Given f(x)=x2, after performing the following transformations: shift upward 60 units and shift 29 units to the right, the new function g(x)

Answer:

[tex]0.23 - 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.172[/tex]

[tex]0.23 + 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.288[/tex]

The 95% confidence interval would be given by (0.172;0.288)

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".  

The population proportion have the following distribution

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

[tex]X=46[/tex] number of vehicles that were not covered by insurance

[tex]n=200[/tex] random sample taken

[tex]\hat p=\frac{46}{200}=0.23[/tex] estimated proportion of vehicles that were not covered by insurance

[tex]p[/tex] true population proportion of vehicles that were not covered by insurance

Solution to the problem

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]0.23 - 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.172[/tex]

[tex]0.23 + 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.288[/tex]

The 95% confidence interval would be given by (0.172;0.288)

Answer: 5/36

Step-by-step explanation:

We assume it's a fair die and the probability of any number coming up is 1/6.

Let's denote first die with it's number as A2, that is first die roled number 2

Let's denote second die with it's number as B4, that is Second die rolled 4.

For us to have the sum of those numbers to be 8, then we have possibilities of rolling the numbers

A2 and B6, A3 and B5, A4 and B4, A5 and B3, A6 and B2

This becomes :

[pr(A2) * pr(B6)] + [pr(A3) * pr(AB5)] + [pr(A4) * pr(B4)] + [pr(A5) * pr(3)] + [pr(A6) * pr(B2)]

Which becomes:

[1/6 * 1/6] + [1/6 * 1/6] + [1/6 * 1/6] + [1/6 * 1/6 + [1/6 * 1/6]

Which becomes :

1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36

Hence, the probability of the sum of the two numbers on the dice being 8 is 5/36

Answer:

Sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.

Step-by-step explanation:

To yield a more accurate estimate of the population mean, margin of error should be minimized.

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

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

for a given confidence level, and the same standard deviation, as the sample size increases, margin of error decreases.

Thus, random sample of 50 people from population A, has smaller margin of error than the sample of 20 people from population B.

Therefore, sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.

Final answer:

In statistics, a larger sample size generally leads to a more accurate estimate of the population mean.

Explanation:

Population A: Sample size = 50, SD = 6.0

Population B: Sample size = 20, SD = 6.0

In this case, the sample mean from Population A (larger sample size) will likely yield a more accurate estimate of its population mean compared to Population B (smaller sample size) due to the larger sample size resulting in a more reliable estimation.

Answer:

3 and 9 are parallelograms

other two are quadrilaterals

Step-by-step explanation:

Answer:

Option a) The confidence interval increases as the standard deviation increases.

Step-by-step explanation:

We have to find the true statements.

Confidence Interval:

[tex]\mu \pm Test_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

a. The confidence interval increases as the standard deviation increases.

As the standard deviation increases, the margin of error increases, thus, the width of the confidence interval increases.

Thus, the given statement is true.

b. The confidence interval increases as the standard deviation decreases.

The given statement is false. The explanation is similar to above part.

c. The confidence interval increases as the sample size increases.

As the sample size increases, the margin of error decreases, thus, the confidence interval width decreases.

Thus, the given statement is false.

d. The t-value for a confidence interval of 95% is smaller than for a confidence interval of 90%

The t-value depend on the level of significance as well as degree of freedom. But for a particular degree of freedom  t-value for a confidence interval of 95% is greater than for a confidence interval of 90%.

Thus, the given statement is false.

In this exercise we have to use our knowledge of statistics to identify the correct alternative that best matches, so we have:

Letter a.

So with the knowledge in statistics topics we can say that:

a. True, as the standard deviation increases, the margin of error increases, thus, the width of the confidence interval increases.

b. False, the confidence interval increases as the standard deviation decreases.

c. False, as the sample size increases, the margin of error decreases, thus, the confidence interval width decreases.

d. False, the t-value depend on the level of significance as well as degree of freedom. But for a particular degree of freedom  t-value for a confidence interval of 95% is greater than for a confidence interval of 90%.

