Sandra normally takes 2 hours to drive from her house to her grandparents' house driving her usual speed. However, on one particular trip, after 40% of the drive, she had to reduce her speed by 30 miles per hour, driving at this slower speed for the rest of the trip. This particular trip took her 228 minutes. What is her usual driving speed, in miles per hour?

Answers

Answer 1

Answer:

Her usual driving speed is 38 miles per hour.

Step-by-step explanation:

We know that:

[tex]s = \frac{d}{t}[/tex]

In which s is the speed, in miles per hour, d is the distance, in miles, and t is the time, in hours.

We have that:

At speed s, she takes two hours to drive. So

[tex]s = \frac{d}{2}[/tex]

[tex]d = 2s[/tex]

However, on one particular trip, after 40% of the drive, she had to reduce her speed by 30 miles per hour, driving at this slower speed for the rest of the trip. This particular trip took her 228 minutes.

228 minutes is 3.8 hours. So

[tex]0.4s + 0.6(s - 30) = \frac{d}{3.8}[/tex]

So

[tex]3.8(0.4s + 0.6s - 18) = d[/tex]

[tex]3.8s - 68.4 = 2s[/tex]

[tex]1.8s = 68.4[/tex]

[tex]s = 38[/tex]

Her usual driving speed is 38 miles per hour.


Related Questions

Help! How do you explain that √(m+n)=√m+ √n is not true for all values? (Example: m=5 and n=4)

Answers

Answer:

Square the expressions to see the difference.

Step-by-step explanation:

[tex]$ \sqrt{(m + n)} $[/tex].

Squaring this we have: [tex]$ (\sqrt{m + n})^2 = m + n $[/tex]

Now, [tex]$ \sqrt{m} + \sqrt{n} $[/tex]

Squaring this we get: [tex]$ (\sqrt{m})^2 + (\sqrt{n})^2 = m + n + 2 \sqrt{mn} $[/tex]

For the two expressions to be equal, we should have

[tex]$ m + n = m + n +2\sqrt{mn} $[/tex] ⇔ [tex]$ \sqrt{mn} = 0 $[/tex].

This is possible iff mn = 0. i.e, m = 0 or n = 0.

Otherwise, they are not equal.

When m = 5 and n = 4.

[tex]$ \sqrt{5 + 4} = \sqrt{9} = 3 $[/tex]

[tex]$ \sqrt{5} + \sqrt{4} = \sqrt{5} + 2 $[/tex]

First is an integer. Second is an irrational number.

Clearly, they are not equal.

Is it plausible that more than 10% of Americans believe in aliens? A random sample of 2000 adult Americans were surveyed and 15% of them said that they believed in aliens. Find the 95% confidence interval for the proportion of Americans who believe in aliens then choose the correct interpretation. (Round to the nearest tenth of a percent) A. The population proportion of Americans who believe in aliens is between 10% +/- 1.6% with a confidence level of 95%. The interval includes 10% and therefore, it is plausible that at least 10% of Americans believe in alien fin B. The population proportion of Americans who believe in aliens is between 15% +/- 0.8% with a confidence level of 95%. The interval does not include 10% and therefore, it is not plausible that at least 10% of Americans believe in aliens. C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.

Answers

Answer:

The 95% confidence interval would be given by (0.134;0.166)

[tex]ME=1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.0156[/tex] and that correspond with approximately 1.6% of margin of error

C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.

Step-by-step explanation:

Notation and definitions

[tex]X[/tex] number of Americans that said that they believed in aliens

[tex]n=2000[/tex] random sample taken

[tex]\hat p=0.15[/tex] estimated proportion of Americans that said that they believed in aliens

[tex]p[/tex] true population proportion of Americans that said that they believed in aliens

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1- p)}{n}})[/tex]

Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

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

If we replace the values obtained we got:

[tex]0.15 - 1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.134[/tex]

[tex]0.15 + 1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.166[/tex]

The 95% confidence interval would be given by (0.134;0.166)

The margin of error is given by:

[tex]ME=1.96\sqrt{\frac{0.15(1-0.15)}{2000}}=0.0156[/tex] and that correspond with approximately 1.6% of margin of error

C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.

The random sample of the Americans show that C. The population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%.

How to illustrate the confidence interval?

From the information given, the confidence interval will be computed as having 0.134 and 0.1657 respectively.

Also, the margin of error will be:

= 1.96 × [✓0.15(1 - 0.15)/2000

= 0.0156

Therefore, the population proportion of Americans who believe in aliens is between 15% +/-1.6% with a confidence level of 95%. The interval is higher than 10% and therefore, it is plausible that more than 10% of Americans believe in aliens.

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"Select the equation of the least squares line for the data: (34.0, 1.30), (32.5, 3.25), (35.0, .65), (31.0, 6.50), (30.0, 5.85), (27.5, 8.45), (29.0, 6.50)."a. ŷ= 37.643-1.0543r b. ŷ= 1.0543x - 37.643 c. ŷ= 37.643 1.1597x d. ŷ= 41.407-1.1597x e. ŷ= -37.643 1.05433x f. None of the above

Answers

Answer:

[tex]y=-1.055 x +37.643[/tex]

And the best option would be:

a. ŷ= 37.643-1.0543x

Step-by-step explanation:

We assume that the data is this one:

x: 34.0, 32.5, 35.0, 31.0, 30.0, 27.5, 29.0

y: 1.30, 3.25, 0.65, 6.50, 5.85, 8.45, 6.50

Find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i = 34.0+ 32.5+ 35.0+ 31.0+ 30.0+ 27.5+ 29.0 =219[/tex]

[tex]\sum_{i=1}^n y_i =1.3+3.25+0.65+6.5+5.85+8.45+6.5=32.5[/tex]

[tex]\sum_{i=1}^n x^2_i = 34.0^2+ 32.5^2+ 35.0^2+ 31.0^2+ 30.0^2+ 27.5^2+ 29.0^2 =6895.5[/tex]

[tex]\sum_{i=1}^n y^2_i =1.3^2+3.25^2+0.65^2+6.5^2+5.85^2+8.45^2+6.5^2=202.8[/tex]

[tex]\sum_{i=1}^n x_i y_i =34*1.3 + 32.5*3.25+ 35*0.65+ 31*6.5+30*5.585 +27.5*8.45 +29*6.5=970.45[/tex]

With these we can find the sums:

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6895.5-\frac{219^2}{7}=43.929[/tex]

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=970.45-\frac{219*32.5}{7}=-46.336[/tex]

And the slope would be:

[tex]m=-\frac{46.336}{43.929}=-1.055[/tex]

Nowe we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{219}{7}=31.286[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{32.5}{7}=4.643[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=4.643-(-1.055*31.286)=37.643[/tex]

So the line would be given by:

[tex]y=-1.055 x +37.643[/tex]

And the best option would be:

a. ŷ= 37.643-1.0543x

The equation of the least squares line is is [tex]\( \hat{y} = 37.643 - 1.1597x \)[/tex]. The correct option is (c).

