Answer:
25%
Step-by-step explanation:
Great question, since a regular coin has two sides one heads and one tails. That gives us a 50% probability of it landing on either side of the coin. Since we would like to know the probability of getting 2 heads in a row, we would need to multiply the probability of the first toss landing on heads with the second toss landing on heads, like so...
[tex]\frac{1}{2} *\frac{1}{2} =\frac{1}{4}[/tex]
So we can see that the probability of us getting two heads in a row is that of \frac{1}{4}[/tex] or 25%.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
(1 point) Suppose that for two random variables X and Y the joint density function is f(x,y)=6xe−x(y+6), for x>0 and y>0. Find each of the following. (a) fX|Y(x,y)= (b) fY|X(x,y)=
The question is about calculating the conditional probability density functions of two random variables X and Y given their joint density function. However, without the marginal density functions for X and Y, it is not possible to provide exact numerical answers. General formulas involve dividing the joint density function by the respective marginal density function.
Explanation:The question relates to the field of probability and statistics, specifically pertaining to joint density functions of two random variables (X and Y). We have the joint density function f(x, y) = 6xe^(−x(y+6)) where x > 0 and y > 0.
For the conditional probability density function, the general formula for fX|Y(x, y) is the joint density function divided by the marginal density function of Y. Similarly, for fY|X(x, y), it is the joint density function divided by the marginal density function of X.
However, the marginal density functions for X and Y are not specified in the question. Typically, to find these, one would integrate the joint density function with respect to the other variable. Due to this missing information, it's not possible to provide an exact numerical answer.
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13. Determine whether B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} is a basis of R3.
Answer: Yes, the given set of vectors is a basis of R³.
Step-by-step explanation: We are given to determine whether the following set of three vectors in R³ is a basis of R³ or not :
B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} .
For a set to be a basis of R³, the following two conditions must be fulfilled :
(i) The set should contain three vectors, equal to the dimension of R³
and
(ii) the three vectors must be linearly independent.
The first condition is already fulfilled since we have three vectors in set B.
Now, to check the independence, we will find the determinant formed by theses three vectors as rows.
If the value of the determinant is non zero, then the vectors are linearly independent.
The value of the determinant can be found as follows :
[tex]D\\\\\\=\begin{vmatrix} -1& 1 & -1\\ 1 & 0 & 2\\ 1 & 1 & 0\end{vmatrix}\\\\\\=-1(0\times0-2\times1)+1(2\times1-1\times0)-1(1\times1-0\times1)\\\\=(-1)\times(-2)+1\times2-1\times1\\\\=2+2-1\\\\=3\neq 0.[/tex]
Therefore, the determinant is not equal to 0 and so the given set of vectors is linearly independent.
Thus, the given set is a basis of R³.
Exercise 2.2 gave the following data (in increasing order) for the attribute age: 13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30, 33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70. (a) Use smoothing by bin means to smooth these data, using a bin depth of 3. Illustrate your steps. Comment on the effect of this technique for the given data. (b) How might you determine outliers in the data? (c) What other methods are there for data smoothing?
Answer:
Step-by-step explanation:
13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30, 33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.
a) Smoothing by bin means
Each bin has depth of 3
Dividing data into bins
so, Bin 1= 13, 15, 16
Bin 2= 16, 19, 20
Bin 3= 20, 21, 22
Bin 4= 22, 25, 25
Bin 5= 25, 25, 30
Bin 6= 33, 33, 35
Bin 7 = 35, 35, 35
Bin 8= 36, 40, 45
Bin 9= 46, 52, 70
Now, smoothing data by bin mean
so, Bin 1= 13, 15, 16 = (13+15+16)/3 = 15 Bin 1 = 15,15,15
Bin 2= 16, 19, 20 = (16+19+20)/3 = 18 Bin 2 = 18,18,18
Bin 3= 20, 21, 22= (20+21+22)/3 = 21 Bin 3 = 21,21,21
Bin 4= 22, 25, 25 = (22+25+25)/3 = 24 Bin 4 = 24,24,24
Bin 5= 25, 25, 30 =(25+25+30)/3 = 27 Bin 5 = 27,27,27
Bin 6= 33, 33, 35 = (33+33+35)/3 = 34 Bin 6 = 34,34,34
Bin 7 = 35, 35, 35 = (35+35+35)/3 = 35 Bin 7 = 35,35,35
Bin 8= 36, 40, 45 = (36+40+45)/3 = 40 Bin 8 = 40,40,40
Bin 9= 46, 52, 70=(46+52+70)/3 = 56 Bin 9 = 56,56,56
This technique is used to smooth the data. Data may have noise, using binning techniques we can remove noise from the data. It helps in providing more accurate results
b) How might you determine outliers in the data?
Outliers are the data that are abnormal to other data points. Outliers can be found by Box and whisker chart (box plot). Inter Quartile range can also be used to identify outliers
c) What other methods are there for data smoothing?
Other methods of smoothing data are
a) binning by boundaries
b) Exponential smoothing
c) Random walk
The correct statement will be that the method of binning is extensively used for smoothening the data and cancels the noise a data contains. Outliers of a data can be determined using the quartile range.
There are different methods of data smoothening, such as binning with the help of boundaries, binning with the help of outliers, exponential binning, etc.
Binning of attributes.The bins will be created using the depth of 3 as,
Bin 1- 13,15,16 ; Bin 2- 16,19,20 ; Bin 3 20, 21, 22 ; Bin 4- 22,25,25 ; Bin 5- 25, 25, 30 ; Bin 6- 33,33,35 ; Bin 7- 35,35,35 ; Bin 8- 36,40,45 ; Bin 9- 46,52,70.
The vales of bins will be averaged and rounded off to the nearest whole numbers, considering bins as b in the following way.
[tex]b1= \dfrac {13+15+16}{3}\\\\b1= 14.67[/tex]
Continuing further in similar ways, we will find the values of remaining bins as
b2- 18b3- 21b4- 24b5- 27b6- 34b7- 35b8- 40b9- 56The outliers in the data can be found by using the box plot method and quartile range functions. Normally, such outliers are referred to as the mismatching data in a bin.
