Answer: Yes, I agree we should study number system in computing.
Step-by-step explanation:
Hexadecimal numbers can be easily converted to binary numbers.
Two cards are randomly selected from an ordinary playing deck.
What is the probability that one of the cards is an ace and the other one is either a ten, a jack, a queen or a king?
Final answer:
To find the probability of selecting one ace and one of either a ten, jack, queen, or king from a deck with replacement, calculate the probability for each draw and account for both possible orders of drawing these cards the final probability is 2 * ((4/52) * (16/52)) to account for both orders.
Explanation:
To calculate the probability that one card is an ace and the other is either a ten, jack, queen, or king when two cards are selected randomly from an ordinary deck, we approach this problem considering the combinations and probabilities of each event happening.
First, let's establish the total possibilities. Since we are replacing the card back into the deck after the first draw, the number of possibilities for each draw remains 52. So, the total number of ways to draw any two cards (with replacement) is 52 * 52.
The probability of drawing an ace (P(Ace)) on the first draw is 4/52 because there are 4 aces in a deck of 52 cards.
The probability of drawing a ten, jack, queen, or king (P(10/J/Q/K)) on the second draw is 16/52, as there are 4 of each of these cards in the deck, totaling 16 cards.
To find the probability of both events happening, you multiply the probabilities of each event: P(Ace) * P(10/J/Q/K). This gives you (4/52) * (16/52).
However, the order in which the ace and the 10/J/Q/K are drawn matters, as the ace could also be drawn second. This means we need to calculate the probability again with the ace being drawn second and the 10/J/Q/K card drawn first, which is the same calculation. Therefore, the final probability is 2 * ((4/52) * (16/52)) to account for both orders.
To calculate the probability of drawing an ace followed by a ten, jack, queen, or king, we find the probabilities of each event separately and multiply them since the events are independent, assuming the card is replaced after the draw.
Explanation:The probability of selecting one ace and another card which is a ten, jack, queen, or king involves understanding the composition of a standard deck and calculating the odds of each draw.
A standard deck has 52 cards with 4 aces and 16 face cards (including the tens). The chance of drawing an ace (event A) is therefore 4/52. For the second draw (event B), with the card being replaced after the first draw, the odds of drawing either a ten, jack, queen, or king from any suit remain identical, since we're back to having a full deck.
To find the overall probability, we calculate the probability of both events occurring, which involves multiplying the probabilities of each individual event, assuming they are independent. This is represented by: P(A and B) = P(A) * P(B).
For the specific example of drawing a four (event A) then a five (event B) when replacing the card back into the deck after each draw, the probability of event A is 1/13 since there are four fours in the deck and the probability of event B is also 1/13 for the same reason. Therefore, the combined probability is (1/13)*(1/13), which simplifies to 1/169.
It is known that 45% of senior citizens are deficient in vitamin D. Let Y be the number of vitamin D efficient individuals in a random sample of n = 15 senior citizens. (a) Calculate P(Y = 5). Also obtain an approximation for this probability using the normal approximation. (b) Calculate P(Y > 7). Also obtain normal approximations for this probability with and without using continuity correction. (c) Calculate P(4
Answer:
a ) 0.1403604645 and 0.1368
b) 0.3464961 and 0.3485
c) 0.802671982 and 0.8018
Step-by-step explanation:
Y~ B (15,0.45)
Y~ N (15*0.45, 15*0.45*0.55) = Y~ N (6.75, 3.7125)
a) P(Y=5) = 15C5 (0.45)^5 * (0.55)^10 = 0.1403604645
For normal approximation
P(Y = 5 ) = P ( 4.5 < Y < 5.5 ) ......... continuity correction
Hence,
[tex]P ( 4.5 < Y < 5.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{5.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.16775 < Z < -0.64875 )[/tex]
The probability P ( 4.5 < Y < 5.5 ) = 0.1368
b) P(Y>7) = 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6 + 15C10 (0.45)^10 * (0.55)^5 + 15C11 (0.45)^11 * (0.55)^4 + 15C12 (0.45)^12 * (0.55)^3 + 15C13 (0.45)^13 * (0.55)^2 + 15C14 (0.45)^14 * (0.55) + (0.45)^15
= 0.3464961
For normal approximation
P(Y > 7 ) = P (Y > 7.5 ) ......... continuity correction
Hence,
[tex]P (Y > 7.5) = P (Z > \frac{7.5-6.75}{\sqrt{3.7125} } ) = P (Z > 0.389249)\\[/tex]
The probability P ( Y>7.5 ) = 0.3485
c) P (4 < Y < 10) = 15C5 (0.45)^5 (0.55)^10 + 15C6 (0.45)^ 6 (0.55)^9 + 15C7 (0.45)^7 (0.55)^8 + 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6
= 0.802671982
For normal approximation
P( 4 < Y < 10 ) = P (4.5< Y < 9.5 ) ......... continuity correction
Hence,
[tex]P ( 4.5 < Y < 9.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{9.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.167748416 < Z < 1.427248064 )[/tex]
The probability P (4.5< Y < 9.5 ) = 0.8018
Find the equations that describe the circle of radius 2 centered at (4, 7, 6) that is parallel to the xy-plane. (Enter your answers as a comma-separated list of equations.)
