Answer:
Step-by-step explanation:
from the word Bi "meaning two"
Bivariate data are two different variables obtained from the same population element. Bivariate data are used if the sampled data cannot be graphically displayed using a single variable.
To form a bivariate data, there are three combinations of variable
1. Both variables are qualitative i.e attribute in nature
2. Both variables are quantitative i.e numerical
3. One of them is qualitative whilee the other is quantitative
Qualitative data are data that are measure of type which could represent a name, colour,symbol. traits or characteristics
Quantitative data are data that can be counted, measured and expressed using numbers/
Question 5 (Fill-In-The-Blank Worth 1 points)
(05.05 MC)
A system of equations is shown below:
6x - 5y = 5 ,
3x + 5y = 4
The x-coordinate of the solution to this system of equations is
Numerical Answers Expected!
Answer:
The x-coordinate of the solution to this system of equations is 1.
Step-by-step explanation:
Given,
[tex]6x - 5y = 5\\\\3x + 5y = 4[/tex]
We have to find out the x-coordinate of the equation.
Solution,
Let [tex]6x-5y=5\ \ \ \ equation\ 1[/tex]
Again let [tex]3x+5y=4\ \ \ \ \ equation \ 2[/tex]
Now using elimination method we will solve the equations.
For this we will add equation 1 and equation 2 and get;
[tex](6x-5y)+(3x+5y)=5+4\\\\6x-5y+3x+5y=9\\\\9x=9[/tex]
Now on dividing both side by '9' we get;
[tex]\frac{9x}{9}=\frac{9}{9}\\\\x=1[/tex]
Hence The x-coordinate of the solution to this system of equations is 1.
1
ur welcome homie
poggers
Philip ran out of time while taking a multiple-choice test and plans to guess on the last 444 questions. Each question has 555 possible choices, one of which is correct. Let X=X=X, equals the number of answers Philip correctly guesses in the last 444 questions. Assume that the results of his guesses are independent.
What is the probability that he answers exactly 1 question correctly in the last 4 questions?
Answer:
There is a 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either it is correct, or it is 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:
There are four questions, so n = 4.
Each question has 5 options, one of which is correct. So [tex]p = \frac{1}{5} = 0.2[/tex]
What is the probability that he answers exactly 1 question correctly in the last 4 questions?
This is [tex]P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{4,1}*(0.2)^{1}*(0.8)^{3} = 0.4096[/tex]
There is a 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.
Answer:
0.41
Step-by-step explanation:
kahn
Which relationship is a direct variation?
Answer:
A relationship is said to have direct variation when one variable changes and the second variable changes proportionally; the ratio of the second variable to the first variable remains constant. For example, when y varies directly as x, there is a constant, k, that is the ratio of y:x.
Suppose the coefficient matrix of a linear system of four equations in four variables has a pivot in each column. Explain why the system has a unique solution. What must be true of a linear system for it to have a unique solution? Select all that apply.
If the coefficient matrix has a pivot in each column, it means that it is shaped like this:
[tex]A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right][/tex]
So, the correspondant system
[tex]Ax = b[/tex]
will look like this:
[tex]\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right][/tex]
This turn into the following system of equations:
[tex]\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}[/tex]
The last equation is solvable for [tex]x_4[/tex]: we easily have
[tex]x_4=\dfrac{b_4}{a_{4,4}}[/tex]
Once the value for [tex]x_4[/tex] is known, we can solve the third equation for [tex]x_3[/tex]:
[tex]x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}[/tex]
(recall that [tex]x_4[/tex] is now known)
The pattern should be clear: you can use the last equation to solve for [tex]x_4[/tex]. Once it is known, the third equation involves the only variable [tex]x_3[/tex]. Once
You are certain to get 3 jacks when selecting 51 cards from a shuffled deck. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.
The question pertains to calculating the probability of drawing exactly 3 jacks from 51 randomly drawn cards from a 52-card deck. While the probability is very high, it's not an absolute certainty. The exact calculations involve complex combinatorial mathematics.
Explanation:The subject of this question pertains to probability in mathematics, specifically to calculate the likelihood of drawing 3 jacks when selecting 51 cards from a shuffled deck of 52 cards.
First off, we need to understand that in a well-shuffled 52-card deck, there are 4 Jacks. Even if you select 51 out of 52 cards, there isn't a guarantee that you will select 3 jacks because the selection is random. The scenario you provided indicates a nearly certain event (since you're pulling nearly all the cards), but it still isn't an absolute certainty.
