Answer:
0
Step-by-step explanation:
1/3m - 1 - 1/2n
1/3(21) - 1 - 1/2(12)
7 - 1 - 6 = 0
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.
Find the equation of the line normal to the curve of y=3cos1/3x, Where x=\pi
Answer:
[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]
Step-by-step explanation:
The equation to the line normal to the curve has the following format:
[tex]y - y(x_{0}) = m(x - x_{0})[/tex]
In whicm m is the derivative of y at the point [tex]x_{0}[/tex]
In this problem, we have that:
[tex]x_{0} = \pi[/tex]
[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]
[tex]y(\pi) = 3\cos{\frac{\pi}{3}} = \frac{3}{2}[/tex]
The derivative of [tex]\cos{ax}[/tex] is [tex]a\sin{ax}[/tex]
So
[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]
[tex]y'(x) = 3*\frac{1}{3}\sin{\frac{x}{3}} = \sin{\frac{x}{3}}[/tex]
[tex]m = \sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{3}[/tex]
The equation of the line normal to the curve of y=3cos1/3x is:
[tex]y - y(x_{0}) = m(x - x_{0})[/tex]
[tex]y - \frac{3}{2} = \frac{\sqrt{2}}{3}(x - \pi)[/tex]
[tex]y = \frac{\sqrt{2}}{3}(x - \pi) + \frac{3}{2}[/tex]
[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]
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.
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.
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
A pizza lover wants to compare the average delivery times for four local pizza restaurants. Over the course of a few weeks, he orders a number of pizzas from each restaurant, and he records the time it takes for each pizza to be delivered.
a) When performing an ANOVA with this data, what is the alternative hypothesis?O All of the restaurants have different mean delivery timesO At least two of the restaurants have different mean delivery timesO Two of the restaurants have different mean delivery timesO One of the restaurants has a different mean delivery time than the others
Answer:
At least two of the restaurant have different mean delivery time.
Step-by-step explanation:
The ANOVA known as Analysis of variance involves the hypothesis testing of means on the basis of variances. The null hypothesis in ANOVA is taken as the equality mean i.e. H0: All means are equal. Whereas alternative hypothesis consists of at least two means are not equal.
The problem states that a pizza lover is comparing average delivery time. The null and alternative hypothesis in this case would be
H0: All restaurant have equal mean delivery time.
H1: At least two restaurant have different mean delivery time.
In an ANOVA comparing four local pizza restaurants' delivery times, the alternative hypothesis is that at least two restaurants have different mean delivery times.
Explanation:When performing an ANOVA with data comparing the average delivery times for four local pizza restaurants, the alternative hypothesis is that at least two of the restaurants have different mean delivery times. This is because ANOVA tests the null hypothesis that all group means are equal against the alternative that at least one group mean is different. Specifically, the null hypothesis for a one-way ANOVA test with four groups is μ1 = μ2 = μ3 = μ4, where each μ represents the mean delivery time of a different restaurant. The alternative hypothesis, which the student is inquiring about, is Ha: μi ≠ μj for some i ≠ j.
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If a 1% level of significance is used to test a null hypothesis, there is a probability of __________ of rejecting the null hypothesis when it is true.
Answer:
If a 1% level of significance is used to test a null hypothesis, there is a probability of ____ less than 1%______ of rejecting the null hypothesis when it is true
Step-by-step explanation:
Given that a hypothesis testing is done.
Level of significance used is 1%
i.e. alpha = 1%
When we do hypothesis test, we find out test statistic Z or t suitable for the test and find p value
If p value is < 1% we reject null hypothesis otherwise we accept null hypothesis.
So p value can be atmost 1% only for accepting null hypothesis.
So the answer is 1%
If a 1% level of significance is used to test a null hypothesis, there is a probability of ____less than 1%______ of rejecting the null hypothesis when it is true.
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)
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 is the minimum number of binary bits needed to represent each of the following unsigned decimal integers? a. 65
Answer: N >/= 7 bits
Minimum of 7 bits
Step-by-step explanation:
The minimum binary bits needed to represent 65 can be derived by converting 65 to binary numbers and counting the number of binary digits.
See conversation in the attachment.
65 = 1000001₂
65 = 7 bits :( 0 to 2^7 -1)
The number of binary digits is 7
N >/= 7 bits
To represent the unsigned decimal integer 65 in binary format, we need a minimum of 7 binary bits.
Explanation:This question is related to the conversion of decimal numbers to binary format. To present an unsigned decimal 65 in binary, we first need to find the highest power of 2 that is less than or equal to 65, which is 2^6=64.
