1. (1010 0110)₂ = (166)₁₀
2. (145)₁₀ = (1001 0001)₂
3. (101 000 111 100)₂ = (5074)₈
4. (1111 1100 0011 0110)₂ = (FC36)₁₆
1. Convert the 8-binary binary expansion (1010 0110)₂ to a decimal expansion.
In order to solve this problem we have to use the expansion:
n = aₓbˣ + aₓ₋₁bˣ⁻¹ + ... + a₁b¹ + a₀
where b = 2, x = 8 - 1 = 7 due is a 8-binary
(1010 0110)₂ = 1 x 2⁷ + 0 x 2⁶ + 1 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2⁰
(1010 0110)₂ = 128 + 0 + 32 + 0 + 0 + 4 + 2 + 0
(1010 0110)₂ = (166)₁₀
2. Convert the following decimal expansion (145)₁₀ to an 8-bit binary expansion.
To solve this problem we have to use the divide by 2 process.
Since we are dividing by 2, when the dividend is an even number, the remainder will be 0, and when the dividend is an odd number the binary residual will be 1.
145 ---------------> 1 Less significant bit
145/2 = 72 -----> 0
72/ 2 = 36 -----> 0
36/2 = 18 ------> 0
18/2 = 9 -------> 1
9/2 = 4 --------> 0
4/2 = 2 ---------> 0
2/2 = 1 ----------> 1 Most significant bit
Then we order from the most significant bit to the less significant bit (from the bottom to the top) to obtain the 8-binary number:
(145)₁₀ = (1001 0001)₂
3. Convert the following hexadecimal expansion (A3C)₁₆ to an octal expansion.
To convert a hexadecimal expansion to an octal expansion we have to convert from hexadecimal to binary and then to octal using the table hexadecimal to binary and binary to octal.
Converting from hexadecimal to binary:
(A3C)₁₆
A = 1010, 3 = 0011 and C = 1100
(A3C)₁₆ = (1010 0011 1100)₂
Converting from binary to octal:
To convert binary to octal we have to order the binary expansion into group of 3-bits and use the table to convert binary to octal.
(1010 0011 1100)₂ = (101 000 111 100)₂
101 = 5, 000 = 0, 111 = 7 and 100 = 4
(101 000 111 100)₂ = (5074)₈
4. Convert the following binary expansion (1111 1100 0011 0110)₂ to a hexadecimal expansion.
To solve this exercise we have to use the binary to hexadecimal table.
(1111 1100 0011 0110)₂
1111 = F, 1100 = C, 0011 = 3 and 0110 = 6
(1111 1100 0011 0110)₂ = (FC36)₁₆
Determine whether the vectors (2, 3, l), (2, -5, -3), (-3, 8, -5) are linearly dependent or linear independent. If the vectors are linearly dependent, express one as a linear combination of the others. (Solutions of homogeneous differential equations form a vector space: it is necessary to confirm whether given functions/vectors are linearly dependent or linearly independent, chapter 4).
Answer:
So the vectors are linearly independent.
Step-by-step explanation:
So if they are linearly independent then the following scalars in will have the condition a=b=c=0:
a(2,3,1)+b(2,-5,-3)+c(-3,8,-5)=(0,0,0).
We have three equations:
2a+2b-3c=0
3a-5b+8c=0
1a-3b-5c=0
Multiply last equation by -2:
2a+2b-3c=0
3a-5b+8c=0
-2a+6b+10c=0
Add equation 1 and 3:
0a+8b+7c=0
3a-5b+8c=0
-2a+6b+10c=0
Divide equation 3 by 2:
0a+8b+7c=0
3a-5b+8c=0
-a+3b+2c=0
Multiply equation 3 by 3:
0a+8b+7c=0
3a-5b+8c=0
-3a+9b+6c=0
Add equation 2 and 3:
0a+8b+7c=0
3a-5b+8c=0
0a+4b+13c=0
Multiply equation 3 by -2:
0a+8b+7c=0
3a-5b+8c=0
0a-8b-26c=0
Add equation 1 and 3:
0a+0b-19c=0
3a-5b+8c=0
0a-8b-26c=0
The first equation tells us -19c=0 which implies c=0.
If c=0 we have from the second and third equation:
3a-5b=0
0a-8b=0
0a-8b=0
0-8b=0
-8b=0 implies b=0
We have b=0 and c=0.
So what is a?
3a-5b=0 where b=0
3a-5(0)=0
3a-0=0
3a=0 implies a=0
So we have a=b=c=0.
So the vectors are linearly independent.
Final answer:
To find out if the vectors (2, 3, l), (2, -5, -3), and (-3, 8, -5) are linearly dependent or independent, set up a linear system with the vectors and look for non-trivial solutions.
Explanation:
To determine whether the vectors (2, 3, l), (2, -5, -3), and (-3, 8, -5) are linearly dependent or linearly independent, we set up the equation a(2, 3, l) + b(2, -5, -3) + c(-3, 8, -5) = (0, 0, 0), where a, b, and c are scalars.
If only the trivial solution exists, where a = b = c = 0, then the vectors are linearly independent. If a non-trivial solution exists, then the vectors are linearly dependent.
Let's solve the system of linear equations generated from the above equation:
2a + 2b - 3c = 0,3a - 5b + 8c = 0,al - 3b - 5c = 0.Using the methods for solving systems of linear equations, such as Gaussian elimination, we can determine whether a unique solution exists.
