Answer:
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
Step-by-step explanation:
The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.
We have the following system:
[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]
[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]
To make the reductions easier, i am going to swap the first two lines. So
[tex]L1 <-> L2[/tex]
Now the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]
Now we reduce the first row, doing the following operations
[tex]L2 = L2 - 2L1[/tex]
[tex]L3 = L3 - L1[/tex]
So, the matrix is:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]
Now we divide L2 by 3
[tex]L2 = \frac{L2}{3}[/tex]
So we have
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]
Now we have:
[tex]L3 = 3L2 + L3[/tex]
So, now we have our row reduced matrix:
[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]
We start from the bottom line, where we have:
[tex]2x_{3} = -8[/tex]
[tex]x_{3} = \frac{-8}{2}[/tex]
[tex]x_{3} = -4[/tex]
At second line:
[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]
[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]
[tex]x_{2} = 0[/tex]
At the first line
[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]
[tex]x_{1} - 4 = -3[/tex]
[tex]x_{1} = 1[/tex]
The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]
Proof by contradiction!
Prove that the following statement is true: the sum of two odd numbers is an even number
Answer:
The proposition: The sum of two odd numbers is an even number is true.
Step-by-step explanation:
A proof by contradiction is a proof technique that is based on this principle:
To prove a statement P is true, we begin by assuming P false and show that this leads to a contradiction; something that always false.
Facts that we need:
Any even number has the form 2nAny odd number has the form 2n + 1Proposition. The sum of two odd numbers is an even number
Proof. Suppose this proposition is false in this case we assume that the sum of two odd numbers is not even. (That would mean that there are two odd numbers out there in the world somewhere that'll give us an odd number when we add them.)
Let a, b be odd numbers. Then there exist numbers m, n, such that a = 2m + 1, b = 2n + 1 .Thus a + b = (2m + 1) + (2n + 1) = 2(m + n + 1) which is even. This contradicts the assumption that the sum of two odd numbers is not even.
Use the augmented matrix method to solve the following system of equations. Your answers may be given as decimals or fractions.
x - 2y - z = 2
x + 3y - 2z = 4
-x + 2y + 3z = 2
x =
y =
z =
Answer:
[tex]x\ =\ \dfrac{21}{5}[/tex]
[tex]y\ =\ \dfrac{3}{5}[/tex]
z = 2
Step-by-step explanation:
Given equations are
x - 2y - z = 2
x + 3y - 2z = 4
-x + 2y + 3z = 2
from the given equations the augmented matrix can be written as
[tex]\left[\begin{array}{ccc}1&-2&-1:2\\1&3&-2:4\\-1&2&3:2\end{array}\right][/tex]
[tex]R_2=>R_2-R_1\ and\ R_3=>R_3+R_1[/tex]
[tex]=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&5&-1:2\\0&0&2:4\end{array}\right][/tex]
[tex]R_2=>\dfrac{R_2}{5}[/tex]
[tex]=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\0&0&2:4\end{array}\right][/tex]
[tex]R_1=>R_1+2.R_2[/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&-1-\dfrac{2}{5}:2+\dfrac{4}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right][/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right][/tex]
[tex]R_3=>\dfrac{R_3}{2}[/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&1:2\end{array}\right][/tex]
[tex]R_1=>R_1+\dfrac{7}{5}R_3\ and\ R_2+\dfrac{1}{5}R_3[/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&0:\dfrac{14}{5}+\dfrac{7}{5}\\\\0&1&0:\dfrac{2}{5}+\dfrac{1}{5}\\\\0&0&1:2\end{array}\right][/tex]
So, from the above augmented matrix, we can write
[tex]x\ =\ \dfrac{21}{5}[/tex]
[tex]y\ =\ \dfrac{3}{5}[/tex]
z = 2
When the piper increased his volume, the number of rats increased 160 percent. If he ended up with 6578 rats, how many rats did he have before the volume was increased?
Answer:
i think the answer is 1,000
Step-by-step explanation:
Before the piper increased his volume, there were approximately 2530 rats. The final total (6578 rats), represents a 160 percent increase over this original amount.
Explanation:The question is asking about the original number of rats before a 160 percent increase. In this scenario, you must remember that 6578 rats is equal to the original number of rats plus an extra 160 percent of the original number. Hence, 6578 rats signifies 260% of the original quantity of rats because inherently 100% represents the original quantity. To find the original number of rats, we'd divide the final quantity by 2.6. So, the original number of rats would be calculated by: 6578 ÷ 2.6 = 2530 rats approximately. Therefore, before the piper increased his volume, there were approximately 2530 rats.
Learn more about Percent increase here:https://brainly.com/question/5449967
#SPJ2
For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.
x1 − 2x2 + 3x3 + 2x4 + x5 = 10
2x1 − 4x2 + 8x3 + 3x4 + 10x5 = 7
3x1 − 6x2 + 10x3 + 6x4 + 5x5 = 27
Answer:
This is the reduced row-echelon form
[tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&0&2&-3\\0&0&0&1&-4&7\end{array}\right][/tex]
from the augmented matrix
[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\2&-4&8&3&10&7\\3&-6&10&6&5&27\end{array}\right][/tex]
Step-by-step explanation:
To transform an augmented matrix to the reduced row-echelon form we need to follow this steps:
1. Write the system of equations as an augmented matrix.
The augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.