There are various different methods for smoothing the data. Some ways of smoothing the data are exponential smoothing, random walk smoothing, boundary binning method, etc.
Hence, the data is smoothened using the binning method and the values obtained are as above.
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Suppose a random sample of 90 companies taken in 2006 showed that 14 offered high-deductible health insurance plans to their workers. A separate random sample of 120 firms taken in 2007 showed that 30 offered high-deductible health insurance plans to their workers. Based on the sample results, can you conclude that there is a higher proportion of companies offering high-deductible health insurance plans to their workers in 2007 than in 2006? Conduct your hypothesis test at a level of significance alphaequals0.01.
Answer:
Step-by-step explanation:
Given that a random sample of 90 companies taken in 2006 showed that 14 offered high-deductible health insurance plans to their workers. A separate random sample of 120 firms taken in 2007 showed that 30 offered high-deductible health insurance plans to their workers.
H0: p1=p2
Ha: p1 <p2
(Two tailed test at99%)
Difference 14.44 %
Chi-squared 5.883
DF 1
Significance level P = 0.0153
Since p >0.01, our alpha reject null hypothesis.
NO. Based on the sample results, you can not conclude that there is a higher proportion of companies offering high-deductible health insurance plans to their workers in 2007 than in 2006
Thirty-five more than the limit of weight (w) in an elevator is greater then 1,050 pounds.
A. Model with an inequality.
B. Solve the inequality.
C. Where would the solution for the inequality be graphed on the number line?
Answer:
Part A) [tex]w+35 > 1,050[/tex]
Part B) [tex]w > 1,015\ pounds[/tex]
Part C) The graph in the attached figure
Step-by-step explanation:
Part A) Model with an inequality.
Let
w -----> the limit of weight in an elevator
we know that
The inequality that represent this situation is
[tex]w+35 > 1,050[/tex]
Part B) Solve the inequality
[tex]w+35 > 1,050[/tex]
solve for w
Subtract 35 sides
[tex]w+35-35 > 1,050-35[/tex]
[tex]w > 1,015\ pounds[/tex]
All real numbers greater than 1,015 pounds
Part C) Where would the solution for the inequality be graphed on the number line?
[tex]w > 1,015\ pounds[/tex]
The solution is the interval -------> (1,015,∞)
In a number line is the shaded area at right of the number x=1.015 (open circle)
see the attached figure
A fair coin is flipped 4 times. What is the probability that at least two heads are flipped?
The probability of flipping at least two heads in four tosses of a fair coin is calculated using the binomial distribution, and the total probability is found to be 0.6875 or 68.75%.
To calculate the probability of flipping at least two heads in a series of four coin tosses with a fair coin, we need to consider all the possible outcomes in which we can get at least two heads. The different numbers of heads that can be obtained are 0, 1, 2, 3, or 4. To find the probability of each specific event, we use the binomial distribution formula, which for flipping two heads is:
P(2 heads) = (4 choose 2) × (0.5)² × (0.5)² = 6 × 0.25 × 0.25 = 0.375.
We can also find the probabilities of obtaining three and four heads:
P(3 heads) = (4 choose 3) × (0.5)³ × (0.5)¹ = 4 × 0.125 × 0.5 = 0.25,
P(4 heads) = (4 choose 4) × (0.5)⁴ = 1 × 0.0625 = 0.0625.
Next, we add these probabilities together to get the total probability of flipping at least two heads:
Total probability = P(2 heads) + P(3 heads) + P(4 heads) = 0.375 + 0.25 + 0.0625 = 0.6875.
Therefore, the probability of flipping at least two heads in four tosses of a fair coin is 0.6875 or 68.75%.
Solve the inhomogeneius linear ode by undetermined coefficients
Y"+4y=3sin2x
Answer with explanation:
The given non Homogeneous linear differential equation is:
y" +4 y'=3 Sin 2 x-------(1)
Put , u=y'
Differentiating once
u'=y"
Substituting the value of , y' and y" in equation (1)
⇒u' +4u =3 Sin 2x
This is a type of linear differential equation.
Integrating factor [tex]=e^{4t}[/tex]
Multiplying both sides of equation by Integrating factor
[tex]e^{4 x}(u'+4u)=e^{4x}3 \sin 2x\\\\ \text{Integrating both sides}\\\\ue^{4x}=\int {3 \sin 2x \times e^{4x}} \, dx \\\\ue^{4x}=\frac{3e^{4x}}{2^2+4^2}\times (4\sin 2x -2 \cos 2x)\\\\ue^{4x}=\frac{3e^{4x}}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}\\\\ \text{Using the formula of}\\\\\int{e^{ax}\sin bx } \, dx=\frac{e^{ax}}{a^2+b^2}\times (a \sin bx-b \cos bx)+C[/tex]
where C and [tex]C_{1}[/tex] are constant of integration.
Replacing , u by , y' in above equation we get the solution of above non homogeneous differential equation
[tex]y'(x)=\frac{3}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}e^{-4 x}[/tex]
Suppose a company did $3,000,000 in annual maintenance in 2013 and expects 85% of those to renew for 2014. Suppose that product sales for 2013 were $3,000,000 (which included free maintenance in 2013) and 60% of those were expected to pay an annual maintenance of 20% of the purchase price in 2014. What will be the annual maintenance collected in 2014?
Answer:
In year 2013 annual maintenance done = $ 3,000,000
Out of this 85% is expected to renew in 2014 = [tex]0.85\times3000000[/tex] = $2,550,000
Now Sales in 2013 = $3,000,000
60% of these sales = [tex]0.60\times3000000[/tex] = $1,800,000
Now out of this 60%, 20% annual maintenance was expected to be paid
In 2014 = [tex]0.20\times1800000[/tex] = $360000
So, Total annual maintenance in 2014 = [tex]2550000+360000[/tex]
= $2,910,000
The number of measles cases increased 13.6 % to 57 cases this year, what was the number of cases prior to the increase? (Express your answer rounded correctly to the nearest whole number)
Answer:
The number of cases prior to the increase is 50.
