Answer:
The probability that a randomly selected person is not color blind is 0.958
Step-by-step explanation:
Given,
C represents red-green color blindness,
Also, the probability that a randomly selected person is color blind,
P(C) = 0.042,
Thus, probability that a randomly selected person is not color blind,
P(C') = 1 - P(C) = 1 - 0.042 = 0.958
What is the difference between any normal distribution and the standard normal distribution?
Explanation:
"Any normal distribution" may have arbitrary mean and standard deviation. The "standard normal distribution" has a mean of zero and a standard deviation of 1.
Suppose that out of 1500 first-year students at ICU, 350 are taking history, 300 are taking mathematics, and 270 are taking both history and mathematics. How many first- year students are taking history or mathematics?
Step-by-step explanation:
Assuming that the 350 taking history and the 300 taking math each includes the 270 taking both history and math, then:
N(H or M) = N(H) + N(M) − N(H and M)
N = 350 + 300 − 270
N = 380
There are 380 first-year students taking history or mathematics.
12. True/False and explain your answers. a) If A is invertible then det(A)det(A-1) = 1 b) Any matrix with a row of all zeros has a determinant of 1. c) If A is a skew symmetric matrix, AT = -A, and A has size n x n then A must be singular if n is odd.
Answer with explanation:
(A)
It is given that, A is invertible, That is inverse of matrix exist.
[tex]|A|=|A^{-1}|\neq 0[/tex]
That is, [tex]|A|=|A^{-1}|=1[/tex], is incorrect Statement.
False
(B)
If a Matrix has , either any row or column has all entry equal to Zero, then value of Determinant is equal to 0.
Any matrix with a row of all zeros has a determinant of 1 ,is incorrect Statement.
False
(C)
The Meaning of Singular matrix is that , then Determinant of Singular Matrix is equal to Zero.
For, a n×n , matrix, whether n is Odd or even
[tex]A^{T}= -A\\\\|A^{T}|=|-A|=(-1)^n|A|[/tex]
So, the statement, If A is a skew symmetric matrix, [tex]A^{T}= -A[/tex],and A has size n x n then A must be singular if n is odd ,is incorrect Statement.
False
Assume the readings on thermometers are normally distributed with a mean of 0degreesC and a standard deviation of 1.00degreesC. Find the probability that a randomly selected thermometer reads between negative 1.52 and negative 0.81 and draw a sketch of the region.
Answer:
Step-by-step explanation:
Given : The readings on thermometers 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 = -1.52
[tex]z=\dfrac{-1.52-0}{1}=-1.52[/tex]
For x = -0.81
[tex]z=\dfrac{-0.81-0}{1}=-0.81[/tex]
The p-value = [tex]P(-1.52<z<-0.81)=P(z<-0.81)-P(z<-1.52)[/tex]
[tex]0.2089701-0.0642555=0.1447146\approx0.1447[/tex]
Hence, the probability that a randomly selected thermometer reads between negative 1.52 and negative 0.81 = 0.1447
To find the probability, standardize the values using z-scores and find the area under the normal curve between the z-scores.
Explanation:To find the probability that a randomly selected thermometer reads between -1.52 and -0.81, we need to find the area under the normal curve between these two values. First, we need to standardize the values by finding the z-scores for these values using the formula z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation. After finding the z-scores, we can then use the normal distribution table or a calculator to find the area between these z-scores.
The z-score for -1.52 is z = (-1.52 - 0) / 1.00 = -1.52 and the z-score for -0.81 is z = (-0.81 - 0) / 1.00 = -0.81. Using a normal distribution table or a calculator, we can find the area to the left of -1.52 and the area to the left of -0.81. The probability that a randomly selected thermometer reads between -1.52 and -0.81 is the difference between these two areas: P(-1.52 < X < -0.81) = P(X < -0.81) - P(X < -1.52).
Using the normal distribution table or a calculator, we can find that P(X < -0.81) is approximately 0.2123 and P(X < -1.52) is approximately 0.0655. Therefore, the probability that a randomly selected thermometer reads between -1.52 and -0.81 is approximately 0.2123 - 0.0655 = 0.1468, or 14.68%. The sketch of the region would be a shaded area under the standard normal curve between -1.52 and -0.81.