Answer:
[tex](x-4)^2+(y-7)^2=4[/tex]
z=6
Step-by-step explanation:
The equation of a circle is satisfied by every point on it. The equation of a circle of radius r and point (h,k,l) parallel to the xy-plane is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
we can just substitute values into the equation:
[tex](x-4)^2+(y-7)^2=r^2[/tex] where z=6
The radius is 2. Therefore:
[tex](x-4)^2+(y-7)^2=4[/tex]
Final answer:
The equations describing the circle with a radius of 2 that is centered at (4, 7, 6) and parallel to the xy-plane are (x - 4)² + (y - 7)² = 4 and z = 6.
Explanation:
The equation that describes a circle of radius 2 centered at (4, 7, 6) that is parallel to the xy-plane can be found using the general equation of a circle in three dimensions, considering that the circle lies in a plane parallel to the xy-plane, hence the z-coordinate will remain constant. The standard equation for a circle in two dimensions is (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is its radius.
For a circle of radius 2, the equation becomes (x - 4)² + (y - 7)² = 2². Since the circle is parallel to the xy-plane and the center has a z-coordinate of 6, the equation does not change with respect to z, remaining constant at z = 6. Therefore, the complete set of equations describing the circle are (x - 4)² + (y - 7)² = 4 and z = 6.
the ration of 28 hours to 42 hours is
Answer:
2:3
Step-by-step explanation:
the ratio of 28 hours to 42 hours is 28:42
28/42
7 divides both the numerator and denominator into small terms
4/6
2 divides both the numerator and denominator into small terms
2/3 = 2:3
In an albatross colony the wingspan of the birds is approximately normally distributed with standard deviation 0.43 metres. In this colony, 67% of the birds have wingspan less than 3.21 metres. Find the average wingspan of the birds in this colony. Give your answer as a decimal, rounded to two decimal places. Do not include the unit.
To find the average wingspan, the z-score for 67% of the normal distribution was found to be 0.44. Using the formula with the given standard deviation and wingspan less than 3.21 meters, the average wingspan was calculated to be approximately 3.02 meters.
Explanation:In order to find the average wingspan of the birds in the albatross colony given the standard deviation of 0.43 meters and that 67% of the birds have a wingspan less than 3.21 meters, we apply concepts from normal distribution.
From the properties of the normal distribution, we know that 67% corresponds to an area under the curve to the left of the mean. Using the standard normal distribution table (z-table), we can find the z-score that corresponds to 0.67 (67%). The z-score close to this area under the curve is approximately 0.44.
The formula to convert a z-score to an actual score is given as:
X = μ + (z × σ)
Where:
X is the raw scoreμ is the mean (average)z is the z-score corresponding to the desired cumulative probabilityσ is the standard deviation.We are given X (3.21), z (0.44), and σ (0.43), and we need to find the mean μ.
Substituting the known values into the formula, we get:
3.21 = μ + (0.44 × 0.43)
Now we solve for μ:
3.21 = μ + 0.1892
μ = 3.21 - 0.1892
μ ≈ 3.02
So, the average wingspan of the albatrosses is approximately 3.02 meters rounded to two decimal places.
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A randon sample of monthly gasoline bills for a company's 15 sales persons are:$216 $254 $247 $257 $231 $265 $221 $226 $228 $252 $235 $265 $272 $285 $266.What is the mean?What is the median?what is the value of quartile 3?What is the value of the modethe range is equal to?What is the value of the standard deviation?what is the value of the coefficient of variation? do not express the answer as a percent, leave it as a ratio?are the numbers skewed positively, negatively or are they symmetrical? (for your answer enter only one word: positively, negatively or symmetrical.what is the value of the coefficient of skewness using pearson's coefficient of skewness?What is the value of quartile 2?
Answer:Mean = $248
Median = $252
Quartile 3 = $265
Mode = $265
range = $69
Standard deviation= 20.126
Coefficient of variation = 8.115
Negatively
Coefficient of skewness = -0.596
Quartile 2 = $252
Step-by-step explanation:
The detailed explanation can be found in the attached pictures
Nite Time Inn has a toll-free telephone number so that customers can call at any time to make a reservation. A typical call takes about 4 minutes to complete, and the time required follows an exponential distribution.
Find the probability that a call takes:
(a) 3 minutes or less.
(b) 4 minutes or less.
(c) 5 minutes or less.
(d) longer than 5 minutes.
Answer:
Step-by-step explanation:
Let X be the time for any customer to call at any time to make reservation in Nite Time Inn.
Given that X is exponential with mean = 4 minutes
We are to find the probability
(a) 3 minutes or less.
=[tex]P(X\leq 3)=1-e^{-3/4} =0.5276[/tex]
(b) 4 minutes or less.