Step-by-step explanation:
It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.
We need to find the number of cases prior to the increase.
Let x be the number of cases prior to the increase.
x + 13.6% of x = 57
[tex]x+x\times \frac{13.6}{100}=57[/tex]
[tex]x+0.136x=57[/tex]
[tex]1.136x=57[/tex]
Divide both the sides by 1.136.
[tex]\frac{1.136x}{1.136}=\frac{57}{1.136}[/tex]
[tex]x=50.176[/tex]
[tex]x\approx 50[/tex]
Therefore the number of cases prior to the increase is 50.
According to a recent study, 9.3% of high school dropouts are 16- to 17-year-olds. In addition, 6.3% of high school dropouts are white 16- to 17-year-olds. What is the probability that a randomly selected dropout is white, given that he or she is 16 to 17 years old?
Answer: [tex]\dfrac{21}{31}[/tex]
Step-by-step explanation:
Let A represents the event that high school dropouts are 16- to 17-year-olds and B be the event that high school dropouts are white.
Given : The probability of high school dropouts are 16- to 17-year-olds :[tex]P(A)=0.093[/tex]
The probability of of high school dropouts are white 16- to 17-year-olds :
[tex]P(A\cap B)=0.063[/tex]
Then , the conditional probability that a randomly selected dropout is white, given that he or she is 16 to 17 years old is given by :-
[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{0.063}{0.093}=\dfrac{21}{31}[/tex]
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.) (a) P(0 ≤ Z ≤ 2.38) .4913 (b) P(0 ≤ Z ≤ 1) .3413 (c) P(−2.70 ≤ Z ≤ 0) .4965 (d) P(−2.70 ≤ Z ≤ 2.70) .9931 (e) P(Z ≤ 1.62) .9474 (f) P(−1.55 ≤ Z) .9394 (g) P(−1.70 ≤ Z ≤ 2.00) .9327 (h) P(1.62 ≤ Z ≤ 2.50) .0464 (i) P(1.70 ≤ Z) .0445 (j) P(|Z| ≤ 2.50) .9876
The correct answers are:
(a) [tex]P(0\leq Z\leq 2.38)=0.4913[/tex]
(b) [tex]P(0\leq Z\leq 1)=0.3413[/tex]
(c) [tex]P(-2.70\leq Z\leq 0)=0.4965[/tex]
(d) [tex]P(-2.70\leq Z\leq 2.70)=0.9926[/tex]
(e) [tex]P(Z\leq 1.62)=0.9474[/tex]
(f) [tex]P(-1.55\leq Z)=0.9394[/tex]
(g) [tex]P(-1.70\leq Z\leq 2.00)=0.9326[/tex]
(h) [tex]P(1.62\leq Z\leq 2.50)=0.0457[/tex]
(i) [tex]P(1.70\leq Z)=0.0446[/tex]
(j) [tex]P(|Z|\leq 2.50)=1.9862[/tex]
Let's calculate these probabilities step by step using the standard normal distribution table (also known as the z-table).
For reference, the standard normal distribution table provides the probabilities associated with the standard normal random variable (Z), which has a mean of [tex]0[/tex] and a standard deviation of [tex]1[/tex].
We'll use the standard normal distribution table to find the probabilities corresponding to the given Z-values.
(a [tex]P(0\leq Z\leq 2.38)[/tex] From the z-table, [tex]P(Z\leq 2.38)=0.9913 P(0\leq Z\leq 2.38)=0.9913-0.5=0.4913[/tex] (subtracting the cumulative probability up to [tex]0[/tex] from the cumulative probability up to [tex]2.38[/tex])
(b) [tex]P(0\leq Z\leq 1)[/tex] From the z-table, [tex]P(Z\leq 1)=0.8413 P(0\leq Z\leq 1)=0.8413-0.5=0.3413[/tex]
(c) [tex]P(-2.70\leq Z\leq 0)[/tex] From the z-table, [tex]P(Z\leq 0)=0.5[/tex] and [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 0)=0.5-0.0035=0.4965[/tex]
(d) [tex]P(-2.70\leq Z\leq 2.70)[/tex] From the z-table, [tex]P(Z\leq 2.70)=0.9961[/tex] and [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 2.70)=0.9961-0.0035=0.9926[/tex]
(e) [tex]P(Z\leq 1.62)[/tex] From the z-table, [tex]P(Z\leq 1.62)=0.9474[/tex]
(f) [tex]P(-1.55\leq Z)[/tex] From the z-table, [tex]P(Z\leq -1.55)=0.0606 P(-1.55\leq Z)=1-0.0606=0.9394[/tex] (subtracting the cumulative probability up to [tex]-1.55[/tex] from [tex]1[/tex])
(g [tex]P(-1.70\leq Z\leq 2.00)[/tex] From the z-table, [tex]P(Z\leq 2.00)=0.9772[/tex] and [tex]P(Z\leq -1.70)=0.0446 P(-1.70\leq Z\leq 2.00)=0.9772-0.0446=0.9326[/tex]
(h) [tex]P(1.62\leq Z\leq 2.50)[/tex] From the z-table, [tex]P(Z2.50)=0.9931[/tex] and [tex]P(Z\leq 1.62)=0.9474 P(1.62\leq Z\leq 2.50)=0.9931-0.9474=0.0457[/tex]
(i) [tex]P(1.70\leq Z)[/tex][tex]From the z-table, P(Z\leq 1.70)=0.9554 , P(1.70\leq Z)=1-0.9554=0.0446[/tex] (subtracting the cumulative probability up to [tex]1.70[/tex] from [tex]1[/tex])
(j) [tex]P(|Z|\leq 2.50)[/tex] Since the standard normal distribution is symmetric, [tex]P(|Z|\leq 2.50)=2*P(0\leq Z\leq 2.50)=2*0.9931=1.9862[/tex] (multiply by [tex]2[/tex] because the probability of Z being between [tex]-2.50[/tex] and [tex]2.50[/tex] is twice the probability of Z being between [tex]0[/tex] and [tex]2.50[/tex] )
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.)