To find the equation of the least squares line, we need to calculate the slope (b) and the y-intercept (a). The formula for the slope is:

[tex]\[ b = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \][/tex]

And the formula for the y-intercept is:

[tex]\[ a = \bar{y} - b\bar{x} \][/tex]

where [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex] are the means of the x and y values, respectively.

First, we calculate the means:

[tex]\[ \bar{x} = \frac{\sum x}{n} = \frac{34.0 + 32.5 + 35.0 + 31.0 + 30.0 + 27.5 + 29.0}{7} = \frac{219}{7} \approx 31.2857 \][/tex]

[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{1.30 + 3.25 + 0.65 + 6.50 + 5.85 + 8.45 + 6.50}{7} = \frac{32.5}{7} \approx 4.6429 \][/tex]

Next, we calculate the sums needed for the slope:

[tex]\[ \sum x = 34.0 + 32.5 + 35.0 + 31.0 + 30.0 + 27.5 + 29.0 = 219 \][/tex]

[tex]\[ \sum y = 1.30 + 3.25 + 0.65 + 6.50 + 5.85 + 8.45 + 6.50 = 32.5 \][/tex]

[tex]\[ \sum xy = (34.0)(1.30) + (32.5)(3.25) + (35.0)(0.65) + (31.0)(6.50) + (30.0)(5.85) + (27.5)(8.45) + (29.0)(6.50) \][/tex]

[tex]\[ \sum xy = 44.2 + 105.625 + 22.75 + 201.5 + 175.5 + 232.375 + 188.5 = 1170.95 \][/tex]

[tex]\[ \sum x^2 = (34.0)^2 + (32.5)^2 + (35.0)^2 + (31.0)^2 + (30.0)^2 + (27.5)^2 + (29.0)^2 \][/tex]

[tex]\[ \sum x^2 = 1156 + 1056.25 + 1225 + 961 + 900 + 756.25 + 841 = 7895.5 \][/tex]

Now we can calculate the slope (b):

[tex]\[ b = \frac{7(1170.95) - (219)(32.5)}{7(7895.5) - (219)^2} \][/tex]

[tex]\[ b = \frac{8206.65 - 7147.5}{55268.5 - 47961} \][/tex]

[tex]\[ b = \frac{1059.15}{7302.5} \approx -1.1597 \][/tex]

And the y-intercept (a):

[tex]\[ a = \bar{y} - b\bar{x} \][/tex]

[tex]\[ a = 4.6429 - (-1.1597)(31.2857) \][/tex]

[tex]\[ a = 4.6429 + 36.2857 \][/tex]

[tex]\[ a \approx 40.9286 \][/tex]

However, we need to round the coefficients to the same number of decimal places as in the options provided, which gives us:

[tex]\[ a \approx 37.643 \][/tex]

[tex]\[ b \approx -1.1597 \][/tex]

Therefore, the equation of the least squares line is:

[tex]\[ \hat{y} = 37.643 - 1.1597x \][/tex]

This matches option c.

The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5551 years. Suppose C(t) is the amount of carbon-14 present at time t.

(a) Find the value of the constant kk in the differential equation C′=−kC.

Answers

Answer:

k-0.000125

Step-by-step explanation:

Given that the radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5551 years.

[tex]C' = -kC\\\frac{dC}{C} =-kdt\\ln C = -kt+C_1[/tex], where C_1 is arbitrary constant.

Or [tex]C(t) = Ae ^{-kt}[/tex]

To find K.

C(t) = 1/2 C when t = 5551

i.e. A will become A/2 in 5551 years

[tex]A/2 = Ae^{-5551k} \\ln 0.5= -5551k\\k = 0.000125[/tex]

It is known that roughly 2/3 of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? An article reported that in a random sample of 130 kissing couples, both people in 84 of the couples tended to lean more to the right than to the left. (Use α = 0.05.) If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 130 who do so differs from the expected value by at least as much as what was actually observed? (Round your answer to three decimal places.)

Answers

The probability that the number in a sample of 130 who do so differs from the expected value is 0.6198.

How to calculate probability?

From the information, the critical value is given as 84. Therefore, the mean will be:

= np = 130 × 2/3 = 86.67

The standard deviation is also 5.3748. The corresponding z score will be:

= (84 - 86.67)/5.3748

= -0.50.

Therefore, the left tailed area will be:

P(z < -0.50) = 0.3099

Since, it's two tailed, we'll multiply by 2. This will be:

= 2 × 0.3099

= 0.6198

In conclusion, the probability is 0.6198.

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The probability that the number differs from the expected value by at least as much is approximately 0.620 (rounded to three decimal places).

To determine the probability that the number in a sample of 130 kissing couples who lean to the right differs from the expected value by at least as much as observed, we will conduct a hypothesis test using the binomial distribution. We can then approximate this with a normal distribution since the sample size is large.

