Answer:
Everything is verified in the step-by-step explanation.
Step-by-step explanation:
We have the following differential equation:
[tex]x'' + 2x' + x = 0[/tex]
This differential equation has the following characteristic polynomial:
[tex]r^{2} + 2r + 1 = 0[/tex]
This polynomial has two repeated roots of [tex]r = -1[/tex].
Since the roots are repeated, our solution has the following format:
[tex]x(t) = c_{1}e^{-t} + c_{2}te^{-t}[/tex]
This shows that the family of solutions satisfies the equation for all values of the constants. The values of the constants depends on the initial conditions.
Lets solve the system with the initial conditions given in the exercise.
[tex]x(0) = 8[/tex]
[tex]c_{1}e^{0} + c_{2}(0)e^{0} = 8[/tex]
[tex]c_{1} = 8[/tex]
--------------------
[tex]x'(0) = 8[/tex]
[tex]x(t) = c_{1}e^{-t} + c_{2}te^{-t}[/tex]
[tex]x'(t) = -c_{1}e^{-t} + c_{2}e^{-t} - c_{2}te^{-t}[/tex]
[tex]-c_{1}e^{0} + c_{2}e^{0} - c_{2}(0)e^{0} = 8[/tex]
[tex]-c_{1} + c_{2} = 8[/tex]
[tex]c_{2} = 8 + c_{1}[/tex]
[tex]c_{2} = 8 + 8[/tex]
[tex]c_{2} = 16[/tex]
With these initial conditions, we have the following solution
[tex]x(t) = 8e^{-t} + 16te^{-t}[/tex]
Using a formula estimate the body surface area of a person whose height is 5 feet and who weighs 120 pounds.
A.
1.52 m2
B.
0.32 m2
C.
1.13 m2
D.
55.9 m2
Answer:
(A) 1.52 m²
Step-by-step explanation:
As per the given data of the question,
Height of a person = 5 feet
As we know that 1 feet = 30.48 cm
∴ Height = 152.4 cm
Weight of a person = 120 pounds
And we know that 1 pound = 0.453592 kg
∴ Weight = 54.4311 kg
The Mosteller formula to calculate body surface area (BSA):
[tex]BSA(m^{2})=\sqrt{\frac{Height (cm)\times Weight(kg)}{3600}}[/tex]
Therefore,
[tex]BSA=\sqrt{\frac{Height (cm)\times Weight(kg)}{3600}}[/tex]
[tex]BSA=\sqrt{\frac{152.4\times 54.4311}{3600}}[/tex]
[tex]BSA= 1.517 m^{2} = 1.52 m^{2}[/tex]
Hence, the body surface area of a person = 1.52 m²
Therefore, option (A) is the correct option.
Is it possible for a simple, connected graph that has n vertices all of different degrees? Explain why or why not.
Answer:
It isn't possible.
Step-by-step explanation:
Let G be a graph with n vertices. There are n possible degrees: 0,1,...,n-1.
Observe that a graph can not contain a vertice with degree n-1 and a vertice with degree 0 because if one of the vertices has degree n-1 means that this vertice is adjacent to all others vertices, then the other vertices has at least degree 1.
Then there are n vertices and n-1 possible degrees. By the pigeon principle there are two vertices that have the same degree.
Prove the following statement.
The square of any odd integer has the form 8m+1 for some integer m.
As per the question,
Let a be any positive integer and b = 4.
According to Euclid division lemma , a = 4q + r
where 0 ≤ r < b.
Thus,
r = 0, 1, 2, 3
Since, a is an odd integer, and
The only valid value of r = 1 and 3
So a = 4q + 1 or 4q + 3
Case 1 :- When a = 4q + 1
On squaring both sides, we get
a² = (4q + 1)²
= 16q² + 8q + 1
= 8(2q² + q) + 1
= 8m + 1 , where m = 2q² + q
Case 2 :- when a = 4q + 3
On squaring both sides, we get
a² = (4q + 3)²
= 16q² + 24q + 9
= 8 (2q² + 3q + 1) + 1
= 8m +1, where m = 2q² + 3q +1
Now,
We can see that at every odd values of r, square of a is in the form of 8m +1.
Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.
Hence , it is clear that square of an odd positive is in form of 8m +1
Show that the given curve c(t) is a flow line of the given velocity vector field F(x, y, z).
c(t) = (2 sin(t), 2 cos(t), 9et); F(x, y, z) = (y, −x, z)
c'(t) = ?
F(c(t)) = ?