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Answer with Step-by-step explanation:

a.[tex]x=b^p[/tex]

[tex]y=b^q[/tex]

Taking both sides log

[tex]log x=plog b[/tex]

Using identity:[tex]logx^y=ylogx[/tex]

[tex]p=\frac{logx}{log b}=log_b x[/tex]

Using identity:[tex]log_x y=\frac{log y}{log x}[/tex]

[tex]log y=qlog b[/tex]

[tex]q=\frac{log y}{log b}=log_b y[/tex]

b.[tex]xy=b^pb^q[/tex]

We know that

[tex]x^a\cdot x^b=x^{a+b}[/tex]

Using identity

[tex]xy=b^{p+q}[/tex]

c.[tex]log_b(xy)=log_b(b^{p+q})[/tex]

[tex]log_b(xy)=(p+q)log_b b[/tex]

Substitute the values then we get

[tex]log_b(xy)=(log_b x+log_b y)[/tex]

By using [tex]log_b b=1[/tex]

Hence, [tex]log_b(xy)=log_b x+log_b y[/tex]

To prove the property log b(xy) = log bx + log by, we let x = [tex]b^p[/tex]and y = [tex]b^q[/tex], express xy in terms of the base b and the exponents p and q, and then use the properties of logarithms to show the equality.

The student is asking to prove the logarithmic property − log b(xy) = log bx + log by. Here's a step-by-step explanation:

Let x = bp and y = bq. To solve for p and q, take the logarithm base b of both sides. Thus, p = logbx and q = logby.

Using the property of exponents, xy = bp*bq = bp+q.

Now compute logb(xy). According to the logarithmic property, logb(bp+q) = p + q. Since p = logbx and q = logby, then logb(xy) = logbx + logby.

Therefore, we have proven the given logarithmic property using the steps provided in the question.

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Answer:

[tex]z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637[/tex]  

[tex]p_v =P(z>3.637)=0.000138[/tex]  

Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

Step-by-step explanation:

Data given and notation

[tex]\bar X_{1}=35621[/tex] represent the mean for sample 1  

[tex]\bar X_{2}=33498[/tex] represent the mean for sample 2  

[tex]\sigma_{1}=4310[/tex] represent the population standard deviation for 1  

[tex]\sigma_{2}=4310[/tex] represent the population standard deviation for 2

[tex]n_{1}=100[/tex] sample size for the group 2  

[tex]n_{2}=120[/tex] sample size for the group 2  

[tex]\alpha=0.01[/tex] Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for US residents (sample 1) is higher than the mean for sample 2, the system of hypothesis would be :  

Null hypothesis:[tex]\mu_{1}-\mu_{2}\leq0[/tex]  

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}> 0[/tex]  

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

[tex]z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

[tex]z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637[/tex]  

P value  

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

[tex]p_v =P(z>3.637)=0.000138[/tex]  

Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

Answer:

Option C.

Step-by-step explanation:

The given equations are

[tex]3A-B=5[/tex]

[tex]2A+3B=-4[/tex]

We need  isolate one of the variables, such that it appears by itself on one side of the equation.

Isolating a variable in two equations is easiest when one of them has a coefficient 1.

In equation 1, coefficient of B is 1. So, we can easily isolate one of the variables.

The equation is

[tex]3A-B=5[/tex]

Subtract 3A from both sides.

[tex]-B=5-3A[/tex]

Multiply both sides by -1.

[tex]B=3A-5[/tex]

Therefore, the correct option is C.

Answer:

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

Step-by-step explanation:

For this case our variable of interest is the average parking time of its students. And we have 280 values for these times. So then the variable of interest is quantitative.

Which of the following graphs should not be used to display information concerning the students parking times?

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

b.Stem-and-leaf display

That incorrect since the Stem and leaf plot is useful when we want to plot quantitative data.

c. Histogram

That incorrect since the Histogram is ideal when we want to plot quantitative data and analyze the distribution of the data.

d. Box plot

That incorrect since the Box plot is ideal and useful when we want to plot quantitative data and see central tendency measures.

Final answer:

A pie chart should not be used to display the average parking times of students because it cannot effectively represent the distribution or variability of parking times, unlike a histogram, box plot, or stem-and-leaf display.