If the determinant of the coefficients matrix is non-zero, the system has a unique solution, indicating linear independence. Otherwise, a non-unique solution indicates linear dependence, and we can express
Pedro owns 5 7/10 acres of farmland. He grows beets on 1/8 of the land. On how many acres of land does Pedro grow beets?
Pedro owns 5 7/10 acres of farmland and grows beets on 1/8 of it. The calculation to determine the area used for beets is to convert 5 7/10 to an improper fraction (57/10), multiply by 1/8 to get 57/80, which is 0.7125 acres.
Explanation:The question asks us to calculate the amount of farmland Pedro uses to grow beets. Pedro owns 5 7/10 acres of farmland and grows beets on 1/8 of his land. To find out how many acres he uses for beets, we do the following calculation:
Convert the mixed number to an improper fraction: 5 7/10 is equal to 57/10 (5 × 10 + 7).Multiply 57/10 by 1/8 to find the fraction of land used for beets: 57/10 × 1/8 = 57/80.Simplify 57/80: It can't be simplified further, so we keep it as is or convert it to decimal form which is 0.7125 acres.Therefore, Pedro grows beets on 0.7125 acres of land.
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Final answer:
Pedro grows beets on 57/80 acres of his land. First, convert 5 7/10 acres to the improper fraction 57/10, then multiply by 1/8 to find the area for beets.
Explanation:
The question is asking us to calculate the amount of land Pedro uses to grow beets. Since Pedro owns 5 7/10 acres of farmland and grows beets on 1/8 of the land, we need to find what 1/8 of 5 7/10 acres is. To do this, we convert 5 7/10 to an improper fraction, which is 57/10 acres. Then, we multiply 57/10 by 1/8 to find the portion of the land used for beets.
Here is the calculation step by step:
Convert the mixed number 5 7/10 to an improper fraction: 57/10.
Multiply 57/10 by 1/8 to get the fraction of the land used for beets.
57/10 × 1/8 = (57 × 1) / (10 × 8) = 57 / 80
Convert the fraction 57/80 to its decimal form or directly to acres to get the final answer.
Therefore, Pedro grows beets on 57/80 acres of his farmland, which can be converted to a decimal to get an exact measure in acres if needed.
An estimator receives an average quote fora traffic control subcontractor of $1570 for the job duration. If the lowest bid is 4 % under average, and the highest bid is 12% above average, what is the cost difference between lowest and highest bid?
Answer:
The cost difference between lowest and highest bid $ 251.20
Step-by-step explanation:
Given,
The average quote for the traffic control subcontractor = $ 1570,
Also, the lowest bid is 4 % under average,
That is, lowest bid = average quote - 4% average quote
= 1570 - 4% of 1570
[tex]=1570-\frac{4\times 1570}{100}[/tex]
[tex]=1570-\frac{6280}{100}[/tex]
[tex]=1570-62.80[/tex]
[tex]=\$1507.2[/tex]
While, the highest bid is 12% above average,
That is, the highest bid = average quote + 12% average quote
= 1570 + 12% of 1570
[tex]=1570+\frac{12\times 1570}{100}[/tex]
[tex]=1570+\frac{18840}{100}[/tex]
[tex]=1570+188.4[/tex]
[tex]=\$1758.4[/tex]
Hence, the cost difference between lowest and highest bid = $ 1758.4 - $ 1507.2 = $ 251.20
The response to a question has three alternatives: A, B, and C. A sample of 120 responses provides 64 A, 23 B, and 33 C responses. Show the frequency and relative frequency distributions (use nearest whole number for the frequency column and 2 decimal for the relative frequency column).
Frequency--
It is the number of times an outcome occurs while performing an experiment some " n " number of times.
Relative frequency--
It is the ratio of the frequency of an outcomes to the total number of times an experiment is been performed.
Here there are TOTAL : 120 responses and three outcomes A , B and C.
The frequency table is given as follows:
Outcome A B C
Frequency 64 23 33
and the Relative frequency table is given as follows:
Outcome A B C
Relative frequency 64/120 23/120 33/120
i.e. the Relative frequency table is given by:
Outcome A B C
Relative frequency 0.53 0.19 0.28
The frequency distribution for the given sample is: A: 64, B: 23, C: 33. The relative frequency distribution is: A: 0.53, B: 0.19, C: 0.28.
Explanation:To find the frequency distribution, we simply count the number of occurrences of each response. For the given sample of 120 responses, we have:
A: 64 responses
B: 23 responses
C: 33 responses
To find the relative frequency distribution, we divide the frequency of each response by the total number of responses (120). The relative frequencies, rounded to two decimal places, are:
A: 0.53
B: 0.19
C: 0.28
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) F(x) = x4/5(x − 2)2
Answer:
The critical numbers/values are x = 0, 4/7, 2
Step-by-step explanation:
This is a doozy; no wonder you have it up here for help!
The critical numbers of a function are found where the derivative of the function is equal to 0. To find these numbers, you have to factor the deriative or simply solve it for 0. This one is especially difficult since it involves rational exponents that have to be factored. But this is fun, so let's get to it.
First off, I am assuming that the function is
[tex]f(x)=x^{\frac{4}{5}}*(x-2)^2[/tex] which involves using the product rule to find the derivative.