Answer:
a) [tex]c'(t) = (2 Cos(t), -2 Sin(t), 9e^t) [/tex]
b) [tex]c'(t) = (2 Cos(t), -2 Sin(t), 9e^t) [/tex]
Step-by-step explanation:
We are given in the question:
[tex]c(t) = (2 Sin(t), 2 Cos(t), 9e^t)[/tex]
F(x,y,z) = (y, -x, z)
a) [tex]c'(t) [/tex]
We differentiate with respect to t.
[tex]c'(t) = (2 Cos(t), -2 Sin(t), 9e^t) [/tex]
b) F(c(t))
This is a composite function.
[tex]F(c(t)) = F(2 Sin(t), 2 Cos(t), 9e^t)[/tex]
[tex]= (2 Cos(t), -2 Sin(t), 9e^t)[/tex]
On three examinations, you have grades of 85, 78, and 84. There is still a final examination, which counts as one grade In order to get an A your average must be at least 90. If you get 100 on the final, what is your numerical average? (Type an integer or a decimal)
Answer:
The average of the provided grades are 86.75
Step-by-step explanation:
Consider the provided information.
On three examinations, you have grades of 85, 78, and 84. In order to get an A your average must be at least 90.
In the last exam you get 100 marks now calculate the average by using the formula:
[tex]\frac{\text{Sum of observations}}{\text{Number of observations}}[/tex]
[tex]\frac{85+78+84+100}{5}[/tex]
[tex]\frac{347}{5}[/tex]
[tex]86.75[/tex]
86.75 is less than 90 so you will not get A.
The average of the provided grades are 86.75
The width of a rectangle is 4 more than half the length.
If the perimeter of the rectangle is 74, what is the width?
Perimeter of rectangle: P = 2l + 2w
width =
length =
Answer:
Width = 15.
Length = 22.
Step-by-step explanation:
If the length is L then the width W = 1/2L + 4.
The perimeter = 2L + 2W, so
2L + 2(1/2L + 4) = 74
2L + L + 8 = 74
3L = 66
L = 22.
So W = 1/2 *22 + 4 = 11 + 4
= 15.
Company A charges $331.35 per week for a compact car with unlimited miles. Company B charges $175 per week plus $0.53 per mile, for the same car. How many miles must be driven in a week so that company A is a better deal than company B?
Answer:
Company A is a better deal than Company B for the number of miles greater than 295 miles
Step-by-step explanation:
Let
y ----> the charge per week in dollars
x ----> the number of miles
we have
Company A
[tex]y=331.35[/tex] -----> equation A
Company B
[tex]y=0.53x+175[/tex] -----> equation B
Solve the system by substitution
Equate equation A and equation B and solve for x
[tex]331.35=0.53x+175[/tex]
[tex]0.53x=331.35-175\\0.53x=156.35\\x=295\ mi[/tex]
For x=295 miles the charge in Company A and Company B is the same
therefore
Company A is a better deal than Company B for the number of miles greater than 295 miles
3.2.19 Statistics students conducted a test to see if people could taste the difference between Coke and Pepsi. They fill two cups with Coke and a third with Pepsi. They then asked their subjects which tasted different than the other two. Of the 64 people they tested, 22 were able to correctly identify which of the three cups of colas tasted different. Determine a 95% theory-based confidence interval for the population proportion that can correctly identify the cola that is different? What is the margin of error from your interval from part (a)?
The confidence interval for population mean is given by :-
[tex]\hat{p}\pm E[/tex], where [tex]\hat{p}[/tex] is sample proportion and E is the margin of error .
[tex]E=z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
Given : Significance level : [tex]\alpha:1-0.95=0.05[/tex]
Sample size : n= 64
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Sample proportion: [tex]\hat{p}=\dfrac{22}{64}\approx0.344[/tex]
[tex]E=(1.96)\sqrt{\dfrac{0.344(1-0.344)}{64}}\approx0.1164[/tex]
Hence, the margin of error = 0.1164
Now, the 95% theory-based confidence interval for the population proportion will be :
[tex]0.344\pm0.1164\\\\=(0.344-0.1164,\ 0.344+0.1164)=(0.2276,\ 0.4604)[/tex]
Hence, the 99% confidence interval is [tex](0.2276,\ 0.4604)[/tex]
When constructing a 95% theory-based confidence interval for the proportion of people that can correctly identify the different cola, the interval ranges from about 0.225 to 0.463. The margin of error is approximately 0.118.
Explanation:This question pertains to a theory-based confidence interval for the population proportion. In this case, the proportion (p) is the number of people who correctly identified the different cola, which is 22 out of 64, or 0.34375. First, we need to calculate the standard error (SE), which is the square root of [ p(1-p) / n ], where n is the sample size. So, SE = sqrt[ 0.34375(1-0.34375) / 64 ] ≈ 0.0602.
