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.
A calculator can display 10 digits in standard notation. what is the largest whole number value of n that the calculator will display in standard notation for 4^n?
Step-by-step explanation:
The largest whole number can be 9,999,999,999 if it is decimal representation is simple. The device with me supports up to 10 ^ 100 exponents, so that 9.9999999E99 could be a candidate too. If your exponents are not limited, then 9E9999999 is the largest (above what the calculator can demonstrate).
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 thannbsp $ 200 on food nbspin the last month. What is the population in the study?
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.
How to determine population in the studyThe 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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A luxury resort is offering an incentive discount. If you book your room six months in advance, you will save 15% off the price of the room. You decide to take advantage of the offer and book a room for 6 nights. If the original price per night for the room is $382, what is the total discounted cost for your stay? Round your answer to the nearest cent.
Answer:the total discounted cost for your stay is $1948.2
Step-by-step explanation:
If you book your room six months in advance, you will save 15% off the price of the room. You decide to take advantage of the offer and book a room for 6 nights. If the original price per night for the room is $382, this means that the value of the discount would be
15/100 × 382 = 0.15 × 382 = $57.3
The cost of the room for a night would be
382 - 57.3 = $324.7
Since you are staying for six nights,
the total discounted cost for your stay would be
324.7 × 6 = $1948.2
Find the critical value χ R corresponding to a sample size of 15 and a confidence level of 90%. 023.685 06.571 O31.319 O29.141
Answer:
Step-by-step explanation:
i) Confidence level = 90%
ii) therefore significance level = [tex]\alpha[/tex] = [tex]\frac{100 - 90}{100}\hspace{0.1cm} = \hspace{0.1cm} 0.1[/tex]
iii) this problem is for a two tailed test.
iv) therefore significance level = [tex]\frac{\alpha }{2}[/tex] = [tex]\frac{0.1}{2}[/tex] = 0.05
iii) sample size = 15
iv) therefore degrees of freedom = sample size - 1 = 15 - 1 = 14
v) the upper ( or right) chi squared value, [tex]\chi^2_{R}[/tex] from Chi squared critical value tables = 23.685
The correct critical value χR is 23.685.
To find the critical value χR corresponding to a sample size of 15 and a confidence level of 90%, we first determine the degrees of freedom.
The degrees of freedom (df) for a chi-square distribution is calculated as n - 1, where n is the sample size. Here, n = 15, so df = 14.
Next, look up the critical value for the chi-square distribution with 14 degrees of freedom and a 90% confidence level.
In statistical tables, the value corresponding to a 90% confidence level (or equivalently, 10% significance level, which leaves 5% in each tail of the distribution) for 14 degrees of freedom is 23.685.
Therefore, the correct critical value χR is 23.685.
When the null hypothesis is true and n is greater than or equal to 5 per group, the test statistic for the Friedman test is approximately distributed as a
(A)t-distribution
(B)F-distribution
(C)normal distribution
(D)chi-square distribution
Answer:
D
Step-by-step explanation:
How many millimeters are in 0.05 kilometers? Question 13 options: 5,000 500 50 50,000
Answer:
its 5000
Step-by-step explanation:
for it to be 50000 like the other person said it would be asking what is 0.5 not 0.05 like it says in the question
Check all statements that are equivalent to If the sky is not clear, then you don't see the stars ? a. Clear sky is necessary and sufficient for seeing the stars. ? b. If the sky is clear, then you see the stars. c. You don't see the stars unless the sky is clear. ? d. Clear sky is necessary to see the stars. e. The sky is clear only if you see the stars. f. Clear sky is sufficient to see the stars. g. If you don't see the stars, then the sky is not clear. h. If you see the stars, then the sky is clear. i. You see the stars only if the sky is clear.
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.
Explanation: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:
'You don't see the stars unless the sky is clear' - This statement implies the same as the original by asserting that seeing the stars is dependent on the clarity of the sky. 'Clear sky is necessary to see the stars' - This statement also implies the same as the original. It plainly states the necessity of a clear sky for star viewing. 'You see the stars only if the sky is clear' - This statement is also equivalent to the original, insisting that the only condition under which stars can be seen is if the sky is clear.