The exact probability computation for this kind of problem are more complex as they would involve combinatorial calculations. For simplicity, let's consider a similar but simpler scenario. Let's assume you are drawing just 4 cards instead. The probability of getting exactly 3 Jacks would be a combination of the probability of picking a Jack, and the probability of picking a non-Jack card. This would be calculated as (C(4,3) * C(48,1)) / C(52,4), with C representing the combination formula. This gives us how many ways we can draw 3 Jacks and a non-Jack divided by how many ways we can draw any 4 cards.
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Final answer:
The probability of drawing 3 jacks from a standard deck of 52 cards when selecting 51 is 1 (certainty), as it is a guaranteed event given the conditions.
Explanation:
The question asks about the probability of a certain event occurring when dealing with a standard deck of 52 cards. In this case, the event is being certain to get 3 jacks when selecting 51 cards out of 52. Since there are 4 jacks in the deck, and upon drawing 51 cards you're left with only 1 card that is not drawn, it is guaranteed that you'll have the 3 jacks among the drawn cards.
Hence, the probability is 1 (certainty), as there is only one card you're not drawing and 4 chances to have drawn a jack, which means you will always end up with all 3 jacks among the chosen 51 cards.
An SRS of 350 350 high school seniors gained an average of ¯ x = 22.61 x¯=22.61 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 53.63 . σ=53.63. We want to estimate the mean change in score μ μ in the population of all high school seniors. (a) Using the 68 68 – 95 95 – 99.7 99.7 Rule or the z - z- table (Table A), give a 95 % 95% confidence interval ( a , b ) (a,b) for μ μ based on this sample.
Answer: (16.9914, 28.2286).
Step-by-step explanation:
The formula to find the confidence interval for population mean is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\sigma[/tex]= Population standard deviation
n= Sample size.
z* = Critical value.
Let μ be the mean change in score in the population of all high school seniors.
As per given , we have
n= 350
[tex]\overline{x}=22.61[/tex]
[tex]\sigma=53.63[/tex]
The critical z-value for 95% confidence interval is z*= 1.96 [From z-table]
Substitute all the value in formula , we get
[tex]22.61\pm (1.96)\dfrac{53.63}{\sqrt{350}}[/tex]
[tex]=22.61\pm (1.96)\dfrac{53.63}{18.708287}[/tex]
[tex]=22.61\pm (1.96)(2.8666)[/tex]
[tex]=22.61\pm (5.6186)[/tex]
[tex]=(22.61-5.6186,\ 22.61+5.6186) =(16.9914,\ 28.2286)[/tex]
Hence, the 95% confidence interval for [tex]\mu[/tex] is (16.9914, 28.2286).
The following scores represent the results of a midterm exam in Statistics class. 25 35 43 44 47 48 54 55 56 57 59 62 63 65 66 68 69 69 71 72 72 73 74 76 77 77 78 79 80 81 81 82 83 85 89 92 93 94 97 98 a) Find the lower and upper quartiles for the data. b) Find the interquartile range. c) Construct a boxplot for this data set.
Answer:
a.
lower Quartile= 57.5
Upper Quartile=81
b.
23.5
c.
box-plot is attached in excel file
Step-by-step explanation:
The data is arranged in ascending order so, the lower quartile denoted as Q1 can be calculated as under
Q1=((n+1)/4)th score=(41/4)th score=(10.25)th score
Q1=10th score+0.25(11th-10th)score
Q1=57+0.25(59-57)=57+0.5=57.5
Q1=57.5
The data is arranged in ascending order so, the third quartile denoted as Q3 can be calculated as under
Q3=(3(n+1)/4)th score=(3*41/4)th score=(30.75)th score
Q3=30th score+0.75(31th-30th)score
Q3=81+0.75(81-81)=81+0=81
Q3=81
b)
Interquartile range=IQR=Q3-Q1=81-57.5=23.5
IQR=23.5
c)
The box-plot is made in excel and it shows no outlier. The box-plot shows the 5-number summary(minimum-Q1-median-Q3-maximum) as 25-57.5-72-81-98.
The exponential probability distribution is a discrete distribution that is often used to describe time between customer arrivals.