Explanation:

The question revolves around determining which graph would be inappropriate for displaying the average parking times of students at a university. The options are: a Pie chart, a Stem-and-leaf display, a Histogram, and a Box plot. To assess the suitability of each graph, we need to understand what type of data we have and what information we wish to convey with the graph.

A pie chart is typically used to show parts of a whole. However, in this scenario, we're interested in the distribution of parking times, not how a single parking time compares as a fraction of the total. Therefore, a pie chart would not effectively represent the variability or distribution of parking times.

Both a histogram and a box plot are well-suited for displaying the distribution of quantitative data such as parking times, making them good choices for this context. A stem-and-leaf display is also a valid option, especially for providing a quick visual summary that includes actual data points, which can be useful for identifying specific patterns or outliers.

In conclusion, the pie chart is the graph that should not be used to display information concerning the students' parking times. It simply doesn't align with the objective of analyzing the distribution or average of the parking times, which is best visualized through the other mentioned graphical methods.

Answer:

There is a 62% probability that the student will be awarded at least one of the two scholarships.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that the student gets a scolarship from Agency A.

B is the probability that the student gets a scolarship from Agency B.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that the student will get an scolarship from agency A but not from agency B and [tex]A \cap B[/tex] is the probability that the student will get an scolarship from both agencies.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

What is the probability that the student will be awarded at least one of the two scholarships?

This is

[tex]P = a + b + (A \cap B)[/tex]

We have that:

[tex]A = 0.55, B = 0.40[/tex]

If the student is awarded a scholarship from Agency A, the probability that the student will be awarded a scholarship from Agency B is 0.60.

This means that:

[tex]\frac{A \cap B}{A} = 0.6[/tex]

[tex]A \cap B = 0.6A = 0.6*0.55 = 0.33[/tex]

----------

[tex]A = a + (A \cap B)[/tex]

[tex]0.55 = a + 0.33[/tex]

[tex]a = 0.22[/tex]

--------

[tex]B = b + (A \cap B)[/tex]

[tex]0.40 = b + 0.33[/tex]

[tex]b = 0.07[/tex]

Answer:

[tex]P = a + b + (A \cap B) = 0.22 + 0.07 + 0.33 = 0.62[/tex]

There is a 62% probability that the student will be awarded at least one of the two scholarships.

The probability that the student will be awarded at least one of the two scholarships is 0.73.

To calculate the probability that the student will be awarded at least one of the two scholarships, we can use the following formula:

P(at least one scholarship) = 1 - P(no scholarships)

The probability that the student will not be awarded a scholarship from either agency is the product of the probability that the student will not be awarded a scholarship from Agency A and the probability that the student will not be awarded a scholarship from Agency B.

The probability that the student will not be awarded a scholarship from Agency A is

1 - 0.55 = 0.45.

The probability that the student will not be awarded a scholarship from Agency B is

1 - 0.40 = 0.60.

Therefore, the probability that the student will not be awarded a scholarship from either agency is

0.45 * 0.60 = 0.27.

Therefore, the probability that the student will be awarded at least one of the two scholarships is

1 - 0.27 = 0.73.

Another way to calculate this probability is to use the following formula:

P(at least one scholarship) = P(scholarship from Agency A) + P(scholarship from Agency B) - P(scholarship from both agencies)

We already know the probability that the student will be awarded a scholarship from each agency. The probability that the student will be awarded a scholarship from both agencies is the product of the probability that the student will be awarded a scholarship from Agency A and the probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A.

The probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A is 0.60.

Therefore, the probability that the student will be awarded a scholarship from both agencies is

0.55 * 0.60 = 0.33.

Therefore, the probability that the student will be awarded at least one of the two scholarships is

0.55 + 0.40 - 0.33 = 0.73.

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The equivalent statements to 'If the sky is not clear, then you don't see the stars' are 'You don't see the stars unless the sky is clear', 'Clear sky is necessary to see the stars', and 'You see the stars only if the sky is clear'. Other statements make assumptions that are not present in the original.

