Answer:
Horizontal axis
Step-by-step explanation:
By convention, the independent variables is arrayed along the Horizontal axis (abscissa) in a scattergram.
The stcattergram has two dimensions
The X (independent) variable is arrayed along the horizontal axis.The Y(dependent) variable is arrayed along the vertical axis.Each dot scattergram is case in data set.The dot is placed at the intersection cases scores on X and Y.You are tasked with building a tower of blocks exactly 10 cm high. You have two types of block to work with. Each block A is 2 cm high, and each block B is 3 cm high. You may use any number of each type of block, and may stack any kind of block on top of any other kind. How many possible towers are there
Answer:
2 possible towers
Step-by-step explanation:
2 possible towers
Step-by-step explanation:
The height of the tower required is 10 cm, and that height may be obtained by 'x' number of blocks A and 'y' number of blocks, knowing its respective height, so 'x' and 'y' must accomplish the equation:
[tex] 10 = 2x + 3y [/ tex]
The maximum value for 'x' is 5; a higher value will give a higher than 10 cm height as a result. So, the possible answers for 'x' are: (0,1,2,3,4,5). Note that both 'x' and 'y' values must be integers, and the only numbers that satisfy that condition are (x = 2, y = 2) and (x = 5, y = 0). The other options for 'x' will give a noninteger value for 'y' because of the equation above.
So, there are just two options.
find a solution using the power series method
y'+ty=0, y(0)=1
Answer: The required solution is
[tex]y=(t-\dfrac{t^3}{2}+~~.~~.~~.).[/tex]
Step-by-step explanation: We are given to find the solution of the following differential equation using power series method :
[tex]y^\prime+ty=0,~~y(0)=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Let [tex]y=\sum_{n=0}^{\infty}a_nq^n[/tex] be the solution of the given equation.
Then, we have
[tex]y^\prime=\sum_{n=1}^{\infty}a_nnt^{n-1}.[/tex]
From equation (i), we get
[tex]\sum_{n=1}^{\infty}a_nnt^{n-1}+t\sum_{n=0}^{\infty}a_nt^n=0\\\\\\\Rightarrow \sum_{n=1}^{\infty}a_nnt^{n-1}+t\sum_{n=-1}^{\infty}a_nt^{n+1}=0.[/tex]
Comparing the coefficients of [tex]t^n,~t^{n+1},~,~~.~~.~~.[/tex] from both sides of the above, we get
[tex]2a_2+a_0\\\\\Rightarrow a_2=-\dfrac{a_0}{2},\\\\\\3a_3+a_1=0\\\\\Rightarrow a_3=-\dfrac{a_1}{3},\\\\\\\vdots~~~\vdots~~~\vdots[/tex]
Therefore, we get
[tex]y=a_0(1-\dfrac{1}{2}t^2+~~.~~.~~.)+a_1(t-\dfrac{t^3}{2}+~~.~~.~~.).[/tex]
The condition y(0) = 1 gives
[tex]a_0=0.[/tex]
So, the required solution is
[tex]y=t-\dfrac{t^3}{2}+~~.~~.~~.[/tex]
Answer with explanation:
The diiferential equation of first order is,
y'+t y=0
[tex]y=\sum_{n=0}^{\infty}a_{n}x^n\\\\y'=\sum_{n=1}^{\infty}na_{n}x^{n-1}\\\\y'+ty=\sum_{n=0}^{\infty}na_{n-1}x^{n-1}+t\sum_{n=0}^{\infty}a_{n}x^n\\\\=\sum_{n=0}^{\infty}(a_{n-1}\frac{n}{x}+a_{n}t)x^n\\\\\rightarrow y'+t y=0\\\\\rightarrow a_{n-1}\frac{n}{x}+a_{n}t=0\\\\a_{n-1}=\frac{-xta_{n}}{n}[/tex]
For, n=1
[tex]a_{0}=\frac{-xta_{1}}{1}\\\\a_{0}=-txa_{1}\\\\a_{1}=\frac{-a_{0}}{tx}\\\\\text{for}, n=2\\\\a_{1}=\frac{-xta_{2}}{2}\\\\a_{2}=\frac{-2a_{1}}{xt}\\\\=\frac{2a_{0}}{x^2t^2}\\\\\text{for}, n=3\\\\a_{2}=\frac{-xta_{3}}{3}\\\\a_{3}=\frac{-6a_{0}}{x^3t^3}\\\\a_{4}=\frac{24a_{0}}{x^4t^4}\\\\a_{n}=(-1)^n\frac{n!a_{0}}{x^nt^n}[/tex]
So,the series can be written as
[tex]y_{n}=\sum_{n=1}^{\infty}(-1)^n\frac{n!a_{0}}{x^nt^n}[/tex]
An excavation has a volume of 203 cubic feet. The material weighs 1,200 pounds per cubic yard. What is the weight of the exca- vation? a. 2,700 pounds b. 9,000 pounds c. 81,120 pounds d. 243,600 pounds
Answer: Option 'd' is correct.