Answer:
True
Step-by-step explanation:
The time between customer arrivals is called inter-arrival time. According to Queueing Notation, the inter-arrival time can be model based on difference probability distribution. The probability distribution by which the inter-arrival time can be modeled include:
Exponential Distribution or Markov distributionConstant or DeterministicHyper - exponentialArbitrary or General distributionA student earned grades of Upper AA, Upper DD, Upper AA, Upper CC, and Upper BB. Those courses had the corresponding numbers of credit hours 44, 22, 22, 33, and 11. The grading system assigns quality points to letter grades as follows: Aequals=4; Bequals=3; Cequals=2; Dequals=1; Fequals=0. Compute the grade point average (GPA) as a weighted mean and round the result with two decimal places. If the Dean's list requires a GPA of 3.00 or greater, did this student make the Dean's list? The grade point average is nothing. (Round to two decimal places as needed.) Did this student make the Dean's list? A. Yes because at least two of the student grades are B or above B. No because the students GPA is not 4.0 C. NoNo because the student has at least one grade lessless than 3 D. NoNo because the student's GPA is lessless than 3.0
Answer:
The grade point average is 2.92The student didn't make the Dean's list because the student's GPA is less than 3.0Step-by-step explanation:
I take the grades as A,D,A,C,B not AA,DD,AA,CC,BB.I take numbers of credit hours as 4,2,2,3,1 not as 44, 22, 22, 33, and 11.Since quality points to letter grades are A=4; B=3; C=2; D=1; F=0, weighted mean is the sum of the qulity points times corresponding credit hours divided by the total credit hours:
[tex]\frac{(4*4) + (1*2) + (4*2) + (2*3) + (3*1)}{12}[/tex] ≈ 2.92
Since 2.92<3.0, the student is not in Dean's list.
Find the mean amount hospitals had to pay in wrong-site lawsuits. Round your answer to the nearest whole dollar.
Answer:
dont see much information here but as far as lawsuits go id aim for the highest answer
Step-by-step explanation:
_____________________________________
In a sample of 11 men, the mean height was 178 cm. In a sample of 30 women, the mean height was 167 cm. What was the mean height for both groups put together?
Answer:
I'm pretty sure it would be 345, just add the two 178 and 167
The stop-board of a shot-put circle is a circular arc 1.22 m in length. The radius of the circle is 1.06 m. What is the central angle?
Answer:
Central angle= 1.15 radians
Step-by-step explanation:
[tex]Arc\,\,length=s= 1.22\,m\\Radius=r=1.06\,m\\\\Central\,\, angle=\theta=?\\\\Using\\\\ s=r\theta\\\\\theta=\frac{s}{r}\\\\\theta= \frac{1.22}{1.06}\\\\\theta=1.15 \,rad[/tex]
For each part, give a relation that satisfies the condition. a. Reflexive and symmetric but not transitive b. Reflexive and transitive but not symmetric c. Symmetric and transitive but not reflexive
Answer:
For the set X = {a, b, c}, the following three relations satisfy the required conditions in (a), (b) and (c) respectively.
(a) R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)} is reflexive and symmetric but not transitive .
(b) R = {(a, a), (b, b), (c, c), (a, b)} is reflexive and transitive but not symmetric .
(c) R = {(a,a), (a, b), (b, a)} is symmetric and transitive but not reflexive .
Step-by-step explanation:
Before, we go on to check these relations for the desired properties, let us define what it means for a relation to be reflexive, symmetric or transitive.
Given a relation R on a set X,
R is said to be reflexive if for every [tex]a \in X, (a,a) \in R[/tex].
R is said to be symmetric if for every [tex](a, b) \in R, (b, a) \in R[/tex].
R is said to be transitive if [tex](a, b) \in R[/tex] and [tex](b, c) \in R[/tex], then [tex](a, c) \in R[/tex].
(a) Let R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)}.
Reflexive: [tex](a, a), (b, b), (c, c) \in R[/tex]
Therefore, R is reflexive.
Symmetric: [tex](a, b) \in R \implies (b, a) \in R[/tex]
Therefore R is symmetric.
Transitive: [tex](a, b) \in R \ and \ (b, c) \in R[/tex] but but (a,c) is not in R.
Therefore, R is not transitive.
Therefore, R is reflexive and symmetric but not transitive .
(b) R = {(a, a), (b, b), (c, c), (a, b)}
Reflexive: [tex](a, a), (b, b) \ and \ (c, c) \in R[/tex]
Therefore, R is reflexive.
Symmetric: [tex](a, b) \in R \ but \ (b, a) \not \in R[/tex]
Therefore R is not symmetric.