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Find the Laplace transform of f(t) when:
f(t)= 9 , 0 = or < t < 2
f(t)= (t-5)^2 , 2= or < t < 5
f(t)= 2te^6t , t > or = 5
[tex]f(t)=\begin{cases}9&\text{for }0\le t<2\\(t-5)^2&\text{for }2\le t<5\\2te^{6t}&\text{for }t\ge5\end{cases}[/tex]
and presumably 0 for [tex]t<0[/tex]. We can express [tex]f(t)[/tex] in terms of the unit step function,
[tex]u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}[/tex]
[tex]f(t)=9(u(t)-u(t-2))+(t-5)^2(u(t-2)-u(t-5))+2te^{6t}u(t-5)[/tex]
Quick explanation: [tex]9u(t)=9[/tex] for [tex]t\ge0[/tex], and [tex]9u(t-2)=9[/tex] for [tex]t\ge2[/tex]. So subtracting these will cancel the value of 9 for all [tex]t\ge2[/tex] and leave us with the value of 9 over the interval we want, [tex]0\le t<2[/tex]. The same reasoning applies for the other 3 terms.
Recall the time displacement theorem:
[tex]\mathcal L_s\{f(t-c)u(t-c)\}=e^{-sc}\mathcal L_s\{f(t)\}[/tex]
By this property, we have
[tex]\mathcal L_s\{9u(t)\}=\mathcal L_s\{9\}=\dfrac9s[/tex]
[tex]\mathcal L_s\{9u(t-2)\}=e^{-2s}\mathcal L_s\{9\}=\dfrac{9e^{-2s}}s[/tex]
[tex]\mathcal L_s\{(t-5)^2u(t-2)\}=\mathcal L_s\{((t-2)-3)^2u(t-2)\}[/tex]
[tex]=e^{-2s}\mathcal L_s\{(t-3)^2\}=\left(\dfrac2{s^3}-\dfrac6{s^2}+\dfrac9s\right)e^{-2s}[/tex]
[tex]\mathcal L_s\{(t-5)^2u(t-5)\}=e^{-5s}\mathcal L_s\{t^2\}=\dfrac{2e^{-5s}}{s^3}[/tex]
[tex]\mathcal L_s\{2te^{6t}u(t-5)\}=\mathcal L_s\{2e^{30}(t-5)e^{6(t-5)}+10e^{30}e^{6(t-5)}\}[/tex]
[tex]=2e^{30-5s}\mathcal L_s\{te^{6t}+5e^{6t}\}=2e^{30-5s}\left(\dfrac1{(s-6)^2}+\dfrac5{s-6}\right)[/tex]
Putting everything together, we end up with
[tex]\boxed{\mathcal L_s\{f(t)\}=\dfrac{(2-6s)e^{-2s}-2e^{-5s}}{s^3}+\dfrac9s-\dfrac{2e^{30-5s}(29-5s)}{(s-6)^2}}[/tex]
The probability that a grader will make a marking error on any particular question of a multiple-choice exam is 0.10. If there are ten questions and questions are marked independently, what is the probability that no errors are made?The probability that a grader will make a marking error on any particular question of a multiple-choice exam is 0.10. If there are ten questions and questions are marked independently, what is the probability that no errors are made?
Answer:
about 34.9%
Step-by-step explanation:
The probability of not making a marking error is 0.9. The probability of doing that 10 times independently is 0.9^10 ≈ 0.34868 ≈ 34.9%.
(b) dy/dx = (x-y + 1)^2
Answer:
The required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].
Step-by-step explanation:
The given differential equation is
[tex]\frac{dy}{dx}=(x-y+1)^2[/tex]
Substitute u=x-y+1 in the above equation.
[tex]\frac{du}{dx}=1-\frac{dy}{dx}[/tex]
[tex]\frac{dy}{dx}=1-\frac{du}{dx}[/tex]
[tex]1-\frac{du}{dx}=u^2[/tex]
[tex]1-u^2=\frac{du}{dx}[/tex]
Using variable separable method, we get
[tex]dx=\frac{du}{1-u^2}[/tex]
Integrate both the sides.
[tex]\int dx=\int \frac{du}{1-u^2}[/tex]
[tex]x+C=\frac{1}{2}\ln|\frac{1+u}{1-u}|[/tex] [tex][\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\n|\frac{a+x}{a-x}|+C][/tex]
Substitute u=x-y+1 in the above equation.
[tex]x+C=\frac{1}{2}\ln|\frac{1+x-y+1}{1-(x-y+1)}|[/tex]
[tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex]
Therefore the required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].
find the value of solid figure not including hole cutout. Round to two decimal places. Cube figure is 12 feet long with 4 ft cutout. THANK YOU!!