Step-by-step explanation:
Since we have given that
Volume of an excavation = 203 cubic feet
Weight of the material = 1200 pounds per cubic yard
So, the weight of the excavation would be
Volume of an excavation × Weight of the material
[tex]=203\times 1200\\\\=243,600\ pounds[/tex]
Hence, the weight of the excavation is 243,600 pounds.
Therefore, Option 'd' is correct.
Problem 2: Suppose that a 3 Ã 5 coefficient matrix A has a pivot in each of the three rows. If we augment A to represent a linear system, would this system have solutions (be consistent) or have no solutions (be inconsistent)? Explain your answer
Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (3, 5, 0) and perpendicular to both i + j and j + k x(t), y(t), z(t) = The symmetric equations are given by −(x − 3) = y − 5 = z. x + 3 = −(y + 5), z = 0. x − 3 = y − 5 = −z. x + 3 = −(y + 5) = z. x − 3 = −(y − 5) = z.
Answer:(x-3)=-(y-5)=z
Step-by-step explanation:
Given
the point through which line passes is (3,5,0)
so we need a vector along the line to get the equation of line
It is given that line is perpendicular to both i+j & j+k
therefore their cross product will give us the vector perpendicular to both
v=(i+j)\times (j+k)=i-j+k
therefore we get direction vector of line so we can write
[tex]\frac{x-3}{1}=\frac{y-5}{-1} =\frac{z-0}{1}[/tex]=t
i.e.
x=t+3,y=-t+5,z=t
The parametric form of the equation is;
[tex]\rm x=x_1+at, \ x=3+1t, \ x=3+t\\\\y=y_1+at, \ y=5+(-1)t, \ y=5-t\\\\z=z_1+at, \ z=0+a(1), \ z=a[/tex]The symmetric form of the equation is [tex]\rm x - 3 = -(y -5) = z[/tex].
Given
The line through (3, 5, 0) and perpendicular to both i + j and j + k
The symmetric form of the equation of the line is given by;[tex]\rm \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}=t[/tex]
Where the value of [tex]\rm x_1=3, \ y_1=5\ and \ z_1=0[/tex].
To find a, b, c by evaluating the product of ( i + j) and ( j + k ).
[tex]\rm= (i+j)\times (j+k)\\\\= i-j+k[/tex]
The value of a = 1, b = -1 and c = 1.
Substitute all the values in the equation.