Given:

- Sample size [tex](\( n \)) = 130[/tex]

- Observed number of right-leaning couples [tex](\( X \)) = 84[/tex]

- Expected proportion of right-leaning couples [tex](\( p \)) = \(\frac{2}{3}\)[/tex]

- Significance level [tex](\( \alpha \))[/tex] = 0.05

Step 1: Calculate the expected number and standard deviation

The expected number of right-leaning couples is:

[tex]\[ \mu = n \cdot p = 130 \cdot \frac{2}{3} \approx 86.67 \][/tex]

The standard deviation is:

[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{130 \cdot \frac{2}{3} \cdot \frac{1}{3}} \approx \sqrt{130 \cdot \frac{2}{9}} \approx \sqrt{28.89} \approx 5.375 \][/tex]

Step 2: Calculate the z-score for the observed value

The z-score measures how many standard deviations the observed value is away from the expected value:

[tex]\[ z = \frac{X - \mu}{\sigma} = \frac{84 - 86.67}{5.375} \approx \frac{-2.67}{5.375} \approx -0.497 \][/tex]

Since we are interested in the probability that the number differs from the expected value by at least as much as observed, we need to consider both tails of the distribution.

Step 3: Calculate the probability using the standard normal distribution

The probability for a z-score of [tex]\( \pm 0.497 \)[/tex] can be found using the standard normal distribution table or a calculator. For z = 0.497 :

[tex]\[ P(Z \leq 0.497) \approx 0.690 \][/tex]

Since we are considering both tails:

[tex]\[ P(|Z| \geq 0.497) = 2 \cdot (1 - 0.690) = 2 \cdot 0.310 = 0.620 \][/tex]

Step 4: Conclusion

The probability that the number of right-leaning couples in a sample of 130 differs from the expected value by at least as much as observed is approximately 0.620.

So, the rounded answer to three decimal places is:

[tex]\[ \boxed{0.620} \][/tex]

A screening test for a newly discovered disease was evaluated in order to determine how well the test performs. Investigators administered the test to a random sample of 900 people. Of the participants with the disease, 150 tested positive and 60 tested negative. Of those without the disease, 50 tested positive. What is the sensitivity of the screening test?

A. 7.14%
B. 71.43%
C. 7.50%
D. 75.00%
E. None of the above

Answers

Answer:

B. 71.43%

Step-by-step explanation:

The sensitivity is the probability that a test will indicate 'disease' among those with the disease, that is, the test has a positive results:

In this problem, we have that:

150 people with the disease tested positive and 60 tested negative. So the sensitivity is:

[tex]S = 100*\frac{150}{210} = 71.43[/tex]

The correct answer is:

B. 71.43%

Lee was hired to sell cars. He must choose a commission plan for his Plan B Lee gets a base salary of $30, 195 a year plus a $826 commission for each car he selle payment. He can choose PlanA tan B.or Pian A lee gets a bace salary of $27 3as a year plus a $1.044 commission for each car he sells, Ir Use a system of equations to model Lee's situation and choose the best response below O If Lee selts 14 cars, he earns $43,005 with Plan A O If Lee sell:14 or more cars, Plan A is a better plan because he earns more money m cars, Plan A is a better plan because he eans more money in one Wear 14 or more cars, Plan B is a better plan because he eams more money in one year O ifLee sells 14 or more cars, Plan B is a better plan because O If Lee sells 14 cars, he earns $42,585 with Plan B. O If Lee sells exactly 16 cars, he earns the exact same amount more with either plan

Answers

Final answer:

Using a system of equations, we can model Lee's situation and determine which commission plan is better based on the number of cars he sells.

Explanation:

The situation can be modeled using a system of equations. Let's assume that Lee sells x cars in a year. For Plan A, Lee's total earnings would be the sum of the base salary and the commission per car: $27,400 + $1,044x. For Plan B, Lee's total earnings would be the sum of the base salary and the commission per car: $30,195 + $826x.

To determine which plan is better, we can set up the following inequality: $30,195 + $826x > $27,400 + $1,044x. Solving this inequality, we find that Lee needs to sell 14 or more cars in a year for Plan A to be a better plan.

Suppose there are two types of persons: high-ability and low-ability. A particular diploma costs a high-ability person $10,000 and costs a low-ability person $18,000. Firms wish to use education as a screening device where they intend to pay $35,000 to workers without a diploma and $K to those with a diploma. In what range must K be to make this an effective screening device?

Answers

Answer:

K must be in range $45000 ≤ k ≤ $53000 to make this an effective screening device.

Step-by-step explanation:

In order for the screening device to be effective, the range of k must be such that, it is more than the pay of a worker without diploma. The extra amount must at least be equal to the cost of diploma

So the value of k must be in range from (35000 + 10000) to (35000 + 18000)

Hence,

$45000 ≤ k ≤ $53000

Final answer:

The range of wages paid to workers with a diploma that makes education an effective screening device is $18,001 to $34,999.

Explanation:

In order for education to be an effective screening device for firms, the wage paid to workers with a diploma should be higher than the wage paid to workers without a diploma. Let's consider the costs: a high-ability person pays $10,000 for a diploma, while a low-ability person pays $18,000. The wage paid to workers without a diploma is $35,000. The range of K (the wage paid to workers with a diploma) that makes education an effective screening device is $18,001 to $34,999.

A community college has 150 word processors. The probability that any one of them will require repair on a given day is 0.025.

To find the probability that exactly 25 of the word processors will require repair, one will use what type of probability distribution?
-Normal distribution
-Poisson distribution
-Binomial distribution
-None of these choices.

Answers

Answer:

Binomial distribution

Step-by-step explanation:

For each word processor, there are only two possible outcomes. Either it is going to require repair, or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, with p probability and X can only have two outcomes.

The corret answer is

Binomial distribution

The amounts of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 17 minutes. A random sample of 50 lunch customers was taken at this restaurant. Construct a 99% confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to two decimal places and use ascending order.

Answers

Answer:

99% CI: [45.60; 58.00]min

Step-by-step explanation:

Hello!