Answer:
1577.20 ft³
Step-by-step explanation:
Cube of length = 12 ft = a
Hole diameter which is cutout = 4 ft = d
Hole radius which is cutout = 4/2 =2 ft = r
Volume of the cube = a³
⇒Volume of the cube = a×a×a
⇒Volume of the cube = 12×12×12
⇒Volume of the cube = 1728 ft³
The hole cut out will be in the shape of a cylinder
Volume of cylinder = πr²h
⇒Volume of cylinder = π×2²×12
⇒Volume of cylinder = 150.79 ft³
Now volume of the solid figure with hole cut out is
Volume of the cube - Volume of cylinder
=1728 - 150.79
=1577.20 ft³
∴ Volume of solid figure not including hole cutout is 1577.20 ft³
15 Points! Answer asap, please! Which is a correct name for the angle shown? (Image below)
Answer:
<CBA
Step-by-step explanation:
The angle name could be either
<ABC or <CBA
The vertex must be in the middle
Answer:
Yes, the answer is CBA
Step-by-step explanation:
The trace of a square n×n matrix A=(aij) is the sum a11+a22+⋯+ann of the entries on its main diagonal. Let V be the vector space of all 2×2 matrices with real entries. Let H be the set of all 2×2 matrices with real entries that have trace 0. Is H a subspace of the vector space V?
Let [tex]\vec h[/tex] and [tex]\vec\eta[/tex] be two vectors in [tex]H[/tex].
[tex]H[/tex] is a subspace of [tex]V[/tex] if (1) [tex]\vec h+\vec\eta\in H[/tex] and (2) for any scalar [tex]k[/tex], we have [tex]k\vec h\in H[/tex].
(1) True;
[tex]\mathrm{tr}(\vec h+\vec\eta)=\mathrm{tr}(\vec h)+\mathrm{tr}(\vec eta)=0[/tex]
so [tex]\vec h+\vec\eta\in H[/tex].
(2) Also true, since
[tex]\mathrm{tr}(k\vec h)=0k=k[/tex]
Therefore [tex]H[/tex] is a subspace of [tex]V[/tex].
Answer: Yes, H is a subspace of V
Step-by-step explanation:
We know that V is the space of all the 2x2 matrices with real entries.
H is the set of all 2x2 matrices with real entries that have trace equal to 0.
Obviusly the matrices that are in the space H also belong in the space V (because in H you have some selected matrices and in V you have all of them). The thing we need to prove is if H is an actual subspace.
Suppose we have two matrices that belong to H, A and B.
We must see that:
1) if A and B ∈ H, then (A + B)∈H
2) for a scalar number k, k*A ∈ H
lets write this as:
[tex]A = \left[\begin{array}{ccc}a1&a2\\a3&a4\\\end{array}\right] B = \left[\begin{array}{ccc}b1&b2\\b3&b4\\\end{array}\right][/tex]
where a1 + a4 = 0 = b1 + b4
then:
[tex]A + B = \left[\begin{array}{ccc}a1 + b1&a2 + b2\\a3 + b3&a4 + b4\\\end{array}\right][/tex]
the trace is:
a1 + b1 - (a4 + b4) = (a1 - a4) + (b1 - b4) = 0
then the trace is nule, and (A + B) ∈ H
and:
[tex]kA = \left[\begin{array}{ccc}k*a1&k*a2\\k*a3&k*a4\end{array}\right][/tex]
the trace is:
k*a1 - k*a4 = k(a1 - a4) = 0
so kA ∈ H
then H is a subspace of V
How many ways can a committee of five be chosen from 120 employees to interview prospective applicants.
Answer:
190578024 ways.
Step-by-step explanation:
We are asked to find the number of ways in which a committee of 5 be chosen from 120 employees to interview prospective applicants.
We will use combinations to solve our given problem.
[tex]_{r}^{n}\textrm{C}=\frac{n!}{(n-r)!r!}[/tex], where,
n = Total number of items,
r = Number of items being chosen at a time.
Upon substituting our given values in above formula, we will get:
[tex]_{5}^{120}\textrm{C}=\frac{120!}{(120-5)!5!}[/tex]
[tex]_{5}^{120}\textrm{C}=\frac{120!}{115!*5!}[/tex]
[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116*115!}{115!*5*4*3*2*1}[/tex]
[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{5*4*3*2*1}[/tex]
[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{120*1}[/tex]
[tex]_{5}^{120}\textrm{C}=\frac{119*118*117*116}{1}[/tex]
[tex]_{5}^{120}\textrm{C}=\frac{190578024}{1}[/tex]
Therefore, the committee of five can be chosen from 120 employees in 190578024 ways.