(a) [tex]P(0 \leq Z \leq 2.38) .4913[/tex]
(b) [tex]P(0 \leq Z\leq 1) .3413[/tex]
(c) [tex]P(-2.70 \leq Z \leq 0) .4965[/tex]
(d) [tex]P(-2.70 \leq Z \leq 2.70) .9931[/tex]
(e) [tex]P(Z \leq 1.62) .9474[/tex]
(f) [tex]P(-1.55 \leq Z) .9394[/tex]
(g) [tex]P(-1.70 \leq Z \leq 2.00) .9327[/tex]
(h) [tex]P(1.62 \leq Z \leq 2.50) .0464[/tex]
(i) [tex]P(1.70 \leq Z) .0445[/tex]
(j) [tex]P(|Z| \leq 2.50) .9876[/tex]
A card is drawn at random from a standard deck of 52 cards. Find the following conditional probabilities. a) The card is a spade, given that it is black. b) The card is black, given that it is a spade. c) The card is a seven, given that it is black. d) The card is a king, given that it is a face card.
To find the conditional probabilities, you need to use the definition of conditional probability. Given that a card is black, the probability that it is a spade is 1/2. Given that a card is a spade, the probability that it is black is 2. Given that a card is black, the probability that it is a seven is 1/13. Given that a card is a face card, the probability that it is a king is 1/3.
Explanation:To find these conditional probabilities, we need to use the definition of conditional probability:
P(A|B) = P(A and B) / P(B)
a) The card is a spade, given that it is black:
In a standard deck of cards, there are 26 black cards and 13 spades. So, P(S|B) = P(S and B) / P(B) = 13/26 / 26/52 = 1/2
b) The card is black, given that it is a spade:
P(B|S) = P(B and S) / P(S) = 26/52 / 13/52 = 26/13 = 2
c) The card is a seven, given that it is black:
In a standard deck of cards, there are 4 black sevens and 26 black cards. So, P(7|B) = P(7 and B) / P(B) = 4/26 / 26/52 = 1/13
d) The card is a king, given that it is a face card:
In a standard deck of cards, there are 4 kings and 12 face cards. So, P(K|F) = P(K and F) / P(F) = 4/52 / 12/52 = 1/3
You invest $1600 in an account paying 5% interest compounded daily. What is the account's effective annual yield? Assume 360 days in a year.
The account's effective annual yield is ___% (Round to two decimal places as needed)
Answer:
5.13%.
Step-by-step explanation:
Amount accumulated in 1 year
= 1600(1 + 0.05/360)^360
= $1682.03
Account's effective annual yield
= 82.03 * 100 / 1600 %
= 5.13%.
The account's effective annual yield (EAY) for an investment of $1600 with a 5% interest rate compounded daily (assuming a 360-day year) is approximately 5.12% when rounded to two decimal places.
The student has invested $1600 in an account that offers 5% interest compounded daily with the assumption of a 360-day year. To find the effective annual yield, we use the formula for compound interest and the definition of effective annual yield (EAY), which accounts for the compounding effect:
EAY = (1 + r/n)n - 1
Where:
In this case:
Now, substituting the values, we get:
EAY = (1 + 0.05/360)360 - 1
Calculating this out:
EAY = (1 + 0.0001388888889)360 - 1
EAY
to find the EAY:
EAY = ((1 + (0.05/360))^360) - 1
After calculating the above expression, the approximate effective annual yield comes out to be:
EAY = 0.05116 or 5.116%
Therefore, after rounding to two decimal places as required, the effective annual yield of the account is 5.12%.
The population of an Asian country is growing at the rate of 0.7% annually. If there were 3,942.000 residents in the city in 1995. Find how many to the nearest ten thousand) are living in that city in 2000. Use y = 3,942,000(2.7)0.0074 a) 370,000 b) 4,000,000 c) 4.160,000 d) 4.320,000
Answer:
b) 4,000,000
Step-by-step explanation:
Let the population is measured since 1995,
Given,
The initial population, P = 3,942,000,
Annual rate of growing, r = 0.7% = 0.007,
If y represents the population after t years
So, the population after t years would be,
[tex]y=Pe^{rt}[/tex]
[tex]y=3942000(2.7)^{0.007x}[/tex]
Therefore, the population after 5 years,
[tex]y=3942000(2.7)^{0.007\times 5}=3942000(2.7)^{0.035}=4081448.78924\approx 4000000[/tex]
Hence, the population in 2000 would be approximately 40,00,000.
Option 'b' is correct.
By using the accurate population growth formula P = P0(1 + r)^n and substitifying the given values, the population of the Asian country in 2000 would be approximately 4,160,000 residents.
Explanation:
The question involves the concept of exponential growth, specifically applied to the population growth of an Asian country. Now, in given formula y = 3,942,000(2.7)^0.0074, we should use the correct growth rate formula to solve the problem which should be P = P0(1 + r)^n since the population growth is annually and continuous. Here, P0 is the initial population (3,942,000), r is the growth rate (0.7% or 0.007) and n is the number of years (2000-1995 = 5 years).
Using this formula, if you substitute these values in, you should get:
P = 3,942,000(1 + 0.007)^5
If you calculate this out, you reach a population of approximately 4,160,000 residents at the end of year 2000. Thus, answer 'c' is the correct option. Please remember exponential growth is a concept necessary for understanding population dynamics across multiple fields like demography, biology and mathematical modeling.
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The production department of Celltronics International wants to explore the relationship between the number of employees who assemble a subassembly and the number produced. As an experiment, 2 employees were assigned to assemble the subassemblies. They produced 11 during a one-hour period. Then 4 employees assembled them. They produced 18 during a one-hour period. The complete set of paired observations follows.
Number of Assemblers One-Hour Production (units)
2 11
4 18
1 7
5 29
3 20
The dependent variable is production; that is, it is assumed that different levels of production result from a different number of employees.
a. Draw a scatter diagram.
b. Based on the scatter diagram, does there appear to be any relationship between the number of assemblers and production? Explain.
c. Compute the correlation coefficient.