That derivative is
[tex]f'(x)=x^{\frac{4}{5}}*2(x-2)+\frac{4}{5}x^{-\frac{1}{5}}(x-2)^2[/tex] which simplifies down to
[tex]f'(x)=x^{\frac{4}{5}}(2x^{\frac{5}{5}}-4)+\frac{4}{5}x^{-\frac{1}{5}}(x^{\frac{10}{5}}-4x^{\frac{5}{5}}+4)[/tex] and
[tex]f'(x)=2x^{\frac{9}{5}}-4x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}[/tex]
Let's get everything over the common denominator of 5 so we can easily add and subtract like terms:
[tex]f'(x)=\frac{10}{5}x^{\frac{9}{5}}-\frac{20}{5}x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}[/tex]
Combining like terms gives us
[tex]f'(x)=\frac{14}{5}x^{\frac{9}{5}}-\frac{36}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}[/tex]
This, however, factors so it is easier to solve for x. First we will set this equal to 0, then we will factor out
[tex]\frac{2}{5}x^{-\frac{1}{5}}[/tex]:
[tex]0=\frac{2}{5}x^{-\frac{1}{5}}(7x^2-18x+8)[/tex]
By the Zero Product Property, one of those terms has to equal 0 for the whole product to equal 0. So
[tex]\frac{2}{5}x^{-\frac{1}{5}}=0[/tex] when x = 0
And
[tex]7x^2-18x+8=0[/tex] when x = 2 and x = 4/7
Those are the critical numbers/values for that function. This indicates where there is a max value or a min value.
To find the critical numbers of the function F(x) = x^(4/5)(x - 2)^2, take the derivative, set it equal to zero, and check for undefined values.
Explanation:To find the critical numbers of the function F(x) = x4/5(x - 2)2, we need to find the values of x where the derivative of the function is equal to zero or undefined.
Step 1: Find the derivative of F(x) using the product rule and simplify.
Step 2: Set the derivative equal to zero and solve for x.
Step 3: Check if the derivative is undefined at any values of x.
The critical numbers of the function are the values of x where the derivative is equal to zero or undefined.
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Accomplished silver workers in India can pound silver into incredibly thin sheets, as thin as 2.87 × 10-7 m. Find the area of such a sheet that can be formed from 1.00 kg of silver. [silver density = 10500 kg/m3]
Answer:
A=331.84 m2
Step-by-step explanation:
Hello
The density of a object is defined by
[tex]d=\frac{m}{v}[/tex]
d is the density
m is the mass of the object
v es the volume of the object
We have
[tex]thickness= t =2.87 * 10^{-7} m\\\ A=unknown=area\\ Volume=Area*thickness\\m=1 kg \\\\\\d=10500kg/m^{3}[/tex]
[tex]d=\frac{m}{v}\\ v=\frac{m}{d}\\ A*t=\frac{m}{d}\\ A=\frac{m}{d*t}\\ \\A=\frac{1 kg}{10500\frac{kg}{m^{3} }*2.87*10^{-7}m}\\A=331.84m^{2}[/tex]
I hope it helps
At the beginning of 1990, 21.7 million people lived in the metropolitan area of a particular city, and the population was growing exponentially. The 1996 population was 25 million. If this trend continues, how large will the population be in the year 2010
Final answer:
To calculate the population in the year 2010 based on exponential growth from 1990, use a growth rate factor and the known population figures from the given years.
Explanation:
Population Growth Calculation:
Determine the growth rate factor from 1990 to 1996: 25 million / 21.7 million = 1.152
Apply the growth rate to find the population in 2010: 21.7 million * (1.152)¹⁴ (14 years from 1996 to 2010) = 48.9 million
A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops, it will have turned through approximately how many revolutions?
Answer:13 revolution
Step-by-step explanation:
Given data
Wheel initial angular velocity[tex]\left ( \omega \right ) [/tex]=18 rad/s
Contant angular deaaceleration[tex]\left ( \alpha \right )[/tex]=2[tex]rad/s^2[/tex]
Time required to stop wheel completely=t sec
[tex]\omega =\omega_0 + \aplha t[/tex]
0 =18 +[tex]\left ( -2\right )t[/tex]
t=9 sec
Therefore angle turn in 9 sec
[tex]\theta [/tex]=[tex]\omega_{0} t[/tex]+[tex]\frac{1}{2}[/tex][tex]\left ( \alpha\right )t^{2}[/tex]
[tex]\theta [/tex]=[tex]18\times 9[/tex]+[tex]\frac{1}{2}[/tex][tex]\left ( -2\right )\left ( 9\right )^2[/tex]
[tex]\theta [/tex]=81rad
therefore no of turns(n) =[tex]\frac{81}{2\times \pi}[/tex]
n=12.889[tex]\approx [/tex]13 revolution
A box contains 19 large marbles and 10 small marbles. Each marble is either green or white. 8 of the large marbles are green, and 4 of the small marbles are white. If a marble is randomly selected from the box, what is the probability that it is small or white? Express your answer as a fraction or a decimal number rounded to four decimal places.
Answer:
25/29
Step-by-step explanation:
see the attached picture.
The probability that marble is small or white is 25/29.
Probability is the ratio of the number of outcomes in an exhaustive set of equally likely outcomes that produce a given event to the total number of possible outcomes.
