more the car has depreciated, and hence the lower its value will be.

Answer:

(x + 10)^2 + (-186)

Step-by-step explanation:

x^2 + 20x - 86 =

Move the constant term to the right leaving a space between the x-term and the constant term.

= x^2 + 20x           - 86

To complete the square, take half of the x-term coefficient and square it.

Half of 20 is 10. 10 squared is 100. This is the number that completes the square. Add it right after the x-term. Now you need to subtract the same amount at the end.

= x^2 + 20x + 100 - 86 - 100

Since 100 was added and subtracted, the expression has the same value. The first three terms are a perfect square trinomial, so we write it as the square of a binomial.

= (x + 10)^2 - 186

Since you have an addition sign, we write -186 as a sum:

= (x + 10)^2 + (-186)

Answer:

Step-by-step explanation:

+10 and -186 is your answer.

Answer:

We need a sample of at least 82 female white-tailed deer

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 99% confidence level, to within 6 pounds of the actual weight?

We need a sample of size at least n.

n is found when [tex]M = 6, \sigma = 21[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]6 = 2.575*\frac{21}{\sqrt{n}}[/tex]

[tex]6\sqrt{n} = 21*2.575[/tex]

[tex]\sqrt{n} = \frac{21*2.575}{6}[/tex]

[tex](\sqrt{n})^{2} = (\frac{21*2.575}{6})^{2}[/tex]

[tex]n = 81.23[/tex]

Rounding up

We need a sample of at least 82 female white-tailed deer

Answer:

[tex]n=(\frac{2.58(21)}{6})^2 =81.54 \approx 82[/tex]

So the answer for this case would be n=82 rounded up to the nearest integer

Step-by-step explanation:

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.

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)

[tex]\sigma=21[/tex] represent the estimation for the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =6 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.58(21)}{6})^2 =81.54 \approx 82[/tex]

So the answer for this case would be n=82 rounded up to the nearest integer

Final answer:

There are 512 different combinations possible for an 8-number combination lock, as the same number can be reused and each of the three positions has 8 choices.

Explanation:

To calculate the number of different combinations possible for a combination lock with a dial containing 8 numbers (0 through 7), we can use the fundamental counting principle. Since the same number can be reused consecutively, there are no restrictions on the choices for each of the three numbers in the sequence. Therefore, for each position of the sequence (the first, second, and third number), there are 8 possibilities.

The total number of combinations is the product of the number of choices for each position. Thus, the number of combinations for the lock is:

8 (choices for the first number) × 8 (choices for the second number) × 8 (choices for the third number)

Doing the calculation, we get:

8 × 8 × 8 = 512

Therefore, there are 512 different combinations possible for the lock.

Answer:

a) - 0.54

b) No, we expect Andy to lose money each time he plays this game. This game is not a good way to make money.

Step-by-step explanation:

We are told the game costs $2 to play.

In a card game, we have the following:

Total number of cards = 52

Number of number cards (2-10)=36

Number of face cards = 12

Number of ace = 3

Number of ace of clubs = 1

The probability and profits of the game is calculated below.

Probability he gets a number card 2-10 =[tex] \frac{36}{52} = 0.69 [/tex]

Profit = - $2

Probability he gets a face card:

[tex] \frac{12}{52} = 0.23 [/tex]

Profit = $3 - $2 = $1

Probability he gets an ace:

[tex] \frac{3}{52} = 0.06 [/tex]

Profit = $5 - $2 =$3

Probability he gets an ace of clubs:

[tex] \frac{1}{52} = 0.02[/tex]

Profit = $25 - $2 =$23

Andy's expected profit per game wil be given as:

E = probability * profit

= [(0.69 * -2)+(0.23 * 1)+(0.06 * 3)+(0.02 * 23)]

= -0.54

b) No, we expect Andy to lose money each time he plays this game. This game is not a good way to make money.

Andy's expected profit per game (-0.542) is negative, which means he doesn't make any profit per game.