The original statement, 'If the sky is not clear, then you don't see the stars' is a conditional statement that refers to the necessary conditions for seeing the stars. In this logic, the clear sky is a necessity to see the stars.

There are several statements equivalent to the original one, according to the principles of logic namely:

However, the statements 'Clear sky is necessary and sufficient for seeing the stars' and 'Clear sky is sufficient to see the stars' are not necessarily equivalent to the original because they assume that a clear sky is all that's needed to see the stars, ignoring the other conditions like absence of light pollution or the time of the day. Whereas, the original statement does not make this assumption.

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Answer:

(0.084,0.396)

Step-by-step explanation:

The 99% confidence interval for the proportion of customers who use debit card monthly can be constructed as

[tex]p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} } <P<p+z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }[/tex]

[tex]p=\frac{x}{n}[/tex]

[tex]p=\frac{12}{50}[/tex]

[tex]p=0.24[/tex]

[tex]q=1-p=1-0.24=0.76[/tex]

[tex]\frac{\alpha }{2} =\frac{\0.01 }{2}=0.005[/tex]

[tex]p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} } <P<p+z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }[/tex]

[tex]0.24-z_{0.005} \sqrt{\frac{0.24*0.76}{50} } <P<0.24+z_{0.005} \sqrt{\frac{0.24*0.76}{50} }[/tex]

[tex]0.24-2.58(0.0604)<P< 0.24+2.58(0.0604)[/tex]

[tex]0.24-0.155832<P<0.24+0.155832[/tex]

By rounding to three decimal places we get,

[tex]0.084<P<0.396[/tex]

The 99% confidence interval for the proportion of customers who use debit card monthly is (0.084,0.396).

To find a 99% confidence interval for the proportion of customers who use their debit card monthly, you can use the formula for the confidence interval for a proportion. The formula is:

[tex]\[ \text{Confidence Interval} = \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]

Where:

- [tex]\(\hat{p}\)[/tex] is the sample proportion (in this case, 12 out of 50 customers).

- Z is the critical value for a 99% confidence level (you can find this value in a standard normal distribution table or use a calculator, and it's approximately 2.576 for a 99% confidence level).

- n is the sample size (50 customers).

Now, plug in the values:

- [tex]\(\hat{p}\)[/tex] = [tex]\frac{12}{50} = 0.24\) (the sample proportion)[/tex].

- [tex]\(Z = 2.576\) (for a 99% confidence level)[/tex].

- [tex]\(n = 50\) (the sample size)[/tex].

Calculate the standard error:

[tex]\[ \text{Standard Error} = \sqrt{\frac{0.24 \cdot (1-0.24)}{50}} \][/tex]

Now, calculate the margin of error:

[tex]\[ \text{Margin of Error} = Z \cdot \text{Standard Error} = 2.576 \cdot \sqrt{\frac{0.24 \cdot (1-0.24)}{50}} \][/tex]

Finally, calculate the confidence interval:

[tex]\[ \text{Confidence Interval} = 0.24 \pm \text{Margin of Error} \][/tex]

Calculate the upper and lower bounds:

Lower Bound: [tex]\(0.24 - \text{Margin of Error}\)[/tex]

Upper Bound: [tex]\(0.24 + \text{Margin of Error}\)[/tex]

Now, calculate these values:

[tex]\[ \text{Margin of Error} = 2.576 \cdot \sqrt{\frac{0.24 \cdot (1-0.24)}{50}} \][/tex]

[tex]\[ \text{Margin of Error} \approx 0.0944 \][/tex]

Lower Bound: [tex]\(0.24 - 0.0944 \approx 0.1456\)[/tex]

Upper Bound: [tex]\(0.24 + 0.0944 \approx 0.3344\)[/tex]

So, the 99% confidence interval for the proportion of customers who use their debit card monthly is approximately 0.1456 to 0.3344.

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Answer:

Population: All undergraduates who attend a university and live in the United States  

Step-by-step explanation:

We are given the following situation in the question:

 "A polling organization contacts 1835 undergraduates who attend a university and live in the United States and asks whether or not they had spent more than nbsp $200 on food nbsp in the last month."