[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\2&-4&8&3&10&7\\3&-6&10&6&5&27\end{array}\right][/tex]
2. Make zeros in column 1 except the entry at row 1, column 1 (this is the pivot entry). Subtract row 1 multiplied by 2 from row 2 [tex]\left(R_2=R_2-\left(2\right)R_1\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\0&0&2&-1&8&-13\\3&-6&10&6&5&27\end{array}\right][/tex]
3. Subtract row 1 multiplied by 3 from row 3 [tex]\left(R_3=R_3-\left(3\right)R_1\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\0&0&2&-1&8&-13\\0&0&1&0&2&-3\end{array}\right][/tex]
4. Since element at row 2 and column 2 (pivot element) equals 0, we need to swap rows. Find the first non-zero element in the column 2 under the pivot entry. As can be seen, there are no such entries. So, move to the next column. Make zeros in column 3 except the entry at row 2, column 3 (pivot entry). Divide row 2 by 2 [tex]\left(R_2=\frac{R_2}{2}\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\0&0&1&-1/2&4&-13/2\\0&0&1&0&2&-3\end{array}\right][/tex]
5. Subtract row 2 multiplied by 3 from row 1 [tex]\left(R_1=R_1-\left(3\right)R_2\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&0&7/2&-11&59/2\\0&0&1&-1/2&4&-13/2\\0&0&1&0&2&-3\end{array}\right][/tex]
6. Subtract row 2 from row 3 [tex]\left(R_3=R_3-R_2\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&0&7/2&-11&59/2\\0&0&1&-1/2&4&-13/2\\0&0&0&1/2&-2&7/2\end{array}\right][/tex]
7. Make zeros in column 4 except the entry at row 3, column 4 (pivot entry). Subtract row 3 multiplied by 7 from row 1 [tex]\left(R_1=R_1-\left(7\right)R_3\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&-1/2&4&-13/2\\0&0&0&1/2&-2&7/2\end{array}\right][/tex]
8. Add row 3 to row 2 [tex]\left(R_2=R_2+R_3\righ)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&0&2&-3\\0&0&0&1/2&-2&7/2\end{array}\right][/tex]
9. Multiply row 3 by 2 [tex]\left(R_3=\left(2\right)R_3\right)[/tex]
[tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&0&2&-3\\0&0&0&1&-4&7\end{array}\right][/tex]
If sin phi sin theta = 0.2 and sin phi cos theta = -0.3 and sin phi > 0 what is theta ? Repeat for sin phi < 0.
Answer:
θ = -33.69°
Step-by-step explanation:
For Φ>0 and Φ<0 (in general Φ≠nπ where n is an integer), sin(Φ) ≠ 0
Dividing both equations:
[tex]\frac{sin(\phi) sin(\theta)}{sin(\phi)cos(\theta)} = tan(\theta) = 0.2/(-0.3)=-2/3\\[/tex]
Therefore:
arctan(θ) = -2/3
θ = -33.69°
The answer does not depend on the sign of Φ, in fact we just need that the sine does not become zero, which occurs when Φ is equal to an integer times π (radians) or 180 (degrees)
Have a nice day!
To find theta (θ) given that sin phi (φ) sin theta (θ) = 0.2 and sin phi (φ) cos theta (θ) = -0.3 with sin phi (φ) being positive or negative, one must first eliminate sin phi (φ) by manipulating the given equations, then solve for theta (θ) using trigonometric identities and inverse functions based on the signs of sin and cos.
Explanation:We have two equations involving sin φ and θ (theta): sin φ sin θ = 0.2 and sin φ cos θ = -0.3. Also, it is given that sin φ > 0 or sin φ < 0. To find θ, first, we need to derive an equation involving only θ by eliminating sin φ. We can do this by squaring and adding both equations.
∑: (sin φ sin θ)^2 + (sin φ cos θ)^2 = 0.2^2 + (-0.3)^2 = 0.04 + 0.09 = 0.13
Using the Pythagorean identity sin^2 θ + cos^2 θ = 1, we can rewrite ∑ as sin^2 φ = 0.13. To solve for θ, we can take either of the initial equations, say sin φ sin θ = 0.2, and substitute sin^2 φ from ∑ giving sin θ = (0.2 / √0.13) or cos θ = (-0.3 / √0.13). Both positive and negative values of sin φ lead to the calculation for different θ values. The actual values of θ are determined by using the arc functions (arcsin, arccos) for both positive and negative scenarios of sin φ, taking into account the range of θ based on the signs of sin and cos.
You deposit the same $10,000 into a bank account at 4% annual interest. How long will it take for the $10,000 to compound to $30,000?
N= I/Y= PV= PMT= FV= P/Y=
Answer:
time = 28 years
Step-by-step explanation:
Given,
principal amount = $10,000
rate = 4%
total amount = $30,000
According to compound interest formula
[tex]A\ =\ P(1+r)^t[/tex]
where, A = total amount
P = principal amount
r = rate
t = time in years
so, from the question we can write,
[tex]30000\ =\ 10000(1+0.04)^t[/tex]
[tex]=>\ \dfrac{30000}{10000}\ =\ (1+0.04)^t[/tex]
[tex]=>\ 3\ =\ (1.04)^t[/tex]
by taking log on both sides, we will get
=> log3 = t.log(1.04)
[tex]=>\ t\ =\ \dfrac{log3}{log1.04}[/tex]
=> t = 28.01
So, the time taken to get the amount from 10000 to 30000 is 28 years.