Andy expected  profit per game is -0.54.

Andy's expected profit per game (-0.542) is negative, which means he doesn't make any profit per game.

Given that,

The game costs $2 to play.

He draws a card from a deck. If he gets a number card (2-10), he wins nothing.

For any face card ( jack, queen or king), he wins $3. For any ace, he wins $5, and he wins $25 if he draws the ace of clubs.

We have to determine,

Andy's expected profit per game is:

Would you recommend this game to Andy as a good way to make money.

According to the question,

The game costs $2 to play.

Total number of cards = 52

Number of number cards (2 - 10) =36

Number of face cards = 12

Number of ace = 3

Number of ace of clubs = 1

The probability and profits of the game is calculated below.

[tex]Probability \ he \ gets \ a\ number\ card \ 2-10 = \dfrac{36}{52} = 0.69[/tex]

Profit = $2

[tex]Probability\ he \ gets\ a \face\ card= \dfrac{12}{52} = 0.23[/tex]

Profit = $3 - $2 = $1

[tex]Probability\ he \ gets\ an \ace = \dfrac{3}{32 } = 0.06[/tex]

Profit = $5 - $2 =$3

[tex]Probability \ he \ gets \ an\ ace\ of \ clubs: = \dfrac{1}{52} = 0.02[/tex]

Profit = $25 - $2 =$23

Andy's expected profit per game will be given as:

[tex]E = \ Probability \times \ profit\\\\E = [(0.69 \times -2)+(0.23 \times 1)+(0.06 \times 3)+(0.02 \times 23)]\\\\E = -0.54[/tex]

The profit of Andy is -0.54.

Andy's expected profit per game (-0.542) is negative, which means he doesn't make any profit per game.

To know more about Probability click the link given below.

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Answer:

[tex]3.786-2.262\frac{0.187}{\sqrt{10}}=3.652[/tex]    

[tex]3.786+2.262\frac{0.187}{\sqrt{10}}=3.920[/tex]    

So on this case the 95% confidence interval would be given by (3.652;3.920)    

Step-by-step explanation:

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.

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  

Solution to the problem

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)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=3.786[/tex]

The sample deviation calculated [tex]s=0.187[/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=10-1=9[/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. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]

Now we have everything in order to replace into formula (1):

[tex]3.786-2.262\frac{0.187}{\sqrt{10}}=3.652[/tex]    

[tex]3.786+2.262\frac{0.187}{\sqrt{10}}=3.920[/tex]    

So on this case the 95% confidence interval would be given by (3.652;3.920)    

To estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline, calculate the sample mean and the 95% confidence interval.

To estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline, we can calculate the sample mean and the 95% confidence interval.

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Answer:

it is D :)

Step-by-step explanation:

trust me ;)

Final answer:

Option D, which is about asking airline passengers to Orlando about their vacation plans, is an example of a non-random sample, specifically convenience sampling.

Explanation:

An example of a non-random sample is option D: Airline passengers to Orlando, Florida, are asked about vacation plans. This is considered non-random sampling because the passengers already have something in common – they are traveling to a popular vacation destination, which likely influences their vacation plans. This method does not give every individual in the broader population an equal chance of being selected and is thus not a random sample. This type of sampling is referred to as convenience sampling, as it involves selecting individuals who are easily accessible rather than using a process that gives every individual an equal chance of being chosen.

Answer:

[tex]76.1\times 9.6=730.56[/tex]

Step-by-step explanation:

In this case, we need to find the value of [tex]76.1\times 9.6[/tex]. The numbers are in decimal form. It is understood that the final answer will have decimal point after two places from right.

For an instance, multiply remove decimals from both numbers. Now simply multiply 761 and 96 such that we will get 73056.

Now, keep the decimal point after two places from right. So, we will get 730.56.

Final answer:

The answer to the multiplication of 76.1 and 9.6 is 730 when rounded to two significant figures in accordance with the rules of significant figures in multiplication.