[tex]\rm \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}\\\\\dfrac{x-3}{1}=\dfrac{y-5}{-1}=\dfrac{z-0}{1}\\\\(x-3)=-(y-5)=z[/tex]
Therefore,
The parametric form of the equation is;
[tex]\rm x=x_1+at, \ x=3+1t, \ x=3+t\\\\y=y_1+at, \ y=5+(-1)t, \ y=5-t\\\\z=z_1+at, \ z=0+a(1), \ z=a[/tex]
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The daily number in the State Lottery is a three-digit nuber from 000 to 999. To win in "straight play", a ticket must have the exact number. To win in a "box play" , a ticket can have the three digit of the daily number in any order. Find the probabilty of winning in :
a) Straight play
b) Box play
Answer:
a. [tex]P = \frac{1}{1000}[/tex]
b. [tex]P = \frac{3}{500}[/tex]
Step-by-step explanation:
To win the 'straight play' a ticket must have the exact number. Given that there are 1000 numbers between 000 and 999. The probability of winning in striaght play is:
[tex]P = \frac{1}{1000}[/tex]
To win in a box play, a ticket can have the three digit in daily order.
Given three numbers A, B and C. There are 6 ways to arrange the numbers:
ABC, ACB, BAC, BCA, CAB, CBA.
Therefore the probability is: [tex]P = \frac{3}{500}[/tex]
Find a polynomial f(x) of degree 4 that has the following zeros.
0, -4, (multiplicity ), 2
Leave your answer in factored form.
Answer:
[tex]f(x)=(x-2)(x)(x+4)^2[/tex]
Step-by-step explanation:
Since there is a multiplicity at x=-4, that would meant that that part of the factor would have to have an even degree. This means that it would have to be 2.
This would give you [tex]f(x)=(x-2)(x)(x+4)^2[/tex]
Answer:
[tex]f (x) = x (x + 4)^2(x-2)[/tex]
Step-by-step explanation:
The zeros of the polynomial are all the values of x for which the function [tex]f (x) = 0[/tex]
In this case we know that the zeros are:
[tex]x = -4,\ x+4 =0[/tex]
[tex]x = -4,\ x+4 =0[/tex]
[tex]x = 0[/tex]
[tex]x = 2[/tex], [tex]x - 2 = 0[/tex]
Now we can write the polynomial as a product of its factors
[tex]f (x) = x (x + 4)(x+4) (x-2)[/tex]
[tex]f (x) = x (x + 4)^2(x-2)[/tex]
Note that the polynomial is of degree 4 because the greatest exponent of the variable x that results from multiplying the factors of f(x) is 4
Write the polynomial as the product of linear factors.
h(x) = x^2 ? 2x + 10
Answer:
If [tex]h(x) = x^2 + 2x + 10[/tex] then [tex](x+1-3i)*(x+1+3i)[/tex].
If [tex]h(x) = x^2 - 2x + 10[/tex] then [tex](x-1-3i)*(x-1+3i)[/tex].
Step-by-step explanation:
In order to write the polynomial as the product of linear factors, we need to find the roots of the polynomial. A quadratic equation is defined as:
[tex]ax^2+bx+c[/tex]
Because the given polynomial expression is a quadratic equation, we can use the following equations for calculating the roots:
[tex]x1=(-b/2a)+\sqrt{b^2-4ac}/2a[/tex]
[tex]x1=(-b/2a)-\sqrt{b^2-4ac}/2a[/tex]
Since the second term sign is not given, then we can write the expression as:
[tex]x^2+s2x+10[/tex], in which a=1, b=s2 where 's' represents a sign (- or +), and c=10.
Using the equation for finding the roots we obtain:
[tex]x1=(-sb/2a)+\sqrt{b^2-4ac}/2a[/tex]
[tex]x1=(-s2/2*1)+\sqrt{2^2-4*1*10}/(2*1)[/tex]; notice that [tex](sb)^{2} = 2^{2}[/tex]
[tex]x1=(-s1)+\sqrt{-36}/2[/tex]
[tex]x1=-(s1)+6i/2[/tex]
[tex]x1=-(s1)+3i[/tex]
[tex]x2=(-sb/2a)-\sqrt{b^2-4ac}/2a[/tex]
[tex]x2=(-s2/2*1)-\sqrt{2^2-4*1*10}/(2*1)[/tex]; notice that (s2)²=2^2
[tex]x2=(-s1)-\sqrt{-36}/2[/tex]
[tex]x2=-(s1)-6i/2[/tex]
[tex]x2=-(s1)-3i[/tex]
If we consider 's' as possitive (+) the roots are:
[tex]x1=-1+3i[/tex] and [tex]x2=-1-3i[/tex]
Whereas if we consider 's' as negative (-) the roots are:
[tex]x1=1+3i[/tex] and [tex]x2=1-3i[/tex]
The above means that if the equation is [tex]h(x) = x^2 + 2x + 10[/tex], then we can express the polynomial as: [tex](x+1-3i)*(x+1+3i)[/tex].