The 95% confidence interval can be calculated as p ± Z * SE, where Z is the Z-score from the standard normal distribution corresponding to the desired level of confidence. For a 95% confidence interval, Z = 1.96. Plug the values into the equation gives us the interval [0.34375 - 1.96(0.0602), 0.34375 + 1.96(0.0602)] which is approximately [0.225, 0.463].
The margin of error is the difference between the endpoint of the interval and the sample proportion, which can be calculated as Z*SE. So the margin of error = 1.96(0.0602) ≈ 0.118.
Learn more about Confidence Intervals here:https://brainly.com/question/34700241
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How many rows of 10 make 50
Answer:
5 rows.
Step-by-step explanation:
Imagine having 10 in each row. If you had 5 rows that means you have 5 groups of 10.
Answer:
Step-by-step explanation:
10 goes into 50, 5 times. I know this because if you multiply 10x5 it gives you 50!
On a safari, a team of naturalists sets out toward a research station located 8.9 km away in a direction 42° north of east. After traveling in a straight line for 2.1 km, they stop and discover that they have been traveling 21° north of east, because their guide misread his compass. What is the direction (relative to due east) of the displacement vector now required to bring the team to the research station?
Answer:
27.19°
Step-by-step explanation:
According to the picture attached, we can find the distance between the two vectors using cosine law
[tex]a^{2} =b^{2} +c^{2} -2ab*cosA\\a=\sqrt{b^{2} +c^{2} -2ab*cosA} \\\\a=\sqrt{2.1^{2} +8.9^{2} -2(2.1)(8.9)*cos21}\\a=6.98\\\\[/tex]
Then we can get C angle by applying one more time cosine law between a and b
[tex]c^{2} =a^{2} +b^{2} -2ab*cosC\\\\c^{2} -a^{2} -b^{2}= -2ab*cosC\\\\\frac{c^{2} -a^{2} -b^{2}}{-2ab}=cosC\\ \\CosC=\frac{8.9^{2} -6.98^{2} -2.1^{2}}{-2*6.98*2.1}\\ \\CosC=-0.89\\\\ArcCos(-0.89)=C\\\\C=152.81[/tex]
We can see that the C angle is complement of the angle we are looking for, so we take away 180 degrees to get the answer
[tex]180=C+?\\\\180-C=?\\\\180-152.81=C\\\\27.19=C[/tex]
27.19 degrees is our answer!
Find the arc length of the given curve on the specified interval.
(6 cos(t), 6 sin(t), t), for 0 ≤ t ≤ 2π
Answer:
Step-by-step explanation:
Given that
[tex]r(t) = (6cost, 6sint, t), 0\leq t\leq 2\pi\\r'(t) = (-6sint, 6cost, 1),\\||r'(t)||=\sqrt{(-6sint)^2 +(6cost)^2+1} =\sqrt{37}[/tex]
Hence arc length = [tex]\int\limits^a_b {||r'(t)||} \, dt[/tex]
Here a = 0 b = 2pi and r'(t) = sqrt 37
Hence integrate to get
[tex]\int\limits^{2\pi} _0 {\sqrt{37} } \, dt\\ =\sqrt{37} (t)\\=2\pi\sqrt{37}[/tex]
Write a differential equation whose only solution is the trivial solution y = 0. Explain your reasoning
Answer:
[tex]2e^{y'}y=0[/tex]
Step-by-step explanation:
The solution for this differential equation [tex]2e^{y'}y=0[/tex] have to be the trivial solution y=0. Because the function [tex]e^{x}[/tex] always have values different of zero, then the only option is the trivial solution y=0.
Show that Z2[i] = {a + bi | a,b € Z2} is not a field
Step-by-step explanation:
On a field every element different from 0 should have a multiplicative inverse. Let's check that in Z2[i] not ALL nonzero elements have multiplicative inverses.
Z2 is made of two elements: 0 and 1, and so Z2[i] is made of four elements: 0+0i,0+1i, 1+0i, 1+1i (which we can simplify from now on as 0, i, 1, 1+i respectively). Now, let's check that the element 1+i doesn't have a multiplicative inverse (we can do this by showing that no matter what we multiply it by, we're not getting 1, which is the multiplicative identity)
[tex](1+i)\cdot 0 = 0[/tex] (which is NOT 1)
[tex](1+i)\cdot i = i+i^2=i-1=1+i[/tex] (which is NOT 1) (remember -1 and 1 are the same in Z2)
[tex](1+i)\cdot 1 = 1+i[/tex] (which is NOT 1)
[tex](1+i)\cdot (1+i) = 1+i+i+i^2=1+2i-1=0+0i=0[/tex] (which is NOT 1) (remember 2 is the same as 0 in Z2)
Therefore the element 1+i doesn't have a multiplicative inverse, and so Z2[i] cannot be a field.