How to find If a marble is randomly selected from the box, what is the probability that it is small or white?
Given A box contains 19 large marbles and 10 small marbles.
Each marble is either green or white.
8 of the large marbles are green, and 4 of the small marbles are white.
Then P(s or w) = P(s) +P(w)
and P(s)=10/29
P(w)=11+4/29 = 15/29.
So, P(s or w) = P(s) +P(w)
=> P(s or w) = 10+15/29
=> 25/29.
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Professor N. Timmy Date has 31 students in his Calculus class and 17 students in his Discrete Mathematics class.
(a) Assuming that there are no students who take both classes, how many students does Professor Date have?
(b) Assuming that there are five students who take both classes, how many students does Professor Date have?
Answer: a) 48
b) 43
Step-by-step explanation:
Given : The number of students Professor Date has in his Calculus class = 31
The number of students Professor Date has in his Discrete Mathematics class = 17
(a) If we assume that there are no students who take both classes, then the total number of students Professor Date Has = 31+17=48
(b) If we assume that there are five students who take both classes, then the total number of students Professor Date Has = 31+17-5=43
A candy box is made from a piece of cardboard that measures 43 by 23 inches. Squares of equal size will be cut out of each corner. The sides will then be folded up to form a rectangular box. What size square should be cut from each corner to obtain maximum volume?
Answer:
For maximum volume of the box, squares with 4.79 inches should be cut off.
Step-by-step explanation:
A candy box is made from a piece of a cardboard that measures 43 × 23 inches.
Let squares of equal size will be cut out of each corner with the measure of x inches.
Therefore, measures of each side of the candy box will become
Length = (43 - 2x)
Width = (23 - 2x)
Height = x
Now we have to calculate the value of x for which volume of the box should be maximum.
Volume (V) = Length×Width×Height
V = (43 -2x)(23 - 2x)(x)
= [(43)×(23) - 46x - 86x + 4x²]x
= [989 - 132x + 4x²]x
= 4x³- 132x² + 989x
Now we find the derivative of V and equate it to 0
[tex]\frac{dV}{dx}=12x^{2}-264x+989[/tex] = 0
Now we get values of x by quadratic formula
[tex]x=\frac{264\pm \sqrt{264^{2}-4\times 12\times 989}}{2\times 12}[/tex]
x = 17.212, 4.79
Now we test it by second derivative test for the maximum volume.
[tex]\frac{d"V}{dx}= 24x - 264[/tex]
For x = 17.212
[tex]\frac{d"V}{dx}= 24(17.212)-264=413.088-264=149.088[/tex]
This value is > 0 so volume will be minimum.
For x = 4.79
[tex]\frac{d"V}{dx}=24(4.79)-264=-149.04[/tex]
-149.04 < 0, so volume of the box will be maximum.
Therefore, for x = 4.79 inches volume of the box will be maximum.
To find the size of the square to cut from each corner to attain maximum volume, one needs to create a function for the volume based on the size of the cut, derive it, and solve for x when the derivative equals zero. However, this solution might require advanced calculus.
Explanation:In this case, we're dealing with a problem in maximum volume. Let's say the size of the square cut is x. The length, width, and height of the box would then be 43-2x, 23-2x, and x, respectively. The volume of the box will then be (43-2x)(23-2x)*x.
To find the maximum volume, we take the derivative of this function (V'(x)) and find for which value of x it equals zero. But as this is a somewhat complex calculus problem, an alternative approach might be to solve it graphically or computationally, seeking for what value of x the volume becomes maximum.
For more detailed calculus solution, consult a mathematics teacher or resources that delve deeper into maximum and minimum problems within calculus.
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A hacker is trying to guess someone's password. The hacker knows (somehow) that the password is 3 digits long, and that each digit could be a number between 0 and 4. Assume that the hacker makes random guesses. What is the probability that the hacker guesses the password on his first try? Round to six decimal places.
Answer:
.008000
Step-by-step explanation:
The first digit is either 0,1,2,3,4
P( right guess) = 1/5
The second digit is either 0,1,2,3,4
P( right guess) = 1/5
The third digit is either 0,1,2,3,4
P( right guess) = 1/5
Since they are independent
P( right,right,right) = 1/5*1/5*1/5 = 1/125 =.008
To six decimal places = .008000
To solve the problem, let's consider each piece of information step by step:
1. The password is 3 digits long.
2. Each digit can be any number from 0 to 4.
Since there are 5 choices for each digit (0, 1, 2, 3, or 4), we calculate the total number of distinct combinations possible for a 3-digit password where each digit has 5 possibilities.
For each place of the three digits, we have 5 choices, which gives us a total combination count using the Multiplication Principle:
- First digit: 5 choices (0-4)
- Second digit: 5 choices (0-4)
- Third digit: 5 choices (0-4)
To find the total number of different password combinations, we multiply the number of choices for each digit:
Total combinations = 5 (choices for the first digit) × 5 (choices for the second digit) × 5 (choices for the third digit) = \( 5^3 = 125 \) possible password combinations.
Each of these combinations is equally likely if the hacker guesses at random. Hence, the probability that the hacker guesses the correct password on the first try is 1 out of the total number of combinations.