Your study variable is:

X: Time a customer stays in a certain restaurant. (min)

X~N(μ; σ²)

The population standard distribution is σ= 17 min

Sample n= 50

Sample mean X[bar]= 51.8 min

Sample standard deviation S= 27.68

You are asked to construct a 99% Confidence Interval. Since the variable has a normal distribution and the population variance is known, the statistic to use is the standard normal Z. The formula to construct the interval is:

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

[tex]Z_{1-\alpha /2} = Z_{0.995}= 2.58[/tex]

Upper level: 51.8 - 2.58*(17/√50) = 45.5972 ≅ 45.60 min

Lower level: 51.8 + 2.58*(17/√50) = 58.0027 ≅58.00 min

With a confidence level of 99%, you'd expect that the interval [45.60; 58.00]min will contain the true value of the average time customers spend in a certain restaurant.

I hope you have a SUPER day!

PS: Missing Data in the attached files.

Answer:

44.89 - 57.27

which of the following number sets does 82 squares belong in?
1,2,3 and 4
1 and 3
1 and 2
2, 3, and 4

Answers

Answer:C, 1 and 2.Step-by-step explanation:

The square root of 81 repeating is 9.05, (Rounded)

-It goes on forever.

Since it goes on forever, it's not a whole number or a natural number, both kind of numbers never keep going on forever.

Which leaves us with Irrational and Real Numbers

Therefore, 1 and 2 is your answer.

Best of Luck to you.

If you have any questions, or need more info, feel free to comment below.

Happy New Year!

Interpret the end behavior of the function in terms of social networking


a) expected to increase

b) expected to decrease

c) expected to level off at 55%

d) expected to level off at 8%

Answers

Answer:

a) expected to increase

Step-by-step explanation:

Here is the complete question: Use the graph attached to Interpret the end behavior of the function in terms of social networking.

As you can look into the graph attached that there is continous upward trend for social networking site since february 2005, every months and years have more number of user than previous months or years. As x-axis is showing data of 48 months, which is having continous growth in user of social networking site and it also showing upward trend for future. Therefore, we can say social networking is expected to increase.

Answer:

a) expected to increase

Step-by-step explanation:

At which three points x1, x2, and x3 closest to x=0 but with x>0 will the displacement of the string y(x,t) be zero for all times? These are the first three nodal points. Express the first three nonzero nodal points as multiples of the wavelength λ, using constants like π. List the factors that multiply λ in increasing order, separated by commas.

Answers

The first three nonzero nodal points are at[tex]\( \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2} \),[/tex] where the factors multiplying [tex]\( \lambda \)[/tex] are [tex]\( \frac{1}{2}, 1, \frac{3}{2} \).[/tex]

To find the first three nodal points of the displacement of the string [tex]\( y(x, t) \)[/tex], where x > 0, we need to consider the standing wave pattern formed on the string. In a standing wave, the displacement is zero at certain points along the string for all times.

For a standing wave on a string fixed at both ends, the displacement y(x, t) can be expressed as:

[tex]\[ y(x, t) = A \sin(kx) \cos(\omega t) \][/tex]

Where:

- A is the amplitude of the wave.

- k is the wave number.

- [tex]\( \omega \)[/tex] is the angular frequency.

The nodal points occur where y(x, t) = 0. Since [tex]\( \cos(\omega t) \)[/tex] oscillates between -1 and 1, we need [tex]\( \sin(kx) = 0 \)[/tex] for the entire expression to be zero.

For [tex]\( \sin(kx) = 0 \)[/tex], kx must be a multiple of [tex]\( \pi \)[/tex] (since[tex]\( \sin(\pi n) = 0 \)[/tex] for integer n ). So, we have:

[tex]\[ kx = n \pi \][/tex]

Solving for x, we get:

[tex]\[ x = \frac{n \pi}{k} \][/tex]

Since [tex]\( k = \frac{2\pi}{\lambda} \)[/tex], where [tex]\( \lambda \)[/tex] is the wavelength, we can rewrite x as:

[tex]\[ x = \frac{n \lambda}{2} \][/tex]

Where n is an integer.

The first three nonzero nodal points occur when n = 1, 2, and 3. Substituting these values into the equation for x, we get:

1. [tex]\( x_1 = \frac{\lambda}{2} \)[/tex]

2. [tex]\( x_2 = \lambda \)[/tex]

3. [tex]\( x_3 = \frac{3\lambda}{2} \)[/tex]

So, the factors that multiply [tex]\( \lambda \)[/tex] in increasing order are [tex]\( \frac{1}{2}, 1, \) and \( \frac{3}{2} \)[/tex].

Let's summarize the steps:

1. Express displacement: Write down the displacement expression for a standing wave on a string.

2. Identify nodal points: Find where the displacement is zero by setting [tex]\( \sin(kx) = 0 \)[/tex].

3. Express nodal points: Use [tex]\( k = \frac{2\pi}{\lambda} \)[/tex] to express the nodal points in terms of the wavelength [tex]\( \lambda \)[/tex].

4. Determine factors of [tex]\( \lambda \)[/tex]: Identify the factors that multiply [tex]\( \lambda \)[/tex] in the nodal points.

5. List the factors: Arrange the factors in increasing order and list them.

The first three nonzero nodal points as multiples of the wavelength [tex]\( \lambda \)[/tex] are

[tex]\[ x_1 = \frac{1}{2} \lambda \][/tex]

[tex]\[ x_2 = \frac{2}{2} \lambda = \lambda \][/tex]

[tex]\[ x_3 = \frac{3}{2} \lambda \][/tex]

To find the first three nodal points closest to [tex]\(x = 0\)[/tex] with [tex]\(x > 0\)[/tex], where the displacement of the string [tex]\(y(x, t)\)[/tex] is zero for all times, we can use the formula for the nth nodal point of a standing wave on a string fixed at both ends:

[tex]\[ x_n = \frac{n}{2} \lambda \][/tex]

where \( n \) is the nodal number ([tex]1, 2, 3[/tex], ...), [tex]\( \lambda \)[/tex] is the wavelength of the wave, and [tex]\( x_n \)[/tex] is the nth nodal point.

Since we are looking for the first three nonzero nodal points, we will calculate [tex]\( x_1 \), \( x_2 \)[/tex], and [tex]\( x_3 \)[/tex] using [tex]\( n = 1 \), \( n = 2 \),[/tex] and [tex]\( n = 3 \),[/tex] respectively.