For the given situation, we have

Sample size, n = 1835

Sample: 1835 undergraduates who attend a university and live in the United States

Variable: whether or not they had spent more than $200 on food in the last month

Thus, the population for this scenario will be

Population: All undergraduates who attend a university and live in the United States

The population in the study is 1835 undergraduates.

The population in the study is the total number of undergraduates who attend a university and live in the United States.

According to the information given, the polling organization contacted 1835 undergraduates who meet these criteria and asked them about their food expenses.

Therefore, the population in the study is 1835 undergraduates.

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Answer:

the pair of equations y = 3, z = 7 represent the intersection of two plans, The set of points is (0,3,7) and the line is parallel to x axis.

Step-by-step explanation:

Consider the provided equation.

y=3 represents a vertical plane which is in xy plane.

Z=7 represents a horizontal plane which is parallel to xy plane

The both planes are perpendicular to each other and intersect.

y=3 and z=7 is the intersection of two plans, where the value of x is zero y=3 and z=7.

The set of points is (0,3,7) and the line is parallel to x axis.

Final answer:

The pair of equations y = 3 and z = 7 represents a set of points in three-dimensional space where the y-coordinate is always 3 and the z-coordinate is always 7.

Explanation:

The pair of equations y = 3 and z = 7 represents a set of points in three-dimensional space where the y-coordinate is always 3 and the z-coordinate is always 7. In other words, any point that satisfies both equations will have a y-value of 3 and a z-value of 7, regardless of the x-coordinate.

Answer:

Therefore, Δx=5/n, when have n intervals.

Step-by-step explanation:

From exercise we have interval [0,5]. So the length of the given interval is     5-0=5. Since all intervals [x0,x1],[x1,x2],…,[xn−1,xn]  are equal in width.

We know that their width is Δx. We conclude that width of each subinterval Δx in terms of the number of subintervals n is equal 5/n.

Therefore, Δx=5/n, when have n intervals.

Answer:

Step-by-step explanation:

a) {(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)}

[tex]\Omega[/tex]=6*6=36

b)A=(2*3)=6

P(A)=6/36=1/6

c) B=6+5+4+3+2+1=21

P(B)=21/36

d) P(A|B) - an event where either a 3 or 4 is rolled first and is followed by an even number and their sum goes over 7

P(A|B)=3

e) Not always, not sure how to explain, I'm not good with English Math

f) Same as above

The sample space for rolling two die has 36 outcomes. The probability of rolling a three or four, followed by an even number is 1/6, while the probability of the sum of the rolls not exceeding seven is 7/12. These two events are not mutually exclusive, but they are independent.

a. The sample space for rolling two dice consists of 36 possible outcomes, as there are six possible outcomes for the first die and six for the second die, and 6*6=36.

b. Event A happens when we roll a three or four first, followed by an even number. There are 4 such outcomes: (3,2), (3,4), (3,6), (4,2), (4,4), and (4,6). The probability of event A occurring is therefore 6/36 = 1/6.

c. Event B happens when the sum of the two rolls is at most seven. There are 21 outcomes: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(5,1),(5,2),(6,1). So, the probability of event B is 21/36 = 7/12.

d. P(A|B) represents the probability of event A occurring given that event B has already occurred. There are 4 outcomes in B that are also in A: (3,2), (3,4), (4,2), (4,3). Hence, P(A|B) = 4/21.

e. Events A and B are not mutually exclusive, as they can both occur in the same trial (e.g., when the dice rolls are (3,2), (3,4), (4,2), or (4,3)).

f. A and B are independent because the probability of A doesn't change whether B occurs or not, and vice versa. The fact that P(A|B) = P(A) confirms this.

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Answer:

0

Step-by-step explanation:

1/3m - 1 - 1/2n

1/3(21) - 1 - 1/2(12)

7 - 1 - 6 = 0

Answer:

17.32

Step-by-step explanation:

to get TU

sinФ = opposite/hypothenus

sin 60° = TU/20

√3/2=TU/20

( √3/2 )  x 20 = 10√3 = TU = 17.3205080757 = 17.32

Answer:

a) [tex]\frac{28}{349}[/tex] of the prize winners were women.

b) [tex]\frac{321}{349}[/tex] of the prize winners were men.