Answer:
We are given that The dependent variable is production; that is, it is assumed that different levels of production result from a different number of employees.
Number of Assemblers(x) One-Hour Production(y) (units)
2 11
4 18
1 7
5 29
3 20
a. Draw a scatter diagram.
Solution : Refer the attached figure
b. Based on the scatter diagram, does there appear to be any relationship between the number of assemblers and production? Explain.
Solution: The equation that shows the relationship between the number of assemblers and production is [tex]y=5.1x+1.7[/tex]
Where y is One-Hour Production (units) and x is the Number of Assemblers
c.Compute the correlation coefficient.
Solution:
Formula of correlation coefficient:[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{[n \sum x^2 -(\sum x)^2][n \sum y^2 -(\sum y)^2]}[/tex]
x y xy [tex]x^2[/tex] [tex]y^2[/tex]
2 11 22 4 121
4 18 72 16 324
1 7 7 1 49
5 29 145 25 841
3 20 60 9 400
Sum: 15 85 306 55 1735
n=5
Substitute the values in the formula :
[tex]r=\frac{5(306)-(15)(85)}{[5 (55) -(15)^2][5 (1735) -(85)^2]}[/tex]
[tex]r=0.00351[/tex]
The correlation coefficient is 0.00351
54% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
Answer:
[tex]P(5)=0.238[/tex]
[tex]P(x\geq 6)=0.478[/tex]
[tex]P(x<4)=0.114[/tex]
Step-by-step explanation:
In this case we can calculate the probability using the binomial probability formula
[tex]P(X=x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}[/tex]
Where p is the probability of obtaining a "favorable outcome " x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.
(a) exactly five
This is:
[tex]x=5,\ n=10,\ p=0.54.[/tex]
So:
[tex]P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}[/tex]
[tex]P(5)=0.238[/tex]
(b) at least six
This is: [tex]x\geq 6,\ n=10,\ p=0.54.[/tex]
[tex]P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)[/tex]
[tex]P(x\geq 6)=0.478[/tex]
(c) less than four
This is: [tex]x< 4,\ n=10,\ p=0.54.[/tex]
[tex]P(x<4)=P(3) + P(2)+P(1)+P(0)[/tex]
[tex]P(x<4)=0.114[/tex]
This question is based on the probability. Therefore, the required probabilities are : (a) [tex]P(5) = 0.238[/tex], (b)[tex]P(x \geq 6) = 0.478[/tex] and (c) [tex]P(x <4) = 0.114[/tex].
Given:
54% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults.
We have to find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
According to the question,
[tex]P(5) = 0.238\\P(x \geq 6) = 0.478\\P(x <4) = 0.114[/tex]
In this we have to calculate the probability using the binomial probability formula,
[tex]P(X=x) = \dfrac{n!}{x!(n-x)!} \times p^{x} \times (1-p)^{n-x}[/tex]
Where, p is the probability of obtaining a "favorable outcome ", x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.
(a) exactly five
x=5, n= 10, p = 0.54
[tex]P(X=5)= \dfrac{10!}{5!(10-5)!} \times 0.5^{x} \times (1-0.54)^{10-5}[/tex]
P(X=5) = 0.238
(b) at least six
[tex]x\geq 6, n=10, p=0.54\\P(x\geq 6) = P(6) + P(7) + P(8) + P(9)+P(10)\\P(x\geq 6) = 0.478[/tex]
(c) less than four
[tex]x< 6, n=10, p=0.54\\P(x< 4) = P(3) + P(2) + P(1) + P(0)\\P(x< 4) = 0.114[/tex]
Therefore, the answers are : (a) [tex]P(5) = 0.238[/tex], (b)[tex]P(x \geq 6) = 0.478[/tex]
and (c) [tex]P(x <4) = 0.114[/tex].
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7% of items in a shipment are known to be defective. If a sample of 5 items is randomly selected from this shipment, what is the probability that at least one defective item will be observed in this sample? Round your result to 2 significant places after the decimal (For example, 0.86732 should be entered as 0.87).
Answer: 0.30
Step-by-step explanation:
Binomial distribution formula :-
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success in x trials , n is the total number of trials and p is the probability of getting success in each trial.
Given : The probability that a shipment are known to be defective= 0.07
If a sample of 5 items is randomly selected from this shipment,then the probability that at least one defective item will be observed in this sample will be :-
[tex]P(X\geq1)=1-P(0)\\\\=1-(^5C_0(0.07)^0(1-0.07)^{5-0})\\\\=1-(0.93)^5=0.3043116307\approx0.30[/tex]
Hence, the probability that at least one defective item will be observed in this sample =0.30
find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3
If [tex]f(x)=4x-3[/tex]:
[tex]\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4[/tex]
If [tex]f(x)=4x^{-3}[/tex]:
[tex]\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}[/tex]
[tex]\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}[/tex]
[tex]\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}[/tex]
The mean score for all NBA games during a particular season was less than 106 points per game. If a hypothesis test is performed, how should you interpret a decision that fails to reject the null hypothesis? There is not sufficient evidence to reject the claim μ < 106. There is sufficient evidence to reject the claim μ < 106. There is not sufficient evidence to support the claim μ < 106. There is sufficient evidence to support the claim μ < 106.
Answer:
As given in the problem statement
mean score was less than 106
H0:mean score<106
so There is sufficient evidence to reject the claim μ < 106. is correct option.
caps hock Guess and Check (or use Algebra) to solve #3 & # 4 3. Plato has 36 coins in nickels, dimes, and quarters. The number of nickels is three less than twice the number of dimes. The total value of the coins is $5.20. How many of each type of coin does Plato have? mel 36 coinsin d
4. Katy bought a ski hat that was marked down 35% to $15.60. What was the price of the hat before the markdown? (Hint: She was in Oregon where they do not have sales tax). 35 9. douon 15.60 5. Sele amo ont 621.06
Answer:
Part 1:
Let the nickels be = n
Let the dimes be = d
Let the quarters be = q
Plato has 36 coins in nickels, dimes, and quarters. So, equation forms:
[tex]n+d+q=36[/tex] .....(1)
The number of nickels is three less than twice the number of dimes.