In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that an adult over 40 years of age is diagnosed as having cancer?
Answer: Our required probability is 0.406.
Step-by-step explanation:
Since we have given that
Probability of selecting an adult over 40 years of age with cancer = 0.05
Probability of a doctor correctly diagnosing a person with cancer as having the disease = 0.78
Probability of incorrectly diagnosing a person without cancer as having the disease = 0.06
Let A be the given event i.e. adult over 40 years of age with cancer. P(A) = 0.05.
So, P(A')=1-0.05 = 0.95
Let C be the event that having cancer.
P(C|A)=0.78
P(C|A')=0.06
So, using the Bayes theorem, we get that
[tex]P(A|C)=\dfrac{P(A).P(C|A)}{P(A).P(C|A)+P(A')P(C|A')}\\\\P(A|C)=\dfrac{0.78\times 0.05}{0.78\times 0.05+0.06\times 0.95}\\\\P(A|C)=0.406[/tex]
Hence, our required probability is 0.406.
In a recent year, 17.7% of household watched the finals of a popular reality series. There are 110.2 million households in the United States. How many households watched the finals?
This is Find the percent of a number.
For this case we propose a rule of three:
110.2 million -------------> 100%
x --------------------------------------> 17.7%
Where the variable "x" represents the number of households (in millions) that watched the finals of a popular reality series.
[tex]x = \frac {17.7 * 110.2} {100}\\x = \frac {1950.54} {100}\\x = 19.5054[/tex]
Thus, a total of 19.5054 million homes watched the finals of a popular reality series.
Answer:
19.5054 million homes watched the finals of a popular reality series.
For the function, f(x) = 8x + 5x, find the following. (a) f(5) D NUD K (b) f(-2) orea (c) f(4.2) muca (d) f(-4.2)
Answer:
(a) 65
(b) -26
(c) 54.6
(d) -54.6
Step-by-step explanation:
(a) [tex]f(5)=8(5)+5(5)=40+25=65[/tex]
(b) [tex]f(-2)=8(-2)+5(-2)=-16-10=-26[/tex]
(c) [tex]f(4.2)=8(4.2)+5(4.2)=33.6+21=54.6[/tex]
(d) [tex]f(-4.2)=8(-4.2)+5(-4.2)=-33.6-21=-54.6[/tex]
Calculate: (Round two decimals places for the final answer):
1880 milliliter (mL)=_____ pints (pts) ?
Answer:
1880 milliliter (mL) = 3.97 pints (pts)
Step-by-step explanation:
This problem can be solved as a rule of three problem.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
Unit conversion problems, like this one, is an example of a direct relationship between measures.
1 milliliter (mL) is equal to 0.002 pints. How many pints are 1880 milliliter (mL)? We have the following rule of three
1 mL - 0.002 pints
1880 mL - x pints
x = 1880*0.002
x = 3.97 pints
There are 3.97 pints in 1880 milliliters.
The dot plot below shows the average fuel efficiency of a number of mid- size sedans for a particular year model.
The variability of each year model's average miles per gallon is 1.68. The difference between the median miles per gallon each year model's line up is approximately how many times the variability?
A. 4
B. 7
C. 6
D. 3
the amount of carbon 14 still present is a sample after t years
is given by the function C(t)=
Coe-.00012t
where co is the initial anong . estimate the age of a sample of
wood discoverd by a arecheologist if the carbon level in the sample
is only 20% of it orginal carbon 14 level.
Answer:
The age of this sample is 13,417 years.
Step-by-step explanation:
The amount of carbon 14 present in a sample after t years is given by the following equation:
[tex]C(t) = C_{0}e^{-0.00012t}[/tex]
Estimate the age of a sample of wood discoverd by a arecheologist if the carbon level in the sampleis only 20% of it orginal carbon 14 level.
The problem asks us to find the value of t when
[tex]C(t) = 0.2C_{0}[/tex]
So:
[tex]C(t) = C_{0}e^{-0.00012t}[/tex]
[tex]0.2C_{0} = C_{0}e^{-0.00012t}[/tex]
[tex]e^{-0.00012t} = \frac{0.2C_{0}}{C_{0}}[/tex]
[tex]e^{-0.00012t} = 0.2[/tex]
[tex]ln e^{-0.00012t} = ln 0.2[/tex]
[tex]-0.00012t = -1.61[/tex]
[tex]0.00012t = 1.61[/tex]
[tex]t = \frac{1.61}{0.00012}[/tex]
[tex]t = 13,416.7[/tex]
The age of this sample is 13,417 years.
Step 3: write an equivalent expression for sin4x that does not contain powers of trigonometric functions greater than 1.