Suppose C is a 3 x 3 matrix such that det (C) = 4. Show that det (C+C) is equal to 32
Step-by-step explanation:
Let's consider C is a matrix given by
[tex]\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right][/tex]
them determinant of matrix C can be written as
[tex]\begin{vmatrix}a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}\ =\ 4.....(1)[/tex]
Now,
[tex]det (C+C)\ =\ \begin{vmatrix}a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}\ +\ \begin{vmatrix}a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}[/tex]
[tex]=\ \begin{vmatrix}2a & 2b & 2c\\ 2d & 2e & 2f\\ 2g & 2h & 2i \end{vmatrix}[/tex]
[tex]=\ 2\times 2\times 2\times \begin{vmatrix}a & b & c\\ d & e & f\\ g & h & i \end{vmatrix}[/tex]
[tex]=\ 8\times 4\ \ \ \ \ \ \ \ from\ eq.(1)[/tex]
= 32
Hence, det (C+C) = 32
Purchase likelihood 18 dash 34 35 dash 44 45 dash 54 55 plus Total More likely 223 373 384 404 1384 Less likely 26 7 26 13 72 Neither more nor less likely 285 210 169 113 777 Total 534 590 579 530 2233 (a) What is the probability that a randomly selected individual is 35 to 44 years of age, given the individual is neither more nor less likely to buy a product emphasized as "Made in our country"? The probability is approximately 0.270 0.270. (Round to three decimal places as needed.) (b) What is the probability that a randomly selected individual is neither more nor less likely to buy a product emphasized as "Made in our country," given the individual is 35 to 44 years of age? The probability is approximately nothing. (Round to three decimal places as needed.)
Answer:
(a) 0.270 . . . . as you know
(b) 0.356
Step-by-step explanation:
(a) p(35-44 | neither) = (35-44 & neither)/(neither total) = 210/777 ≈ 0.270
__
(b) p(neither | 35-44) = (neither & 35-44)/(35-44 total) = 210/590 ≈ 0.356
martha kept track of her hot dog sales. of every 5 hotdogs sold , 4 had mustard. what percent had mustard?
Answer:
80 %
Step-by-step explanation:
Hi there!
To find the percent of hot dogs with mustard we must divide the number of hotdogs with mustard by the number of total hotdogs, and multiply this number by 100:
[tex]P = \frac{N_{withMustard}}{N_{total}}*100= 100*(4/5) = 80[/tex]
Greetings!
For which equations below is x = -3 a possible solution? Select three options.
x = 3
x = -3
|-x1 = 3
|-x) = -3
-la = -3
Answer:
x=-3
|-x| = 3
|x| = 3
Step-by-step explanation:
we know that
If a number is a solution of a equation, then the number must satisfy the equation
Verify each case
case 1) we have
x=3
substitute the value of x=-3
-3=3 -----> is not true
therefore
x=-3 is not a solution of the given equation
case 2) we have
x=-3
substitute the value of x=-3
-3=-3 -----> is true
therefore
x=-3 is a solution of the given equation
case 3) we have
|-x| = 3
substitute the value of x=-3
|-(-3)| = 3
|3| = 3
3=3-----> is true
therefore
x=-3 is a solution of the given equation
case 4) we have
|x| = 3
substitute the value of x=-3
|(-3)| = 3
3=3-----> is true
therefore
x=-3 is a solution of the given equation
case 5) we have
-|x| = 3
substitute the value of x=-3
-|(-3)| = 3
-3=3-----> is not true
therefore
x=-3 is not a solution of the given equation
Find all relative extrema and inflection points for fx)=(2x+7)^4
Answer:
[tex]x=-\frac{7}{2}[/tex] Extrema point.
The function does not have inflection points.
Step-by-step explanation:
To find the extrema points we have:
[tex]f'(x)=0[/tex]
Then:
[tex]f(x)=(2x+7)^4[/tex]
[tex]f'(x)=4(2x+7)^3(2)[/tex]
[tex]f'(x)=8(2x+7)^3[/tex]
Now:
[tex]f'(x)=8(2x+7)^3=0[/tex]
[tex]8(2x+7)^3=0[/tex]
[tex](2x+7)^3=0[/tex]
[tex]2x+7=0[/tex]
[tex]2x=-7[/tex]
[tex]x=-\frac{7}{2}[/tex]
To find the inflection points we need to calculate [tex]f''(x)=0[/tex] but due to that que have just one extrema point, the function does not have inflection points.
Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line
(x + 2)/3 = (2y + 3)/4 = (3z + 4)/5
Answer:
The distance is:
[tex]\displaystyle\frac{3\sqrt{142}}{10}[/tex]
Step-by-step explanation:
We re-write the equation of the line in the format:
[tex]\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}} [/tex]
Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.
In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:
[tex]\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right> [/tex]
Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:
[tex]x=-2+3t\\y=3+2t\\\displaystyle z=-4+\frac{5}{3}t[/tex]
We plug them into the equation of the plane to get the intersection of that line and the plane, since that intersection is the image on the plane of the point (-2, 3, -4) parallel to the given line:
[tex]\displaystyle x+y+z=3\to -2+3t+3+2t-4+\frac{5}{3}t=3[/tex]
Then we solve that equation for t, to get:
[tex]\displaystyle \frac{20}{3}t-3=3\to t=\frac{9}{10}[/tex]
Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:
[tex]\displaystyle x=-2+3\left(\frac{9}{10}\right)=\frac{7}{10}\\\displaystyle y=3+2\left(\frac{9}{10}\right)=\frac{24}{5} \\\displaystyle z=-4+\frac{5}{3}\left(\frac{9}{10}\right)=-\frac{5}{2}[/tex]
Then to find the distance we just use the distance formula:
[tex]\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}[/tex]
So we get:
[tex]\displaystyle d=\sqrt{\left(-2-\frac{7}{10}\right)^2+\left(3-\frac{24}{5}\right)^2+\left(-4 +\frac{5}{2}\right)^2}=\frac{3\sqrt{142}}{10}[/tex]
List all element of the following sets
a. { 1/n ∣ n ∈ { 3 , 4 , 5 , 6 } }
b. {x∈Z ∣ x=x+1}
c. {n∈P ∣ n is a factor of 24 }
Answer:
a) The elements are [tex]\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}[/tex]
b) The elements are (-∞ ...-1,0,1,2,..∞).
c) The elements are 2 and 3.
Step-by-step explanation:
To find : List all element of the following sets ?
Solution :
a) [tex]\{\frac{1}{n}| n\in \{ 3 , 4 , 5 , 6 \} \}[/tex]
Here, The function is [tex]f(n)=\frac{1}{n}[/tex]
Where, [tex]n\in \{ 3 , 4 , 5 , 6 \}[/tex]
Substituting the values to get elements,
[tex]f(3)=\frac{1}{3}[/tex]
[tex]f(4)=\frac{1}{4}[/tex]
[tex]f(5)=\frac{1}{5}[/tex]
[tex]f(6)=\frac{1}{6}[/tex]
The elements are [tex]\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}[/tex]
b) [tex]\{x\in \mathbb{Z} | x=x+1\}[/tex]
Here, The function is [tex]f(x)=x+1[/tex]
Where, [tex]x\in \mathbb{Z}[/tex] i.e. integers (..,-2,-1,0,1,2,..)
For x=-2
[tex]f(-2)=-2+1=-1[/tex]
For x=-1
[tex]f(-1)=-1+1=0[/tex]
For x=0
[tex]f(0)=0+1=1[/tex]
For x=1
[tex]f(1)=1+1=2[/tex]
For x=2
[tex]f(2)=2+1=3[/tex]
The elements are (-∞ ...-1,0,1,2,..∞).
c) [tex]\{n\in \mathbb{P}| \text{n is a factor of 24}\}[/tex]
Here, The function is n is a factor of 24.
Where, n is a prime number
Factors of 24 are 1,2,3,4,6,8,12,24.
The prime factor are 2,3.
The elements are 2 and 3.
Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
This sequence has generating function
[tex]F(x)=\displaystyle\sum_{k\ge0}k^3x^k[/tex]
(if we include [tex]k=0[/tex] for a moment)
Recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k[/tex]
Take the derivative to get
[tex]\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k[/tex]
[tex]\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k[/tex]
Take the derivative again:
[tex]\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k[/tex]
[tex]\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k[/tex]
Take the derivative one more time:
[tex]\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k[/tex]
[tex]\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k[/tex]
so we have
[tex]\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}[/tex]
In the equation g = 312 ÷ α , the variable g can be described best as the 1. number of degrees that a skateboarder turns when making α rotations. 2. total number of groups, g, with α students each that can be made if there are 312 students to be grouped. 3. weight of a bag containing α grapefruits if each piece of fruit weighs 312 grams. 4. total number of goats that can graze on 312 acres if each acre can feed α goats. 5. number of grams of fuel, g, needed to raise the temperature of a solution, α, to a temperature of 312◦F.
Answer:
5) True. G is the Number of grams of fuel, g, needed to raise the temperature of a solution, α, to a temperature of 312◦F.