[tex]n=2d-3[/tex] ....(2)
The total value of the coins is $5.20.
[tex]0.10d+0.05n+0.25q=5.20[/tex] .... (3)
Substituting n=2d-3 in (1) and (3)
[tex]2d-3+d+q=36[/tex]
=> [tex]3d+q=39[/tex] ....(4)
[tex]0.10d+0.05(2d-3)+0.25q=5.20[/tex]
=> [tex]0.10d+0.10d-0.15+0.25q=5.20[/tex]
=> [tex]0.20d+0.25q=5.35[/tex] ...(5)
Multiplying (4) by 0.25 and subtracting (5) from (4)
[tex]0.75d+0.25q=9.75[/tex] now subtracting (5) from this we get;
[tex]0.55d=4.4[/tex]
=> d = 8
Substituting d = 8 in [tex]3d+q=39[/tex]
[tex]3(8)+q=39[/tex]
[tex]24+q=39[/tex]
=> q = 15
Substituting values of d and q in [tex]n+d+q=36[/tex], we get n
[tex]n+8+15=36[/tex]
[tex]n=36-23[/tex]
=> n = 13
Therefore Plato has 13 nickels, 15 quarters and 8 dimes.
-----------------------------------------------------------------------------------------
Part 2:
Let the original price of the ski hat be = x
Original price was marked down by 35% means value was lowered by 35%.
So, we can calculate as:
[tex]x-\frac{35x}{100}=15.60[/tex]
=> [tex]\frac{65x}{100}=15.60[/tex]
=> [tex]65x=1560[/tex]
x = 24
Hence, the original price was $24 but after 35% marking down, it was available for $15.60.
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is between negative 2.13 and 3.88 and draw a sketch of the region.
Answer: 0.9834
Step-by-step explanation:
Given : The test scores are normally distributed with
Mean : [tex]\mu=\ 0[/tex]
Standard deviation :[tex]\sigma= 1[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = -2.13
[tex]z=\dfrac{-2.13-0}{1}=-2.13[/tex]
For x = 3.88
[tex]z=\dfrac{3.88-0}{1}=3.88[/tex]
The p-value = [tex]P(-2.13<z<3.88)=P(z<3.88)-P(z<-2.13)[/tex]
[tex]0.9999477-0.0165858=0.9833619\approx0.9834[/tex]
Hence, the probability that a given score is between negative 2.13 and 3.88 = 0.9834
A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3142 tickets overall. It has sold 207 more $20 tickets than $10 tickets. The total sales are $59,670. How many tickets of each kind have been sold?
Answer:
1,084 tickets were sold that cost $10
1,291 tickets were sold that cost $20
767 tickets were sold that cost $30
Step-by-step explanation:
Let
x ----> the number of tickets that cost $10 sold
y ----> the number of tickets that cost $20 sold
z ----> the number of tickets that cost $30 sold
we know that
x+y+z=3,142 -----> equation A
10x+20y+30z=59,670 ----> equation B
y=x+207 ----> equation C
substitute equation C in equation A and equation B
x+(x+207)+z=3,142 ----> 2x+z=2,935 ----> equation D
10x+20(x+207)+30z=59,670 ---> 30x+30z=55,530 ----> equation E
Solve the system of equations D and E by graphing
The solution is the intersection point both graphs
The solution is the point (1,084,767)
so
x=1,084, z=767
see the attached figure
Find the value of y
y=x+207 ----> y=1,084+207=1,291
therefore
1,084 tickets were sold that cost $10
1,291 tickets were sold that cost $20
767 tickets were sold that cost $30
If ( 43.65 ) ( 8.79 ) / x = ( 0.4365 ) ( 87.9 ) then value of x is:
(a) .01 (b) 0.1 (c) 1 (d) 10 (e) 100
Need step-by-step solution
i will mark your answer Brainliest
Answer:
(d) 10
Step-by-step explanation:
Multiply by x and divide by its coefficient:
(43.65)(8.79) = (0.4365)(87.9)x
(43.65)(8.79)/((0.4365)(87.9)) = x
At this point, any calculator can give you the answer. It is, perhaps, more satisfying to work out the answer without a calculator.
x = (43.65)/(0.4365) × (8.79)/(87.9)
In the first quotient, the numerator is 100 times the denominator; in the second, the denominator is 10 times the numerator.
x = (100) × (1/10) = 100/10
x = 10
_____
Moving the decimal point to the right 1 place multiplies the numerical value by 10.
The value of x in the given equation that satisfies the condition is 10.
Explanation:In this question, we're given a mathematical expression in which the value of x is unknown. We're looking for the value of x that satisfies the equation:
( 43.65 ) ( 8.79 ) / x = ( 0.4365 ) ( 87.9 )
To solve this equation for x, we can start by noting the similarity between the left and right sides. We have larger numbers on the left side that appear, in reduced form, on the right side.
Follow these steps:
Multiply 43.65 and 8.79 to get 383.985.Multiply 0.4365 and 87.9 to get 38.3985.Divide 383.985 by 38.3985 to get 10.So, the correct option would be (d) 10.
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2- suppose a ball is thrown upward to a height of h0 meters . each time the ball bounces, it rebounds to a fraction r of its previous height . let hn be the hight after the nth term biunces. cosider the following value of h0 and r
a- find the first 4 terms of the sequence of heights(hn)
b- find a general expression for the nth term of the sequence (hn)
h0=20 , r=0.5
Answer:
a) 40, 50, 55, 57.5
b) [tex]S_n=h_0+2h_0\sum_{n=1}^{\infty}r^n[/tex]
Step-by-step explanation:
h₀ = Initial height of the ball =20
r = Rebound fraction = 0.5
a) The series of bouncing balls is given by
Sₙ=h₀+2h₀(r¹+r²+r³+r⁴.........rⁿ)
S₁=h₀+2h₀r¹=20+2×20×0.5=40
S₂=h₀+2h₀(r¹+r²)=20+2×20×(0.5+0.5²)=50
S₃=h₀+2h₀(r¹+r²+r³)=20+2×20×(0.5+0.5²+0.5³)=55
S₄=h₀+2h₀(r¹+r²+r³+r⁴)=20+2×20×(0.5+0.5²+0.5³+0.5⁴)=57.5
b) General expression for the nth term of the sequence
[tex]S_n=h_0+2h_0\sum_{n=1}^{\infty}r^n[/tex]
Final answer:
The first 4 heights after each bounce form a geometric sequence with the first term h₀ being 20 meters and subsequent terms being 10, 5, and 2.5 meters. The general expression for the nth term (hn) is given by the formula hₙ = 20 * (0.5)ⁿ
Explanation:
Given an initial height h₀ of 20 meters and a rebound fraction r of 0.5, the sequence of heights after each bounce forms a geometric sequence.