Explanation:

The student asked for the product of 76.1 and 9.6. To compute this, you multiply the two numbers. Since 76.1 has three significant figures and 9.6 has two significant figures, our final answer should be reported with two significant figures, owing to the least number of significant figures in the given numbers.

Now, calculating the product:

76.1 times 9.6 = 730.56

When rounding to two significant figures, the answer is 730.

In calculations involving significant figures, it is important to report your final answer with the correct number of significant figures. In multiplication and division, the number of significant figures in the final answer should be the same as the least number of significant figures in any of the numbers being calculated.

Answer:

(9)[tex]\frac{1}{12}[/tex]  (10) [tex]\frac{1}{12}[/tex]  (11)[tex]\frac{5}{12}[/tex]  (12)[tex]\frac{1}{4}[/tex]  (13)[tex]\frac{1}{6}[/tex] 14)[tex]\frac{5}{36}[/tex] (15)[tex]\frac{1}{12}[/tex]  (16)0

Step-by-step explanation:

The sample Space of the single die rolled twice is presented below:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.

n(S)=36

(9)Probability of getting two numbers whose sum is 9.

The possible outcomes are:  (3, 6), (4, 5),  (5, 4)

[tex]P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}[/tex]

10) Probability of getting two numbers whose sum is 4.

The possible outcomes are:  (1, 3),(2, 2),(3, 1),

[tex]P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}[/tex]

11.)Find the probability of getting two numbers whose sum is less than 7.

The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4),  (3, 1), (3, 2), (3, 3),  (4, 1), (4, 2),  (5, 1)

[tex]P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}[/tex]

12.Probability of getting two numbers whose sum is greater than 8

The possible outcomes are:(4, 5), (4, 6),  (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)

[tex]P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}[/tex]

(13)Probability of getting two numbers that are the same (doubles).

The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4),  (5, 5), (6, 6)

[tex]P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}[/tex]

14.Probability of getting a sum of 7 given that one of the numbers is odd.

The possible outcomes are: (2, 5),  (3, 4), (4, 3), (5, 2),  (6, 1)

[tex]P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}[/tex]

(15)Probability of getting a sum of eight given that both numbers are even numbers.

The possible outcomes are: (2, 6), (4, 4), (6, 2)

[tex]P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}[/tex]

16.Probability of getting two numbers with a sum of 14.

[tex]P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0[/tex]

Answer:

The numerical limits for a B grade is between 81 and 89.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 79.7, \sigma = 8.4[/tex]

B: Scores below the top 13% and above the bottom 56%

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.127 = \frac{X - 79.7}{8.4}[/tex]

[tex]X - 79.7 = 8.4*1.127[/tex]

[tex]X = 89[/tex]

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.15 = \frac{X - 79.7}{8.4}[/tex]

[tex]X - 79.7 = 8.4*0.15[/tex]

[tex]X = 81[/tex]

The numerical limits for a B grade is between 81 and 89.

Answer:

Test: [tex]\mu[/tex] < 20 with normal distⁿ

Hypothesis test:

H0:  [tex]\mu \ge[/tex] 20                         (Null Hypothesis, H₀ : [tex]\mu[/tex] = 20 is also correct)

H1: [tex]\mu[/tex] < 20                             (Alternative Hypothesis, also called H₁)

This is lower tailed test.

Since sample is large, sample standard deviation can be taken as an  approximation of population standard deviation.

x = 19.3

[tex]\sigma[/tex] = 11.1

n = 444

significance level, [tex]\alpha[/tex] = 0.05 (If no value is given, we take level of 0.05)

Test statistic [tex]z* = \frac{x-\mu}{\sigma/\sqrt{n}}[/tex]

                         [tex]= \frac{19.3-20}{11.1/\sqrt{444}}[/tex]

                       = - 1.33

The attached files contains additional information

Answer:

1.97

Step-by-step explanation:

The null hypothesis is:

[tex]H_{0} = 250[/tex]

The alternate hypotesis is:

[tex]H_{1} \neq 250[/tex]

Our test statistic is:

[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the hypothesis tested(null hypothesis), [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

In this problem, we have that:

[tex]X = 251.6, \mu = 250, \sigma = 5.4, n = 44[/tex]

So

[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]t = \frac{251.6 - 250}{\frac{5.4}{\sqrt{44}}}[/tex]

[tex]t = 1.97[/tex]

Answer:

50 is the initial value

Step-by-step explanation:

The equation is of the form

y = ab^x  where a is the initial value and b is the growth/decay value

 b>1 is growth and b<1 means decay

   y = 50(1.6)^x  

50 is the initial value and 1.6 means growth since it is greater than 1

1.6 -1 = .6 so it has a 60% growth rate        

Answer:

99% CI for the variance [tex]\sigma^{2}[/tex] , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.

Step-by-step explanation:

We are given that the following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for Wire electrical-discharge machining (WEDM) : 21, 16, 29, 35, 42, 24, 24, 25

Firstly, the pivotal quantity for 99% confidence interval for the population variance is given by;

                          P.Q. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex]  ~ [tex]\chi^{2}__n_-_1[/tex]

where,  [tex]s^{2}[/tex] = sample variance = [tex]\frac{\sum (X-\bar X)^{2} }{n-1}[/tex]  = 67.43

             n = sample of observations = 8

            [tex]\sigma^{2}[/tex] = population variance

Here for constructing 99% confidence interval we have used chi-square test statistics.

So, 99% confidence interval for the population variance, [tex]\sigma^{2}[/tex] is ;

P(0.9893 < [tex]\chi^{2}_7[/tex] < 20.28) = 0.99  {As the critical value of chi-square at 7

                                          degree of freedom are 0.9893 & 20.28}  

P(0.9893 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 20.28) = 0.99

P( [tex]\frac{0.9893 }{(n-1)s^{2} }[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{20.28 }{(n-1)s^{2} }[/tex] ) = 0.99

P( [tex]\frac{(n-1)s^{2} }{20.28 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1)s^{2} }{0.9893 }[/tex] ) = 0.99

99% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2} }{20.28 }[/tex] , [tex]\frac{(n-1)s^{2} }{0.9893 }[/tex] ]

                                                   = [ [tex]\frac{7\times 67.43 }{20.28 }[/tex] , [tex]\frac{7\times 67.43 }{0.9893 }[/tex] ]

                                                   = [23.275 , 477.115]

99% confidence interval for [tex]\sigma[/tex]  = [ [tex]\sqrt{23.275}[/tex] , [tex]\sqrt{477.115}[/tex] ]

                                                  = [4.824 , 21.843]

Therefore, 99% CI for the variance [tex]\sigma^{2}[/tex] , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.

Answer:

4/52

Step-by-step explanation:

4 of each card 52 cards

Answer:

[tex] 7p + 5 \le 50 [/tex]

Step-by-step explanation:

The number of pizzas is p.

One pizza costs $7.

p number of pizzas cost 7p.

The total cost is the cost of the pizzas plus the $5 tip.

The total cost is 7p + 5

He can spend $50 or less, so the total cost must be less than or equal to $50.

Answer: [tex] 7p + 5 \le 50 [/tex]

Answer:

[tex]217.636-1.96\frac{20}{\sqrt{11}}=205.817[/tex]    

[tex]217.636+1.96\frac{20}{\sqrt{11}}=229.455[/tex]    

So on this case the 95% confidence interval would be given by (205.82;229.46)    

Step-by-step explanation:

Assuming the following data: Weight 150 169 170 196 200 218 219 262 269 270 271

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.