Step-by-step explanation:
Hi!
Let's examine better this equation: [tex]g=\frac{312}{a}[/tex]
What we have here 312 is a dependent variable, and it is inversely proportional to a. The more a increases the more g decreases.
1) Number of degrees that a skateboarder turns when making "α" rotations
[tex]g=\frac{312}{a}[/tex]
1 rotation ----------- 312°
2 rotation ----------- 156°
Here we have a problem. The skateboarder must necessarily and randomly turn 312°, and its fractions. But in a circle, the rotation cannot follow this pattern.
False
2) The total number of groups, g, with "α" students each that can be made if there are 312 students to be grouped.
[tex]g=\frac{312}{a}[/tex]
1 group --------------- 312 students
2 groups ------------ 156 students
5 groups -------------62.4 students
Even though 312 is divisible for 1,2,3,4 it is not for 5,7,9, and the group is a countable, natural category.
False
3) Weight of a bag containing "α" grapefruits if each piece of fruit weighs 312 grams
[tex]g=\frac{312}{a}[/tex]
g=1 bag with 1 grapefruit-------------- 312 g
g=1 bag with 2 grapefruits ---------- 156 g
That doesn't make sense, since for this description. The best should be g=312a and not g=312/a.
False
4) The total number of goats that can graze on 312 acres if each acre can feed "α" goats.
Since there's a relation
1 acre can feed ----------------- 1 goat
312 acres can feed ----------------g
g= 312/1 = 312 acres can feed 1 goat (1 acre for 1 goat)
g=312/2= since 312 acres can feed 156 goats (1 acre for 2 goats)
g =312/3 = 312 acres can feed 104 (1 acre for 3 goats)
Clearly, this function g=312/a does not describe this since the ratio is not the same, as long as we bring more goats to graze on those 312 acres.
False
5) Number of grams of fuel, g, needed to raise the temperature of a solution, α, to a temperature of 312◦F
g= number of grams of a fuel
a= initial temperature of a solution
g=312/a
Let's pick a=100 F initial temperature
g=312/100
g=3.12 grams
Let's now pick 200F as our initial temperature.
g=312/200 g=1.56 grams of solution
The more heat needed to raise, the more fuel necessary. Then True
If the interest rate is 3% and a total of $4,370.91 will be paid to you at the end of 3 years, what is the present value of the sum
Answer:
The present value (or initial investment) is $4000.00
Step-by-step explanation:
I'm going to assume that the correct formula here is
[tex]A(t)=P(1+r)^t[/tex]
and we are looking to solve for P, the principle investment. We know that A(t) is 4370.91; r is .03 and t is 3:
[tex]4370.91=P(1+.03)^3[/tex] and
[tex]4370.91=P(1.03)^3[/tex] and
4370.91 = 1.092727P so
P = 4000.00
(a) Find all points where the function f(z) = (x^2+y^2-2y)+i(2x-2xy) is differentiable, and compute the derivative at those points.
Answer:
The given function is differentiable at y = 1.
At y = 1, f'(z) = 0
Step-by-step explanation:
As per the given question,
[tex]f(z)\ = (x^{2}+y^{2}-2y)+i(2x - 2xy)[/tex]
Let z = x + i y
Suppose,
[tex]u(x,y) = x^{2}+y^{2}-2y[/tex]
[tex]v(x,y) = 2x - 2xy[/tex]
On computing the partial derivatives of u and v as:
[tex]u'_{x} =2x[/tex]
[tex]u'_{y}=2y -2[/tex]
And
[tex]v'_{x} =2-2y[/tex]
[tex]v'_{y}=-2x[/tex]
According to the Cauchy-Riemann equations
[tex]u'_{x} =v'_{y} \ \ \ \ \ \ \ and\ \ \ \ \ \ u'_{y} = -v'_{x}[/tex]
Now,
[tex](u'_{x} =2x) \neq (v'_{y}=-2x)[/tex]
[tex](u'_{y}=2y -2) \ = \ (- v'_{x} =-(2-2y) =2y-2)[/tex]
Therefore,
[tex]u'_{y}=- v'_{x}[/tex] holds only.
This means,
2y - 2 = 0
⇒ y = 1
Therefore f(z) has a chance of being differentiable only at y =1.