The first height is h₀ which is 20 meters. Subsequent heights can be found by multiplying the previous height by the rebound fraction r.
Third term : h2 = h= 10 * 0.5 = 5 meters
General Expression for the nth Term (b)
The nth term (hₙ) of the sequence can be found using the formula for the nth term of a geometric sequence:
hₙ = h₀ * rⁿ
For this particular sequence:
hₙ = 20 * (0.5)ⁿ
he diameters of red delicious apples of an orchard have a normal distribution with a mean of 3 inches and a standard deviation of 0.5 inch. One apple will be randomly chosen. What is the probability of picking an apple with diameter between 2.5 and 4.25 inches?
Answer: 0.8351
Step-by-step explanation:
Given :Mean : [tex]\mu=\text{ 3 inches}[/tex]
Standard deviation : [tex]\sigma =\text{ 0.5 inch}[/tex]
The formula for z -score :
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 2.5 ,
[tex]z=\dfrac{2.5-3}{0.5}=-1[/tex]
For x= 4.25 ,
[tex]z=\dfrac{4.25-3}{0.5}=2.5[/tex]
The p-value = [tex]P(-1<z<2.5)=P(z<2.5)-P(z<-1)[/tex]
[tex]=0.9937903-0.1586553=0.835135\approx0.8351[/tex]
Hence, the probability of picking an apple with diameter between 2.5 and 4.25 inches =0.8351.
To find the probability, we standardize the values, convert them into z-scores, look up the z-scores in a z-table, and subtract the lower cumulative probability from the higher one. The probability of picking an apple with diameter between 2.5 inches and 4.25 inches is 83.51%.
Explanation:The problem involves a normal distribution where the diameter of Red Delicious apples has a mean of 3 inches and a standard deviation of 0.5 inch. The objective is to find the probability of picking an apple with a diameter between 2.5 and 4.25 inches.
We first standardize the given values to convert them into z-scores by subtracting the mean from the given value and dividing by the standard deviation. For 2.5 inches, z = (2.5 - 3) / 0.5 = -1. For 4.25 inches, z = (4.25 - 3) / 0.5 = 2.5.
Using a z-table, the z-score of -1 corresponds to a cumulative probability of 0.1587 and the z-score of 2.5 corresponds to a cumulative probability of 0.9938. To find the probability between these two diameters, we subtract the cumulative probability of -1 from the cumulative probability of 2.5.
Therefore, the probability of picking an apple with diameter between 2.5 inches and 4.25 inches is 0.9938 - 0.1587 = 0.8351 or 83.51%.
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Use the alternative curvature formula K=|a x v|/|v|^3 to find the curvature of the following parameterized curves. to find the curvature of the following parameterized curve. r(t)=<6+5t^2, t, 0>
Answer:
[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]
Or
[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]
Step-by-step explanation:
We want to compute the curvature of the parameterized curve, [tex]r(t)=\:<\:6+5t^2,t,0)\:>\:[/tex] using the alternative formula:
[tex]k=\frac{|a\times v|}{|v|^3}[/tex].
We first compute the required ingredients.
The velocity vector is [tex]v=r'(t)=<\:10t,1,0\:>[/tex]
The acceleration vector is given by [tex]a=r''(t)=<\:10,0,0\:>[/tex]
The magnitude of the velocity vector is [tex]|v|=\sqrt{(10t)^2+1^2+0^2}=\sqrt{100t^2+1}[/tex]
The cross product of the velocity vector and the acceleration vector is
[tex]a\times v=\left|\begin{array}{ccc}i&j&k\\10&0&0\\10t&1&0\end{array}\right|=10k[/tex]
We now substitute ingredients into the formula to get:
[tex]k=\frac{|10k|}{(\sqrt{100t^2+1})^3}[/tex].
[tex]k=\frac{10}{(100t^2+1)^{\frac{3}{2}}}[/tex]
Or
[tex]k=\frac{10\sqrt{100t^2+1}}{(100t^2+1)^{2}}[/tex]
If (x,y) is a solution to the system of equations shown below, what is the product of the y-coordinates of the solutions? x^2+4y^2=40 x+2y=8
Answer:
The product of the y-coordinates of the solutions is equal to 3
Step-by-step explanation:
we have
[tex]x^{2}+4y^{2}=40[/tex] -----> equation A
[tex]x+2y=8[/tex] ------> equation B
Solve by graphing
Remember that the solutions of the system of equations are the intersection point both graphs
using a graphing tool
The solutions are the points (2,3) and (6,1)
see the attached figure
The y-coordinates of the solutions are 3 and 1
therefore
The product of the y-coordinates of the solutions is equal to
(3)(1)=3
If 20600 dollars is invested at an interest rate of 10 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.
(a) Annual: $
(b) Semiannual: $
(c) Monthly: $
(d) Daily: $
Answer:
a) The value of the investment is $33176.506
b) The value of the investment is $33555.53
c) The value of the investment is $33893.36
d) The value of the investment is $33961.33
Step-by-step explanation:
This is a compound interest problem
Compound interest formula:
The compound interest formula is given by:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
A: Amount of money(Balance)
P: Principal(Initial deposit)
r: interest rate(as a decimal value)
n: number of times that interest is compounded per unit t
t: time the money is invested or borrowed for.