[tex]=P(X\leq 4)\\=1-e^{-4/4} =0.6321[/tex]
(c) 5 minutes or less.
[tex]=P(X\leq 5)\\=1-e^{-5/4} =0.7135[/tex]
(d) longer than 5 minutes.
=1-P(X≤5) = 0.2865
(a) identify the independent variable and the dependent variable of each equation (use 't' for the independent variable if an independent variable is not given explicitly): (b) give the order of each differential equation (enter '1' for first order. '2' for second order and so on: do not include the quotes); and (C) state whether the equation is linear or nonlinear If your answer to (C) is nonlinear, make sure that you can explain why this is true
Answer:
You didn't give the differential equations, but I'll explain how to identify the independent variable, dependent variable, how to know the order, linearity, and nonlinearity of a differential equation.
Step-by-step explanation:
DIFFERENTIAL EQUATION
This is any equation that involves differential coefficients. It is a relationship between an independent variable, x, a dependent variable, y, and one or more derivatives of y with respect to x.
Examples
(1) xd²y/dx² + 7dy/dx = 0
(2) y²dy/dx + 2x = 0
(3) xd³y/dx³ = y½ + 1
(4) 2xy'' - 3y' + 5y = 0
(5) (y''')² + 30xy = 0
Note that the dependent variable is always the numeratior, and the independent, denominator, in a different coefficient. In the case of our examples, y is the dependent variable, and x is the independent.
Example (4) is another way of writing a differential coefficient, y' (read as y-prime) is the same as dy/dx (read as dee-y dee-x). In some cases when the independent variable is time t, it is written as ÿ, which is the same as d²y/dt² (read as dee-two-y dee-t-squared)
ORDER
This is the order of the highest derivative in a differential equation. You need not consider other derivatives, just the highest.
In the examples, the orders are
(1) two
(2) one
(3) three
(4) two
(5) three
LINEAR DIFFERENTIAL EQUATION
This is the kind of differential equation 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.
NONLINEAR DIFFERENTIAL EQUATION
If any condition for linearity is not met, then it is nonlinear.
(1) Linear
(2) Nonlinear because y is the dependent variable, and y² is nonlinear, and even still, it multiplies a derivative.
(3) Nonlinear because y½ is nonlinear
(4) Linear
(5) Nonlinear because (y''')² in nonlinear.
Understanding this, you can determine the order, linearity or nonlinearity of any differential equation. Cheers!
Equations 1 and 3 are nonlinear due to their respective terms, while Equation 2 is linear. The orders are first for Equations 1 and 2, and second for Equation 3.
Let's analyse each of the given differential equations step by step.
Equation 1: y' = y - x²
Independent Variable: xEquation 2: xy' = 2y
Independent Variable: xEquation 3: x'' + 5x = e-x
Independent Variable: t (since no specific variable is given, it's common practice to use 't')Dependent Variable: xOrder: 2 (Second order, because x'' involves the second derivative of x)Linearity: This equation is nonlinear because the term e-x is an exponential nonlinear function of x.Thus,
1. [tex]\( y' = y - x^2 \)[/tex]
(a) Independent variable: x, Dependent variable: y
(b) First order
(c) Nonlinear because of the term -x².
2. xy' = 2y
(a) Independent variable: x, Dependent variable: y
(b) First order
(c) Nonlinear because of the product xy.
3. [tex]\( x'' + 5x = e^{-x} \)[/tex]
(a) Independent variable: t (not explicitly given), Dependent variable: x
(b) Second order
(c) Linear because it has no products or powers of x other than x and x''.
Complete question: In problems below
(a) identify the independent variable and the dependent variable of each equation (use 't' for the independent variable if an independent variable is not given explicitly): (b) give the order of each differential equation (enter '1' for first order. '2' for second order and so on: do not include the quotes); and (C) state whether the equation is linear or nonlinear If your answer to (C) is nonlinear, make sure that you can explain why this is true
Equation:
1. y' = y-x²
2. xy' = 2y
3. [tex]\( x'' + 5x = e^{-x} \)[/tex]
Which equation best represent Fallon’s earning in term of Donald’s earnings?
Answer:
C. y=x+7
Step-by-step explanation:
If you add 7 to the values on the left, you'll get the values on the right.
Answer: I think c is the answer
Step-by-step explanation:
What is the slope of the line ? Pls help .
let P(-2,0) & Q(0,2)
slope = 2-0/0+2
= 1
the slope is 1 which will be 45°
Answer: slope = 1
Step-by-step explanation:
The formula for finding slope is given as :
slope = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]x_{1}[/tex] = -5
[tex]x_{2}[/tex] = 3
[tex]y_{1}[/tex] = -3
[tex]y_{2}[/tex] = 5
substituting the values into the formula , we have :
slope = [tex]\frac{5-(-3)}{3-(-5)}[/tex]
slope = [tex]\frac{5+3}{3+5}[/tex]
slope = [tex]\frac{8}{8}[/tex] = 1
Therefore : the slope of the line is 1
Evaluate the function at the given value:
f(a)=-3a^3+10a^2-3a-7 at a=2
Btw this unit is on the division of polynomials
Answer:
51
Step-by-step explanation:
Substitute the value of a
3(2)³+10(2)²-3(2)-7
3(8)+10(4)-3(2)-7
24+40-6-7
64-13
51
While on a walk in the country, you pass a field of goats and chickens. After a quick count, you determine there are 38 heads and 108 feet in the field. How many of each animal are there
Answer:there are 16 goats and 22 chickens.