But, if the equation is [tex]h(x) = x^2 - 2x + 10[/tex], then we can express the polynomial as: [tex](x-1-3i)*(x-1+3i)[/tex].
As men and women age, their risk of death increases. There are also health issues that can greatly impair older citizens' abilities to participate in activities and travel. Assume you are planning a 60th high school reunion and are wondering how many of the older adults would be able to attend. If you know that 22 women and 18 men are still alive, and are told that exactly 8 of them will attend, what is the probability that you would have exactly four men and exactly four women attend?
Answer:
The total number of men and women is 42.
So the number of ways of selecting 8 people from 42 is C(42,8) = 118030185.
The number of ways of selecting 4 men from 19 and 4 women from 23 is C(19,4)*C(23,4) = 3876*8855 = 34321980
Therefore, the required probability is 34321980 / 118030185 = 0.2907
Step-by-step explanation:
To determine the probability of exactly four men and four women attending a 60th high school reunion given certain conditions, we use the combination formula. The probability is found by dividing the product of the combinations of 4 men from 18 and 4 women from 22 by the total combinations of 8 people from 40.
Explanation:The question is asking us to find the probability that exactly four men and exactly four women would attend the high school reunion. Considering that there are 40 (22 women + 18 men) people still alive, we are told that exactly 8 will attend the reunion.
We utilise the formula for combinations, to calculate this problem.
So, the possible combinations of 4 men attending from 18 is calculated as C(18,4).
And the possible combinations of 4 women attending from 22 is calculated as C(22,4).
Finally, the total combinations of 8 people attending from 40 is calculated as C(40,8).
Probability is calculated as the desired outcomes over total possibilities. Therefore, the desired probability will be [C(18,4) * C(22,4)] / C(40,8).
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If a set of sample measurements has a mean of 100, a normal distribution, a standard deviation of 2, and control limits of 94 and 106, what percentage of the samples are expected to be between 94 and 106? Explain your answer.
Answer: 99.73%
Step-by-step explanation:
Given : Mean : [tex]\mu=100[/tex]
Standard deviation : [tex]\sigma=2[/tex]
Let X be the random variable that represents the data values.
Formula for Z-score : [tex]z=\dfrac{X-\mu}{\sigma}[/tex]
For x=94, we have
[tex]z=\dfrac{94-100}{2}=-3[/tex]
For x=106, we have
[tex]z=\dfrac{106-100}{2}=3[/tex]
The probability that the samples are between 94 and 106:-
[tex]P(-3<x<3)=1-2P(z<-3)\\\\=1-2(0.0013499)=0.9973002\approx0.9973=99.73\%[/tex]
Hence, the percent of the samples are expected to be between 94 and 106 = 99.73%
Approximately 95% of the samples are expected to be between 94 and 106.
Explanation:The percentage of samples that are expected to be between 94 and 106 can be determined using the empirical rule, also known as the 68-95-99.7 rule. According to this rule, approximately 68% of the samples will fall within one standard deviation of the mean, and approximately 95% of the samples will fall within two standard deviations of the mean.
In this case, the control limits are 94 and 106, which are one standard deviation below and above the mean, respectively. Therefore, 95% of the samples are expected to be between 94 and 106.
Note that this rule applies to data that follow a bell-shaped, symmetric distribution, which is assumed to be the case for this set of sample measurements.