Answer: I don't know if you wanted to write sin(4x) or [tex]sin^{4} (x)[/tex] , but here we go:
ok, sin(4x) = sin(2x + 2x), and we know that:
sin (a + b) = sin(a)*cos(b) + sin(b)*cos(a)
then sin (2x + 2x) = sin(2x)*cos(2x) + cos(2x)*sin(2x) = 2cos(2x)*sin(2x)
So 2*cos(2x)*sin(2x) is equivalent of sin(4x)
If you writed [tex]sin^{4} (x)[/tex] then:
[tex]sin^{4} (x) = sin^{2} (x)*sin^{2} (x)[/tex]
and using that: [tex]sins^{2} (x) = \frac{1-cos(2x)}{2}[/tex]
we have: [tex]sin^{2} (x)*sin^{2} (x) = \frac{(1-cos(2x))*(1-cos(2x))}{4} = \frac{1-2cos(2x) + cos^{2}(2x) }{4}[/tex]
and using that: [tex]cos^{2} (x) = \frac{1 + cos(2x)}{2}[/tex]
[tex]\frac{1-2cos(2x) + cos^{2}(2x) }{4} = \frac{1-2cos(2x) + \frac{1+cos(4x)}{2} }{4}[/tex]
You can keep simplifying it, but there is your representation of [tex]sin^{4} (x)[/tex] that does not contain powers of trigonometric functions greater than 1.
To write an equivalent expression for sin(4x) without powers of trigonometric functions greater than 1, use the trigonometric identity sin(2x) = 2sin(x)cos(x). Apply this identity twice to get 4sin(x)cos(x)(cos^2(x) - sin^2(x)).
Explanation:To write an equivalent expression for sin(4x) that does not contain powers of trigonometric functions greater than 1, we can use the trigonometric identity sin(2x) = 2sin(x)cos(x). By applying this identity twice, we get:
sin(4x) = sin(2(2x)) = 2sin(2x)cos(2x)Using the identity sin(2x) = 2sin(x)cos(x) again, we have 2sin(2x)cos(2x) = 2(2sin(x)cos(x))(cos^2(x) - sin^2(x))Simplifying further, we obtain 2(2sin(x)cos(x))(cos^2(x) - sin^2(x)) = 4sin(x)cos(x)(cos^2(x) - sin^2(x))Therefore, an equivalent expression for sin(4x) that does not contain powers of trigonometric functions greater than 1 is 4sin(x)cos(x)(cos^2(x) - sin^2(x)).
Learn more about Trigonometric identities here:https://brainly.com/question/24377281
#SPJ11
Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order. Each permutation is equally likely. Define the following events: A: The letter b falls in the middle (with three before it and three after it) B: The letter c appears to the right of b, although c is not necessarily immediately to the right of b. For example, "agbdcef" would be an outcome in this event. C: The letters "def occur together in that order (e.g. "gdefbca") Calculate the probability of each individual event. That is, calculate p(A), P(B), and p(c). What is p(AIC)? (c) What is p(BIC)? What is p(AIB)? (e) Which pairs of events among A, B, and C are independent? Feedback?
Answer:
P(A)=1/7
P(B)=1/2
P(C)=1/42
P(A|C)=1/10
P(B|C)=1/10
P(A|B)=1/7
A and B are independent
A and C aren't independent
B and C aren't independent
Step-by-step explanation:
A="b falls in the middle"
- b can fall in seven possible places, but only one is the middle. So, P(A)=1/7
B="c falls to the right of b"
X=i means "b falls in the i-th position"
Y=j means "c falls in the j-th position"
if b falls in the first place, c can fall in the 2nd, 3rd, 4th, 5th, 6th or 7th place.
if b falls in the 2nd place, c can fall in the 3rd, 4th, 5th, 6th or 7th place
...
If b falls in the 6th place, c can fall in the 7th place
then:
[tex][tex]P(B)=\displaystyle\sum_{i=1}^{6}( P(X=i)\displaystyle\sum_{j=i+1}^{7} P(Y=j))=\displaystyle\sum_{i=1}^{6}( \frac{1}{7}\displaystyle\sum_{j=i+1}^{7} \frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}[/tex][/tex]
- if d falls in the 1st place, e falls in the 2nd and f in the 3rd place
- if d falls in the 2nd place, e falls in the 3rd and f in the 4th place
- if d falls in the 3rd place, e falls in the 4th and f in the 5th place
- if d falls in the 4th place, e falls in the 5th and f in the 6th place
- if d falls in the 5th place, e falls in the 6th and f in the 7th place
X=i means "d falls in the i-th position"
Y=j means "e falls in the j-th position"
Z=k means "f falls in the k-th position"
[tex]P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}[/tex]
P(A|C)=P(A∩C)/P(C)=?
A∩C:
- d falls in the 1st place, e in the 2nd, f in the 3rd and b in the 4th place
- b falls in the 4th place, d in the 5th place, e in the 6th, f in the 7th place
P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420
P(A|C)=(1/420)/(1/42)=1/10
P(B|C)=P(B∩C)/P(C)=?
X=i means "d falls in the i-th position"
Y=j means "e falls in the j-th position"
Z=k means "f falls in the k-th position"
V=k means "b falls in the k-th position"
W=k means "c falls in the k-th position"
[tex]P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)[/tex]
[tex]P(B\cap C)=\displaystyle\sum_{i=1}^{3} \frac{1}{7}\times\frac{1}{6}\times\frac{1}{5}(\displaystyle\sum_{j=i+3}^{6}\frac{1}{4}\times\frac{1}{3})=\frac{1}{2520}\displaystyle\sum_{i=1}^{3} \displaystyle\sum_{j=i+3}^{6}1=\frac{1}{420}[/tex]
P(B|C)=(1/420)/(1/42)=1/10
P(A|B)=P(B∩A)/P(B)=?