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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The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party, and then returned the purchase? This is called a one-time fling. About 5% of all adults deliberately do a one-time fling and feel no guilt about it! In a group of nine adult friends, what is the probability of the following? (Round your answers to three decimal places.) Answer the following:
A. no one has done a one time fling
B. at least one person has done a one time fling
c. no more than two people have done a one time fling
Answer:
a) There is a 63.02% probability that no one has done an one time fling.
b) There is a 36.98% probability that at least one person has done a one time fling
c) There is a 99.15% pprobability that no more than two people have done a one time fling.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they have done an one time fling, or they have not. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 9, p = 0.05[/tex]
A. no one has done a one time fling
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.05)^{0}.(0.95)^{9} = 0.6302[/tex]
There is a 63.02% probability that no one has done an one time fling.
B. at least one person has done a one time fling
Either no one has done a one time fling, or at least one person has. The sum of the probabilities of these events is decimal 1.
So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - `(X = 0) = 1 - 0.6302 = 0.3698[/tex]
There is a 36.98% probability that at least one person has done a one time fling
c. no more than two people have done a one time fling
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 1) = C_{9,1}.(0.05)^{1}.(0.95)^{8} = 0.2985[/tex]
[tex]P(X = 2) = C_{9,2}.(0.05)^{2}.(0.95)^{7} = 0.0628[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6302 + 0.2985 + 0.0628 = 0.9915[/tex]
There is a 99.15% pprobability that no more than two people have done a one time fling.
In a group of 9 adults, the probability that no one has done a 'one-time fling' is 0.630, the probability that at least one person has done it is 0.370, and the probability that no more than two people have done it is approximately 0.853.
Explanation:The question here involves probability given certain conditions. The probability of an event happening can be found by dividing the number of times the event occurs by the total number of occurrences. Here we have the probability of the 'one-time fling' happening as 5% or 0.05, the complement of which (it not happening) is 1 - 0.05 = 0.95.
A. The probability of none out of nine friends having done a one-fling time is (0.95)^9 = 0.630. So, there’s a 0.630 chance that none of them have done a one-time fling.
B. The probability of at least one person having done a one-time fling would be the complement of no one having done it, which equals 1 - (the probability of no one having done it), 1 - 0.630 = 0.370.
C. The probability of no more than two people having done a one-time fling is found by adding the probabilities of exactly zero, one, and two persons having done it. Using the binomial probability formula, this would amount to 0.630 (for 0 person) + 9(0.05)*(0.95)^8 (for one person) + 36*(0.05)^2*(0.95)^7 (for two people) = 0.853, approximately.
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Express the confidence interval 0.333< p < 0.999 in the form p±E.
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.
Explanation: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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Use the following steps to prove that log b(xy)- log bx+ log by.
a. Letx-bp and y= bq. Solve these expressions for p and q, respectively
b. Use the property b*b b to express xy in terms of b, p, and q
c. Compute log b(xy) and simplify
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.
Explanation: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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Which set of integers is a Pythagorean triple and are the side lengths of a right triangle?
*PLEASE PROVIDE HOW YOU GOT YOUR ANSWER
A. 20,23,28
B. 18, 26, 44
C. 9, 40, 41
D. 8, 20, 32
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.
A bank sampled its customers to determine the proportion of customers who use their debit card at least once each month. A sample of 50 customers found that only 12 use their debit card monthly. Find a 99% confidence interval for the proportion of customers who use their debit card monthly.
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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Which represents the solution(s) of the equation x2 = 36?
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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Solve for the right triangle
Angle V 51
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 °
What does the pair of equations y = 3, z = 7 represent? In other words, describe the set of points (x, y, z) such that y = 3 and z = 7.
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.
A box contains 3 coins, one fair, one weighted with a 2 3 chance of coming up heads, and one weighted with a 2 3 chance of coming up tails. You take a coin at random from the box, flip it 20 times, and get 11 heads. What is the probability you have the fair coin?
Answer:
The probability that we have a fair coin is 0.2717
Step-by-step explanation:
Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)
We also let H be the event that the coin which is flipped lands (which is head)
Therefore, the problem arise:
That: P(H/C1) = 1/2.............fair coin
p(H/C2) = 2/3..........Chance of head
p(H/C3) = 2/3............Chance of tail
Now, noting that the coin was picked at random,
We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3
We can then say that or calculate thus:
P( C2 | H ) = P (H ∩ C2) ÷ P(H)
Where P (H ∩ C2) means the probability of event intersection
= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)
= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)
= (1 / 2) (1 / 3) ÷ (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)
0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)
0.15 ÷ 0.15 + 0.201 + 0.201
=0.15 / 0.552
= 0.2717
The idea of a large, stable middle class (defined as those with annual household incomes in 2010 between $42,000 and $126,000 for a family of three), is central to America's sense of itself. But the U.S. middle class shrank steadily from 61% of all adults in 1971 (t = 0) to 51% in 2011 (t = 4), where t is measured in decades. Research has shown that this decline was approximately linear.† (a) Find a linear function f(t) giving the percentage of middle-income adults in decade t, where t = 0 corresponds to 1971.