Now we can compute the derivative
[tex]f'(z)=\frac{1}{2}[(u'_{x}+iv'_{x})-i(u'_{y}+iv'_{y})][/tex]
[tex]f'(z) =\frac{1}{2}[(2x+i(2-2y))-i(2y-2+i(-2x))][/tex]
[tex]f'(z) = i(2-2y)[/tex]
At y = 1
f'(z) = 0
Hence, the required derivative at y = 1 , f'(z) = 0
All the fourth-graders in a certain elementary school took a standardized test. A total of 81% of the students were found to be proficient in reading, 74% were found to be proficient in mathematics, and 64% were found to be proficient in both reading and mathematics. A student is chosen at random.(a) What is the probability that student is proficient in mathematics but not in reading?(b) What is the probability that student is proficient in reading but not in mathematics?
Answer:
The probability that a student is proficient in mathematics, but not in reading is, 0.10.
The probability that a student is proficient in reading, but not in mathematics is, 0.17
Step-by-step explanation:
Let's define the events:
L: The student is proficient in reading
M: The student is proficient in math
The probabilities are given by:
[tex]P (L) = 0.81\\P (M) = 0.74\\P (L\bigcap M) = 0.64[/tex]
[tex]P (M\bigcap L^c) = P (M) - P (M\bigcap L) = 0.74 - 0.64 = 0.1\\P (M^c\bigcap L) = P (L) - P (M\bigcap L) = 0.81 - 0.64 = 0.17[/tex]
The probability that a student is proficient in mathematics, but not in reading is, 0.10.
The probability that a student is proficient in reading, but not in mathematics is, 0.17
Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, set x3 and solve for x1 and x2. x1-3x3 =-5 3x1 X2 2x34 2x1 + 2x2 + x3 = 7 Need Help? Read It Talk to a Tutor Submit Answer Save Progress Practice Another Version 1 points LarLinAlg8 1.2.033 Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x, x2, and x3 in terms of the parameter t.) My Notes Ask Your Teach 2x1+ 4x1-3x2 + 7x3 = 2 8x1 - 9x2 15x3 12 3x33 (x1, x2, x3) -
Therefore, the solution to the system of equations is:
x=0
y=-3t+4
z=t
We can solve the system of equations using Gaussian elimination with back-substitution. Here's how:
Steps to solve:
1. Eliminate x from the second and fourth equations:
x+y+3z=4
0=0 (2x+5y+15z=20)-(x+2y+6z=8)
3y+9z=12
-x+2y+6z=8
2. Eliminate y from the fourth equation:
x+y+3z=4
0=0
3y+9z=12
3y+9z=12 (3y+9z=12)-(3y+9z=12)
0=0
3. Since the last equation is always true, we can ignore it.
4. Solve the remaining equations:
x+y+3z=4
0=0
3y+9z=12
From the second equation, we know that y=-3z+4. Substituting this into the first equation, we get:
x+(-3z+4)+3z=4
x+4=4
x=0
Now that we know x=0, we can substitute it back into the third equation to solve for z:
3(-3z+4)+9z=12
-9z+12+9z=12
12=12
This equation is always true, so there are infinitely many solutions. We can express x, y, and z in terms of the parameter t as follows:
x=0
y=-3z+4
z=t
Complete Question:
Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x, y, and z in terms of the parameter t.)
3x + 3y + 9z = 12
x + y + 3z = 4
2x + 5y + 15z = 20
-x + 2y + 6z = 8
(x, y, z) =
Find an equation of a line passing through the point (8,9) and parallel to the line joining the points (2,7) and (1,5).
Answer:
2x - y - 7 = 0
Step-by-step explanation:
Since the slope of parallel line are same.
So, we can easily use formula,
y - y₁ = m ( x ₋ x₁)
where, (x₁, y₁) = (8, 9)
and m is a slope of line passing through (x₁, y₁).
and since the slope of parallel lines are same, so here we use slope of parallel line for calculation.
and, Slope = m = [tex]\dfrac{y_{b}-y_{a}}{x_{b}-x_{a}}[/tex]
here, (xₐ, yₐ) = (2, 7)
and, [tex](y_{a},y_{b}) = (1, 5 )[/tex]
⇒ m = [tex]\dfrac{5-7}{1-2}[/tex]
⇒ m = 2
Putting all values above formula. We get,
y - 9 = 2 ( x ₋ 8)
⇒ y - 9 = 2x - 16
⇒ 2x - y - 7 = 0
which is required equation.
Answer:
y=2x-8
Step-by-step explanation:
In order to solve this you first have to calculate the slope of the parallel line, since that would be equal to the slope of our line:
[tex]Slope=\frac{y2-y1}{x2-x1}[/tex]
Now we insert the values into the formula:
[tex]Slope=\frac{y2-y1}{x2-x1}\\Slope=\frac{5-7}{1-2}\\Slope= \frac{-2}{-1}\\ Slope:2[/tex]
And remember that the formula for general line is:
[tex]Y-y1= M(x-x1)\\y-9=2(x-8=\\y=2x-16+9\\y=2x-7[/tex]
So the equation for the line passing through point 8,9 and parallel to the line joining 2,7 and 1,5 would be y=2x-7
Show that if a, b e Z, then a^2 - 4b =/ 2
Step-by-step explanation:
Proposition If a, b [tex]\in[/tex] [tex]\mathbb{Z}[/tex], then [tex]a^{2}-4b \neq2[/tex]
You can prove this proposition by contradiction, you assume that the statement is not true, and then show that the consequences of this are not possible.