B∩A:
- b falls in the 4th place and c in the 5th
- b falls in the 4th place and c in the 6th
- b falls in the 4th place and c in the 7th
P(B∩A)=3*(1/7*1/6)=1/14
P(A|B)=(1/14)(1/2)=1/7
If one event is independent of another, P(X∩Y)=P(X)P(Y)
So:
P(A∩B)=1/14=(1/7)*(1/2)=P(A)P(B), A and B are independent
P(A∩C)=1/420≠(1/7)*(1/42)=1/294=P(A)P(C), A and C aren't independent
P(B∩C)=1/420≠(1/2)*(1/42)=1/84=P(A)P(C), B and C aren't independent
using a directrixof y = -3 and a focus of (2, 1), what quadratic funtion is created
Answer:
The function is a parabola described by: [tex]y = 1 +\frac{(x-2)^{2}}{8}[/tex]
Step-by-step explanation:
Since a directrix and a focus are given, then we know that they are talking about a parabola.
We know that the distance from any point in a parabola (x, y), to the focus is exactly the same as the distance to the directrix. Therefore, in order to find the required equation, first we will compute the distance from an arbitrary point on the parabola (x, y) to the focus (2, 1):
[tex]d_{pf} = \sqrt{(x-2)^2+(y-1)^2}[/tex]
Next, we will find the distance from an arbitrary point on the parabola (x, y) to the directrix (y= -3). Since we have an horizontal directrix, the distance is easily calculated as:
[tex]d_{pd}=y-(-3)=y+3[/tex]
Finally, we equate the distances in order to find the parabola equation:
[tex]\sqrt{(x-2)^2+(y-1)^2}=y+3[/tex]
[tex](x-2)^2+(y-1)^2=(y+3)^2[/tex]
[tex](x-2)^{2} +y^{2}-2y+1=y^{2}+6y+9[/tex]
[tex](x-2)^{2}-8=8y[/tex]
[tex]y = \frac{(x-2)^{2}}{8}-1[/tex]
A patient is to receive 2 tablets po ACHS for 30 days. How many tablets should you dispense? DO NOT include units
Answer: We should dispense 60 tablets.
Step-by-step explanation:
Given : A patient is to receive 2 tablets po ACHS for 30 days.
i.e. Dose for each day = 2 tablets
Number of days = 30
If a patient takes 2 tablets each day , then the number of tablets he require for 30 days will be :-
[tex]2\times30=60[/tex] [Multiply 2 and 30]
Therefore, the number of tablets we should dispense = 60
The Powerball lottery is decided every Wednesday and Saturday night by drawing five white balls out of a drum with 69 balls and one red ball out of a drum with 26 red balls. The Powerball jackpot is won by matching all five white balls in any order and the red Powerball. Each ticket costs $2. A ticket is defined as a set of five white and 1 red number. a. [2 pts] How many possible different Powerball tickets can be purchased? b. [1 pts] How many possible different winning Powerball tickets are there?
There are 11,238,513 ways Powerball tickets can be purchased.
There are 292, 201, 338 ways to win Powerball tickets.
Given
Number of white balls = 69
Number of white balls drawn = 5
Number of red balls = 26
Number of red balls drawn = 1
What is the combination?The combination is the way to select the number of objects from a group.
The formula is used to select the number of the object is;
[tex]\rm = \ ^nC_r \\\\\rm = \dfrac{n!}{(n-r)!r!}[/tex]
Where n is the total number of objects and r is the number of selected objects.
1. How many possible different Powerball tickets can be purchased?
The number of ways Powerball tickets can be purchased is;
[tex]\rm = \ ^{69}C_5\\\\= \dfrac{69!}{(69-5)!. 5!}\\\\= \dfrac{69!}{64!.5!}\\\\= 11238513[/tex]
There are 11,238,513 ways Powerball tickets can be purchased.
2. How many possible different winning Powerball tickets are there?
A number of ways to win Powerball tickets are there is;
[tex]\rm = \ ^{69}C_5 \times ^{26}C_1\\\\= \dfrac{69!}{(69-5)!. 5!} \times \dfrac{26!}{(26-1)!\times 1!}\\\\= \dfrac{69!}{64!.5!} \times \dfrac{26!}{25!.1!}\\\\= 11238513 \times 26\\\\= 292, 201, 338 ways[/tex]
There are 292, 201, 338 ways to win Powerball tickets.
To know more about Combination click the link given below.
https://brainly.com/question/25351212
The total number of possible different Powerball tickets is 292,201,338, calculated by multiplying 11,238,513 combinations of white balls with the 26 possible red balls. There is only one winning combination for the Powerball jackpot, so there is only one possible winning Powerball ticket.
Explanation:To calculate the total number of possible different Powerball tickets, we must consider the combination of white balls and the selection of the red Powerball. There are 5 white balls drawn from a set of 69, and this is a combination because the order does not matter. Therefore, the number of combinations of white balls is calculated using the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of balls to choose from and k is the number of balls chosen. So, the calculation is C(69, 5) = 69! / (5!(69-5)!) = 11,238,513 combinations of white balls.