Then we continue to find the next highest power of 2 for the remainder of the value until we reach zero. This process would require a total of 7 bits. Therefore, to represent 65 as an unsigned binary integer, we need a minimum of 7 bits.
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Find the probability choosing 7 numbers that match 7 randomly selected balls numbered 1-34.
Answer:72
Step-by-step explanation:
You are given the equation 27bx7•cx8•7xd=455x27. If b and c are both integers greater than 1 and b less than cb
Answer:
The question is not so clear and complete
Step-by-step explanation:
But for questions like this, since the equation has been given, what is expected is for us to make comparison, compare the RHS with the LHS or by method of comparing coefficients.
We follow the stated conditions since we are told that b and c are both integers which are greater than 1 and b is less than the product of cb. from these conditions, we can compare and get the values of b , c and d.
Another approach is to assume values, make assumptions with the stated conditions, however, our assumptions must be valid and correct if we substitute the assumed values of b, c and d in the equation, it must arrive at the same answer for the RHS. i.e LHS = RHS
Let v1,....., vk be vectors, and suppose that a point mass of m1,....., mk is located at the tip of each vector. The center of mass for this set of point masses is equal to: v = [(m1v1 +.....+ mkvk)/m] where m = m1 +.....+ mk. Determine how to divide a total mass of 11 kg among the vectors u1 = (−1, 3), u2 = (3, −2), and u3 = (5, 2) so that the center of mass is (21/11, 6/11).
Answer: m1 = 4
m2 = 5
m3 = 2
Step-by-step explanation:
given (21/11, 6/11) = m1 (-1/3) + m2 (3, -2) + m3 (5, 2)
= (-m1 + 3m2 + 5m3) / 11 = 21/11
= (3m1 + (-2)m2 + 2m3) / 11 = 6/11
so that m1 + m2 +m3 = 11
-m1 + 3m2 + 5m3 = 21
3m1 - 2m2 + 2m3 = 6
from this, we get the augmented matrix as
\left[\begin{array}{cccc}-1&1&1&11\\-1&3&5&21\\3&-2&2&6\end{array}\right]
= \left[\begin{array{cccc}-1&1&1&11\\0&4&6&32\\0&-5&-1&-27\end{array}\right] \left \{ {{R2=R2 + R1} \atop {R3=R3 -3R1 }]} \right.
= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&-5&-1&-27\end{array}\right]
= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&0&13/2&13\end{array}\right]
(R3 = R3 + 5R2)
this gives m1 + m2 + m3 = 11
m2 + 3/2 m3 = 8
13/2 m3 = 8
13/2 m3 = 13
m3 = 2
m2 = 8 -3/2 (2) = 5
= m1 = 11- 5 - 2 = 4
this gives
m1 = 4
m2 = 5
m3 = 2
To achieve the given center of mass, 4 kg of mass should be allocated to the vector u1, 6 kg to the vector u2, and 1 kg to the vector u3.
Explanation:The provided equation for the center of mass expresses a weighted average of the position vectors, with each mass acting as the weight on its respective vector. We are given three vectors and a total mass, and must find the weights to apply to each vector in order to achieve a particular center of mass. Setting up equations for each component of the center of mass, we get two equations:
21/11 = (-1m1 + 3m2 + 5m3) / 11 and 6/11 = (3m1 - 2m2 + 2m3) / 11.
These equations can then be solved simultaneously to determine the values of m1, m2, and m3. Solving these equations gives m1 = 4 kg, m2 = 6 kg, and m3 = 1 kg. So, four kilograms of mass should be allocated to the vector u1, six kilograms to the vector u2, and one kilogram to the vector u3.
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"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).
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
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]
Consider the following cumulative relative frequency distribution.
Class Cumulative
Relative
Frequency
150 up to 200 0.15
200 up to 250 0.23
250 up to 300 0.62
300 up to 350 1.00
a-1. Construct a relative frequency distribution. (Round your answers to 2 decimal places.)
Class Relative
Frequency
150 up to 200
200 up to 250
250 up to 300
300 up to 350
Total
a-2. What percent of the observations are at least 300 but less than 350?
Percent of observations
Answer:
Step-by-step explanation:
The relative frequencies are 0.15, 0.08, 0.39, and 0.38 for the respective classes while 38% of the observations are at least 300 but less than 350.