Transitive: [tex](a, a), (a, b) \in R[/tex] and [tex](a, b) \in R[/tex].
Therefore, R is transitive.
Therefore, R is reflexive and transitive but not symmetric .
(c) R = {(a,a), (a, b), (b, a)}
Reflexive: [tex](a, a) \in R[/tex] but (b, b) and (c, c) are not in R
R must contain all ordered pairs of the form (x, x) for all x in R to be considered reflexive.
Therefore, R is not reflexive.
Symmetric: [tex](a, b) \in R[/tex] and [tex](b, a) \in R[/tex]
Therefore R is symmetric.
Transitive: [tex](a, a), (a, b) \in R[/tex] and [tex](a, b) \in R[/tex].
Therefore, R is transitive.
Therefore, R is symmetric and transitive but not reflexive .
Relation from the set of two variables is subset of certain product. The relation for the condition are,
[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]
[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]
[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]
Relation-Relation from the set of two variables is subset of certain product. Relation are of three types-
ReflexiveSymmetricTransitive1) Reflexive and symmetric but not transitive -
Let a data set as,
[tex]X=1,2,3[/tex]
For the data set the relation can be given as,
[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]
[tex]R_1[/tex] is reflexive as it can be represent as [tex]R_1(a,a)[/tex] for,
[tex]a=1,2,3, \;\;\;\;\; [/tex]
[tex]a[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is symmetric as it can be represent as [tex]R_1(a,b)[/tex] for,
[tex]a,b \;\;\;\;(1,2) (2,1)[/tex]
[tex]a,b[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is not transitive as it can be represent as [tex]R_1\neq (a,c)[/tex] .
[tex]a,c\neq \;\;\;\;(1,3) (3,1)[/tex]
2) Reflexive and transitive but not symmetric
Let a data set as,
[tex]X=1,2,3[/tex]
For the data set the relation can be given as,
[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]
[tex]R_2[/tex] is reflexive as it can be represent as [tex]R_2(a,a)[/tex] for,
[tex]a=1,2,3, \;\;\;\;\; [/tex]
[tex]a[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is transitive as it can be represent as [tex]R_1(a,c)[/tex] for,
[tex]a,c \;\;\;\;(1,3) (3,1)[/tex]
[tex]a,c[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is not symmetric as it can be represent as [tex]R_1\neq (a,b)[/tex] .
[tex]a,b\neq \;\;\;\;(1,2) (2,1)[/tex]
3) Symmetric and transitive but not reflexive
Let a data set as,
[tex]X=1,2,3[/tex]
For the data set the relation can be given as,
[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]
[tex]R_1[/tex] is symmetric as it can be represent as [tex]R_3(a,b)[/tex] for,
[tex]a,b=(1,2),(2,1) \;\;\;\;\; [/tex]
[tex]a,b[/tex] ∈ [tex]X[/tex]
[tex]R_3[/tex] is transitive as it can be represent as [tex]R_3(a,c)[/tex] for,
[tex]a,c \;\;\;\;(1,3) (3,1)[/tex]
[tex]a,c[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is not reflexive as it can be represent as [tex]R_3\neq (a,a)[/tex] .
[tex]a,a\neq \;\;\;\;(1,1) [/tex]
Thus the relation for the condition are,
[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]
[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]
[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]
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Samples of skin experiencing desquamation are analyzed for both moisture and melanin content. The results from 100 skin samples are as follows: melanin content high low moisture high 13 10 content low 47 30 Let A denote the event that a sample has low melanin content, and let B denote the event that a sample has high moisture content. Determine the following probabilities. Round your answers to three decimal places (e.g. 98.765).
a) P(A)
b) P(B)
c) P (A|B)
d) P (BA)
Answer: a. 0.40 b. 0.23 c . 0.435 d . 0.25
Step-by-step explanation:
melanin content Total
high low
moisture high 13 10 23
content low 47 30 77
Total 60 40 100
Let A denote the event that a sample has low melanin content, and let B denote the event that a sample has high moisture content.