For the red Powerball, there is 1 ball drawn from a set of 26, and since there is only one ball, there are 26 possible outcomes. To find the total number of possible Powerball tickets, multiply the number of white ball combinations by the number of possible red balls: 11,238,513 × 26 = 292,201,338 possible Powerball tickets.
For part b, considering that there is only one winning combination, there is just one possible winning Powerball ticket.
A manufacturer has a monthly fixed cost of $110,000 and a production cost of $14 for each unit produced. The product sells for $20 per unit.
(a) What is the cost function?
C(x) =
(b) What is the revenue function?
R(x) =
(c) What is the profit function?
P(x) =
(d) Compute the profit or loss corresponding to a production level of 12,000 and 23,000 units. (Input a negative value to indicate a loss.)
at 12,000 units $ ______
at 23,000 units $______
Answer:
Cost function C(x) == FC + VC*Q
Revenue function R(x) = Px * Q
Profit function P(x) =(Px * Q)-(FC + VC*Q)
P(12000) = -38000 Loss
P(23000) = 28000 profit
Step-by-step explanation:
Total Cost is Fixed cost plus Variable cost multiplied by the produce quantity.
(a)Cost function
C(x) = FC + vc*Q
Where
FC=Fixed cost
VC=Variable cost
Q=produce quantity
(b)
Revenue function
R(x) = Px * Q
Where
Px= Sales Price
Q=produce quantity
(c) Profit function
Profit = Revenue- Total cost
P(x) =(Px * Q)-(FC + vc*Q)
(d) We have to replace in the profit function
at 12,000 units
P(12000) =($20 * 12,000)-($110,000 + $14*12,000)
P(12000) = -38000
at 23,000 units
P(x) =($20 * 23,000)-($110,000 + $14*23,000)
P(23000) = 28000
For the statement "if Fury is the director of SHIELD then Hill and Coulson are SHIELD agents" (a) Write the contrapositive (b) Write the converse (c) Write the inverse (d) Write the negation
Step-by-step explanation:
Consider the provided information.
If Fury is the director of SHIELD then Hill and Coulson are SHIELD agents"
For the condition statement [tex]p \rightarrow q[/tex] or equivalent "If p then q"
The rule for Converse is: Interchange the two statements. [tex]q \rightarrow p[/tex]The rule for Inverse is: Negative both statements. [tex]\sim p \rightarrow \sim q[/tex]The rule for Contrapositive is: Negative both statements and interchange them. [tex]\sim q \rightarrow \sim p[/tex]The rule for Negation is: If p then q" the negation will be: p and not q. [tex]p \rightarrow q=\sim p\vee q=p\vee \sim q[/tex]Part (A) Write the contrapositive.
Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
Contrapositive: If Hill and Coulson are not SHIELD agents, then Fury is not the director of SHIELD.
Part (b) Write the converse.
Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
Converse: If Hill and Coulson are SHIELD agents, then Fury is the director of SHIELD.
Part (c) Write the inverse.
Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
Inverse: If Fury is not the director of SHIELD then Hill and Coulson are not SHIELD agents
Part (D) Write the negation.
Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
Negation: Fury is the director of SHIELD and Hill and Coulson are not SHIELD agents"
Step-by-step explanation:
Consider the provided information.
If Fury is the director of SHIELD then Hill and Coulson are SHIELD agents." For the condition statement or equivalent "If p then q"
The rule for Converse is: Interchange the two statements.
The rule for Inverse is: Negative both statements.
The rule for Contrapositive is: Negative both statements and interchange them.
The rule for Negation is: If p then q" the negation will be: p and not q.
- Part (A) Write the contrapositive.
.Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
Contrapositive: If Hill and Coulson are not SHIELD agents, then Fury is not the director of SHIELD.
- Part (b) Write the converse.
.Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
.Converse: If Hill and Coulson are SHIELD agents, then Fury is the director of SHIELD.
- Part (c) Write the inverse.
.Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
.Inverse: If Fury is not the director of SHIELD then Hill and Coulson are not SHIELD agents
- Part (D) Write the negation.
.Here p is Fury is the director of SHIELD, and q is Hill and Coulson are SHIELD agents.
.Negation: Fury is the director of SHIELD and Hill and Coulson are not SHIELD agents"
Compare the numbers 26^39 and 39^26. Which one is bigger? Are they the same? No calculators.
Answer:
[tex]26^{39}[/tex] is greater.
Step-by-step explanation:
Given numbers,
[tex]26^{39}\text{ and }39^{26}[/tex]
∵ HCF ( 26, 39 ) = 13,
That is, we need to make both numbers with the exponent 13.
[tex]26^{39}=((13\times 2)^3)^{13}=(13^3\times 2^3)^{13}=(13^3\times 8)^{13}=(13^2\times 104)^{13}[/tex]
[tex](\because (a)^{mn}=(a^m)^n\text{ and }(ab)^m=a^m.a^n)[/tex]
[tex]39^{26}=((13\times 3)^2)^{13}=(13^2\times 3^2)^{13}=(13^2\times 9)^{13}[/tex]
Since,
[tex]13^2\times 104>13^2\times 9[/tex]
[tex]\implies (13^2\times 104)^{13} > (13^2\times 9)^{13}[/tex]
[tex]\implies 26^{39} > 39^{26}[/tex]
Can 5/8 be estimated to 1/2
Answer:
Step-by-step explanation:
Estimated because there is no number that can go into 5 and 8 evenly
A set contains eleven elements. How many subsets can be formed from this set?