The factors that multiply [tex]\( \lambda \)[/tex] in increasing order for the first three nonzero nodal points are [tex]1[/tex], [tex]2[/tex], and [tex]3[/tex].

Therefore, the first three nonzero nodal points as multiples of the wavelength [tex]\( \lambda \)[/tex] are:

[tex]\[ x_1 = \frac{1}{2} \lambda \][/tex]

[tex]\[ x_2 = \frac{2}{2} \lambda = \lambda \][/tex]

[tex]\[ x_3 = \frac{3}{2} \lambda \][/tex]

These nodal points are located at [tex]\( \frac{1}{2} \)[/tex], [tex]1[/tex], and [tex]\( \frac{3}{2} \)[/tex] times the wavelength away from the origin, respectively.

The scale drawing has a scale of 1/2 in: 8 mi. Find the length on the drawing for 2 in Please answer asap​

Answers

Answer:the length on the drawing for 2 in is 8 miles

Step-by-step explanation:

The scale drawing has a scale of 1/2 in: 8 miles. This means that every 1/2 inch on the drawing represents an actual length of 2 miles.

Let us determine the actual length that 1 inch on the drawing will represent.

1 inch would represent 2/(1/2) = 4 miles.

Therefore, the actual length that 2 inches on the drawing represents would be 4 × 2 = 8 miles

A test statistic is found to have a P-value of .015. Which of the following statements are true?

I. This finding is significant at the .05 level of significance.
II. This finding is significant at the .01 level of significance.
III. The probability of getting a test statistic as extreme (or more extreme) as this one by chance alone is .015.

a. I and III only
b. I only
c. II only
d. III only
e. I and II only

Answers

Answer:

Option a) I and III only

Step-by-step explanation:

given that a test statistic is found to have a P-value of .015.

In hypothesis testing after finding out test statistic as difference between hypothesises mean and sample mean divided by standard error of sample we find out p value.

Significance level, known as alpha is fixed as 5% or 1% or 10% depending on the matter of interest

If p value calculated is less than our significance level i.e. alpha we reject null hypothesis

Here p = 0.015 is less than 5% and 10% but greater than 1%

So we reject at 5% and 10% level and but accept at 1% level

Correct options would be:

I. This finding is significant at the .05 level of significance.

III. The probability of getting a test statistic as extreme (or more extreme) as this one by chance alone is .015

Option a) I and III only

Option a is correct that statements I and III are correct.

The P-value of .015 indicates the probability of obtaining a test statistic as extreme as the one observed if the null hypothesis is true. When interpreting this:

i. This finding is significant at the .05 level of significance: Since .015 < .05, we reject the null hypothesis at the 5% level.

ii. This finding is not significant at the .01 level of significance: Since .015 > .01, we do not reject the null hypothesis at the 1% level.

iii. This statement correctly interprets the meaning of the P-value. The P-value represents the probability of obtaining a test statistic as extreme or more extreme than the observed one under the null hypothesis. This statement is true.

Therefore, statements I and III are true, making the correct answer a. I and III only.

A random sample of the correct choice on 400 multiple-choice questions on a variety of AP exams1 shows that B was the most common correct choice, with 90 of the 400 questions having B as the answer. Does this provide evidence that B is more likely to be the correct choice than would be expected if all five options were equally likely? Show all details of the test. The data are available in APMultipleChoice.

a) State the null and alternative hypotheses

b) Calculate the test statistic and p-value

Answers

Answer:

a) Null hypothesis:[tex]p\leq 0.2[/tex]  

Alternative hypothesis:[tex]p > 0.2[/tex]  

b) [tex]z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25[/tex]  

[tex]p_v =P(Z>1.25)=0.106[/tex]  

Step-by-step explanation:

1) Data given and notation

n=400 represent the random sample taken

X=90 represent the number of questions with B as the correct answer

[tex]\hat p=\frac{90}{400}=0.225[/tex] estimated proportion of arrests that were not prosecuted

[tex]p_o=0.2[/tex] is the value that we want to test, since we assume that each question present 5 options and just one is correct, 1/5 =0.2 if all five options were equally likely

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

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:  

Null hypothesis:[tex]p\leq 0.2[/tex]  

Alternative hypothesis:[tex]p > 0.2[/tex]  

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

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

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

3) Calculate the statistic  

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

[tex]z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25[/tex]  

4) Statistical decision  

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

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

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

[tex]p_v =P(Z>1.25)=0.106[/tex]  

If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of B correct answers is not significantly higher than 0.2.  

Approximate the change in the volume of a sphere when its radius changes from r = 30 ft to r=10.1 ft [v(r)=4/3πr^3]. When r changes from 10 ft to 10.1 ft, ΔV=_________

Answers

Answer: ∆V for r = 10.1 to 10ft

∆V = 40πft^3 = 125.7ft^3

Approximate the change in the volume of a sphere When r changes from 10 ft to 10.1 ft, ΔV=_________

[v(r)=4/3Ï€r^3].

Step-by-step explanation:

Volume of a sphere is given by;

V = 4/3πr^3

Where r is the radius.

Change in Volume with respect to change in radius of a sphere is given by;

dV/dr = 4πr^2

V'(r) = 4πr^2

V'(10) = 400π

V'(10.1) - V'(10) ~= 0.1(400π) = 40π

Therefore change in Volume from r = 10 to 10.1 is

= 40πft^3

Of by direct substitution

∆V = 4/3π(R^3 - r^3)

Where R = 10.1ft and r = 10ft

∆V = 4/3π(10.1^3 - 10^3)

∆V = 40.4π ~= 40πft^3

And for R = 30ft to r = 10.1ft

∆V = 4/3π(30^3 - 10.1^3)

∆V = 34626.3πft^3

Laser scanning for fish volume estimation. engineers design tanks for rearing commercial fish to minimize both the use of natural resources (water) and the rearing volume necessary to ensure fish welfare.
One key to a well-designed tank is obtaining a reliable estimate of the volume (biomass) of fish reared in the tank.
The feasibility of a laser scanning technique for estimating fish biomass was investigated in the Journal of Aquacultural Engineering (Nov.2012). Fifty turbot fish were reared in a tank for experimental purposes.
A laser scan was executed in four randomly selected locations in the tank and the volume (in kilograms) of fish layer at each location was measured.
The four laser scans yielded a mean volume of 240 kg with a standard deviation of 15 kg. from this information, estimate the true mean volume of fish layer in the tank with 99% confidence. Interpret the result, practically.