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5. Find the general solution to y'''-y''+4y'-4y = 0
For any equation,
[tex]a_ny^(n)+\dots+a_1y'+a_0y=0[/tex]
assume solution of a form, [tex]e^{yt}[/tex]
Which leads to,
[tex](e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0[/tex]
Simplify to,
[tex]e^{yt}(y^3-y^2+4y-4)=0[/tex]
Then find solutions,
[tex]\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}[/tex]
For non repeated real root y, we have a form of,
[tex]y_1=c_1e^t[/tex]
Following up,
For two non repeated complex roots [tex]y_2\neq y_3[/tex] where,
[tex]y_2=a+bi[/tex]
and,
[tex]y_3=a-bi[/tex]
the general solution has a form of,
[tex]y=e^{at}(c_2\cos(bt)+c_3\sin(bt))[/tex]
Or in this case,
[tex]y=e^0(c_2\cos(2t)+c_3\sin(2t))[/tex]
Now we just refine and get,
[tex]\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}[/tex]
Hope this helps.
r3t40
To find the general solution for the differential equation y''' - y'' + 4y' - 4y = 0, we solve the characteristic equation r³ - r² + 4r - 4 = 0, and find the roots r = 1 and r = ±2i. Thus, the general solution is y(x) = C1eˣ + C2cos(2x) + C3sin(2x).
The given differential equation is:
y''' - y'' + 4y' - 4y = 0
First, we find the characteristic equation corresponding to this differential equation. Replace derivatives with powers of r:This is the general solution to the differential equation y''' - y'' + 4y' - 4y = 0.
Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the range, interquartile range, variance, and standard deviation (to a maximum of 2 decimals, if decimals are necessary).
So we have a set of integers,
[tex]X=\{15,20,25,25,27,28,30,34\}[/tex]
And we need to compute some stats.
First the range,
The range of data set is difference between the maximum and the minimum of the data set.
So the minimum is 15 and maximum is 34 therefore the range is 34 - 15 = 19.
Second the interquartile range,
The interquartile range of data set is the difference of the first and third quartiles.
First quartile is the value separating the lower quarter and higher three - quarters of the set. The first quartile is computed by taking median of the lower half of the sorted set. This is 15, 20, 25, 25 and it's median is 22.5
Third quartile is therefore the median of 27, 28, 30, 34 which is 29.
Next up, variance.
The variance of set measures how much data is spread out. For data set [tex]x_1,\dots, x_n[/tex] with an average [tex]a[/tex],
[tex]\mathsf{var}(X)=\Sigma_{i=1}^{n}\dfrac{(x_i-a)^2}{n-1}[/tex]
So first compute the average value which is actually a mean,
[tex]a=\dfrac{1}{2}\Sigma_{i=1}^{n}a_i[/tex]
The sum [tex]\Sigma_{i=1}^{n}a_i[/tex] of numbers in set is 204.
Divide this by number of elements in the set (8).
[tex]a=\dfrac{204}{8}=25.5[/tex]
Then compute the variance,
[tex]\mathsf{var}(X)=\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}\approx\boxed{34.57}[/tex]
And finally standard deviation,
Since have computed variance the standard deviation is nothing too hard. It's defined as a square root of variance,
[tex]\mathsf{sde}(X)=\sqrt{\dfrac{\Sigma_{i=1}^{n}(x_i-a)^2}{n-1}}=\sqrt{34.57}\approx\boxed{5.88}[/tex]
Hope this helps.
r3t40
The values of the statistical measures for the distribution are :
Range = 19Interquartile range = 6.5Variance = 34.57Standard deviation = 5.88Given the data :
27, 25, 20, 15, 30, 34, 28, 25.
Rearranging the values :
15, 20, 25, 25, 27, 28, 30, 34
A.)
The range = Maximum - Minimum
Maximum = 34 ; Minimum = 15
Range = 34 - 15 = 19
B.)
The interquartile range = Upper quartile - Lower quartile
Upper quartile (Q3) = 3/4(n+1)th term
n = number of values = 8
Q3 = 3/4(9) = 6.75 th term = (28+30)/2 = 29
Lower quartile (Q1) = 1/4(n+1)th term
Q1 = 1/4(9) = 2.25th term = (20+25)/2 = 22.5
Interquartile range = 29 - 22.5 = 6.5
C.)