a) Total skin samples has low melanin content = 10+30=40
P(A)=[tex]\dfrac{40}{100}=0.40[/tex]
b) Total skin samples has high moisture content = 13+10=23
P(B) =[tex]\dfrac{23}{100}=0.23[/tex]
c) A ∩ B = Total skin samples has both low melanin content and high moisture content =10
P(A ∩ B) =[tex]\dfrac{10}{100}=0.10[/tex]
Using conditional probability formula , [tex]P (A|B)=\dfrac{P(A\cap B)}{P(B)}[/tex]
[tex]P (A|B)=\dfrac{0.10}{0.23}=0.434782608696\approx0.435[/tex]
d) [tex]P (B|A)=\dfrac{P(A\cap B)}{P(A)}[/tex]
[tex]P (B|A)=\dfrac{0.10}{0.40}=0.25[/tex]
10- [6-2•2 + (8-3)]•2
Answer:
10-[6-4+(5)]×2
10-[2+5]×2
10-(7)×2
10-14= -4
how many ways are there to list the digits {1,2,2,3,4,5,6} so that identical digits are not in consecutive position?
Answer: 2520 ways
Step-by-step explanation:
7!/2!
Answer:
no
Step-by-step explanation:
indicate if the following systems are lineare or non linjear systems d^2x/dt+5dx/dt+10x = 0
Answer: You have only provided one Differential Equation (DE), it looks like you intended listing more.
The equation you wrote contains an incorrect d²x/dt, it is likely to be d²x/dt² + 5dx/dt + 10x = 0, which is linear. Unless it is (dx/dt)² + 5dx/dt + 10x = 0, then it is nonlinear.
Not to worry though, I will explain what linear and nonlinear DE's are.
Step-by-step explanation:
LINEAR DE: This is the kind of DE in which the functions of the dependent variable are linear. There are no powers of the dependent variable and/or its derivatives, there are no products of the dependent variable and its derivative, there are no functions of the dependent variable like cos, sin, exp, etc.
Example:
* 5d²x/dt² + dx/dt - x = 2t
This is linear, as it satisfies all the conditions.
NONLINEAR DE: If any condition explained for linear DE is not satisfied, then it is called nonlinear.
Example:
* d²x/dt² - sinx = 0
This is nonlinear because of the presence of sinx.
* d²x/dt² + xdx/dt = 0
This is nonlinear because of the product of the dependent variable, x, and its derivative, dx/dt.
* d²x/dt² + x² = 0
This is nonlinear because a function of the dependent variable is not linear. You shouldn't have x².
* (dx/dt)³ + 3dx/dt = 0 is equally nonlinear. You can't have nonlinear functions of the dependent variable or its derivatives.
I hope this helps answer the remaining parts of your question.
PLEASE HELP!!!
Carol paid $0.78 per liter for gas while driving across Canada. Find the cost per gallon to the nearest cent.
Please give a step by step
Answer:
2.95 cent
Step-by-step explanation:
1 gallon = 231 cubic inches
1 litre = 1000ml = 61.0237 cubic inches
1 galloon = 231 / 61.0237 = 3.7854118 liters
if Carol paid $0.78 per litre
1 galloon = 0.78 x 3.7854118 = 2.952621204 ≅ 2.95 cent
Evaluate the limit using the appropriate Limit Law(s). (If an answer does not exist, enter DNE.) lim x→8 1 + 3 x 5 − 6x2 + x3
Answer:
[tex] [tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)[/tex]=[tex]1-384 +512+3\sqrt{8} -18(8)^{5/2} +3 (8)^{7/2} =1223.601[/tex]
And the limit on this case exists.
Step-by-step explanation:
We want to find the following limit:
[tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)[/tex]
First we can distribute the polynomials like this:
[tex] lim_{x \to 8} (1-6x^2 +x^3+3\sqrt{x} -18 x^{5/2} +3x^{7/2})[/tex]
And Now we can use the distributive property for the limit and we got:
[tex] lim_{x \to 8} 1 - 6 lim_{x \to 8} x^2 + lim_{x \to 8} x^3 +3 lim_{x \to 8} \sqrt{x} -18 lim_{x \to 8} x^{5/2} + 3 lim_{x \to 8} x^{7/2}[/tex]
And now we can evaluate the limit and we got:
[tex] [tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)[/tex]=[tex]1-384 +512+3\sqrt{8} -18(8)^{5/2} +3 (8)^{7/2} =1223.601[/tex]
And the limit on this case exists.
To solve limit problems in mathematics, limit laws are often very useful. In this specific case, as the function is a polynomial and defined for all real number values, a direct substitution of x=8 into the function is sufficient. Therefore, the limit as x approaches 8 for function 1 + 3x5 - 6x2 + x3 is calculable.