You start with 5000.00. Simple interest and yields .37% APR How much interest earned in one year?
Answer:
$1850.
Step-by-step explanation:
We are asked to find amount of interest earned in one year on an amount of $5000.
We will use simple interest formula to solve our given problem.
[tex]I=Prt[/tex], where,
I = Amount of interest earned,
r = Annual interest rate in decimal form.
t = Time in years.
Let us convert our given interest rate in decimal form.
[tex]37\%=\frac{37}{100}=0.37[/tex]
Upon substituting our given value in simple interest formula, we will get:
[tex]I=\$5000\times 0.37\times 1[/tex]
[tex]I=\$5000\times 0.37[/tex]
[tex]I=\$1850[/tex]
Therefore, an amount of $1850 is earned as interest in one year.
Let A {x ∈ N : 3 ≤ x ≤ 13}, B {x ∈ N : x is even}, and C {x ∈ N : x is odd}.
(a) Find A ∩ B.
(b) Find A ∪ B.
(c) Find B ∩ C.
(d) Find B ∪ C. 3. Find an example of sets A and B such that A ∩ B {3, 5} and A ∪ B {2, 3, 5, 7, 8}.
The answer for the sets corresponding to the given conditions is as follows:
a) A ∩ B = {4, 6, 8, 10, 12}.
b) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
c) B ∩ C = {}.
d) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8...}.
e) Set A = {3, 5, 7} and Set B = {2,4, 6, 8, 10, 12}
Given:
Set A = {x ∈ N : 3 ≤ x ≤ 13}
Set B = {x ∈ N : x is even}
Set C = {x ∈ N : x is odd}.
Solve each option:
(a) Find A ∩ B (the intersection of sets A and B):
Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.
The intersection of A and B includes even numbers that are between 3 and 13: A ∩ B = {4, 6, 8, 10, 12}.
(b) Find A ∪ B (the union of sets A and B):
Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.
The union of A and B includes all numbers from both sets: A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
(c) Find B ∩ C (the intersection of sets B and C):
Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.
Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.
The intersection of B and C is the empty set, as there are no numbers that are both even and odd.
(d) Find B ∪ C (the union of sets B and C):
Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.
Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.
The union of B and C includes all natural numbers: B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...}.
(e) For the given example:
Set A = {3, 5, 7}
Set B = {2,4, 6, 8, 10, 12}
This example satisfies the conditions A ∩ B = {3, 5} and A ∪ B = {2, 3, 5, 7, 8}
The intersection and Unioun of all the sets is found from A and B.
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The intersection of sets A and B is {4, 6, 8, 10, 12}. The union of sets A and B is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. The intersection of sets B and C is {}. The union of sets B and C is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
Explanation:(a) To find the intersection of sets A and B, we need to identify the elements that are common to both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. The elements that are common to both sets are: {4, 6, 8, 10, 12}. Therefore, A ∩ B = {4, 6, 8, 10, 12}.
(b) To find the union of sets A and B, we need to combine all the elements from both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
(c) To find the intersection of sets B and C, we need to identify the elements that are common to both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. The elements that are common to both sets are: {}. Therefore, B ∩ C = {}.
(d) To find the union of sets B and C, we need to combine all the elements from both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, B ∪ C = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
Solve Using Dirac Deltla/discontinuous forcing
Consider the following scenario. A salt tank that initially contains 50 gallons of pure water. A brine solution containing 1/5 lb/gal of salt flows in to the tank at a rate of 5 gal/min. Brine flows out of the tank at the same rate. At time t = 15 minutes the mechanism regulating salt flow in to the tank breaks and 20 pounds of salt is dumped instantaneously in to the tank. Although water continues to flow in to the tank at the original 5 gal/min, there is no salt in the water.
1. Write an IVP describing the amount of salt in the tank at time t.
2. Solve the IVP from Problem 1.
3. Plot the solution found in Problem 2 and explain the results.
Let [tex]A(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex]. We're told that [tex]A(0)=0[/tex].
For [tex]0\le t\le15[/tex], the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for [tex]t>15[/tex]. We can capture this in terms of the unit step function [tex]u(t)[/tex] and Dirac delta function [tex]\delta(t)[/tex] as
[tex]\text{rate in}=u(t)-u(t-15)+20\delta(t-15)[/tex]
(in lb/min)
The salt from the mixed solution flows out at a rate of
[tex]\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}[/tex]
Then the amount of salt in the tank at time [tex]t[/tex] changes according to
[tex]\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}[/tex]
Let [tex]\hat A(s)[/tex] denote the Laplace transform of [tex]A(t)[/tex], [tex]\hat A(s)=\mathcal L_s\{A(t)\}[/tex]. Take the transform of both sides to get
[tex]s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)[/tex]
Solve for [tex]\hat A(s)[/tex], then take the inverse of both sides.