Explanation:The cumulative relative frequency distribution given can be used to construct a relative frequency distribution. To find the relative frequency for each class, subtract the cumulative frequency of the previous class from the cumulative frequency of the current class. Therefore:
For the class 150 up to 200, its relative frequency would be 0.15 because it's the first class.For the class 200 up to 250, its relative frequency, obtained by subtracting the cumulative frequency of the previous class from that of the present one, would be 0.23 - 0.15 = 0.08.For the class 250 up to 300, the relative frequency would be 0.62 - 0.23 = 0.39.For the class 300 up to 350, the relative frequency would be 1.00 - 0.62 = 0.38.The Total relative frequency is the sum of all the relative frequencies and should equal 1 (or almost equal to 1 due to rounding differences)
For the second part of your question: 'what percent of the observations are at least 300 but less than 350?', since this is the relative frequency distribution, it means that class 300 to 350 has already been expressed as a percentage (since relative frequency is a proportion and can be expressed as a percentage). So, 38% of the observations are at least 300 but less than 350.
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PLZ HELP ASAP 30 POINTS !!!! Which is a zero of the quadratic function f(x) = 9x2 – 54x – 19?
x= 1/3
x= 3 1/3
x= 6 1/3
x= 9 1/3
Answer: x =6 1/3
Step-by-step explanation:
Whichever value of x from the given option that make f(x) = 0 is the zero of f(x). x = 6 1/3 satisfies this, you can check by replacing 6 1/3 by x in the equation. 6 1/3 can be written as 19/3
f(19/3) = 9(19/3)² - 54(19/3) - 19
= 9(361/9) - 54(19/3) - 19
= 361 - 342 - 19
= 0.
The remaining values don't give 0.
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).
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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Based upon market research, the Hawthone Company has determined that consumers are willing to purchase 121 units of their portable media player each week when the price is set at s96.90 per unit. At a unit price f 40.20,cosumers are wiling to buy 310 units per week (a) Determine the weekdy denund equation for this product, ansuming price, p, and quantity, x, are inearly related b) Determine the weekdy revenue fanction A(x) of wnits consumers will demand weekly when the price is $50.70 per portable media player d Determine the mumber of ts consumers will demand weekly when the revenue is maximieed )Detormine the price of each unit when the revene is maximized
Answer:
Step-by-step explanation:
Given that price, p, and quantity, x, are inearly related
We are given two points on this line as (p,x) = (96.90, 121) and (40.20,310)
Using two point formula we find linear equation as
[tex]\frac{y-121}{310-121} =\frac{0-96.90}{40.20-96.90} \\(x-121)(-56.70)=189(p-96,.90)\\189p+56.70x =25174.80[/tex]
a)[tex]189p+56.70x =25174.80[/tex] is the linear equation
b) A(x)
Substitute to get
189(50.70)+56.70x = 25174.80
x=275 units
c) This is linear function hence no local maxima or minima
d) No maximia or minima
System of Linear Equations In Exercises 25–38, solve the system using ei ther Gaussian elimination with back-substitution or Gauss-Jordan elimination.
Hi, you haven't provided the system of linear equations that you need to solve. Therefore, I'll just explain how to use Gauss-Jordan in a system of equations and you can apply the same method to the system of equations you have.
Answer with explanation and step by step solution:
1. For the system of equations:
[tex]4X_{1} + 8X_{2} + 12X_{3} = 36\\8X_{1} + 10X_{2} + 12X_{3} = 48\\4X_{1} + 14X_{2} + 24X_{3} = 60\\[/tex]
2. We can represent it as a matrix by placing every number of the equation as follow:
[tex]\left[\begin{array}{ccccc}4&8&12&|&36\\4&5&6&|&24\\2&7&12&|&30\end{array}\right][/tex]
3. As you can see all the coefficients in the equation are divisible by two, so we can express the system of equations as follow:
[tex]\left[\begin{array}{ccccc}2&4&6&|&18\\4&5&6&|&24\\2&7&12&|&30\end{array}\right][/tex]
4. Gauss-Jordan method solves the system of equations by applying simple operations to the Matrix: Multiplication by non-zero numbers, adding a multiple of one row to another and swapping rows.
Step by step solution:
Divide both sides of equation one by two:
[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\4&5&6&|&24\\2&7&12&|&30\end{array}\right][/tex]
Subtract two times the equation two to the equation three:
[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\4&5&6&|&24\\-6&-3&0&|&-18\end{array}\right][/tex]
Divide equation number three by minus three and subtract two times the equation one to equation two:
[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\2&1&0&|&6\\2&1&0&|&6\end{array}\right][/tex]
Subtract the equation two to the equation three:
[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\2&1&0&|&6\\0&0&0&|&0\end{array}\right][/tex]
Because now we have two equations for three unknown values X1, X2 and X3 the system has an infinite number of solutions.