Therefore, the probability is:
\( P(\text{correct on first try}) = \frac{1}{125} \)
Let's convert this probability to a decimal and then round it to six decimal places:
\( P(\text{correct on first try}) = \frac{1}{125} = 0.008 \)
When rounded to six decimal places, the probability is:
\( P(\text{correct on first try}) \approx 0.008000 \)
So, the probability that the hacker guesses the password correctly on the first try is approximately 0.008000.
find the gcd and lcm of 20 and 56
By gcd, I think you mean gcf ( Greatest Common Factor).
To find the gcf find all the factors of each number:
Factors of 20: 1, 2, 4,5 ,10 , 20
Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56
The largest common factor is 4.
LCM = Least Common Multiple
This is the smallest number that both numbers divide into evenly
Find the prime factors of each number:
Prime factors of 20: 2 * 2 * 5
Prime factors of 56: 2 * 2 * 2 * 7
To find the LCM, multiply all prime factors the most number of times they occur.
In the prime factor of 56, 2 appears 3 times and 7 appears once.
In the prime factor of 20 5 appears once.
LCM = 2 * 2 *2 * 7 * 5 = 280
Graph the line and the parabola and find the points of intersection: y= 3-x ; y=x(squared)+x-12
Answer:
Step-by-step explanation:
We have to graph a line y = 3 - x which has the slope = -1 and y intercept 3.
We will select two points where line intersects at x = 0 and y = 0
The given line will intersect x-axis at (3, 0) and at y- axis (0, 3).
Joining these two points we can draw a straight line showing y = -x + 3
Now we will draw the parabola given by equation y = x² + x - 12
We will convert this equation in vertex form first to get the vertex and line of symmetry.
Standard equation of a parabola in vertex form is
y = (x - h)² + k
Where (h, k) is the vertex and x = h is the line of symmetry.
y = x² + x - 12
y = x² + 2(0.5)x + (0.5)²- (0.5)²-12
y = (x + 0.5)² - 12.25
Therefore, vertex will be (-0.5, -12.25) and line of symmetry will be x = 0.5
For x intercept,
0 = (x + 0.5)² - 12.25
x + 0.5 = ±√12.25
x = -0.5 ± 3.5
x = -4, 3
For y- intercept,
y = (0+0.5)² - 12.25
= 0.25 - 12.25
y = -12
So the parabola has vertex (-0.5, - 12.25), line of symmetry x = 0.5, x intercept (4, 0), (and y-intercept (0, -12).
Now we have to find the points of intersection of the given line and parabola.
For this we will replace the values of y
3 - x = x² + x - 12
x² + 2x - 15 = 0
x² + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
x = 3, -5
For x = 3
y = 3- 3 = 0
For x = -5
y = 3 + 5 = 8
Therefore, points of intersection will be (3, 0) and (-5, 8)
There is a probability of 20% that a milk container is underweight throughout of packaging line. Suppose milk containers are shipped to retail outlets in boxes of 10 containers. What is the probability that at least nine milk containers in a box are properly filled?
Answer: 0.3758
Step-by-step explanation:
Given : The probability that a milk container is underweight throughout of packaging line: [tex]p = 0.20[/tex]
The number of containers : n= 10
The formula binomial distribution formula :-
[tex]^nC_rp^{n-r}(1-p)^r[/tex]
The probability that at least nine milk containers in a box are properly filled is given by :-
[tex]P(X\geq9)=P(9)+P(10)\\\\=^{10}C_9(0.2)^{10-9}(1-0.20)^9+^{10}C_{10}(0.2)^{10-10}(1-0.2)^{10}\\=10(0.2)(0.8)^9+(1)(0.8)^{10}\\=0.3758096384\approx0.3758[/tex]
Seven trucks are filled equally from a gasoline tank and 1/3 of gasoline is still in the tank. The capacity of each truck is what part of tank:
a) 1/10 b) 2/15 c) 3/20 d) 2/21 e) 4/15
Answer:
2/21
Step-by-step explanation:
We start out with a full tank. Once the trucks take from it, it is down to 1/3 of a tank. Therefore,
[tex]\frac{3}{3} -\frac{1}{3} =\frac{2}{3}[/tex]
So the trucks took 2/3 of the gas.
If there were 7 trucks and we need to know how much of that 2/3 was taken by each truck, we divide 2/3 by 7:
[tex]\frac{\frac{2}{3} }{7}[/tex]
When dividing fractions, we bring up the lower fraction and flip it and multiply:
[tex]\frac{2}{3}*\frac{1}{7}=\frac{2}{21}[/tex]
The capacity of each truck is 2/21 of the total gasoline tank, which is calculated by dividing the used part of the gasoline tank (2/3) by the number of trucks (7).
Explanation:Let's denote the total gasoline tank volume as one unit, or 1. Seven trucks share 2/3 of the gasoline tank capacity (since 1/3 is still left); each truck capacity could be gotten by dividing this 2/3 equally among the seven trucks. By dividing 2/3 by 7, we get each truck's capacity as 2/21 of the total gasoline tank capacity. Therefore, the correct answer is d) 2/21.
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The functions f and g are defined as follows.
f (x) = 3x^2 - 3x g (x) = 3x -1
Find f(-4) and g(-6)
Simplify your answers as much as possible.