Finite Math question.
To determine the number of subsets that can be formed from a set containing eleven elements, we use the formula [tex]2^n,[/tex] yielding [tex]2^11[/tex] = 2048 possible subsets.
Explanation:The question is about finding the number of subsets that can be formed from a set containing eleven elements. To determine this, we employ the principle that for any set with n elements, the number of possible subsets is 2n. This includes both the empty set and the set itself as subsets.
Therefore, for a set with 11 elements, the number of possible subsets is 211 = 2048. This calculation reveals that one can form 2048 different subsets from a set of eleven elements. This includes all possible combinations of elements within the set, ranging from choosing no elements (the empty set) to choosing all eleven elements (the set itself).
The Hoover Dam in the U.S produces 2 x 10 ^9 Watts of electricity and is composed of 7 x 10^9 kg of concrete. Concrete requires 1MJ of energy to produce 1 kg
1) How much energy did it take to produce the dam?
2) How long is the "energy payback time" for the dam?
Show your work.
Answer:
1) [tex]7 x 10^{15} J[/tex]
2) [tex]40.5[/tex] days
Step-by-step explanation:
1) First of all we use [tex]1MJ=1x10^{6}J[/tex] so the total energy will be [tex]7x10^{9} kg * 1x10^{6} \frac{J}{kg} =7x10^{15}J[/tex].
2)Then we use [tex]1W=1\frac{J}{s}[/tex] and [tex]Time=\frac{7 x 10^{15} J }{2x10^{9}J/s} =3.5x10^{6}s[/tex] or [tex]3.5x10^{6}s*\frac{1h}{3600s}*\frac{1day}{24h} =40.5 days[/tex]
Use a definition, postulate, or theorem to find the value of x In the figure described
SV is an angle bisector of ZRST. IfmZRSV = (2x +9)° and mZRST = (6x - 26)°, find x.
Select each definition, postulate, or theorem you will use.
A
Angle Addition Postulate
definition of midpoint
C
definition of angle bisector
D
Linear Palr Theorem
The solution is x =
Write the base number for each expression: Please I need with steps
1. 5^12
2. 1.2^2
3. (1/3)^4
Answer:
1) Base - 5
2) Base - 1.2
3) Base - [tex]\frac{1}{3}[/tex]
Step-by-step explanation:
To find : Write the base number for each expression ?
Solution :
Base number is defined as the number written in exponent form [tex]a^n[/tex]
which tells you to multiply a by itself, so a is the base of power n.
1) [tex]5^{12}[/tex]
In number [tex]5^{12}[/tex] the base is 5 as the power is 12.
2) [tex]1.2^{2}[/tex]
In number [tex]1.2^{2}[/tex] the base is 1.2 as the power is 2.
3) [tex](\frac{1}{3})^{2}[/tex]
In number [tex](\frac{1}{3})^{2}[/tex] the base is [tex]\frac{1}{3}[/tex] as the power is 4.
The base number for 5^12 is 5, for 1.2^2 is 1.095, and for (1/3)^4 is 0.577.
Explanation:To find the base number for 5^12, we need to determine what number raised to the 12th power gives us 5. Since 5 is a prime number, it cannot be expressed as a perfect power of another number. Therefore, the base number for 5^12 is 5.To find the base number for 1.2^2, we need to determine what number raised to the 2nd power gives us 1.2. We can rewrite 1.2 as 6/5. Taking the square root of 6/5 gives us approximately 1.095. Therefore, the base number for 1.2^2 is 1.095.To find the base number for (1/3)^4, we need to determine what number raised to the 4th power gives us 1/3. We can rewrite 1/3 as 3^(-1). Taking the 4th root of 3^(-1) gives us approximately 0.577. Therefore, the base number for (1/3)^4 is 0.577.Learn more about Exponents here:https://brainly.com/question/5497425
#SPJ12
Sixty percent of the eligible voting residents of a certain community support the incumbent candidate. If eight of the residents are selected at random, find the probability that at least five of the eight support the candidate.
By the binomial theorem we know that
[tex]1 = (.4 + .6)^8 \\ = {8 \choose 0} (.4)^{8} (.6)^{0} + {8 \choose 1} (.4)^{7} (.6)^{1} +{8 \choose 2} (.4)^{6} (.6)^{2} + {8 \choose 3} (.4)^{5} (.6)^{3} + {8 \choose 4} (.4)^{4} (.6)^{4} \\ + \quad {8 \choose 5} (.4)^{3} (.6)^{5} + {8 \choose 6} (.4)^{2} (.6)^{6} + {8 \choose 7} (.4)^{1} (.6)^{7} + {8 \choose 8} (.4)^{0} (.6)^{8}[/tex]
The probability that exactly 5 of 8 support the incumbent is the term
[tex]{8 \choose 5} (.4)^{3} (.6)^{5}[/tex]
So at least five of eight support is the sum of this term and beyond,
[tex]p={8 \choose 5} (.4)^{3} (.6)^{5} + {8 \choose 6} (.4)^{2} (.6)^{6} + {8 \choose 7} (.4)^{1} (.6)^{7} + {8 \choose 8} (.4)^{0} (.6)^{8}[/tex]
No particularly easy way of calculating that except popping it into Wolfram Alpha which reports
[tex]p = \dfrac{ 46413}{78125}[/tex]
Shouldn't half the terms work out to .6 ? Interestingly it's not exactly .6 but pretty close at .594.