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)

[tex]\sigma= \sqrt{400}= 20[/tex] represent the population standard deviation

n=11 represent the sample size  

Solution to the problem

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

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

In order to calculate the mean we can use the following formula:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

The mean calculated for this case is [tex]\bar X=217.636[/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. The excel command would be: "=NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

Now we have everything in order to replace into formula (1):

[tex]217.636-1.96\frac{20}{\sqrt{11}}=205.817[/tex]    

[tex]217.636+1.96\frac{20}{\sqrt{11}}=229.455[/tex]    

So on this case the 95% confidence interval would be given by (205.82;229.46)    

8x² - 24 can be written in factorized form as  8 (x² - 3).

Step-by-step explanation:

Given expression is

8x² - 24

It can be factorized by taking the common factors as,

Since 8 is the common factor for both the terms and the expression can be written as,

8x² - 24

It can be expanded as,

= 8x² - 8×3

Now both the terms has 8, so it can be taken out and the expression can be written as,

= 8 (x² - 3)

So it can be written in factorized form as  8 (x² - 3).

Answer:

carmen winsted

Step-by-step explanation:

Answer:

3/8

Step-by-step explanation:

Since we have given that

Measure of an angle = 135°

As we know that

Total angle formed by a circle = 360°

So, Fraction of a circle when the an angle 135° turn is given by

135/360=

Answer:

The probability that 10 squared centimetres of dust contains more than 10150 particles is 0.067.

Step-by-step explanation:

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 1,000.

If X follows Poisson (λ) and λ > 1,000 then the distribution of X can be approximated but he Normal distribution.

The mean of the approximated distribution of X is:

μ = λ

The standard deviation of the approximated distribution of X is:

σ = √λ

Thus, if λ > 1,000, then [tex]X\sim N(\mu=\lambda,\ \sigma^{2}=\lambda)[/tex].

Let X = number of asbestos particles in a sample of 1 squared centimetre of dust.

The random variable X follows a Poisson distribution with mean, μ = 1000.

Then the average number of asbestos particles in a sample of 10 squared centimetre of dust will be, [tex]\lambda = 10\times \mu=10\times 1000=10,000[/tex].

Compute the probability that 10 squared centimetres of dust contains more than 10150 particles as follows:

[tex]P(X>10150)=P(\frac{X-\mu}{\sigma}>\frac{10150-10000}{\sqrt{10000}})[/tex]

                       [tex]=P(Z>1.50)\\=1-P(Z<1.50)\\=1-0.93319\\=0.06681\\\approx0.067[/tex]

*Use a z-table for the probability.

Thus, the probability that 10 squared centimetres of dust contains more than 10150 particles is 0.067.

We use the Poisson distribution and normal approximation to calculate the probability of observing more than 10150 particles in 10 square centimeters of dust, given the average particle count per centimeter. The normal approximation is necessary due to the large area being considered.

The question pertains to the calculation of a probability using the Poisson distribution in conjunction with the normal approximation. Since the average number of asbestos particles in a square centimeter of dust is 1000, for 10 square centimeters, the expected number of particles will be 10 * 1000 = 10000. However, when considering a larger area, the number of events (in this case, asbestos particles) might not follow the Poisson distribution strictly, which necessitates the use of the normal approximation.

Normal approximation of a Poisson variable involves transforming the Poisson variable to a normally distributed variable. The mean (μ) of this normal distribution stays the same (μ = λ = 10000 for 10 square centimeters), and the standard deviation (σ) becomes the square root of the mean (σ = √λ = √10000 = 100).

To calculate the probability of getting more than 10150 particles (denoted as X) in 10 square centimeters of dust, we use the normal cumulative distribution function (normalcdf). We want P(X > 10150), which with the normalcdf function, becomes 1 - P(X <= 10150) = 1 - normalcdf(-∞, 10150, 10000, 100). The result of the calculation using a standard normal table or calculator gives the probability, rounded to three decimal places.

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Answer:

Step-by-step explanation:

Systolic. Calculate the mean

118+128+158+96+156+122+116+136+126+120+

=1276÷10 = 127.6.

To calculate the median arrange the values from the lowest to the highest.

96,116,118,120,122,126,128,136,156,158.

122+126+2=124.