Step-by-step explanation:

Given:

Prize awarded to women = 56

Prize awarded to men = 642

we need to find;

a. fraction of the prize winners were​ women

b. fraction were​ men

Solution:

First we will find the Total prize awarded.

Total Prize awarded is equal to sum of Prize awarded to women and Prize awarded to men.

framing in equation form we get;

Total Prize awarded = [tex]56+642 = 698[/tex]

Now we need to find the fraction of prize winners were women.

to find the fraction of prize winners were women we will divided Prize awarded to women from Total Prize awarded we get;

fraction of the prize winners were​ women = [tex]\frac{56}{698} = \frac{2\times28}{2\times 349}=\frac{28}{349}[/tex]

a) Hence [tex]\frac{28}{349}[/tex] of the prize winners were women.

Now we can say that;

to find the fraction of prize winners were men we will divided Prize awarded to men from Total Prize awarded we get;

fraction of the prize winners were​ men = [tex]\frac{642}{698}= \frac{2\times321}{2\times 349} = \frac{321}{349}[/tex]

b) Hence [tex]\frac{321}{349}[/tex] of the prize winners were men.

Answer:

a = L

b = MT^(-1)

c = LT^(-1)

k = MT^(-2)

f = MLT^(-2)

S = T^(-1)

Step-by-step explanation:

x (0) = a

x is denoted by displacement in vibration analysis hence attains units of x.

Hence, a = L

b is the damping coefficient:

[tex]b = \frac{F}{\frac{dx}{dt} } \\= MLT^(-2) / LT^(-1)\\= MT^(-1)[/tex]

x'(0) = c

dx/dt = velocity hence c attains the units of velocity

c = LT^(-1)

Coefficient k is the stiffness:

[tex]k = \frac{F}{x} = \frac{MLT^(-2)}{L} = MT^(-2)[/tex]

Coefficient f is the magnitude of the exciting force

[tex]F = m*acceleration = MLT^(-2)[/tex]

Coefficient S is the angular frequency

angular frequency is displacement in radians per seconds; hence,

S = T^(-1)

Answer:

Substitution of x+1

Step-by-step explanation:

We are given that

[tex]\int \frac{1}{x^2+2x+2}dx[/tex]

[tex]\int\frac{1}{(x^2+2x+1)+1}dx[/tex]

[tex]\int\frac{1}{(x+1)^2+1^2}dx[/tex]

By using identity

[tex](a+b)^2=a^2+b^2+2ab[/tex]

Substitute x+1=t

Differentiate w.r.t x

dx=dt

Substitute the values

[tex]\int\frac{1}{t^2+1^2}dx[/tex]

[tex]\frac{1}{1}tan^{-1}\frac{t}{1}+C[/tex]

By using formula :[tex]\int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}\frac{x}{a}+C[/tex]

[tex]tan^{-1}(x+1)+C[/tex]

To complete the square for the quadratic polynomial x^2 + 2x + 2, we obtain (x + 1)^2 + 1. Substituting u = x + 1 simplifies the integral, which is then recognized as the derivative of arctan(u), resulting in the solution arctan(x + 1) + C.

The student asked to complete the square for the integrand x2 + 2x + 2 and to propose a substitution to compute the integral ∫ 1 / (x2 + 2x + 2) dx.

First, we complete the square for the quadratic polynomial x2 + 2x + 2. That gives us:

(x + 1)2 + 1 = x2 + 2x + 1 + 1 = x2 + 2x + 2.

We have now written our quadratic in the form of a perfect square plus a constant, which simplifies our integral:

∫ 1 / ((x + 1)2 + 1) dx.

We can substitute u = x + 1, which implies du = dx.

This transforms our integral into:

∫ 1 / (u2 + 1) du,

which can be recognized as the derivative of arctan(u), hence the integral is:

arctan(u) + C, where C is the constant of integration.