Variance :
Σ[(X - mean)²] / (n-1)
ΣX/ n
Mean = 204 / 8 = 25.5
[(15-25.5)² + (20-25.5)² + (25-25.5)² + (25+25.5)² + (27-25.5)² + (28-25.5)² + (30-25.5)² + (34 - 25.5)²] ÷ (8 - 1)
Variance = 242 / 7
Variance = 34.57
Standard deviation = √variance
Standard deviation = √34.57
Standard deviation = 5.879
Standard deviation = 5.88 (2 decimal place).
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A property contained ten acres. How many lots of not less than 50 feet by 100 feet can be subdivided from the property if 26,000 square feet were dedicated for roads?
Answer:
81 lots of 5,000 square feet fits in the property.
Step-by-step explanation:
The first step is calculate the square feet of the minimum area:
[tex]50 \: feets \times 100 \: feet = 5,000 \:square \: feet[/tex]
Second, from the ten acres we subtract the 26,000 feets of roads:
[tex]1 \: acre = 60 \: feet \times 660\:feet = 43,560\:square\:feet\\43,560 \times 10 = 435,600\\435,600 - 26,000 = 409,600 \:square \: feet\:available[/tex]
Third, we divide the lot area over the available square feet
[tex]\frac{409,600}{5,000} = 81.92[/tex]
81 lots of 5,000 square feet fits in the property.
Suppose a bank offers a CD that earns 1% interest compounded 353 times per year. You invest $1,282 today. How much will you have (in dollars and cents) after 8 years?
Answer:
1388 dollars 77 cents.
Step-by-step explanation:
Since, the amount formula in compound interest is,
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
Where, P is the principal amount,
r is the annual rate,
n is the number of periods in a year,
t is the number of years,
Given,
P = $ 1,282,
r = 1 % = 0.01,
n = 353,
t = 8 years,
Hence, the amount after 8 years would be,
[tex]A=1282(1+\frac{0.01}{353})^{2824}[/tex]
[tex]=\$ 1388.77244711[/tex]
[tex]\approx \$ 1388.77[/tex]
= 1388 dollars 77 cents.
The height of a photograph is 40 cm greater than the width, and the area of the photograph is 896 cm2what is the height of the photograph?
Answer:
Height = 56 cm
Step-by-step explanation:
let the width of photograph be = 'w' cm
let the Height of photograph be = 'h' cm
Now it is given that height is 40 cm greater than width
=> h = w + 40 cm.............................(i)
Now it is given that area of photograph = [tex]896cm^{2}[/tex]
We know that area = [tex]Width^{}[/tex]x[tex]Height^{}[/tex]
Thus we have [tex]h^{}[/tex] x [tex]w^{}[/tex]=[tex]896cm^{2}[/tex]
Applying value of 'h' from equation i we get
[tex](w+40^{})[/tex]x[tex]w^{}[/tex]=[tex]896cm^{2}[/tex]
[tex]w^{2} +40w=896cm^{2}[/tex]
This is a quadratic equation in 'w' whose solution in standard form is given by
w=[tex]\frac{-b\mp\sqrt{b^{2}-4ac}}{2a}[/tex]
upon comparing with standard equation we see that
a =1
b=40
c=896
applying values in the formula we get
w=[tex]\frac{-40\mp\sqrt{40^{2}-4\times 1\times- 896}}{2 \times1 }[/tex]
w1 = 16 cm
w2 = -56 cm
We discard -56 cm since length cannot be negative thus
width = 16 cm
Height = 40+16 cm = 56 cm
Alvin is 15 years older than Elga. The sum of their ages is 89. What is Elga's age?
Answer:
Elga is 37 years old.
Step-by-step explanation:
Let 'x' be the age of elga.