Answer:
f(-4)=60 and g(-6)=17
Step-by-step explanation:
f(x)Plug in -4 for x-values
3(-4)^2 - 3(-4)Square -4
3(16) - 3(-4)Multiply 3 by 16 and -3 by -4 then solve
48+12=60Simplify
f(-4)=60g(x)Plug in -6 for x
-3(-6)-1Multiply -3 by -6
-3(-6)=18Subtract 1
18-1=17Simplify
g(-6)=17For this case we have the following functions:
[tex]f (x) = 3x ^ 2-3x\\g (x) = 3x-1[/tex]
We must find the value of the function [tex]f (x)[/tex] when [tex]x = -4[/tex], then:
[tex]f (-4) = 3 (-4) ^ 2-3 (-4)\\f (-4) = 3 * 16 12\\f (-4) = 48 12\\f (-4) = 60[/tex]
We must find the value of the function g (x) when [tex]x = -6[/tex], then:
[tex]g (-6) = 3 (-6) -1\\g (-6) = - 18-1\\g (-6) = - 19[/tex]
Answer:
[tex]f (-4) = 60\\g (-6) = - 19[/tex]
Find the time required for an investment of 5000 dollars to grow to 6400 dollars at an interest rate of 7.5 percent per year, compounded quarterly. Your answer is t = _____
Answer:
t= 3.322 years
Step-by-step explanation:
investment made= $5000 (Principal)
amount obtained after a specific time= $6400
rate %= 7.5% per year compounded quarterly which means
r= 7.5/(100*4)= 0.01875
time = 4t ( compounded quarterly)
we know that Amount obtained is given by
[tex]A= P(1+r)^{4t}[/tex]
[tex]6400= 5000(1+0.01875)^{4t}[/tex]
[tex](1.01875)^{4t}=1.28[/tex]
taking log on both sides and solving we get
t= 3.322 years
hence my answer t= 3.322 years
Verify that y1 = x and y2 = x ln x are solutions to x 2y ′′ − xy′ + y = 0. b) Use the Wronskian to show that y1 and y2 are linearly independent. c) Find the particular solution to the differential equation with initial conditions y(1) = 7, y′ (1) = 2
a. Substitute the given solutions and their derivatives into the ODE.
[tex]y_1=x\implies {y_1}'=1\implies{y_1}''=0[/tex]
[tex]x^2y''-xy'+y=-x+x=0[/tex]
[tex]y_2=x\ln x\implies{y_1}'=\ln x+1\implies{y_1}''=\dfrac1x[/tex]
[tex]x^2y''-xy'+y=x-x(\ln x+1)+x\ln x=0[/tex]
Both solutions satisfy the ODE.
b. The Wronskian determinant is
[tex]\begin{vmatrix}x&x\ln x\\1&\ln x+1\end{vmatrix}=x(\ln x+1)-x\ln x=x\neq0[/tex]
so the solutions are indeed independent.
c. The ODE has general solution [tex]y(t)=C_1x+C_2x\ln x[/tex]. Then with the given initial conditions, the constants satisfy
[tex]y(1)=7\implies 7=C_1[/tex]
[tex]y'(1)=2\implies2=C_1+C_2\implies C_2=-5[/tex]
So the ODE has the particular solution,
[tex]\boxed{y(t)=7x-5x\ln x}[/tex]
Final answer:
The functions y1 = x and y2 = x ln x are verified as solutions to the differential equation x^2y'' - xy' + y = 0. They are confirmed to be linearly independent through a non-zero Wronskian. Lastly, the particular solution is found to be y = 7x - 5x ln x using given initial conditions.
Explanation:
To verify that y1 = x and y2 = x ln x are solutions to the differential equation x2y'' - xy' + y = 0, we need to substitute each function into the equation and show that the left-hand side reduces to zero.
For y1 = x, its derivatives are y1' = 1 and y1'' = 0. Substituting these into the equation gives x2(0) - x(1) + x = 0, which simplifies to 0, confirming that y1 is a solution.
For y2 = x ln x, its first derivative is y2' = ln x + 1, and the second derivative is y2'' = 1/x. Substituting these into the equation gives x2(1/x) - x(ln x + 1) + x ln x = 0, which also simplifies to 0, confirming that y2 is a solution.
To demonstrate that y1 and y2 are linearly independent, we must calculate the Wronskian, W(y1,y2), and show that it is non-zero. The Wronskian is:
W(y1,y2) = y1y2' - y1'y2 = x(ln x + 1) - (x ln x) = x.
Since the Wronskian is not zero for all x
e 0, y1 and y2 are linearly independent.
For the particular solution of the differential equation with initial conditions y(1) = 7, y'(1) = 2, we express y as a linear combination of y1 and y2:
y = c1y1 + c2y2 = c1x + c2x ln x.
Applying the initial conditions, we get two equations:
1) y(1) = c1(1) + c2(1 ln 1) = 7
2) y'(1) = c1 + c2(ln 1 + 1) = 2
Simplifying these equations gives us c1 = 7 and c2 = -5, therefore the particular solution is y = 7x - 5x ln x.
Please help me I don't know how to do these at all.