Suppose the proposition If a, b [tex]\in[/tex] [tex]\mathbb{Z}[/tex], then [tex]a^{2}-4b \neq2[/tex] is false. Thus there exist integers If a, b [tex]\in[/tex] [tex]\mathbb{Z}[/tex] for which [tex]a^{2}-4b=2[/tex]
From this equation you get [tex]a^{2}=4b+2=2(2b+1)[/tex] so [tex]a^{2}[/tex] is even. Since [tex]a^{2}[/tex] is even, a is even, this means [tex]a=2d[/tex] for some integer d. Next put [tex]a=2d[/tex] into [tex]a^{2}-4b=2[/tex]. You get [tex] (2d)^{2}-4b=2[/tex] so [tex]4(d)^{2}-4b=2[/tex]. Dividing by 2, you get [tex]2(d)^{2}-2b=1[/tex]. Therefore [tex]2((d)^{2}-b)=1 [/tex], and since [tex](d)^{2}-b[/tex] [tex]\in[/tex] [tex]\mathbb{Z}[/tex], it follows that 1 is even.
And that is the contradiction because 1 is not even. In other words, we were wrong to assume the proposition was false. Thus the proposition is true.
Find the period and amplitude of the function. y-2sin 6x Give the exact values, not decimal approximations. Period: 2 Amplitude:
Answer:
Amplitude=2
Period=[tex]\frac{\pi}{3}[/tex]
Step-by-step explanation:
We are given that [tex]y=2sin6x[/tex]
We have to find the value of period and amplitude of the given function
We know that [tex]y=a sin(bx+c)+d [/tex]
Where a= Amplitude of function
Period of sin function =[tex]\frac{2\pi}{\mid b \mid}[/tex]
Comparing with the given function
Amplitude=2
Period=[tex]\frac{2\pi}{6}=\frac{\pi}{3}[/tex]
Hence, period of given function=[tex]\frac{\pi}{3}[/tex]
Amplitude=2
Twenty girls (ages 9-10) competed in the 50-meter freestyle event at a local swim meet. The mean time was 43.70 seconds with a standard deviation of 8.07 seconds. The median time was 40.15 with an IQR of 4.98 seconds. Without looking at a graphical display, what shape would you expect the distribution of swim times to have?
The distribution of swim times from the data given would most likely be right-skewed, as the mean is larger than the median and the Interquartile Range (IQR) suggests the data is concentrated towards the middle.
Explanation:From the given data about the 50-meter freestyle event, one can deduce probable distribution shape of the swim times. Notably, the mean of 43.70 seconds significantly exceeds the median of 40.15 seconds. This fact suggests a possible right skewed distribution, with longer swim times occurring less frequently but affecting the mean more strongly due to their higher values. It's called right-skewed because the 'tail' of the distribution curve extends more towards the right.
We can also examine the Interquartile Range (IQR), which measures spread in the middle 50% of the data. This is found by subtracting the lower quartile (first 25% of data) from the upper quartile (last 25% of data). An IQR of 4.98 seconds signifies much of the data is bunched in the middle of the distribution rather than at the ends, reinforcing the notion of a skewed distribution.
Thus, without a graphical representation, the swim times would be expected to exhibit a right-skewed distribution, presenting a positive skewness in the data.
Learn more about Data Distribution here:https://brainly.com/question/18150185
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The expected shape of the distribution of swim times would be right-skewed.
To determine the expected shape of the distribution, we compare the mean and median of the swim times:
- The mean time is 43.70 seconds, which is greater than the median time of 40.15 seconds.
- The standard deviation is 8.07 seconds, which is relatively large compared to the mean, indicating a wide spread of times.
- The interquartile range (IQR) is 4.98 seconds, which is relatively small compared to the standard deviation, suggesting that the middle 50% of the data is more tightly clustered.
In a perfectly symmetric distribution, the mean and median would be equal. However, when the mean is greater than the median, it suggests that there are some outliers or a longer tail on the right side of the distribution, pulling the mean up. The relatively large standard deviation in comparison to the IQR reinforces this idea, as it indicates there are some times that are significantly higher than the majority of the times, which are more closely packed around the median.