[tex]\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}[/tex]
[tex]\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}[/tex]
A problem is given below. Instead of trying to solve that problem, state a simpler problem and solve it instead Thirteen people are being honored for their work in reducing pollution. In how many ways can we line up these people for a picture? Which of the following is a simpler version of the given problem? O A. In how many ways can fifteen people be lined up for a picture? O B. In how many ways can three people be selected from a group of thirteen people? O C. In how many ways can three people be lihed up for a picture? O D. In how many ways can thirteen people be selected from a group of fifteen people? What is the solution to this simpler problem? ways Click to select your answeris) here to search Q G
Answer:
d
Step-by-step explanation:
13 cant be divided equally nor cubed because its not an even number u can try to give all thirteen of then
The simpler version of the initial problem is arranging three people in a line. There are three choices for the first spot, two for the second, and one for the third, which results in a total of six possible arrangements. This involves the principle of permutation in combinatorics.
Explanation:The subject of the given problem can be defined as permutations. If we're looking for a simpler version of it, we should choose a problem which still involves line-up or arrangement of a smaller number of people. Hence, the best option is: 'In how many ways can three people be lined up for a picture?'
To solve this simpler problem, we consider the number of available spots for each person in the line. For the first spot, there are 3 people that could be selected. After the first person is chosen, there are only 2 people left for the second spot. Lastly, there is only 1 person left for the third spot. So, the total number of ways we can line up 3 people for a picture is 3*2*1 = 6 ways.
This is a basic principle called permutation in combinatorics which is a fundamental concept in mathematics that deals with counting, both as a means and an end in obtaining results.
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A baker pours 108 oz of batter into 36 muffin tins, pouring the same amount in each. How much batter is in each tin?
Answer: Each muffin tin contains 3 oz of batter.
Step-by-step explanation: Given that a baker pours 108 oz of batter into 36 muffin tins such that each tin has same amount of batter.
We are to calculate the quantity of batter in each tin.
We will be using the UNITARY method to solve the given problem.
Quantity of batter in 36 muffin tins = 108 oz.
Therefore, the quantity of batter in 1 muffin tin is given by
[tex]Q_t=\dfrac{108}{36}=3~\textup{oz}.[/tex]
Thus, each muffin tin contains 3 oz of batter.
By dividing 108 oz of batter by 36 muffin tins, you find that each tin contains 3 oz of batter. This simple division problem helps distribute the batter evenly. Each tin thus gets exactly 3 oz.
To find out how much batter is in each muffin tin, you need to divide the total amount of batter by the number of muffin tins.
Here are the steps:
Take the total amount of batter, which is 108 oz.Divide this amount by the number of muffin tins, which is 36.Perform the division:108 ÷ 36 = 3 oz per muffin tin.So, there are 3 oz of batter in each muffin tin.
A ramp is 10 feet in length. The ramp is lifted 4 feet off the ground to the truck door. What is the distance across the ground from the bottom of the ramp to the ground underneath the truck door? Approximate to the nearest hundredth. ≈ 3.46 feet ≈ 5.29 feet ≈ 9.17 feet ≈ 10.77 feet
Answer:
The distance is 9.17 feet.
Step-by-step explanation:
The ramp, vertical distance it is lifted, and the ground form a right triangle, whose hypotenuse the ramp, and whose base and perpendicular are the ground and the lifted distance respectively.
Thus we have a triangle whose hypotenuse [tex]H[/tex] is 10 feet, the perpendicular [tex]P[/tex] is 4 feet, and a base [tex]B[/tex] feet.
The Pythagorean theorem gives:
[tex]H^2=P^2+B^2[/tex]
We substitute the values [tex]H=10[/tex], [tex]P =4[/tex] and solve for B:
[tex]B=\sqrt{H^2-P^2} =\sqrt{10^2-4^2} =9.17.[/tex]
Thus the distance is 9.17 feet.
Answer:
the Answer is ≈ 9.17 feet
Step-by-step explanation:
it is correct on edge 2020
If your navigator tells you your position is 0.7 nautical miles from the West Point light, how far is that in feet? And how far is it in meters?