Substituting back our original variable, the solution is:

arctan(x + 1) + C.

Answer:

Step-by-step explanation:

Triangle BCD is a right angle triangle.

From the given right angle triangle

BC represents the hypotenuse of the right angle triangle.

Taking 45 degrees as the reference angle,

BC represents the adjacent side of the right angle triangle.

BD represents the opposite side of the right angle triangle.

To determine BC, we would apply trigonometric ratio

Cos θ = adjacent side/hypotenuse side. Therefore,

Cos 45 = BC/2√2

√2/2 = BC/2√2

BC = 2√2 × √2/2

BC = 2

Answer:

d. Cluster

Step-by-step explanation:

Random: Random is asking a group of people from a population. For example, to estimate the proportion of Buffalo residents who are Bills fans, you ask 100 Buffalo residents and estimate to the entire population.

Systematic: Similar to random. For example, you want to estimate something about a population, and your sample is every 5th people you see on the street.

Cluster:Divides the population into groups, with geographic characteristics.. Each element is the groups is used. Suppose you want to study the voting choices of Buffalo Bills players. You can divide into offense, defense and special teams, and ask each player of these 3 groups.

Stratified: Done on a group of clusters, that is, from each cluster(group), a number of people are selected.

In this problem, we have that:

All employees at three stores of a large retail chain were asked to fill out a survey.

Divided by clusters(stores).

So the orrect answer is:

d. Cluster

The type of sampling described in the question, where all employees at three selected stores are surveyed, is an example of cluster sampling. This method divides the population into groups, or 'clusters', and includes all members from selected clusters in the study.

The scenario described seems to be an example of cluster sampling. In cluster sampling, all subjects (in this case, employees at the stores) within selected groups (the three stores in this case) are studied. The selection is not random or systematic, but based on grouping. It is also not a stratified sample as we are not choosing samples proportionally from subgroups within the population.

In this case, the whole population is divided into groups (stores). Each group is called a cluster. All the members (employees) of the selected groups (stores) are included in the study. It is important to note that unlike stratification, where samples are taken from each group, in cluster sampling, we study the entire group.

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Answer:

e = 0.6342

S = 0.808

γsat = 126.207 lb/ft³

Step-by-step explanation:

Given

VT = 0.320 ft³

WT = 38.9 lb

w = 19.2% = 0.192

Gs = 2.67

then we apply

γ = WT / VT

γ = 38.9 lb / 0.320 ft³

γ = 121.5625 lb/ft³

then we get γdry as follows

γdry = γ / (1 + w)

γdry = 121.5625 lb/ft³ / (1 + 0.192)

γdry = 101.982 lb/ft³

the void ratio (e) can be obtained applying this equation

γdry = Gs*γw / (1 + e)

101.982 = 2.67*62.42 / (1 + e)

⇒ e = 0.6342

We can get the degree of saturation (S) as follows

S*e = Gs*w

S = Gs*w / e

S = 2.67*0.192 / 0.6342

S = 0.808

The saturated unit weight (γsat) will be obtained applying this formula

γsat = (Gs + e)*γw / (1 + e)

γsat = (2.67 + 0.6342)*62.42 lb/ft³/ (1 + 0.6342)

γsat = 126.207 lb/ft³

Answer:

A. $45.83

Step-by-step explanation:

This problem can be solved by a simple rule of three.

Each year has 12 months. The company will pay $550 per year for your health insurance. This means that in 12 months, the company will pay $550.

What is the monthly amount they will pay?

This is how much they are going to pay in one month. So:

1 month - $x

12 months - $550

[tex]12x = 550[/tex]

[tex]x = \frac{550}{12}[/tex]

[tex]x = 45.83[/tex]

So the correct answer is:

A. $45.83

Answer:

Step-by-step explanation:

x2 = 36?

solution

x^2-(36)=0

Factoring:  x2-36

Check : 36 is the square of 6

Check :  x2  is the square of  x1  

Factorization is :       (x + 6)  •  (x - 6)

(x + 6) • (x - 6)  = 0

x+6 = 0

x = -6

x-6 = 0

add 6 to both side

x = 6

x = -6

The solutions of an equation are the true values of the equation.