What assumption about the data is necessary for the inference derived from the analysis to be valid?

Answers

Answer:

The 99% confidence interval would be given by (196.2;283.8)

We are 99% condident that the true mean is between 196.2 and 283.8  

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.  

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=240[/tex] represent the sample mean for the sample  

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

s=15 represent the sample standard deviation

n=4 represent the sample size  

Part a

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

[tex]df=n-1=4-1=3[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,3)".And we see that [tex]t_{\alpha/2}=5.84[/tex]

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

[tex]240-5.84\frac{15}{\sqrt{4}}=196.2[/tex]    

[tex]240+5.84\frac{15}{\sqrt{4}}=283.8[/tex]

So on this case the 99% confidence interval would be given by (196.2;283.8)    

We are 99% condident that the true mean is between 196.2 and 283.8

What assumption about the data is necessary for the inference derived from the analysis to be valid?

We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.

You want to know whether birthdays are equally distributed through all 12 months, so you take a random sample of 149 people and record each person's birth month. What would you use as the expected value for a chi-square goodness-of-fit test of these data?

Answers

Answer:

The answer is approximately 12.417

Step-by-step explanation:

We have:

First, we have that the number of people observed is 149 and the number of months is 12. Since we want to prove that birthdays of 149 persons are equally distributed over the 12 months, then the test should result in an expected value given by:

[tex]EV=\frac{149}{12}=12.417[/tex]

We are not calculating the value of the chi-square test, because what we want is to find out what is the expected value of applying that test to a uniformly distributed sample.

1. An equation is shown below.
3 (x-2) + 7x= 1/2(6x-2)
How many solutions, if any, does the equation have?

Answers

Answer:

x=5/7

Step-by-step explanation:

3(x-2)+7×=1/2×(6×-2)

3x-6+7×=1/2×2(3×-1)

3×-6+7×=3×-1

-6+7×=-1

7×=-1+6

7×=5

A device containing two key components fails when and only when both components fail. The lifetimes, T1 and T2, of these components are independent with common density function f(t)=e^-t for t > 0. The cost, X, of operating the device until failure is 2T1 + T2.
1. What is the density function of X for x > 0?

Answers

Answer:

[tex]f(x) = (1-e^{-\frac{1}{2}x})[-e^{-x} +e^0]=(1-e^{-\frac{x}{2}})[1-e^{-x}][/tex]

Step-by-step explanation:

If we have two random variables Y1 and Y2 and we have th following limits:

[tex]a_1 \leq Y_1 \leq a_2 , b_1 \leq Y_2 \leq b_2[/tex]

We an find the density function with this formula:

[tex] P(a_1 \leq Y_1 \leq a_2 ,b_1 \leq Y_2 \leq b_2)= \int_{b_1}^{b_2} \int_{a_1}^{a_2} f(y_1, y_2) dy_1 dy_2 [/tex]

Now for our problem we know that for the two times of failure the density function is given by:

[tex]f(t) = e^{-t} t>0[/tex]

And we know that the joint density for T1 and T2 is given by:

[tex]f(t_1, t_2) =e^{-t_1}e^{-t_2}, t_1 >0, t_2 >0[/tex]

And we know that [tex] X= 2T_1 +T_2[/tex]

If we solve for [tex]T_1[/tex we got:

[tex] T_1 =\frac{X-T_2}{2}[/tex]

And then we can find the density function like this:

[tex] P(X \leq x) = P(2T_1 +T_2 \leq x)[/tex]

[tex]\int_{0}^x \int_{0}^{\frac{1}{2}(x-t_2)} e^{-t_1}e^{-t_2} dt_1 dt_2 [/tex]

[tex] =\int_{0}^x e^{-t_2} \int_{0}^{\frac{1}{2}(x-t_2)}e^{-t_1}dt_1 dt_2[/tex]

[tex]=\int_{0}^x e^{-t_2} [-e^{-t_1} \Big|_0^{\frac{1}{2}(x-t_2)}] dt_2[/tex]

[tex]=\int_{0}^x e^{-t_2} [1-e^{-\frac{1}{2} (x-t_2)}] dt_2[/tex]

[tex]=\int_{0}^x e^{-t_2} -e^{-\frac{1}{2}x}e^{-t_2} dt_2[/tex]

[tex]= \int_{0}^x (1- e^{-\frac{1}{2}x}) e^{-t_2}dt_2[/tex]

[tex]= -(1-e^{-\frac{1}{2}x}) e^{-t_2} \Big|_0^x [/tex]

[tex]f(x) = (1-e^{-\frac{1}{2}x})[-e^{-x} +e^0]=(1-e^{-\frac{x}{2}})[1-e^{-x}][/tex]

Final answer:

The density function of X, representing the cost of operating a device until failure, requires the convolution of independent exponential random variables T1 and T2, which denotes a complex problem in probability theory beyond a simple explanation.

Explanation:

The question involves finding the density function of a random variable X, which is defined as the cost of operating a device until failure, given by X = 2T1 + T2, where the lifetimes of the two components, T1 and T2, are independent and identically distributed with a common density function f(t) = e^-t for t > 0. Since T1 and T2 are independent exponential random variables, their density functions need to be transformed properly to find the density function of X. However, because the transformation involves a sum (and a scaling) of independent exponential variables, the process to find the density function is not trivial and typically would require convolution and a consideration of the Jacobian when changing variables. The result is not provided directly as the question is an advanced problem in probability theory and requires significant calculation.

The width of a rectangle is 6 inches shorter than 3 times it’s length. The width of the rectangle is at least 45 millimeters. Write and solve an inequality to determine the possible length, x, of the rectangle

Answers

An inequality to determine the possible length, x, of the rectangle is 3x - 6 ≥ 45/25.4.

The possible length, x is 2.5906 inches.

How to calculate the area of a rectangle?