Answer:
that would be 4253.28 feet
that would be 1296.4 m
Step-by-step explanation:
I think that's the answer and I hope it helps :)
Assume that an airline’s flights for miami leave every 33 minutes and flights from dallas leave every 45 minutes. If the flights to miami and dallas have just departed, how many minutes will it be before this happen again?
Answer:
495
Step-by-step explanation:
To find this you have to find the LCM of the two times which in this case is 33 and 45. The LCM of those two is 495.
The minutes it will be before this happens again is 495.
What is the unitary method?The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.
Assume that an airline’s flights for Miami leave every 33 minutes and flights from Dallas leave every 45 minutes.
To find the LCM of the two times which in this case is 33 and 45.
Factor;
33 = 3 x 11
45 = 5 x 3 x 3
Thus, The LCM of those two is 495.
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find a nonzero vector x perpendicular to the vector v= [-2,-8,-7,2] u= [6,7,-2,8] x= [answer,answer,answer,answer]
Non zero vector x perpendicular to u and v : x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]
Given, v= [-2,-8,-7,2] u= [6,7,-2,8]
Let the vector be x = [[tex]x_1 , x_2 , x_3, x_4[/tex]]
Now x is non xero vector perpendicular to vector 'v' and 'u' .
So,
x . v = 0
[tex]-2x_1 - 8x_2 - 7x_3 + 2x_4 = 0[/tex] .........1
x . u = 0
[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex] .........2
Solve 1 and 2 to eliminate [tex]x_4[/tex] .
Multiply 1 with 4 to make the coefficients of [tex]x_4[/tex] same .
[tex]-8x_1 - 32x_2 - 28x_3 + 8x_4 = 0[/tex]
[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex]
Subtract two equations,
[tex]-14x_1 -39x_2 -26x_3 = 0[/tex]
[tex]-14x_1 = 39x_2 + 26x_3[/tex]
[tex]x_1 = \frac{-39}{14} x_2 - \frac{26}{14} x_3[/tex]
From equation 1,
[tex]x_4 = x_1 + 4x_2 + \frac{7}{2} x_3[/tex]
[tex]x_4 = \frac{-39}{14} x_2 - \frac{26}{14} x_3+ 4x_2 + \frac{7}{2} x_3\\\\x_4 = \frac{17}{14}x_2 + \frac{23}{14} x_3[/tex]
Thus x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]
[tex]x_2 = [-39/14 , 1 , 0 , 17/14] + x_3[-26/14, 0 , 1 , 23/14 ][/tex]
[tex]x_1 , x_3[/tex] are arbitrary .
For every value of [tex]x_2 , x_3[/tex] vector x is obtained.
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To find a vector x that is perpendicular to vectors v and u, we can use the cross product.
Explanation:To find a vector x that is perpendicular to vectors v and u, we can use the cross product. The cross product of two vectors is a vector that is perpendicular to both of them. To find the cross-product, we can use the formula:
x = (v2u3 - v3u2, v3u1 - v1u3, v1u2 - v2u1)
Plugging in the values, we get:
x = (-8(-2) - (-7)(7), (-7)(6) - (-2)(-2), (-2)(8) - (-8)(6)) = (1, 52, -32)
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5. Convert 11010 to base ten. A) 22 B) 26 C) 11 D) 9
Answer:
26
Step-by-step explanation:
Converting 11010 to base 10.
1*24=16
1*23=8
0*22=0
1*21=2
0*20=0
Adding all to get Ans=26_10
Step2 converting 26_10 to 10
The equation calculation formula for 26_10 number to 10 is like this below.
10|26
10|2|6
10|2|2
Ans:26_10
Assuming the given number is in base 2, we have
[tex]11010_2=2^4+2^3+2^1=16+8+2=26_{10}[/tex]
We have 7 boys and 3 girls in our church choir. There is an upcoming concert in the local town hall. Unfortunately, we can only have 5 youths in this performance. This performance team of 5 has to be picked randomly from the crew of 7 boys and 3 girls. What is the probability that exactly 4 boys are picked in this team of 5?
Answer:
105/252 = 0.41666...
Step-by-step explanation:
There are (7C4)(3C1) = (35)(3) = 105 ways to choose exactly 4 boys. There are 10C5 = 252 ways to choose 5 youths, so the probability that a randomly chosen team will consist of exactly 4 boys is ...
105/252
_____
nCk = n!/(k!(n-k!))
Answer:
There is a 41.67% probability that exactly 4 boys are picked in this team of 5.