Explanation:In the field of mathematics, limit laws are used quite frequently for evaluating limits. In this case, we want to calculate the limit as x approaches 8 for the function 1 + 3x5 - 6x2 + x3.
For a given polynomial function like this one, an easy and very straightforward approach is to substitute the value x is approaching (in this scenario, x = 8) directly into the polynomial function.
So, after substitution, our function becomes: 1 + 3*(8)^5 - 6*(8)^2 + (8)^3. Simplifying it further, the limit as x approaches 8 of this function gives us a definite numeric value.
Always remember while applying limit laws, you might at times need the limit laws to evaluate complex limit problems but in this given scenario, direct substitution works perfectly fine because this polynomial function is defined for all real number values of X.
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Equations - Item 2829
The circumference (C) of a circle is 16 cm. Which formula can you use to find the
diameter (d) if you know that C = 3.14d?
The formula is used to find the diameter of circle is: [tex]d = \frac{C}{3.14}[/tex]
The diameter of circle is 5.1 cm
Solution:
Given that,
Circumference (C) of a circle is 16 cm
The formula for circumference of circle when diameter is given is:
[tex]C = \pi d\\\\\pi \text{ is a constant equal to 3.14}\\\\C = 3.14d[/tex]
Rearrange the formula to get "d"
Divide both sides by 3.14
[tex]d = \frac{C}{3.14}[/tex]
The above formula is used to find the diameter of circle
Given that, circumeference = C = 16 cm
Substituting we get,
[tex]d = \frac{16}{3.14}\\\\d = 5.095 \approx 5.1[/tex]
Thus diameter of circle is 5.1 cm
Assume that about 30% of all U.S. adults try to pad their insurance claims. Suppose that you are the director of an insurance adjustment office. Your office has just received 140 insurance claims to be processed in the next few days. What is the probability that from 45 to 47 of the claims have been padded?
a. 0.222
b. 0.167
c. 0.119
d. 0.104
e. 0.056
Answer:
For x=45
sample proportion=45/140=0.321
z=(0.321-0.30)/sqrt(0.3*(1-0.3)/140)
z=0.54
For x=47
sample proportion=47/140=0.336
z=(0.336-0.30)/sqrt(0.3*(1-0.3)/140)
z=0.93
Now,
P(0.54<z<0.93)=P(z<0.93)-P(z<0.54)
=0.8238-0.7054
=0.118
So,correct option is 0.119
Find the vector representing the area of the triangle ABC where A=(4,5,6), B=(6,4,5) and C=(5,4,6) oriented so that it faces upward.
Answer: Area of triangle is √3 / 2
Step-by-step explanation:
The explanation can be found in the attached in picture
A student who has created a linear model is disappointed to find that herR2 value is a very low 13%. a) Does this mean that a linear model is not appropriate? Explain. b) Does this model allow the student to make accurate predictions? Explain.
Answer:
a) No it doesn't mean that linear model is inappropriate
b) No. The prediction using this model will not be accurate.
Step-by-step explanation:
a)
For answering this part, firstly consider the concept of [tex]R^{2}[/tex]
The [tex]R^{2}[/tex] also known as coefficient of determination is used to determine the amount of variability in dependent variable is explained by the linear model. Lower [tex]R^{2}[/tex] depicts that less variation of dependent is explained by the independent variable using the linear model. The linearity of model is determined by scatter plot. Thus, if the [tex]R^{2}[/tex] is lower, it doesn't mean that linear model is inappropriate.
b)
The predictions made by the model having lower [tex]R^{2}[/tex] value are erroneous. The model is used for prediction if the linear model explains the larger portion of variability in dependent variation. If the predictions made from the model that have lower [tex]R^{2}[/tex] value then the predicted values will not be close to the actual value and thus residuals will not be minimum as residuals are the difference of actual and predicted values.
An airplane has a front nad a rear door that are bother openedto allow passengers to exit when the plane lands. the planehas 100 passengers seated. the number of passengers exitingthrought the front door shougl have
a) a binomial distribution with mean 50
b) a binomial distribution with 100 trials but successprobability not equal to .5
c)a normal didtribution with a standard deviation of5
d) none of the above
Answer:
a) a binomial distribution with mean 50
Step-by-step explanation:
Given that an airplane has a front nad a rear door that are bother opened to allow passengers to exit when the plane lands. the plane has 100 passengers.