Disastolic

80+76+74+52+90+88+58+64+72+82= 736÷10=73.6

To calculate the median arrange the values from the lowest to the highest.

52,58,64,72,74,76,80,82,88,90

Median= 74+76=150÷2=75.

Comparison

The Systolic and diastolic blood pressure measures different things, so comparing them is of no use.

It will be good if the relationship between the blood pressure can be investigated because the data are in pairs.

Compute the mean and median for both the systolic and diastolic measurements. Compare the results to identify skewness in the data. The mean and median do not represent data variability, so also consider using the Standard Deviation or Interquartile Range for a broad view.

First, let's begin by calculating the mean or average of both systolic and diastolic data sets. For this, you'll add up all the measurements and divide by the number of measurements.

For the median, you'll arrange the measurements in ascending order. If the total number is odd, the median is the middle number. If it is even, the median will be the average of the two middle numbers.

Comparing the two data sets would involve looking at the calculated mean and median values and determine whether they're close in value or not. If they are close, it indicates a lack of skewness in the data. If they're not, there would appear to be skewness, with the mean being influenced by particularly high or low values.

The mean and median are measures of central tendency that provide some insight into the data, however, they do not show the spread or variability of the data. Therefore, using additional statistics like the Standard Deviation or Interquartile Range might be good options to understand dispersion in the data set.

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Answer:

B=22.348 Inches per minutes

Step-by-step explanation:

A snail is traveling along a straight path. The snail’s velocity can be modeled by [tex]v(t)=1.4ln(1+t^2)[/tex] inches per minute for 0 ≤ t ≤ 15 minutes.

If the snail's velocity is [tex]v(t)=1.4ln(1+t^2)[/tex] per minute, its displacement for 0 ≤ t ≤ 15 minutes is given by the integral:

[tex]\int_{0}^{15}1.4ln(1+t^2)dt=76.04307[/tex]

The ant travels with a constant acceleration of 2 Inches per minute.

Therefore, the velocity of the ant will be:

[tex] \int 2 dt=2t+c,[/tex] inches per minutes, for some constant c.

For the interval, 12≤t≤15, the displacement of the ant is:

[tex] \int_{12}^{15}(2t+c)dt=t^2+ct|_{12}^{15}=81+3c[/tex]

Since the snails displacement and that of the ant are equal in 12≤t≤15.

81+3c=76.04307

3c=76.04307-81

3c=-4.95693

c=-1.65231

The velocity of the ant at t=12 is therefore:

2t+c=2(12)-1.65231=22.348 Inches per minutes

B=22.348 Inches per minutes

The value of B to the nearest whole number, given all of the factors enumerated above, is 22.4 inches/Min. (For the full answer, please see the attached.)

This simply refers to the pace or rate at which an object or a person changes their position in relation to a frame of reference. It is also a function of time.

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Answer:

0.9%

Step-by-step explanation:

To determine the percentage of Earth's surface affected by a meteor strike, we calculate the surface area of Earth and the area of impact, then divide the area of impact by Earth's total surface area and multiply by 100 to get a percentage. Using the given Earth's radius of 3959 miles and the meteor's diameter of 750 miles, the affected area is approximately 0.224% of Earth's surface.

To calculate the percentage of the Earth's surface affected by a meteor strike, we can use the formula for the surface area of a sphere and the formula for the surface area of a circle, which represents the affected area.

The total surface area of the Earth (AEarth) is given by 4πr2, where r is the radius of the Earth. The radius of the Earth is 3959 miles.

The surface area affected by the meteor (AMeteor) can be approximated by the surface area of a circle, πd2/4, where d is the diameter of the meteor. The diameter of the meteor is 750 miles. Therefore, the affected area is approximately π * (7502)/4.

Now we can calculate both the total surface area of the Earth and the affected area:

To find the percentage of the surface affected, we divide the affected area by the total surface area and then multiply by 100:

Percentage = (AMeteor / AEarth) * 100

Percentage = (441,963 / 197,000,000) * 100 = 0.224% of the Earth's surface is affected by the meteor strike.