Step-by-step explanation:
Let x represent the number of goats in the field.
Let y represent the number if chicken in the field.
After a quick count, you determine there are 38 heads and 108 feet in the field. A goat has one head. A chicken also has one head. It means that
x + y = 38
A goat has four legs. A chicken has 2 legs. It means that
4x + 2y = 108 - - - - - - - - - - - 1
Substituting x = 38 - y into equation 1, it becomes
4(38 - y) + 2y = 108
152 - 4y + 2y = 108
- 4y + 2y = 108 - 152
- 2y = - 44
y = - 44/ - 2
y = 22
x = 38 - y = 38 - 22
x = 16
PLZ help 40 points!!
Classify each scenario as a linear or non-linear function.
1. The number of pages a copy machine can print over time, expressed by n=30m.
2. The amount of money an hourly worker makes over time, expressed by a=15h.
3. The temperature outside at noon each day for a year.
4. The height increase of a child over 3 years.
Answer:
1. Linear relation
2. Linear relation
3. Non Linear relation
4. Non Linear relation
Step-by-step explanation:
i) the number of pages a copy machine can print over time, expressed by n = 30m is a linear relation where the value of 'n' is directly proportional to 'm'.
ii) The amount of money an hourly worker makes over time, expressed by a = 15h, is a linear relation where the value of 'a' is directly proportional to 'h'.
iii) The temperature outside at noon each day for a year is a non-linear relation because the temperature will be different every day and not by the same amount.
iv) The height increase of a child over 3 years is also a non-linear relation as the height increase will vary from year to year and not by the same amount.
Customers filter into a record shop at an average of 1 per minute (exponential interarrivals) where the service rate is 15 per hour (exponential service times). What is the minimum number of servers needed to keep the average time in the system under 6 minutes?
A minimum of 10 servers is needed to keep the average time in the system under 6 minutes.
What is the utilization factor?The utilization factor is the ratio of the power station's maximum demand to its rated capacity. The time that equipment is in use./ The total time that it could be in use.
Given, In a record store with a service rate of 15 per hour, customers typically enter at a rate of 1 per minute (exponential interarrivals) (exponential service times).
Customers Per Minute into record shop = 1 per minute
Per Hour = 1 * 60 = 60
Number of Customers > Per Hour/(Server)
6 > 60/ server
server > 10
Therefore, as per the utilization factor, a minimum of 10 servers should be installed to keep the average time under 6 minutes.
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The given scenario can be modeled as a queuing system, where the arrival rate (\(\lambda\)) and service rate (\(\mu\)) are given.
From the information provided:
- The arrival rate, \(\lambda\), is 1 customer per minute, hence \(\lambda = 1\).
- The service rate, \(\mu\), is 15 customers per hour. Since we are working with minutes, we need to convert this to minutes. There are 60 minutes in an hour, so \( \mu = \frac{15}{60} = 0.25 \) customers per minute.
The goal is to keep the average time in the system, \( W \), under 6 minutes. The average time in the system for a M/M/c queue (which is the kind we're dealing with since both arrival and service rates are exponentially distributed) can be calculated using the formula for the average time in a system with \(c\) servers:
\[ W = \frac{W_q + 1}{\mu} \]
where:
- \(W_q\) is the average time a customer spends waiting in the queue.
- \(\mu\) is the service rate per server.
- The term \( \frac{1}{\mu} \) is the average service time.
The queue time \(W_q\) depends on the number of servers \(c\) and the utilization factor \(\rho = \frac{\lambda}{c\mu}\). The queue time increases with the utilization factor. The formula for \( W_q \) in an M/M/c queue is complex as it involves Erlang B formulas and iterative methods to solve for different values of \(c\).
However, for the practical purposes of this problem, we can use the approximation for \( W \) without explicitly calculating \( W_q \) as:
\[ W = \frac{1}{\mu - \lambda} \]
provided that this system is stable, which occurs only if \( \lambda < c \cdot \mu \).