Answer:
The quotient is (-x³ + 4x² + 4x - 8) and the remainder is 0
Step-by-step explanation:
Look to the attached file
Solving Quadratic Equations by completing the square:
z^2 - 3z - 5 = 0
Answer:
[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]
Step-by-step explanation:
We are given the following quadratic equation by completing the square:
[tex]z^2 - 3z - 5 = 0[/tex]
Rewriting the equation in the form [tex]x^2+2ax+a^2[/tex] to get:
[tex]z^2 - 3z - 5+(-\frac{3}{2} )^2-(-\frac{3}{2} )^2[/tex]
[tex]z^2-3z+(-\frac{3}{2} )^2=(z-\frac{3}{2} )^2[/tex]
Completing the square to get:
[tex] ( z - \frac{ 3 } { 2 } )^ 2 - 5 - ( - \frac { 3 } { 2 } ) ^ 2[/tex]
[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]
Answer: [tex]z_1=4.19\\\\z_2=-1.19[/tex]
Step-by-step explanation:
Add 5 to both sides of the equation:
[tex]z^2 - 3z - 5 +5= 0+5\\\\z^2 - 3z = 5[/tex]
Divide the coefficient of [tex]z[/tex] by two and square it:
[tex](\frac{b}{2})^2= (\frac{3}{2})^2[/tex]
Add it to both sides of the equation:
[tex]z^{2} -3z+ (\frac{3}{2})^2=5+ (\frac{3}{2})^2[/tex]
Then, simplifying:
[tex](z- \frac{3}{2})^2=\frac{29}{4}[/tex]
Apply square root to both sides and solve for "z":
[tex]\sqrt{(z- \frac{3}{2})^2}=\±\sqrt{\frac{29}{4} }\\\\z=\±\sqrt{\frac{29}{4}}+ \frac{3}{2}\\\\z_1=4.19\\\\z_2=-1.19[/tex]
Ethan is playing in a soccer league that has 6 teams (including his team). Each team plays every other team twice during the regular season. The top two teams play in a final championship game after the regular season. In this league, how many soccer games will be played in all? 7.
Answer:
There are going to be 31 matches played in the soccer league.
Step-by-step explanation:
The soccer league has 6 teams, so if every team plays against the others twice, there are going to be played 30 matches:
-Team 1: v Team 2 (2), v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)
-Team 2: v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)
-Team 3: v Team 4 (2), v Team 5 (2), v Team 6 (2)
-Team 4: v Team 5 (2), v Team 6 (2)
-Team 5: v Team 6 (2)
-Team 6: -
If there is a final championship game after the 30 regular season matches, there are going to be 31 matches played in the league.
___________is the use of EHRs in a meaningful manner.
A. Interoperability
B. Meaningful use
C. Integration
Answer:
B. Meaningful use
Step-by-step explanation:
Meaningful use is the use of EHRs in a meaningful manner.
QUESTIONS The average price of wheat per metric ton in 2012 was $30575. Demand in millions of metric tons) in 2012 was 672. The average price of wheat per metric ton in 2013 was $291.56, while the demand was 700. Calculate the elasticity and classify as elastic or inelastic -0.04 elastic -0.9, elastic -0.9; inelastic -3.84, elastic -0.04 inelastic
Answer: -0.9 ; inelastic
Explanation:
Given:
The average price of wheat per metric ton in 2012 = $305.75
Demand (in millions of metric tons) in 2012 = 672
The average price of wheat per metric ton in 2013 = $291.56
Demand (in millions of metric tons) in 2013 = 700
We will compute the elasticity using the following formula:
ε = [tex]\frac{\frac{(Q_{2} - Q_{1})}{\frac{(Q_{2} +Q_{1})}{2}}}{\frac{(P_{2} - P_{1})}{\frac{(P_{2} +P_{1})}{2}}}[/tex]
ε = [tex]\frac{\Delta Q}{\Delta P}[/tex]
Now , we'll first compute [tex]\Delta Q[/tex]
i.e. [tex]\frac{\Delta Q}{\Delta P}[/tex] = [tex]\frac{(700 - 672)}{\frac{(700 +672)}{2}}[/tex]
[tex]\Delta Q[/tex] = 0.04081
Similarly for [tex]\Delta P[/tex]
i.e. [tex]\Delta P[/tex] = [tex]\frac{(291.56 - 305.75)}{\frac{(261.56 +305.75)}{2}}[/tex]
[tex]\Delta P[/tex] = -0.0475
ε = [tex]\frac{0.04081}{-0.0475}[/tex]
ε = -0.859 [tex]\simeq[/tex] -0.9
[tex]\because[/tex] we know that ;
If, ε > 1 ⇒ Elastic
ε < 1 ⇒ Inelastic
ε = 1 ⇒ unit elastic
[tex]\because[/tex] Here , ε = -0.859 [tex]\simeq[/tex] -0.9
Therefore ε is inelastic.
3500 to purchase a government bonds
Pays 4.89% annual simple interest.
How much will you have in 3 years?
Answer:
4013.45
Step-by-step explanation:
Given,
Purchased amount of the bond, P = 3500,
Annual rate of simple interest, r = 4.89% = 0.0489,
Time ( in years ), t = 3,
Since, the total amount of a bond that earns simple interest is,
[tex]A=P(1+r\times t)[/tex]
By substituting values,
The amount of the bond would be,
[tex]=3500(1+0.0489\times 3)[/tex]
[tex]=3500(1.1467)[/tex]
[tex]=4013.45[/tex]
Express the weight of the main axle of the 1893 Ferris wheel in kilograms.