Equivalente system (From matrix to equation notation):
[tex]1X_{1} + 2X_{2} + 3X_{3} = 9\\2X_{1} + 1X_{2} = 6\\[/tex]
Conclusion:
For whatever system you have you need to convert the system into a matrix notation and using the basic operations, described here, reduce the complexity of the system until:
You have a solution, you discover that the system has an infinite number of solutions or the system of equation is inconsistent.
Example of inconsistency
If after making the basic operations to your system you get a result like this
[tex]\left[\begin{array}{ccccc}7&0&4&|&9\\2&1&0&|&6\\0&0&0&|&-1\end{array}\right][/tex]
You can say that the system is inconsistent because zero is not equal to minus one.
Example of solution
If after making the basic operations to your system you get a result like this
[tex]\left[\begin{array}{ccccc}1&0&0&|&9\\0&1&0&|&-6\\0&0&1&|&-1\end{array}\right][/tex]
You can say that the system have a solution in which X1 = 9, X2 = -6 and X3 = -1
To solve a system of linear equations using Gaussian elimination with back-substitution or Gauss-Jordan elimination, follow the steps of writing the equations in matrix form, performing row operations, creating a diagonal of 1's, and using back-substitution to find the variable values.
Explanation:To solve a system of linear equations using Gaussian elimination with back-substitution or Gauss-Jordan elimination, follow these steps:
Write the system of equations in matrix form.Perform row operations to create zeros below the diagonal elements.Continue row operations to create a diagonal of 1's.Perform back-substitution to find the values of the variables.Example: Solve the system of equations 2x + 3y = 8 and 4x - 2y = 10 using Gaussian elimination with back-substitution.
Write the augmented matrix:The solution to the system of equations is x = 23/8 and y = 3/4.
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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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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.
At most restaurants, the more food you order, the more money you have to pay. This is true whether you go to an overpriced "gourmet" restaurant where the plates are tiny but very expensive, or to a cafeteria where the portions are huge and the prices cheap. If you collect data at any ONE restaurant, the correlation between the amount of food served per person and the price paid for it should be:
A.positive and fairly strong
B.negative and fairly strong
C.fairly weak
Answer:
A.positive and fairly strong
Step-by-step explanation:
Since an increase in food quantity usually means an increase in price, quantity and price are directly proportional, which configures a positive correlation.
Since it is stated that this relationship in observed at most restaurants, it can be concluded that there is a fairly strong correlation between the amount of food served per person and the price paid for it.
Therefore, the answer is, A.positive and fairly strong
Answer:
The answer is A which is Positive and fairly strong.
Step-by-step explanation:
To properly do justice to the selected answer above, we describe a simple scenario.
Imagine yourself, content and perhaps a bit over-full after a lovely meal at a local restaurant. Then, the extravagant bill arrives.
Does this high cost affirm your belief that this meal was valuable and thereby influence your reordering of it?
Or does the cost of the meal overshadow your enjoyment of it and leave you wishing you had chosen a simple meal at a better price point?
What features must an expensive restaurant provide you over a bargain one to justify the extra cost?
Based of the question asked in respect to the scenario above, researchers went on a an experiment collecting different data from a restaurant and used the Latent Dirichlet Allocation (LDA) model to classify each review of the data and key words like ( e.g. food, service, price, ambiance, anecdotes, miscellaneous) were paid attention to.
After analysis, it was perceived that the value is such a tricky parameter to measure, marketers, restaurateurs and economists often overlook it, instead focusing on objective restaurant price and quality’s effect on customer satisfaction.
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 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
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%.
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A researcher studies length of time in college, first through fourth year, and its relation to academic motivation. To get the most detail out of her measures, she assesses each student in both the fall and spring semesters of each of their four years in school. She finds that students have increasingly higher motivation from their first semester to their seventh semester (the start of their fourth year), with a trailing off in the last semester. What is the independent variable in this study?
Answer:
A researcher studies length of time in college, first through fourth year, and its relation to academic motivation. To get the most detail out of her measures, she assesses each student in both the fall and spring semesters of each of their four years in school. She finds that students have increasingly higher motivation from their first semester to their seventh semester (the start of their fourth year), with a trailing off in the last semester. The independent variable are:
Semester in school Year in schoolStep-by-step explanation:
The dependent variable in the given scenario is academic motivation as she is researching on academic motivation so this factor is influencing the results.The semester and year in school are independent variables as these two factors are not influencing the results or research.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