These 100 passengers can select either back door or front door with equal probability (assuming)
so probability for selecting front door = 0.5
No of passengers =100
Each passenger is independent of the other
Hence X no of passengers exiting through the front door is binomial with
p =0.5 and n =100
Mean of the variable X = np = 100(0.5) = 50
Variance of X = 100(0.5)(0.5)
Hence std dev = 10(0.5) = 5
So correct answers are
a) a binomial distribution with mean 50
A general 2x2 diagonal matrix has the form(a00b). Thus the two unknown real numbers a b are needed to specify each 2x2 diagonal matrix. In Exercises 11 16, how many unknown real numbers are needed to specify each of the given matrices
1. An upper triangular 2x2 matrix?
2.) An m × n matrix?
Answer:
1. 3, and 2. m x n
Step-by-step explanation:
1. for an upper triangular 2x2 matrix i.e. (a,0,c,d), three (03) unknown elements a, c, and d are needed to be specified.
2. for m x n matrix, m*n elements are needed to be specified.
Final answer:
To specify an upper triangular 2x2 matrix, 3 unknown real numbers are needed. For an m × n matrix, m × n unknown real numbers are required.
Explanation:
The question asks how many unknown real numbers are needed to specify each of the given matrices: an upper triangular 2x2 matrix, and an m × n matrix.
1. An Upper Triangular 2x2 Matrix
An upper triangular matrix has the form:
(a, b)
(0, c)
Thus, to specify an upper triangular 2x2 matrix, 3 unknown real numbers are needed: a, b, and c.
2. An m × n Matrix
An m × n matrix has m rows and n columns. To specify such a matrix, one needs m × n unknown real numbers, representing each element in the matrix.
Find the equation of the plane that is parallel to the vectors left angle 3 comma 0 comma 3 right angle and left angle 0 comma 1 comma 3 right angle, passing through the point (2 comma 0 comma negative 1 ).
Answer:
[tex]x + 3y -z - 3 = 0[/tex]
Step-by-step explanation:
We have to find the equation of plane that is parallel to the vectors
[tex]\langle 3,0,3\rangle, \langle0,1,3\rangle[/tex]
The plane also passes through the point (2,0,-1).
Hence, the equation of plane s given by:
[tex]\displaystyle\left[\begin{array}{ccc}x-2&y-0&z+1\\3&0&3\\0&1&3\end{array}\right]\\\\=(x-2)(0-3) - (y-0)(9-0) + (z+1)(3-0)\\=-3(x-2)-9y+3(z+1)\\\Rightarrow -3x + 6 - 9y + 3z + 3 = 0\\\Rightarrow 3x + 9y -3z -9 = 0\\\Rightarrow x + 3y -z - 3 = 0[/tex]
It is the required equation of plane.
Which represents a quadratic function?
f(x) = −8x3 − 16x2 − 4x
f (x) = three-quarters x 2 + 2x − 5
f(x) = StartFraction 4 Over x squared EndFraction minus StartFraction 2 Over x EndFraction + 1
f(x) = 0x2 − 9x + 7
Answer:
The answer to your question is the second option
Step-by-step explanation:
A Quadratic function is a polynomial of degree two. That means that the higher exponent is 2.
a) This option is incorrect because the highest power is 3 not two.
b) This option is the right answer, the highest power is 2, so, it is a quadratic function.
c) This option is incorrect, the highest power is -2.
d) This option is incorrect, the highest option is 1.
Answer:
Option 2 is the correct answer
Step-by-step explanation:
A quadratic function is a function in which the highest power to which the variable is raised is 2
1) f(x) = −8x3 − 16x2 − 4x
The given function is a cubic function because the highest power
to which the variable,x is raised is 3
2) f(x) = 3x²/4 + 2x - 5
The given function is a quadratic function because the highest power
to which the variable,x is raised is 2
3) f(x) = 4/x² - 2/x + 1
It can be rewritten as
f(x) = 4x^-2 - 2x^-1 + 1
The given function is not a quadratic function because the highest power to which the variable,x is raised is - 2
4) f(x) = 0x2 − 9x + 7
It can be rewritten as
f(x) = - 9x + 7
The given function is not a quadratic function because the highest power to which the variable,x is raised is 1
1. Suppose the coefficient matrix of a linear system of four equations in four variables has a pivot in each column. Explain why the system has a unique solution.
2. What must be true of a linear system for it to have a unique solution?
Select all that apply.