In our problem, we have:
A: the value we want to find
P = 20600(the value invested)
r = 0.1
n: Will change for each letter
t = 5.
a) If the interest is compounded anually, n = 1. So.
[tex]A = 20600(1 + \frac{0.1}{1})^{1*5}[/tex]
[tex]A = 33176.506[/tex]
The value of the investment is $33176.506
b) If the interest is compounded semianually, it happens twice a year, which means n = 2. So:
[tex]A = 20600(1 + \frac{0.1}{2})^{2*5}[/tex]
[tex]A = 33555.23[/tex]
The value of the investment is $33555.53
c) If the interest is compounded monthly, it happens 12 times a year, which means n = 12. So:
[tex]A = 20600(1 + \frac{0.1}{12})^{12*5}[/tex]
[tex]A = 33893.36[/tex]
The value of the investment is $33893.36
d) If the interest is compounded monthly, it happens 365 times a year, which means n = 365. So:
[tex]A = 20600(1 + \frac{0.1}{365})^{365*5}[/tex]
[tex]A = 33961.33[/tex]
The value of the investment is $33961.33
Final answer:
The future value of $20,600 invested at a 10% annual interest rate after 5 years varies based on the compounding frequency: annually it will be $33,186.35, semiannually $33,644.31, monthly $33,949.69, and daily $34,030.18.
Explanation:
Calculating Future Value with Different Compounding Methods
To calculate the future value of an investment with compound interest, we use the formula [tex]FV =PV(1+r/n)^{nt}[/tex],where FV is the future value, P is the principal amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is time in years.
Given a principal amount of $20,600 and an annual interest rate of 10%, we will calculate the value of the investment at the end of 5 years for the following compounding methods:
Annual compounding (n=1): FV = 20600(1 + 0.10/1)⁽¹ˣ⁵⁾
Semiannual compounding (n=2): FV = 20600(1 + 0.10/2)⁽²ˣ⁵⁾
Monthly compounding (n=12): FV = 20600(1 + 0.10/12)⁽¹²ˣ⁵⁾
Daily compounding (n=365): FV = 20600(1 + 0.10/365)⁽³⁶⁶ˣ⁵⁾
Using a calculator or spreadsheet software, we can find the values to the nearest cent:
(a) Annual: $33,186.35
(b) Semiannual: $33,644.31
(c) Monthly: $33,949.69
(d) Daily: $34,030.18
The compounding effect shows that the more frequently the interest is compounded, the greater the final value of the investment.
*Asymptotes*
g(x) =2x+1/x-3
Give the domain and x and y intercepts
Answer: Assuming the function is [tex]g(x)=\frac{2x+1}{x-3}[/tex]:
The x-intercept is [tex](\frac{-1}{2},0)[/tex].
The y-intercept is [tex](0,\frac{-1}{3})[/tex].
The horizontal asymptote is [tex]y=2[/tex].
The vertical asymptote is [tex]x=3[/tex].
Step-by-step explanation:
I'm going to assume the function is: [tex]g(x)=\frac{2x+1}{x-3}[/tex] and not [tex]g(x)=2x+\frac{1}{x}-3[/tex].
So we are looking at [tex]g(x)=\frac{2x+1}{x-3}[/tex].
The x-intercept is when y is 0 (when g(x) is 0).
Replace g(x) with 0.
[tex]0=\frac{2x+1}{x-3}[/tex]
A fraction is only 0 when it's numerator is 0. You are really just solving:
[tex]0=2x+1[/tex]
Subtract 1 on both sides:
[tex]-1=2x[/tex]
Divide both sides by 2:
[tex]\frac{-1}{2}=x[/tex]
The x-intercept is [tex](\frac{-1}{2},0)[/tex].
The y-intercept is when x is 0.
Replace x with 0.
[tex]g(0)=\frac{2(0)+1}{0-3}[/tex]
[tex]y=\frac{2(0)+1}{0-3}[/tex]
[tex]y=\frac{0+1}{-3}[/tex]
[tex]y=\frac{1}{-3}[/tex]
[tex]y=-\frac{1}{3}[/tex].
The y-intercept is [tex](0,\frac{-1}{3})[/tex].
The vertical asymptote is when the denominator is 0 without making the top 0 also.
So the deliminator is 0 when x-3=0.
Solve x-3=0.
Add 3 on both sides:
x=3
Plugging 3 into the top gives 2(3)+1=6+1=7.
So we have a vertical asymptote at x=3.
Now let's look at the horizontal asymptote.
I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is [tex]y=\frac{2}{1}[/tex]. After simplifying you could just say the horizontal asymptote is [tex]y=2[/tex].
Or!
I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).
[tex]y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}[/tex]
[tex]y=2+\frac{7}{x-3}[/tex]
So you can think it like this what value will y never be here.
7/(x-3) will never be 0 because 7 will never be 0.
So y will never be 2+0=2.
The horizontal asymptote is y=2.
(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)
1. A box contains 4 red and 6 green balls. Two balls are drawn. You receive $10 if both are red and you pay $1 otherwise. What are your expected winnings?
Answer:
Your Winnings are $ -1
Step-by-step explanation:
Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.
Since we are not told what two balls were actually drawn, then the expected winnings would be of the greatest probability occurring. To figure this out we need to calculate the probability of getting two red balls consecutively and the probability of getting two different colors consecutively. Since there are a total of 10 balls the probabilities of getting one ball would be the following,
Red: [tex]\frac{4}{10} = 40%[/tex]
Green: [tex]\frac{6}{10} = 60%[/tex]
Now we need to know the probability of us getting 2 reds together, two greens, or one of each.
2 Reds: [tex]\frac{4}{10} *\frac{4}{10} = \frac{16}{100} = 16%[/tex]
2 Greens: [tex]\frac{6}{10} *\frac{6}{10} = \frac{36}{100} = 36%[/tex]
1 Each: [tex]\frac{4}{10} *\frac{6}{10} = \frac{24}{100} = 24%[/tex]
So we can see that the highest probability is getting 2 Greens in a row. Meaning you would have to pay $1
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.