The probability that the sample proportion will differ from the population proportion by more than 5% is 0.0108 or 1.08%.

To solve this problem, we need to find the probability that the sample proportion of defective cell phones will differ from the population proportion (19%) by more than 5%.

Given information:

- Population proportion of defective cell phones = 0.19 (or 19%)

- Sample size = 399

Calculate the standard error of the proportion.

[tex]\[\text{Standard error of the proportion} = \sqrt{\frac{p \times (1 - p)}{n}}\]Substituting the given values:\[\text{Standard error of the proportion} = \sqrt{\frac{0.19 \times (1 - 0.19)}{399}}\]\[\text{Standard error of the proportion} = \sqrt{\frac{0.19 \times 0.81}{399}}\]\[\text{Standard error of the proportion} = \sqrt{\frac{0.1539}{399}}\]\[\text{Standard error of the proportion} = 0.0196 \text{ or } 1.96\%\][/tex]

Calculate the maximum allowable difference from the population proportion.

Maximum allowable difference = 0.05 (or 5%)

Calculate the z-score for the maximum allowable difference.

z-score = (Maximum allowable difference - Population proportion) / Standard error of the proportion

z-score = (0.05 - 0.19) / 0.0196

z-score = -2.55

Find the probability using the standard normal distribution table or calculator.

The z-score of -2.55 corresponds to a probability of 0.0054 (or 0.54%) in the standard normal distribution table.

Since the question asks for the probability that the sample proportion will differ from the population proportion by more than 5%, we need to find the probability of both tails.

Probability of both tails = 2 × 0.0054 = 0.0108 or 1.08%

Therefore, the probability that the sample proportion will differ from the population proportion by more than 5% is 0.0108 or 1.08%.

The probability that the sample proportion will differ from the population proportion by more than 5% is 0.0108 or 1.08%.

To solve this problem, we need to find the probability that the sample proportion of defective cell phones will differ from the population proportion (19%) by more than 5%.

Given information:

- Population proportion of defective cell phones = 0.19 (or 19%)

- Sample size = 399

Calculate the standard error of the proportion.

[tex]\[\text{Standard error of the proportion}[/tex]=[tex]\sqrt{p \times (1 - p))/(n)[/tex]

Substituting the given values:[tex]\[\text{Standard error of the proportion} = \sqrt{(0.19 \times (1 - 0.19))/(399)[/tex]

[tex]\[\text{Standard error of the proportion} = \sqrt{(0.19 \times 0.81)/(399)}\\\text{Standard error of the proportion} = \sqrt{(0.1539)/(399)}\\\text{Standard error of the proportion} = 0.0196 \text{ or } 1.96\%\][/tex]

Calculate the maximum allowable difference from the population proportion.

Maximum allowable difference = 0.05 (or 5%)

Calculate the z-score for the maximum allowable difference.

z-score = (Maximum allowable difference - Population proportion) / Standard error of the proportion

z-score = (0.05 - 0.19) / 0.0196

z-score = -2.55

Find the probability using the standard normal distribution table or calculator.

The z-score of -2.55 corresponds to a probability of 0.0054 (or 0.54%) in the standard normal distribution table.

Since the question asks for the probability that the sample proportion will differ from the population proportion by more than 5%, we need to find the probability of both tails.

Probability of both tails = 2 × 0.0054 = 0.0108 or 1.08%

Therefore, the probability that the sample proportion will differ from the population proportion by more than 5% is 0.0108 or 1.08%.

Answer:

Option A  - ∑Fx and ∑τz and ∑Fy

Step-by-step explanation:

All the force will be in x and y plane only

So Torque will be in z plane

These 3 quantities should be 0

∑Fx and ∑τz and ∑Fy

For static equilibrium in a plane, the net forces in the x and y directions, and the net torque about the z-axis, all need to be zero. This is true for a ladder leaning against a wall (a planar equilibrium situation). Hence, the correct set of quantities is ∑Fx, ∑Fy, and ∑τz.