Step-by-step explanation:
The order is not important, so we use the combinations formula.
[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Number of desired outcomes.
Four boys and one girl: So
[tex]C_{7,4}*C_{3,1} = \frac{7!}{4!(7-4)!}*\frac{3!}{1!(3-1)!} = 35*3 = 105[/tex]
Number of total outcomes:
Combination of five from a set of 10.
So
[tex]C_{10,5} = \frac{10!}{5!(10-5)!} = 252[/tex]
What is the probability that exactly 4 boys are picked in this team of 5?
[tex]P = \frac{105}{252} = 0.4167[/tex]
There is a 41.67% probability that exactly 4 boys are picked in this team of 5.
Find an implicit solution to the ODE. ("Homogeneous") y' = y^2x/y^3 +x^3 + y/x
I'm going to guess that you meant to include parentheses somewhere, so that the ODE is supposed to be
[tex]y'=\dfrac{y^2x}{y^3+x^3}+\dfrac yx[/tex]
Then substitute [tex]y(x)=xv(x)[/tex] so that [tex]y'(x)=xv'(x)+v(x)[/tex]. Then
[tex]xv'+v=\dfrac{x^3v^2}{x^3v^3+x^3}+v[/tex]
[tex]xv'=\dfrac{v^2}{v^3+1}[/tex]
which is separable as
[tex]\dfrac{v^3+1}{v^2}\,\mathrm dv=\dfrac{\mathrm dx}x[/tex]
Integrate both sides: on the left,
[tex]\displaystyle\int\frac{v^3+1}{v^2}\,\mathrm dv=\int\left(v+\frac1{v^2}\right)\,\mathrm dv=\dfrac12v^2-\dfrac1v[/tex]
The other side is trivial. We end up with
[tex]\dfrac12v^2-\dfrac1v=\ln|x|+C[/tex]
Solve in terms of [tex]y(x)[/tex]:
[tex]\boxed{\dfrac{y^2}{2x^2}-\dfrac xy=\ln|x|+C}[/tex]
I have two bags. Bag 1 contains 10 blue marbles, while Bag 2 contains 15 blue marbles. I pick one of the bags at random, and throw 6 red marbles in it. Then I shake the bag and choose 5 marbles (without replacement) at random from the bag. If there are exactly 2 red marbles among the 5 chosen marbles, what is the probability that I have chosen Bag 1?
The probability that has been chosen Bag 1 is 0.2087.
Given that, bag 1 contains 10 blue marbles, while bag 2 contains 15 blue marbles.
Here we have;
Bag 1 contains 10 blue marbles
Bag 2 contains 15 blue marbles
Chosen a bag at random and throw 5 red marbles in it.
[tex]Required Probability = P(\frac{Bag 1}{2 red and 3 blue marbles})[/tex]
= [tex]\frac{P(bag 1)\cap(2 Red \ and \ 3 blue)}{P(2 \ red \ and \ 3 \ blue \ marbles)}[/tex]
= [tex]\frac{\frac{1}{2}\times ^6C_2\times^{10}C_3}{\frac{1}{2}\times^6C_2\times^{10}C_3+\frac{1}{2}\times^6C_2\times^{15}C_3}[/tex]
= 0.2087
Therefore, the probability that has been chosen Bag 1 is 0.2087.
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To find the probability of choosing Bag 1 given there are 2 red marbles among the 5 chosen marbles, we can use Bayes' theorem to calculate the probability.
Explanation:To solve this problem, we can use Bayes' theorem to find the probability that Bag 1 was chosen given there are exactly 2 red marbles among the 5 chosen marbles. Let's denote Bag 1 as event A and Bag 2 as event B.
The probability of choosing Bag 1 is 1/2, since we picked one of the bags at random.The probability of choosing 6 red marbles from Bag 1 is (10+6) choose 6 / (10+15+6) choose 6.The probability of choosing 2 red marbles and 3 non-red marbles from Bag 1 is (10 choose 2) * (21 choose 3) / (31 choose 5).Putting all the probabilities together, we can calculate the probability of choosing Bag 1 given that there are exactly 2 red marbles among the 5 chosen marbles.The answer to the question is the probability of choosing Bag 1 given there are exactly 2 red marbles among the 5 chosen marbles.