We need to find the minimum number of servers \( c \) such that the average time in the system \( W \) is less than 6 minutes:
\[ W < 6 \]
Using the approximate formula:
\[ \frac{1}{c\mu - \lambda} < 6 \]
Therefore:
\[ c\mu - \lambda > \frac{1}{6} \]
Solving for the number of servers \( c \):
\[ c > \frac{\lambda + \frac{1}{6}}{\mu} \]
Plugging in the values of \(\lambda\) and \(\mu\):
\[ c > \frac{1 + \frac{1}{6}}{0.25} \]
\[ c > \frac{\frac{6}{6} + \frac{1}{6}}{{0.25}} \]
\[ c > \frac{\frac{7}{6}}{0.25} \]
\[ c > \frac{7}{6} \cdot \frac{1}{0.25} \]
\[ c > \frac{7}{6} \cdot 4 \]
\[ c > \frac{7 \cdot 2}{3} \]
\[ c > \frac{14}{3} \]
As we cannot have a fraction of a server, we need to round up to the nearest whole number:
\[ c > 4.67 \]
So we need a minimum of 5 servers to keep the average time in the system under 6 minutes.
A hole 2 inches in radius is drilled out of a solid sphere of radius 5 inches. Find the volume of the remaining solid.
Answer:
The volume of remaining sphere is 489.84 cubic inches.
Step-by-step explanation:
We are given the following in the question:
A hole 2 inches in radius is drilled out of a solid sphere of radius 5 inches.
Radius of sphere = 5 inches
Radius of hole = 2 inches
Volume of sphere =
[tex]\dfrac{4}{3}\pi r^3[/tex]
where r is the radius of sphere.
Volume of sphere =
[tex]\displaystyle\frac{4}{3}\pi (5)^3\\\\=\frac{4}{3}\times 3.14\times (5)^3\\\\=523.33\text{ cubic inches}[/tex]
Volume of hole =
[tex]\displaystyle\frac{4}{3}\pi (2)^3\\\\=\frac{4}{3}\times 3.14\times (2)^3\\\\=33.49\text{ cubic inches}[/tex]
Volume of remaining solid =
Volume of sphere - Volume of hole
[tex]=523.33 - 33.49\\=489.84\text{ cubic inches}[/tex]
The volume of remaining sphere is 489.84 cubic inches
Michael gets test grades of 73, 77, 82, and 86. He gets a 93 on his final exam. Find the weighted mean if the tests each count for 15% and the final exam counts for 40% of the final grade. Round to one decimal place.
Answer:
84.9
Step-by-step explanation:
The weighted mean is given by the sum of the products of each grade by its respective weight. If the first four grades correspond to 15% of the final grade each, and the final exam is equivalent to 40% of the final grade, Michael's final grade is:
[tex]G= (73+77+82+86)*0.15+(93*0.4)\\G= 84.9[/tex]
Michael's final weighted mean is 84.9.
Michael's final grade, calculated using the weighted mean formula, taking into consideration the individual weights of each test (15%) and the final exam (40%), is approximately 85.4.
Explanation:The subject of this problem is weighted mean, which is used quite often in the field of statistics. The formula to calculate the weighted mean is *(w1*x1 + w2*x2 + ... + wn*xn) / (w1 + w2 + ... + wn), where w represents the weights and x represents the values.
In Michael's case, his weights for the test grades are 15% for each test, and 40% for the final exam. Thus, the calculation will be as follows:
(0.15 * 73 + 0.15 * 77 + 0.15 * 82 + 0.15 * 86 + 0.4 * 93) / (0.15 + 0.15 + 0.15 + 0.15 + 0.4)
After calculation, the weighted mean, or the final grade, is approximately 85.4, rounded to one decimal place.
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Excel is used to find the mean and standard deviation of a discrete probability distribution and the results are asâ follows: mu equals 2.0 and sigma equals negative 3.5 . Can these results beâ correct? Explain. Choose the correct answer below. A. No. The mean must be negative if the standard deviation is negative. B. No. The standard deviation cannot be negative. C. Yes. The results can be correct. D. No. The standard deviation is too large.
Answer:
Option B) No. The standard deviation cannot be negative.
Step-by-step explanation:
We are given the following in the question:
[tex]\mu = 2.0\\\sigma = -3.5[/tex]
The above results are not possible.
The mean of data can be positive or negativeThe standard deviation of data can never be negative.Standard deviation is defined as the square root of squares of difference from the mean divide by the number of observations.It tells us about variation of data.The standard deviation may be smaller or greater than the mean but it cannot have a negative value.Thus, the correct answer isOption B) No. The standard deviation cannot be negative.
Final answer:
The results from Excel indicating that the standard deviation of a discrete probability distribution is negative are incorrect, as B. standard deviation cannot be negative.
Explanation:
When analyzing a discrete probability distribution in Excel, we expect certain characteristics from the mean (mu) and standard deviation (sigma). The mean represents the expected value or average of the distribution, while the standard deviation measures how much the values in the distribution vary from the mean. In the context of a probability distribution, a standard deviation cannot be negative as it represents the square root of variance, which is a squared value and therefore nonnegative.
Given the Excel results showing mu equals 2.0 and sigma equals negative 3.5, we can immediately identify an error. A negative standard deviation does not make sense in the real world since it is a measure of dispersion. Therefore, the correct answer to whether these results can be correct is: B. No. The standard deviation cannot be negative.