Answer:
40514.837 kg
Step-by-step explanation:
The weight of the main axle of the 1893 ferris wheel built by George Washington Gale Ferris Jr. in Chicago, USA was 89,320 pounds (lb).
1 kg=2.20462 pounds (lb)
[tex]\Rightarrow 1 lb=\frac{1}{2.20426}[/tex]
[tex]\Rightarrow 1 lb=0.453592 kg[/tex]
[tex]\Rightarrow 89320 lb=89320\times 0.453592[/tex]
[tex]\therefore 89320 lb=40514.837 kg[/tex]
If the profit is $8000 and the profit % is 4%, what are net sales?
Answer:
8000/4*100 = $200'000
Step-by-step explanation:
An urn contains 11 numbered balls, of which 6 are red and 5 are white. A sample of 4 balls is to be selected. How many samples contain at least 3 red balls?
Answer:
The total number of samples that contain at least 3 red balls is 115.
Step-by-step explanation:
Total number of balls = 11
Total number of red balls = 6
Total number of white balls = 5
A sample of 4 balls is to be selected that contain at least 3 red. It means either 3 out of 4 balls are red or 4 out of 4 ball are red.
[tex]\text{Total ways}=\text{Three balls are red}+\text{Four balls are red}[/tex]
[tex]\text{Total ways}=^6C_3\times ^5C_1+^6C_4\times ^5C_0[/tex]
Combination formula:
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
Using this formula we get
[tex]\text{Total ways}=\frac{6!}{3!(6-3)!}\times \frac{5!}{1!(5-1)!}+\frac{6!}{4!(6-4)!}\times \frac{5!}{0!(5-0)!}[/tex]
[tex]\text{Total ways}=20\times 5+15\times 1[/tex]
[tex]\text{Total ways}=115[/tex]
Therefore the total number of samples that contain at least 3 red balls is 115.
Using the combination formula, it is found that 115 samples contain at least 3 red balls.
The balls are chosen without replacement, which is why the combination formula is used.
Combination formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, the outcomes with at least 3 red balls are:
3 red from a set of 6 and 1 white from a set of 5.4 red from a set of 6.Hence:
[tex]T = C_{6,3}C_{5,1} + C_{6,4} = \frac{6!}{3!3!}\frac{5!}{1!4!} + \frac{6!}{4!2!} = 20(5) + 15 = 100 + 15 = 115[/tex]
115 samples contain at least 3 red balls.
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The root of the equation f(x) = 0 is found by using the Secant method. The initial guesses are x-1 = 3.6 and x0 = 1.5. Given that f(3.6) = 7.1 and f(1.5) = 3.9, the angle the secant line makes with the x axis is ___ (Report your answer in in degrees ; keep 4 decimal places.)
Answer:
Angle made by secant line equals[tex]56.7251^{o}[/tex]
Step-by-step explanation:
Solpe of a line joining points [tex](x_{1},y_{1}),(x_{2},y_{2})[/tex] is given by
[tex]tan(\theta)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
where [tex]y_{i}=f(x_{i})[/tex]
Applying values we get
[tex]tan(\theta)=\frac{7.1-3.9}{3.6-1.5}\\\\\theta =tan^{-1}\frac{32}{21}\\\\\theta=56.7251^{o}[/tex]
A ball is shot out of a cannon at ground level, We know that its height H in feet after t sec given by the function H(t) 144t-16t Com is a. Find H(3), H(6), H(4), and H(5). Why are some of the outputs equal? H(3) feet
Answer:
H(3) = 288 feet
H(4) = 320 feet
H(5) = 320 feet
H(6) = 288 feet
Step-by-step explanation:
A ball is shot out of a cannon at ground level so the ball will follow a parabolic path.
Since height H and time t of the ball have been described by a function H(t) = 144t - 16t²
Then we have to find the values of H(3), H(4), H(5) and H(6).
H(3) = 144×3 - 16(3)²
= 432 - 144
= 288 feet
H(4) = 144×4 - 16(4)²
= 576 - 256
= 320 feet
H(5) = 144×5 - 16(5)²
= 720 - 400
= 320 feet
H(6) = 144×6 - 16(6)²
= 864 - 576
= 288 feet
Here we are getting the value like H(3), H(6) and H(4), H(5) are same because in a parabolic path ball first increase in the height above the ground then after the maximum height it decreases.
Therefore, after t = 3 and t = 6 heights of the canon ball are same. Similarly after t = 4 and t = 5 heights of the canon above the ground are same.
The heights of the ball at different times are calculated by substituting the times into the quadratic function H(t) = 144t - 16t². Some heights are equal because the ball reaches the same height on its way up and on its way down due to the parabolic path of the projectile motion.
Explanation:The height function for the ball being shot out of a cannon is H(t) = 144t - 16t². This is a quadratic function, which models projectile motion. To find the heights at specific times we substitute these times into the function.
H(3) = 144(3) - 16(3²) = 288 feetH(6) = 144(6) - 16(6²) = 0 feetH(4) = 144(4) - 16(4²) = 320 feetH(5) = 144(5) - 16(5²) = 320 feetNotice that H(4) = H(5) = 320 feet. This is because the path of the ball follows parabolic motion. The ball reaches the same height of 320 feet on its way up (at 4 seconds) and on its way down (at 5 seconds), which is why some output values are equal.
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