Answer:
[tex]f(t) = -0.025t + 0.61[/tex]
In which t is measured in decades.
Step-by-step explanation:
A linear function f(t) has the following format:
[tex]f(t) = at + b[/tex]
We are given two points of the function, so we can solve a system of equations to find the values for a and b.
61% of all adults in 1971 (t = 0)
This means that [tex]f(0) = 0.61[/tex]
So
[tex]f(t) = at + b[/tex]
[tex]0.61 = a*(0) + b[/tex]
[tex]b = 0.61[/tex]
51% in 2011 (t = 4)
This means that [tex]f(4) = 0.51[/tex]
So
[tex]f(t) = at + 0.61[/tex]
[tex]0.51 = 4a + 0.61[/tex]
[tex]4a = -0.10[/tex]
[tex]a = -0.025[/tex]
So the linear function is:
[tex]f(t) = -0.025t + 0.61[/tex]
In which t is measured in decades.
"is it appropriate to use the normal approximation for the sampling distribution of"
Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.
Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.
1. Binomial distribution: Let the sampling distribution be the binomial distribution [tex]B(n,p)[/tex], where [tex]n[/tex] is the number of trials and [tex]p[/tex] is the probability of success. It can be approximated by the Normal distribution with the mean of [tex]np[/tex] and the variance of [tex]np(1-p)[/tex], denoted by [tex]N(np,np(1-p))[/tex] if the following condition is met:
[tex]n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)[/tex]
2. Poisson distribution: Let the sampling distribution be the Poisson distribution [tex]P(\lambda)[/tex] where [tex]\lambda[/tex] is its mean. It can be approximated by the Normal distribution with the mean [tex]\lambda[/tex] and the variance [tex]\lambda[/tex], denoted by [tex]N(\lambda,\lambda)[/tex] when [tex]\lambda[/tex] is large enough, say [tex]\lambda>1000[/tex] (however, different sources may give different lower value for [tex]\lambda[/tex] but the greater it is, the better the approximation).
A prize was awarded to 56 women and 642 men.
a. What fraction of the prize winners were women?
b. What fraction were men?
SIMPLIFY EACH plz
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.
Two dice are rolled and someone indicates that the two numbers that come up are different. Find the probability that the sum of the two numbers is 8.
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
Convert the following equation to polar coordinates. y = 6/x The polar from of y = 6/x is
Answer:
[tex]r=\sqrt{6/sinucosu}[/tex]
Step-by-step explanation:
To convert to polar form:
[tex]x=rcosu[/tex]
[tex]y=rsinu[/tex]
We substitute into our function:
[tex]y=6/x[/tex]
[tex]rsinu=6/rcosu[/tex]
multiply both sides by r:
[tex]r^2sinu=6/cosu[/tex]
solve for r by dividing by sinu:
[tex]r=\sqrt{6/sinucosu}[/tex]
Complete the square and give a substitution (not necessarily trigonometric) which could be used to compute the integral. ∫1x2+2x+2dx
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.
Explanation: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.
A coin is tossed three times. Consider the following events
A: Heads on the first toss.
B: Tails on the second.
C: Heads on the third toss.
D: All three outcomes the same (HHH or TTT).
E: Exactly one head turns up.
(a) Which of the following pairs of these events are independent? i. A, B ii. A, D iii. A, E iv. D, E
(b) Which of the following triples of these events are independent? i. A, B, C ii. A, B, D iii. C, D, E
Consider a fair coin flipped a number of times.
(a) Which event is more likely HHHHHHHHH or HHHHHHHHT?
(b) Calculate the probability of each event.
(c) If someone thinks the first event is more likely, they are suffering from what?
(d) If someone thinks the second even is more likely, they are suffering from what?
1 a) Pairs of Independent Events are A and B, A and D
b) Triples of Independent Events are A, B, C
2 a) Both HHHHHHHHH and HHHHHHHHT are equally likely.
b) Probability of each event:
HHHHHHHHH: 1/512
HHHHHHHHT: 1/512
c) The person is suffering from gambler's fallacy
d) The person is suffering from misconception of randomness.
1) Events
A: Heads on the first toss.
B: Tails on the second toss.
C: Heads on the third toss.
D: All three outcomes the same (HHH or TTT).