If a meter measures the total amount of fluid that has passed a designated point, then what is it most likely measuring?"
A. Laminar flow
B. Turbulent flow
C. Total flow
D. Flow rate
Answer:
C. Total flow
Step-by-step explanation:
The total amount of fluid that has passed a designated point is designated as the total flow.
The flow rate is the rate at which fluid passes through a certain point, not the total amount of fluid.
Laminar and turbulent flow are different types of fluid flow patterns and don't characterize a volume of fluid.
Therefore, the correct answer is C. Total flow.
People's scores on a test of extraversion are likely to be most strongly correlated with the number of social conversations they initiate during the course of a single a hour. b semester. c month. d day. e week.
Answer:
MARK ME BRAINIEST OR ILL BREAK MY KEYBOARD
Step-by-step explanation:
BBBBBBBBBBBBB
Choose a method to solve the following system of equations. Explain why you chose that method.
Solve.
y = 2x + 7
y = -2x - 5
Answer:
Step-by-step explanation:
The given simultaneous equations are expressed as
y = 2x + 7 - - - - - - - - - -1
y = -2x - 5 - - - - - - - - - - 2
We would apply the method of substitution.
The first step is to equate equation 1 to equation 2. It becomes
2x + 7 = - 2x - 5
Next step is to add 2x to the left hand side and the right hand side of the equation. It becomes
2x + 2x + 7 = - 2x - 2x - 5
4x + 7 = - 5
Next step is to subtract 7 from the left hand side and the right hand side of the equation. It becomes
4x + 7 - 7 = - 5 - 7
4x = - 12
Next step is to divide the left hand side and the right hand side of the equation by 4. It becomes
4x/4 = -12/4
x = - 3
Substituting x = - 3 into equation 1, it becomes
y = 2 × - 3 + 7 = - 6 + 7
y = 1
Find parametric equations for the following curve. Include an interval for the parameter values. The line that passes through the points (- 4, 3) and (2, - 5). oriented in the direction of increasing x.
Choose the correct set of parametric equations and interval below.
A. x = -4 + 3t, y = 3 - 4t: - 2 lessthanorequalto t lessthanorequalto 3
B. x = - 4 + 3t, y = 3 - 4t: - infinity < t < infinity
C. x = 3 - 4t, y = - 4 + 3t: - infinity < t < infinity
D. x = 3 - 4t, y = - 4+ 3t: -2 lessthanorequalto t lessthanorequalto 3
Answer:
A. x = -4 + 3t, y = 3 - 4t: - 2 lessthanorequalto t lessthanorequalto 3
Step-by-step explanation:
Given that a line in two dimension passes through (-4,3) and (2,-5) oriented in the direction of increasing x
We can write the line equation in parametric form as
[tex]\frac{x-x_1}{x_2-x_1} =\frac{y-y_1}{y_2-y_1} \\\frac{x+4}{6} =\frac{y-3}{-8} =t\\x=-4+6t\\y = 3-8t[/tex]
The values of t when x=-4 is 0 and when x =2 is 1
So t varies from 0 to 1
If instead of t we give t' say which is t +2
then we have
t he parametric equations as
x =-4+3t and y = 3-4t
For x=-4, t =2 and for x = 2 , t =3
So option A is right.
The rates of return on the investments in a large portfolio had mound shapeddistribution, with a mean of 20% and a standard deviation of 10%.
a. What proportion of the investments had a return of between 10% and 30%?b. What proportion of investments had a return that was either less than 10% or morethan 30%?
Answer:
a) 68% of the investments had a return of between 10% and 30%.
b) 32% of investments had a return that was either less than 10% or morethan 30%.
Step-by-step explanation:
We can use the Empirical Rule to solve this question:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean. This also means that 32% of the measures are more than 1 standard deviation from the mean.
95% of the measures are within 2 standard deviation of the mean. This also means that 5% of the measures are more than 2 standard deviations from the mean.
99.7% of the measures are within 3 standard deviations of the mean. This also means that 0.3% of the measures are more than 3 standard deviations from the mean.
In this problem, we have that:
Mean = 20%.
Standard deviation = 10%.
a. What proportion of the investments had a return of between 10% and 30%?
10 is the mean subtracted by 1 one standard deviation
30 is one standard deviation added to the mean.
So 10 and 30 are within 1 standard deviation of the mean. So 68% of the investments had a return of between 10% and 30%.
b. What proportion of investments had a return that was either less than 10% or morethan 30%?
This is the proportion of investments that were farther than one standard deviation of the mean.
By the Empirical Rule, 32% of investments had a return that was either less than 10% or morethan 30%.
In octagon $ABCDEFGH$, every side is perpendicular to each of its adjacent sides. What is the perimeter of $ABCDEFGH$?