To achieve static equilibrium in any system, the sum of all external forces and torques acting on the body must be zero. This is true for both linear and rotational movements. In the case of the ladder against the wall, we're considering a planar equilibrium condition (in the xy-plane), where the spatial extension of the object (ladder) and the effect of z-axis can be disregarded.

Thus, for static equilibrium, the following conditions need to be satisfied: The net force in the x-direction (∑Fx) is zero, the net force in the y-direction (∑Fy) is zero, and the net torque about the z-axis (∑τz) is also zero. This implies that the ladder neither slides along the ground nor falls away from the wall, and does not rotate about its center. Therefore, the correct set of quantities is option A) ∑Fx, ∑Fy, and ∑τz.

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Answer:

95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Step-by-step explanation:

We are given that a sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

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

where, [tex]\bar X[/tex] = sample mean meat consumption = 3 pounds

             s = population standard deviation = 1.4 pounds

            n = sample of people = 2092

            [tex]\mu[/tex] = population mean consumption of meat

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                  of significance are -1.96 & 1.96}  

P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P( [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]

                 = [ [tex]3-1.96 \times {\frac{1.4}{\sqrt{2092} } }[/tex] , [tex]3+1.96 \times {\frac{1.4}{\sqrt{2092} } }[/tex] ]

                 = [2.9 pounds , 3.1 pounds]

Therefore, 95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Answer:

We are 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

Step-by-step explanation:

Let p = proportion of people who believe that the reality TV shows are either "totally made up" or "mostly distorted".

A random sample of n = 1006 adults are selected. Of these adults 86% believes that the reality TV shows are either "totally made up" or "mostly distorted".

The (1 - α)% confidence interval for the population proportion is:

[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

Compute the critical value of z for 95% confidence level as follows:

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

*Use a z-table.

Compute the 95% confidence interval for the population proportion as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

    [tex]=0.86\pm 1.96\times \sqrt{\frac{0.86(1-0.86)}{1006}}\\=0.86\pm 0.0214\\=(0.8386, 0.8814)\\\approx (0.839, 0.881)[/tex]

The 95% confidence interval for the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is (0.839, 0.881).

This confidence interval implies that:

We are 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

95% confident that the proportion of all adults who believe that the shows are either  made up" or "mostly distorted" is within 83.9% and 88.1%.

Given that,

An article included the following statement: "Few people believe there's much reality in reality TV:

Total of 86% said the shows are either 'totally made up' or 'mostly distorted.'" This statement was based on a survey of 1006 randomly selected adults

We have to determine,

Compute a bound on the error (based on 95% confidence) of estimation for the reported proportion of 0.86

According to the question,

Let, p = proportion of people who believe that the reality TV shows are either "totally made up" or "mostly distorted".

A random sample of n = 1006 adults are selected.

These adults 86% believes that the reality TV shows are either "totally made up" or "mostly distorted".

The (1 - α)% confidence interval for the population proportion is:

[tex]C.I. = p\ \pm z_\frac{\alpha}{2} \times \sqrt{\frac{p(1-p)}{n} }[/tex]

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

[tex]z_\frac{\alpha}{2} = z_\frac{0.05}{2} = z_0._0_2_5 = 1.96[/tex]

To calculate the critical value of z for 95% confidence level as follows:

Compute the 95% confidence interval for the population proportion as follows:

[tex]C.I. = p \pm z_\frac{\alpha}{2} \sqrt{\frac{p(1-p)}{n} } \\\\C.I. = 0.86\pm 1.96\sqrt{\frac{0.86(1-0.86)}{1006} }\\\\C.I. = 0.86 \pm 0.0214\\\\C.I. = (0.8366, 0.8814)\\\\C.I = (0.839, 0.88)[/tex]

The 95% confidence interval for the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is (0.839, 0.881).

Hence, 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

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