The expression that represents the solution(s) of [tex]\mathbf{x^2 = 36}[/tex] is [tex]\mathbf{x = \pm6}[/tex]

The equation is given as:

[tex]\mathbf{x^2 = 36}[/tex]

Take square roots of both sides

[tex]\mathbf{x = \pm\sqrt{36}}[/tex]

Express the square root of 36 as 6.

So, we have

[tex]\mathbf{x = \pm6}[/tex]

So, the expression that represents the solution(s) of [tex]\mathbf{x^2 = 36}[/tex] is [tex]\mathbf{x = \pm6}[/tex]

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Answer:

Therefore, the final form

p+/-E = 0.666+/-0.333

Step-by-step explanation:

Given:

Confidence interval = 0.333 < p < 0.999

To express the confidence interval in the forn p+/-E, where;

p is the midpoint of the confidence interval

E is the error.

The midpoint of the confidence interval is

p = (0.333+0.999)/2 = 1.332/2

p = 0.666

The error can be calculated using the formula:

Error = interval width/2

E = (0.999-0.333)/2 = 0.666/2

E = 0.333

Therefore, the final form

p+/-E = 0.666+/-0.333

The confidence interval 0.333< p <0.999 is expressed in the form p ± E as 0.666 ± 0.333. The midpoint of the interval is calculated by adding the two bounds and dividing by 2, and the distance from this point to either end of interval is calculated by subtracting the lower limit from the midpoint.

The confidence interval 0.333< p <0.999 can be expressed in the form p ± E by calculating the middle point of the interval (p), and the distance from the middle point to either end of the interval (E). To calculate the midpoint, add the two bounds and divide the result by 2. Thus, p = (0.999 + 0.333) / 2 = 0.666. Then, calculate E by subtracting the lower limit from p. So, E = 0.666 - 0.333 = 0.333. So, the confidence interval can be written as p ± E, or 0.666 ± 0.333.

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Answer:

Step-by-step explanation:

Triangle VWX is a right angle triangle.

From the given right angle triangle

VX represents the hypotenuse of the right angle triangle.

With 39 degrees as the reference angle,

WX represents the adjacent side of the right angle triangle.

VW represents the opposite side of the right angle triangle.

1) To determine VX, we would apply trigonometric ratio

Sin θ = opposite side/hypotenuse Therefore,

Sin 39 = 4/VX

VX Sin39 = 4

VX = 4/Sin39 = 4/0.6293

VX = 6.4

2) To determine WX, we would apply trigonometric ratio

Tan θ = opposite side/adjacent side. Therefore,

Tan 39 = 4/WX

WX Tan39 = 4

WX = 4/Tan 39 = 4/0.8089

WX = 4.9

3) the sum of the angles in a triangle is 180 degrees. Therefore

∠V + 90 + 39 = 180

∠V = 180 - (90 + 39)

∠V = 51 °

Answer:

D

Step-by-step explanation:

Answer: the integers that is a Pythagorean triple and are the side lengths of a right triangle is

C. 9, 40, 41

Step-by-step explanation:

A Pythagorean triple is a set of three numbers which satisfy the Pythagoras theorem. The Pythagoras theorem is expressed as

Hypotenuse^2 = opposite side^2 + adjacent side^2

Let us try each set of numbers.

A. 20,23,28

28^2 = 20^ + 23^2

784 = 400 + 529 = 929

Since both sides of the equation are not equal, the set of numbers is not a Pythagoras triple.

B. 18, 26, 44

44^2 = 18^ + 26^2

1936 = 324 + 676 = 1000

Since both sides of the equation are not equal, the set of numbers is not a Pythagoras triple.

C. 9, 40, 41

41^2 = 9^ + 40^2

1681 = 81 + 1600 = 1681

Since both sides of the equation are equal, the set of numbers is a Pythagoras triple.

D. 8, 20, 32

32^2 = 20^ + 8^2

1024 = 400 + 64 = 464

Since both sides of the equation are not equal, the set of numbers is not a Pythagoras triple.