K is the midpoint, in the line JL, equally. The space between segment JK is 2k-5. The space between segment KL is 3x-8. What is JL
Answer:
JL=2 units
Step-by-step explanation:
we know that
If k is the midpoint in the line JL
then
JL=JK+KL
JK=KL
substitute the given values
2x-5=3x-8
Solve for x
3x-2x=-5+8
x=3
so
JK=2x-5=2(3)-5=1 units
KL=3x-8=3(3)-8=1 units
therefore
JL=JK+KL=1+1=2 units
What is 2 to the power of three halves equal to?
Final answer:
2 to the power of three halves is equivalent to the square root of 2 cubed, which is approximately 2.83.
Explanation:
In mathematics, when we raise a number to a fraction exponent, we are essentially taking the root of that number. In this case, 2 to the power of three halves is equivalent to the square root of 2 cubed.
2 to the power of three halves = [tex]\sqrt{(2^3)}[/tex] = [tex]\sqrt{8}[/tex] = 2.83
Find the solution of the given initial value problem:
y''- y = 0, y(0) = 2, y'(0) = -1/2
Answer: The required solution of the given IVP is
[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]
Step-by-step explanation: We are given to find the solution of the following initial value problem :
[tex]y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.[/tex]
Let [tex]y=e^{mx}[/tex] be an auxiliary solution of the given differential equation.
Then, we have
[tex]y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.[/tex]
Substituting these values in the given differential equation, we have
[tex]m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.[/tex]
So, the general solution of the given equation is
[tex]y(x)=Ae^x+Be^{-x},[/tex] where A and B are constants.
This gives, after differentiating with respect to x that
[tex]y^\prime(x)=Ae^x-Be^{-x}.[/tex]
The given conditions implies that
[tex]y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
and
[tex]y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Adding equations (i) and (ii), we get
[tex]2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.[/tex]
From equation (i), we get
[tex]\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.[/tex]
Substituting the values of A and B in the general solution, we get
[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]
Thus, the required solution of the given IVP is
[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]
ydx+(y-x)dy=0
Please be as thorough as possible when explaining this, I'm struggling very much trying to solve ODE's
Answer: The required solution of the given differential equation is
[tex]x+y\log y=Cy.[/tex]
Step-by-step explanation: We are given to solve the following ordinary differential equation :
[tex]ydx+(y-x)dy=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We will be using the following formulas for integration and differentiation :
[tex](i)~d\left(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{y^2},\\\\\\(ii)~\int\dfrac{1}{y}dy=\log y.[/tex]
From equation (i), we have
[tex]ydx+(y-x)dy=0\\\\\Rightarrow ydx+ydy-xdy=0\\\\\\\Rightarrow \dfrac{ydx+ydy-xdy}{y^2}=\dfrac{0}{y^2}~~~~~~~~~~~~~~~~~~~~[\textup{dividing both sides by }y^2]\\\\\\\Rightarrow \dfrac{ydx-xdy}{y^2}+\dfrac{1}{y}dy=0\\\\\\\Rightarrow d\left(\dfrac{x}{y}\right)+d(\log y)=0.[/tex]
Integrating the above equation on both sides, we get
[tex]\int d\left(\dfrac{x}{y}\right)+\int d(\log y)=C~~~~~~~[\textup{where C is the constant of integration}]\\\\\\\Rightarrow \dfrac{x}{y}+\log y=C\\\\\Rightarrow x+y\log y=Cy.[/tex].
Thus, the required solution of the given differential equation is
[tex]x+y\log y=Cy.[/tex].
The side of a triangle with 3 equal sides is 8 inches shorter than the side of a square. The perimeter of the square is 46 inches more than the perimeter of the triangle. Find the length of a side of the square.
Answer:
The length of a side of the square is 22 inches.
Step-by-step explanation:
Let each side of square be = s
Let each side of triangle be = s - 8
Perimeter of square, p₁ = 4s
Perimeter of triangle = p₂ = 3s
= 3(s-8)
= 3s - 24
Therefore, according to the question
p₁ - p₂ = 46
4s - (3s - 24) = 46
4s - 3s + 24 = 46
s = 46 - 24
s = 22
The length of a side of the square is 22 inches.
Evaluate the expression:
v ⋅ w
Given the vectors:
r = <8, 8, -6>; v = <3, -8, -3>; w = <-4, -2, -6>
Answer:
v.w = 22
Step-by-step explanation:
We are given
r = <8, 8, -6>; v = <3, -8, -3>; w = <-4, -2, -6>
and we need to evaluate v.w
Using the formula: v.w = vxwx+vywy+vzwz
Putting values and solving:
v.w = 3(-4)+(-8)(-2)+(-3)(-6)
v.w = -12+16+18
v.w = 22
So, v.w = 22