E: Exactly one head turns up.
a) To check if two events are independent, we see if [tex]P(A \cap B) = P(A) \times P(B)[/tex]
Calculating Individual Probabilities
P(A) = 1/2 (Heads on the first toss)
P(B) = 1/2 (Tails on the second toss)
P(C) = 1/2 (Heads on the third toss)
P(D) = P(HHH) + P(TTT) = 1/8 + 1/8 = 1/4
P(E) = 3/8 (Exactly one head: H, T, T; T, H, T; T, T, H)
Checking Independence
i. A and B
P(A ∩ B): Probability of A and B happening together (H on first and T on second):
There are 4 outcomes: HTH, HTT, HHH, HHT. Only HTT has T on second.
P(A ∩ B) = 1/4.
P(A) × P(B) = 1/2 × 1/2 = 1/4.
They are independent
ii. A and D
P(A ∩ D): Probability of A and D (first is H and all are the same):
Only HHH satisfies this.
P(A ∩ D) = 1/8.
P(A) × P(D) = 1/2 × 1/4 = 1/8.
They are independent
iii. A and E
P(A ∩ E): Probability of A and E (first is H and only one head):
Possible outcomes: H, T, T (1 possibility).
P(A ∩ E) = 1/8.
P(A) × P(E) = 1/2 × 3/8 = 3/16.
They are not independent.
iv. D and E
P(D ∩ E): Probability of D and E (all the same and exactly one head):
Impossible since D can only be HHH or TTT, and E requires one head.
P(D ∩ E) = 0.
P(D) × P(E) = 1/4 × 3/8 = 3/32.
They are not independent
b) To check if three events are independent, we check if P(A ∩ B ∩ C) = P(A) × P(B) × P(C)
i. A, B, C
P(A ∩ B ∩ C): Probability of H on first, T on second, H on third:
Outcome: HTH.
P(A ∩ B ∩ C) = 1/8.
P(A) × P(B) × P(C) = 1/2 × 1/2 × 1/2 = 1/8.
They are independent
ii. A, B, D
P(A ∩ B ∩ D): Cannot have T on second and all the same
P(A ∩ B ∩ D) = 0.
P(A) × P(B) × P(D) = 1/2 × 1/2 × 1/4 = 1/16.
They are not independent
iii. C, D, E
P(C ∩ D ∩ E): Cannot have all the same (D) and exactly one head (E).
P(C ∩ D ∩ E) = 0.
P(C) × P(D) × P(E) = 1/2 × 1/4 × 3/8 = 3/64.
They are not independent.
2a) Both events are equally likely since they are specific sequences of tosses.
b) Probability of each event:
Probability of HHHHHHHHH
9 heads in a row:
Probability = (1/2)^9 = 1/512.
Probability of HHHHHHHHT
8 heads followed by 1 tail:
Probability = (1/2)^9 = 1/512.
c) They are suffering from the gambler's fallacy, which is the incorrect belief that past events influence independent events.
d) They may be suffering from the misconception of randomness, believing that certain sequences are more likely due to patterns they perceive.
g Which statement is true below for the size of the confidence interval for µ? Group of answer choices a. The confidence interval increases as the standard deviation increases. b. The confidence interval increases as the standard deviation decreases. c. The confidence interval increases as the sample size increases. d. The t-value for a confidence interval of 95% is smaller than for a confidence interval of 90%
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%.
See more about statistics at brainly.com/question/10951564
1/3m - 1 - 1/2n when m=21 and n=12
Answer:
0
Step-by-step explanation:
1/3m - 1 - 1/2n
1/3(21) - 1 - 1/2(12)
7 - 1 - 6 = 0
Isolating a variable in two equations is easiest when one of them has a coefficient 1.
Let's say we have the two equations:
3A−B = 5
2A+3B = −4
and want to isolate one of the variables, such that it appears by itself on one side of the equation.
Which of the following is an equation with one of the above variables isolated?
a) 3B = -2A-4
b) B = 5-3A
c) B = 3A-5
d) 2A = -3B-4
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.
According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance (CNBC, February 23, 2006). Sample results, consistent with the CBNC report, showed 46 out of 200 vehicles were not covered by insurance. Develop a 95% confidence interval for the population proportion.
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)
The company will pay $550 per year for your health insurance. What is the monthly amount they will pay?
A. $45.83
B. $46.50
C. $47.65
D. $48.26
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
state the most specific name for each figure.
WHAT DO I DO
Answer:
3 and 9 are parallelograms
other two are quadrilaterals
Step-by-step explanation:
Please help in this IXL for geometry!!
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