The total probability of at least five residents supporting the candidate, denoted as P(X≥5), is calculated by summing the probabilities of exactly five, six, seven, and eight residents supporting the candidate.
To find the probability of at least five out of eight randomly selected residents supporting the incumbent candidate when 60% of the community supports them, calculate and sum the binomial probabilities for exactly five to eight residents supporting the candidate.
The student is asking for the probability of at least five out of eight randomly selected community members supporting the incumbent candidate, given that 60% of the eligible voting residents support the candidate. This is a binomial probability problem because each selection is a Bernoulli trial with only two possible outcomes (support or do not support) and the probability of a resident supporting the candidate is constant (60%).
To calculate this probability, we will sum the probabilities of exactly five, six, seven, and eight residents supporting the candidate:
Calculate the probability of exactly 5 residents supporting the candidate using the binomial probability formula: P(X = 5) = (8 choose 5) * (0.6)^5 * (0.4)^3.
Repeat the process for P(X = 6), P(X = 7), and P(X = 8).
Finally, sum these probabilities to get the total probability of at least five residents supporting the candidate: P(X \\u2265 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8).
This sum provides the required probability.
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange remainder
Answer:
As consequence of the Taylor theorem with integral remainder we have that
[tex]f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt[/tex]
If we ask that [tex]f[/tex] has continuous [tex](n+1)[/tex]th derivative we can apply the mean value theorem for integrals. Then, there exists [tex]c[/tex] between [tex]a[/tex] and [tex]x[/tex] such that
[tex] \int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x[/tex]
Hence,
[tex] \int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .[/tex]
Thus,
[tex] \int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} [/tex]
and the Taylor theorem with Lagrange remainder is
[tex] f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}[/tex].
Step-by-step explanation:
How do you express sin x + cos x in terms of sine only?
Answer:
[tex]\sin x + \sqrt{1-\sin^2x}[/tex]
Step-by-step explanation:
Given: sin x + cos x
To change the given trigonometry expression in term of sine only.
Trigonometry identity:-
[tex]\sin^2x+\cos^2x=1[/tex][tex]\cos x=\sqrt{1-\sin^2x}[/tex]Expression: [tex]\sin x+\cos x[/tex]
We get rid of cos x from expression and write as sine form.
Expression: [tex]\sin x + \sqrt{1-\sin^2x}[/tex] [tex]\because \cos x=\sqrt{1-\sin^2x}[/tex]
Hence, The final expression is only sine function.
Line segment DF is dilated from the origin to create line segment D’F’ at D’ (0, 10) and F’ (7.5, 5). The original coordinates for line segment DF are D (0, 4) and F (3, 2). What scale factor was line segment DF dilated by? A: 0.5 B: 0.75 C: 1.5 D: 2.5
Answer: D. 2.5
Step-by-step explanation:
Given : Line segment DF is dilated from the origin to create line segment D’F’ at D' (0, 10) and F' (7.5, 5).
The original coordinates for line segment DF are D(0, 4) and F(3, 2).
We know that the scale factor(k) is the ratio of the coordinates of the image points and the original points.
Then, [tex]k=\dfrac{\text{ y-coordinate of D'}}{\text{y-coordinate of D}}\\\\=\dfrac{10}{4}=\dfrac{5}{2}=2.5[/tex]
Hence, the scale factor was line segment DF dilated by = 2.5
Answer:
The answer to this is 2.5
Step-by-step explanation:
Well 1 i took the test and 2 if you mulitply the otginial coordinates by 2.5 then you will get the new cordiates
An account invested in a money market fund grew from $73,377.34 to $73,677,48 in a month What was the interest rate to the nearest tenth? What was the interest rate?
Answer:
The rate is 4.9%.
Step-by-step explanation:
The compound interest formula is :
[tex]A=p(1+\frac{r}{n})^{nt}[/tex]
A = 73677.48
p = 73377.34
r = ?
n = 12
t = 1/12
Putting the values in formula we get;
[tex]73677.48=73377.34(1+\frac{r}{12})^{12*1/12}[/tex]
=> [tex]73677.48=73377.34(1+\frac{r}{12})^{1}[/tex]
=> [tex]\frac{73677.48}{73377.34} =(1+\frac{r}{12})[/tex]
=> [tex]1.004090=(\frac{12+r}{12})[/tex]
=> [tex]12.04908=12+r[/tex]
=> [tex]r = 12.04908-12[/tex]
r = 0.04908
or r = 4.9%
Therefore, the interest rate is 4.9%.
Final answer:
Explaining how to calculate the interest rate for an investment question.
Explanation:
To find the interest rate, subtract the initial amount from the final amount: $73,677.48 - $73,377.34 = $300.14. This represents the interest earned in one month.
Next, calculate the interest rate using the formula we get the :
Interest Rate = (Interest Earned / Initial Amount) x 100%.
Plugging in the values: Interest Rate = ($300.14 / $73,377.34) x 100% ≈ 0.409% or 0.4% to the nearest tenth.