Answer:
The perimeter is 48 units
Step-by-step explanation:
The picture of the question in the attached figure
we know that
The perimeter of the octagon is the sum of its length sides
so
[tex]P=AB+BC+CD+DE+EF+FG+GH+HA[/tex]
we have
[tex]BC=10\ units\\CD=6\ units\\EF=4\ units\\GH=8\ units[/tex]
substitute
[tex]P=AB+10+6+DE+4+FG+8+HA[/tex]
Combine like terms
[tex]P=AB+DE+FG+HA+28[/tex]
we know that
[tex]BC=DE+FG+HA[/tex] ---> by segment addition postulate
[tex]BC=10\ units[/tex]
so
[tex]DE+FG+HA=10\ units[/tex]
substitute in the expression of perimeter
[tex]P=AB+(DE+FG+HA)+28[/tex]
[tex]P=AB+10+28\\P=AB+38[/tex]
Since
[tex]DC= 6\ units[/tex]
and
[tex]EF = 4\ units[/tex]
The distance between F and line BC must be
[tex]6-4=2\ units[/tex]
so
[tex]AB = HG + 2 = 10\ units[/tex]
substitute
[tex]P=AB+38\\P=10+38=48\ units[/tex]
Is (5, 2) a solution to this system of equations?
16x + y = 9
17x + 3y = 20
Here are the number of hours that 9 students spend on the computer on a typical day: 1 6 7 6 8 11 6 12 15 The data from the above 9 students form what type of distribution? Unimodal Ultimodal Bimodal
Answer:
Unimodal Distribution
Mode = 6
Step-by-step explanation:
We are given the data of hours that 9 students spend on the computer on a typical day:
1, 6, 7, 6, 8, 11, 6, 12, 15
Sorted data: 1, 6, 6, 6, 7, 8, 11, 12, 15
Mode is the most frequent observation in the given data.
Here, 6 repeats itself most frequently that is 3 times. No other observation repeats itself three times.
Thus, the mode of the data is 6.
Since, there is a single mode for the given data, the distribution of students is unimodal.
Unimodal Distribution
Mode = 6
The calculation is as follows:
The sorted data is 1, 6, 6, 6, 7, 8, 11, 12, 15
Since 6 repeats 3 times so the mode should be 6
Also, there should be the single mode so the type of distribution is unimodal
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You roll a 4-sided die repeatedly. On your odd-numbered rolls (1st,3rd,5th, etc.) you are victorious if you get a 4. On your even-numbered rolls, you are victorious if you get a 3 or 4. You stop once you are victorious. Let Y be the number of times you roll.
Find E[Y].
Answer:
E (Y) = 3
Step-by-step explanation:
If a 4-sided die is being rolled repeatedly; and the odd-numbered rolls (1st 3rd,5th, etc.)
The probability of odd number roll will be, p(T) = [tex]\frac{1}{2}[/tex]
However, on your even-numbered rolls, you are victorious if you get a 3 or 4. Also, the probability of even number roll, p(U) = [tex]\frac{1}{2}[/tex]
In order to calculate: E (Y); We can say Y to be the number of times you roll.
We know that;
E (Y) = E ( Y|T ) p(T) + E ( Y|U ) p(U)
Let us calculate E ( Y|T ) and E ( Y|U )
Y|T ≅ geometric = [tex]\frac{1}{4}[/tex]
Y|U ≅ geometric = [tex]\frac{1}{2}[/tex]
also; x ≅ geometric (p)
∴ E (x) = [tex]\frac{1}{p}[/tex]
⇒ [tex]\frac{Y}{T}[/tex] = 4 ; also [tex]\frac{Y}{U}[/tex] = 2
E (Y) = 4 × [tex]\frac{1}{2}[/tex] + 2 ×
= 2+1
E (Y) = 3
For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distributed with parameters ???? and ????. Formulate ???? as a random sum and show that ???? has a binomial distribution with parameters p???? and ????.
Answer:
See the proof below.
Step-by-step explanation:
Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."
Solution to the problem
For this case we can assume that we have N independent variables [tex]X_i[/tex] with the following distribution:
[tex] X_i Bin (1,p) = Be(p)[/tex] bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:
[tex] Z = \sum_{i=1}^N X_i[/tex]
From the info given we know that [tex] N \sim Bin (M,q) [/tex]
We need to proof that [tex] Z \sim Bin (M, pq)[/tex] by the definition of binomial random variable then we need to show that:
[tex] E(Z) = Mpq[/tex]
[tex] Var (Z) = Mpq(1-pq)[/tex]
The deduction is based on the definition of independent random variables, we can do this:
[tex] E(Z) = E(N) E(X) = Mq (p)= Mpq[/tex]
And for the variance of Z we can do this:
[tex] Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2 [/tex]
[tex] Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2[/tex]
And if we take common factor [tex]Mpq [/tex] we got:
[tex] Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq][/tex]
And as we can see then we can conclude that [tex] Z \sim Bin (M, pq)[/tex]
To show that X has a binomial distribution with parameters p*n and q*m, we can formulate X as a random sum of indicator random variables Zi = 1 if the i-th trial is a success and 0 otherwise.