A. The system has no free variables.
B. The system has one more equation than free variable.
C. The system is inconsistent.
D. The system is consistent. Your answer is correct.
E. The system has at least one free variable.
F. The system has exactly one free variable.
Answer:its A
Step-by-step explanation:it was
Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial)2x^2 + 6x - 362(x^2 + 3x - 18)
Answer:
2(x+6)(x-3)
Step-by-step explanation:
Factor the GCF out of the trinomial on the left side of the equation.
[tex]2x^2 + 6x - 36 =2(x^2 + 3x - 18)[/tex]
Greatest common factor of 2, 6, 18 is 2
so GCF is 2
divide each term when we take out GCF 2
so [tex]2(x^2 + 3x - 18)[/tex]
now factor the trinomial
product is -18 and sum is +3
6 times -3 is -18 and 6-3=3
[tex]2(x^2+3x-18)\\2(x+6)(x-3)[/tex]
Determine which matrices are in reduced echelon form and which others are only in echelon form. a. [Start 3 By 4 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 0 4st Column 0 2nd Row 1st Column 0 2nd Column 2 3rd Column 0 4st Column 0 3rd Row 1st Column 0 2nd Column 0 3rd Column 1 4st Column 1 EndMatrix ]1 0 0 0 0 2 0 0 0 0 1 1 b. [Start 3 By 4 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 1 4st Column 1 2nd Row 1st Column 0 2nd Column 1 3rd Column 1 4st Column 1 3rd Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 0 EndMatrix ]1 0 1 1 0 1 1 1 0 0 0 0 c. [Start 4 By 4 Matrix 1st Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 0 2nd Row 1st Column 1 2nd Column 3 3rd Column 0 4st Column 0 3rd Row 1st Column 0 2nd Column 0 3rd Column 1 4st Column 0 4st Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 1 EndMatrix ]
Answer:
Step-by-step explanation:
Check the attachment for the solution
Answer:
Echelon form.Reduced Echelon form.Neither.Step-by-step explanation:
The objective is to determine which of the following matrices are in reduced echelon form and which others are only in echelon form. The given matrices are
[tex]\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0& 2 & 0 & 0 \\ 0& 0 & 1 & 1 \end{bmatrix}[/tex], [tex]\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0& 1& 1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}[/tex] and [tex]\begin{bmatrix} 0& 0 & 0 & 0 \\ 1& 3 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0& 0 & 0 & 1 \end{bmatrix}[/tex].
First, recall what is an echelon and reduced echelon form of a matrix.
A matrix is said to be in a Echelon form if
If there is any zero rows, all nonzero rows are placed above them;Each first non-zero entry in a row, which is the leading entry, is placed to the right of the leading entry of the row above it;All elements below the leading entry must be equal to zero in each column.A matrix is said to be in a Reduced Echelon form if
In each non-zero row, the leading entry is 1.In its column, each leading 1 is actually the only non-zero element.A column that contains a leading 1 which is the only non-zero element is called a pivot column.
Now, let's have a look at the first matrix
[tex]\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0& 2 & 0 & 0 \\ 0& 0 & 1 & 1 \end{bmatrix}[/tex]
As we can see, it doesn't have any zero rows. Each leading entry in a row is placed to the right of the leading entry from the row above and all elements below the leading entries in all columns are equal to zero. Therefore, this matrix is in an Echelon form.
In the second row, the leading entry is 2, not 1, so because of the first property of the Reduced Echelon form, it is not in a Reduced Echelon form.
Notice that it can be transformed to the Reduced Echelon form by multiplying the second row by [tex]\frac{1}{2}.[/tex]
The second matrix is
[tex]\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0& 1& 1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}[/tex]
There is a zero row, and all non-zero rows are placed above it. Each leading entry in a row, which is the first non-zero entry, is placed to the right of the entry of the row above it and all elements below the leading entry are equal to zero in each column, so it is in the Echelon form.
It is also in the Reduced Echelon form, since all non-zero rows the leading entry is 1 and it is the only non zero element in each column.
The least given matrix is
[tex]\begin{bmatrix} 0& 0 & 0 & 0 \\ 1& 3 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0& 0 & 0 & 1 \end{bmatrix}[/tex]
This matrix doesn't satisfy the condition that if there is any zero-row, it must be below all other non-zero rows, so it is not in Echelon form.
A matrix that is not in an Echelon form, it is not in an Reduced Echelon form either.
Therefore, this matrix is not in an Reduced Echelon form.