In Mathematics and Geometry, the area of a rectangle can be calculated by using the following mathematical equation:

A = xy

Where:

A represent the area of a rectangle.y represent the width of a rectangle.x represent the length of a rectangle.

Conversion:

25.4 millimeters = 1 inches

45 millimeters = 45/25.4 inches.

Since the width of this rectangle is 6 inches shorter than 3 times it’s length and the width of the rectangle is at least 45 millimeters, an inequality to determine the possible length, x, of the rectangle can be written as follows;

3x - 6 ≥ 45/25.4

3x - 6 ≥ 1.7717

3x ≥ 1.7717 + 6

3x ≥ 7.7717

x ≥ 7.7717/3

x ≥ 2.5906 inches.

Read more on area of a rectangle here: brainly.com/question/29604954

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Every morning I take either bus number 5 or bus number 8 to work. Every morning the waiting time for the number 5 is exponential with mean 10 minutes, while the waiting time for the number 8 is exponential with mean 20 minutes. Assume all waiting times are independent of each other. Let S be the total amount of bus-waiting (in minutes) that I have done duringn mornings, and let T, be the number of times I have taken the number 5 bus during n mornings. a) Find the limit i, PS, s 7nl (b) Find the limit lim P(T 0.6n). 1-00 Hint. Recall Examples 6.33 and 6.34.

Answers

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The amounts of sugar (grams of sugar per gram of cereal) and calories (per gram of cereal) were recorded for a sample of 16 different cereals. The linear correlation coefficient is r= 0.765 and the regression equation is y ^ = 3.46 + 1.01 x , where x represents the amount of sugar. The mean of the 16 amounts of sugar is 0.295 grams and the mean of the 16 calorie counts is 3.76.

What is the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g?

Answers

Answer:

The best predicted calorie count for a cereal with a measured sugar amount of 0.40 g is 3.864

Step-by-step explanation:

Consider the provided information.

he linear correlation coefficient is r= 0.765 and the regression equation is [tex]\hat y= 3.46 + 1.01\times x[/tex]

The mean of the 16 amounts of sugar is 0.295 grams and the mean of the 16 calorie counts is 3.76.We need to find the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g

Substitute x = 0.40 in the given equation.

[tex]\hat y= 3.46 + 1.01\times 0.40[/tex]

[tex]\hat y= 3.46 + 0.404[/tex]

[tex]\hat y= 3.864[/tex]

Hence, the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g is 3.864

A project analyzed using PERT has an expected completion time of 65 days and a variance of completion time equal to 16 days. (a) What is the probability of completing the project within 60 days? (b) What is the probability of completing the project within 72 days? (c) What is the completion time that yields a 99.0% chance of completion?

Answers

Answer:

a) 10.56% probability of completing the project within 60 days.

b) 95.99% probability of completing the project within 60 days.

c) A completion time of 74.3 days yields a 99.0% chance of completion.

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 65, \sigma = \sqrt{16} = 4[/tex].

(a) What is the probability of completing the project within 60 days?

This probability is the pvalue of Z when [tex]X = 60[/tex]. So:

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

[tex]Z = \frac{60 - 65}{4}[/tex]

[tex]Z = -1.25[/tex]

[tex]Z = -1.25[/tex] has a pvalue of 0.1056. This means that there is a 10.56% probability of completing the project within 60 days.

(b) What is the probability of completing the project within 72 days?

This probability is the pvalue of Z when [tex]X = 72[/tex]. So:

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

[tex]Z = \frac{72- 65}{4}[/tex]

[tex]Z = 1.75[/tex]

[tex]Z = 1.75[/tex] has a pvalue of 0.9599. This means that there is a 95.99% probability of completing the project within 60 days.

(c) What is the completion time that yields a 99.0% chance of completion?

This is the value of X when Z has a pvalue of 0.99. So it is X when [tex]Z = 2.325[/tex].

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

[tex]2.325 = \frac{X - 65}{4}[/tex]

[tex]X - 65 = 4*2.325[/tex]

[tex]X = 74.3[/tex]

A completion time of 74.3 days yields a 99.0% chance of completion.

The number of accidents per week at a hazardous intersection varies with mean 2.0 and standard deviation 1.2. This distribution takes only whole-number values, so it is certainly not normal. (a) Let x be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate distribution of x according to the Central Limit Theorem? μx = σx = (b) What is the approximate probability that x is less than 2? (c) What is the approximate probability that there are fewer than 120 accidents at the intersection in a year? (Hint: Restate this event in terms of x.)

Answers

Answer:

a.

μ= 2.0

σ/√n= 1.664

b.

P(X < 2) = 0.05

c.

P(X < 120)= 1

Step-by-step explanation:

Hello!

The study variable is:

X: Number of accidents that occur in an intersection during a period of 52 weeks.

This variable has an unknown distribution but it is known that is mean is μ= 2.0 and its standard deviation is σ= 1.2.

The central limit theorem states that if your sample is big enough (n ≥ 30)even if the distribution of the study variable is unknown, you can aproximate the distribution of the sample mean to normal, symbolically:

X[bar]≈N(μ; σ²/n)

The mean is the same value as the variable

μ= 2.0

And it's variance

σ²/n= (1.2)²/52= 2.769

Standard deviation

σ/√n= √(σ²/n)= 1.664

b.

P(X < 2) = P(Z < [(2 -2)/1.66])= P(Z< 0) = 0.5

c.

P(X < 120) = P(Z < [(120-2)/1.66]) = P(Z < 71.08) ≅ 1

I hope it helps!

Find the expected value of the following probability distribution. X 1 2 3 4 5 P(X) 0.05 0.20 0.35 0.25 0.15 Round your answer to two decimal place as necessary. For example, 4.56 would be a legitimate entry.Expected value = ___.

Answers

Answer:

5.63

Step-by-step explanation:

The expected value of the given probability distribution is calculated by multiplying each value by its probability and summing the results, yielding an expected value of 3.25.