Explanation:Let X be a binomial random variable with parameters p and n, and let Y be a binomial random variable with parameters q and m. To show that X has a binomial distribution with parameters p*n and q*m, we can formulate X as a random sum of indicator random variables Z1, Z2, ..., Zn, where Zi = 1 if the i-th trial is a success and 0 otherwise. Then X can be expressed as X = Z1 + Z2 + ... + Zn. Since each Zi is independent and follows a Bernoulli distribution with parameter p, the sum of n independent Bernoulli random variables is a binomial random variable with parameters p and n. Therefore, X has a binomial distribution with parameters p*n and q*m.
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the circumference (C) of a swimming pool is 56 feet. which formula can you use to find the diameter (d) if you know that c=
The formula used is: [tex]d = \frac{C}{ \pi }[/tex]
The diameter is 17.83 feet
Solution:
Given that,
Circumference of pool = C = 56 feet
To find: diameter (d)
The circumference of circle is given as:
[tex]C = 2 \pi r[/tex]
Where "r" is the radius of circle
We know, that diameter is twice the radius
[tex]d = 2r[/tex]
Thus the formula becomes,
[tex]C = \pi d[/tex]
Rearrange for "d"
[tex]d = \frac{C}{ \pi }[/tex]
Substituting the values we get,
[tex]d = \frac{56}{3.14} = 17.83[/tex]
Thus the diameter is 17.83 feet
A tank in the shape of a right rectangular prism has a height of 5 feet, width of 3 feet, and length of 2 feet. It is full of a liquid weighing 50 pounds per cubic foot. How much work does it take to pump all of the liquid out of the top of the tank
Answer:
Step-by-step explanation:
Given
height of tank [tex]h=5\ ft[/tex]
Width of tank [tex]w=3\ ft[/tex]
length of tank [tex]L=2\ ft[/tex]
suppose a layer of water at height h of thickness dh from bottom needed to be pump out
So distance moved by this layer to come out of tank is [tex]\Delta h=5-h[/tex]
weight density of water [tex]\rho =50\ pounds/ft[/tex]
Force required to hold this layer up [tex]F_s=2\times 3\times \Delta h\times 50=300\Delta h[/tex]
Work done to remove the water
[tex]W=\int_{0}^{5}300\Delta hdh[/tex]
[tex]W=\int_{0}^{5}300\left ( 5-h\right )dh[/tex]
[tex]W=3750\ Pound-ft[/tex]
The work does it take to pump all of the liquid out of the top of the tank is 3750-pound feet.
Given that
A tank in the shape of a right rectangular prism has a height of 5 feet, width of 3 feet, and length of 2 feet.
It is full of a liquid weighing 50 pounds per cubic foot.
We have to determineHow much work does it take to pump all of the liquid out of the top of the tank?
According to the questionA tank in the shape of a right rectangular prism has a height of 5 feet, width of 3 feet, and length of 2 feet.
The distance moved by this layer to come out of the tank is;
[tex]\rm \triangle h = 5-h[/tex]
The force required to hold this layer up is;
[tex]\rm Force = \triangle Height \times Width \times length \times liquid \ weighing\\ \\ Force = 2 \times 3 \times (5-h) \times 50\\ \\ Force = 300 (5-h)[/tex]
The work does it take to pump all of the liquid out of the top of the tank is calculated by;
[tex]\rm Work = \int\limits^5_0 {300(5-h)} \, dh\\ \\ Work = 300(5\int\limits^5_0 {} \, dh - \int\limits^5_0 {h} \, dh) \\ \\ Work = 300(5[h]^5_0- [\dfrac{h^2}{2}]^5_0)\\ \\ Work = 300(5(5-0)-\dfrac{5^2}{2}-\dfrac{0^2}{2})\\ \\ Work = 300(25-\dfrac{25}{2})\\ \\ Work = 300\times \dfrac{50-25}{2}\\ \\ Work = 300 \times \dfrac{25}{2}\\ \\ Work = 300 \times (12.5)\\ \\ Work = 3750 \ pounds \ feet[/tex]
Hence, The work does it take to pump all of the liquid out of the top of the tank is 3750-pound feet.
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The sum of three consecutive integers is 237. Find the integers.
Answer:
78,79,80
Step-by-step explanation:
use algebraic values. in this case, x, x+1, x+2
Final answer:
The three consecutive integers that sum up to 237 are 78, 79, and 80. To find these, an algebraic equation is set up and solved step by step to identify the smallest integer followed by the next two consecutive ones.
Explanation:
The sum of three consecutive integers is 237. To find these integers, you can set up an algebraic expression for the problem. Let the smallest integer be x. Then the next two consecutive integers would be x + 1 and x + 2. The sum can be written as:
x + (x + 1) + (x + 2) = 237
Combine like terms to solve for x:
3x + 3 = 237
3x = 234
x = 78
Now we have the smallest integer. The next two consecutive integers are:
x + 1 = 79
x + 2 = 80
Therefore, the three consecutive integers are 78, 79, and 80.