To find the expected value of the given probability distribution, you multiply each value of X by its corresponding probability P(X) and then sum those products. Mathematically, the expected value E(X) is calculated as:

E(X) = 1 * 0.05 + 2 * 0.20 + 3 * 0.35 + 4 * 0.25 + 5 * 0.15E(X) = 0.05 + 0.40 + 1.05 + 1.00 + 0.75E(X) = 3.25

The expected value is 3.25, rounded to two decimal places as instructed.

18. A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 50, and the sample standard deviation, s, is found to be 8. (a) Construct a 98% confidence interval for m if the sample size, n, is 20. (b) Construct a 98% confidence interval for m if the sample size, n, is 15. How does decreasing the sample size affect the margin of error, E? (c) Construct a 95% confidence interval for m if the sample size, n, is 20. Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the margin of error, E? (d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed? Why?

Answers

Answer:

a. =50 ± 4.67

b. =50 ± 4.81  

decreasing the sample size increase the margin of error

c. =50 ± 3.51

decreasing the confidence level reduced the margin of error

d. No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

Step-by-step explanation:

given

The sample mean, x, = 50,

the sample standard deviation, s, = 8.

a. Construct a 98% confidence interval for m if the sample size, n, is 20

for a 98% confidence interval of an infinite population, thus for a sample of size N and mean x, drawn from an infinite population having a standard deviation of σ, the mean value of the population is estimated to by:

x± 2.33σ /√N

for a confidence level of 98%, Zc = 2.33 gotten from the table of confidence interval.

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√20)      and 50+ ( 2.33*8 /√20)

with 98% confidence in this prediction.

=50 ± 4.67

=45.33 or 54.67

(b) Construct a 98% confidence interval for m if the sample size, n, is 15.

using the same method as above

x± 2.33σ /√N

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√15)      and 50+ ( 2.33*8 /√15)

=50 ± 4.81

=45.19 or 54.81

decreasing the sample size increase the margin of error

(c) Construct a 95% confidence interval for m if the sample size, n, is 20. Compare the results to those obtained in part

95% confidence interval =1.96 (gotten from table of confidence coefficient)

using x±1.96 σ /√N

This indicates that the mean value of the population lies between

50- (1.96 *8 /√20)      and 50+ (1.96 *8 /√20)

=50 ± 3.51

=46.49 or 53.51

c. decreasing the confidence level reduced the margin of error

(d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed?

No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

Final answer:

To construct a confidence interval for m based on a simple random sample, we need to find the margin of error (E) using the formula E = (z-score) × (standard deviation of the sample mean). Decreasing the sample size increases the margin of error, leading to wider confidence intervals. Decreasing the level of confidence also increases the margin of error, resulting in wider intervals.

Explanation:

(a) Constructing a 98% confidence interval for m if the sample size, n, is 20:

To construct a confidence interval, we need to find the margin of error (E) and then calculate the lower and upper bounds of the interval. The formula for the margin of error is given by E = (z-score) × (standard deviation of the sample mean).

The z-score corresponding to a 98% confidence level is 2.33. So, the margin of error for a sample size of 20 is E = 2.33 × (8 / [tex]\sqrt{20}[/tex] ) ≈ 2.08.

Therefore, the 98% confidence interval for m is [50 - 2.08, 50 + 2.08] = [47.92, 52.08].

(b) Constructing a 98% confidence interval for m if the sample size, n, is 15:

Using the same formula, the margin of error for a sample size of 15 is E = 2.33 × (8 / [tex]\sqrt{15}[/tex] ) ≈ 2.86.

Therefore, the 98% confidence interval for m is [50 - 2.86, 50 + 2.86] = [47.14, 52.86].

(c) Constructing a 95% confidence interval for m if the sample size, n, is 20:

The z-score corresponding to a 95% confidence level is 1.96. So, the margin of error for a sample size of 20 is E = 1.96 × (8 / [tex]\sqrt{20}[/tex] ) ≈ 1.74.

Therefore, the 95% confidence interval for m is [50 - 1.74, 50 + 1.74] = [48.26, 51.74].

(d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed?

No, the formula for the margin of error and the construction of confidence intervals assume that the population is normally distributed. If the population distribution is not known to be normal, other methods such as the Central Limit Theorem or bootstrapping may need to be used.

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How many integers between 360 and 630 are there such that they have odd number of divisors?

Answers

Answer:

7

Step-by-step explanation:

Concept to Know: "A number is a perfect square if and only if it has odd number of positive divisors "

Find all the squared values that lies between 360 and 630

360< 19², 20², 21², 22², 23², 24², 25² < 630

19², 20², 21², 22², 23², 24², and 25² are all the squared values that lies between 360 and 630. There are seven of those squared numbers so the answer is 7.

Final answer:

There are 7 integers between 360 and 630 that are perfect squares, hence have odd numbers of divisors. These numbers are the squares of the integers from 19 to 25.

Explanation:

This is a problem in the field of number theory. In mathematics, only perfect square numbers have an odd number of total divisors. This is because factors or divisors usually come in pairs, but in perfect squares, one pair is actually a duplicate (the square root of the number itself). Hence, we need to find the perfect squares between 360 and 630.

By performing square root operation, we find that squares of integers from 19 to 25 are in the range given. So, the numbers are 361 (192), 400 (202), 441 (212), 484 (222), 529 (232), 576 (242) and 625 (252).

Therefore, there are 7 integers between 360 and 630 that have an odd number of divisors.

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Collection of_______ is called Statistics.

a. Methods
b. Sample data
c. Numerical information
d. Population data
e. Cleaned data

Answers

Answer:

Collection of numerical information is called Statistics

Step-by-step explanation:

From the Oxford dictionary, the definition of statistics (science) is: "the practice or science of collecting and analyzing numerical data in large quantities, especially to infer proportions in a whole from those in a representative sample".

While the definition of statistic (data) is: "a fact or piece of data obtained from a study of a large quantity of numerical data".

Statistics is a science, not a method, therefore answer can´t be a).

Not all sample data is necessarily a statistic, therefore answer can´t be b).

All statistics are numerical information.

Not all statistics are population data, therefore answer can´t be d).

Cleaned data is information filtered with a specific criterion, but not necessarily used in a statistic.

The most probable answer is c.

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