Alvin's age is: x + 15
Given that the sum of their ages is 89, we have that:
x + 15 + x = 89
2x = 89 - 15
2x = 74
x = 74/2 = 37
Elga is 37 years old.
Answer:
Elga is 37 years old.
Step-by-step explanation:
Alvin (A) is 15 years older than Elga (E).
The sum their anges (A and E) is 89.
So the system we have is:
A=15+E
A+E=89
We are going to input 15+E for A in A+E=89 since first equation says A+15+E.
A+E=89 with A=15+E
(15+E)+E=89
15+E+E=89
15+2E=89
Subtract 15 on both sides:
2E=89-15
2E=74
Divide both sides by 2:
2E/2=74/2
E=74/2
E=37
Elga is 37 years old.
A=15+E=15+37=52 so Alvin is 52 yeard old.
52+37=89.
and
52=15+37.
y''-4y=cosx find general solution.
Answer:
Step-by-step explanation:
[tex]y''-4y=cosx[/tex]
The auxialary equation and solutions are as follows:
[tex]m^2-4=0\\m=2,-2[/tex]
General solution is
[tex]y=Ae^{2x} +Be^{-2x}[/tex]
Right side =cosx
Particular Integral
= [tex]\frac{cosx}{-1-4} =\frac{-cosx}{5}[/tex]
Hence full solution is
What is the measure of ∠X?
69°
111°
180°
21°
How many different four-letter passwords can be formed from the letters Upper A comma Upper B comma Upper C comma Upper D comma Upper E comma Upper F comma and Upper G if no repetition of letters is allowed?
Answer with explanation:
Number of Letters from which we have to make four-letter passwords
={A,B,C,D,E,F,G}=7
Since repetition of digit is not allowed.In Permutation order of arrangement is Important,while in combination order of arrangement is not Important.
Out of seven letters we have to select four letters, keeping in mind the order of Alphabets.So, we will use the concept of Permutation.
Number of ways of arrangement,that is to make four letter password from 7 Alphabets
[tex]_{4}^{7}\textrm{P}\\\\=\frac{7!}{(7-4)!}\\\\=\frac{7!}{3!}\\\\=4\times 5\times 6 \times 7=840[/tex]
So, total number of ways of making password from 7 alphabets=840 ways
Second Method⇒
Since repetition of alphabets is not allowed, first Alphabet can be chosen in 7 ways , second alphabet can be chosen in 6 ways, third alphabet can be chosen in 5 ways, fourth alphabet can be chosen from four alphabets.
So, total number of ways of making four letter password
=7×6×5×4
= 840 ways
We can form 840 unique four-letter passwords with no repeats from the letters A, B, C, D, E, F, G. This is a problem in combinatorics, calculating permutations of positions in the password.
Explanation:We are trying to determine the number of unique four-letter passwords that can be created using the letters A, B, C, D, E, F, G with no repetition of letters. This is a problem of permutations, a concept in combinatorics, a branch of mathematics that deals with counting.
In this case, for the first position of the password, we have 7 choices. Once we've picked this first letter, we then have 6 remaining letters to choose from for the second position. Similarly, there are 5 choices remaining for the third position and then 4 for the last position.
The total number of password combinations will then be a product of the number of choices for each of the positions. Mathematically, this can be represented as 7*6*5*4.
Consequently, there are 840 different four-letter passwords that can be created from the given letters without repeating any letter.
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a parking lot is filled with 260 cars. 1/2 of the vehicles are suvs. 1/3 are sub compact cars. the rest are luxury cars. how many cars in the lot are luxury cars?
Answer:
The total number of luxury cars are 43.
Step-by-step explanation:
Consider the provided information.
It is given that there are 260 cars in the parking lot out of 1/2 are SUVs.
Total number of SUV cars = [tex]260\times{\frac{1}{2}=130}[/tex]
Out of 260 cars 1/3 are sub compact cars.
Total number of sub compact cars = [tex]260\times{\frac{1}{3}}=86.66\approx{87}[/tex]
Therefore the total number of luxury cars are:
260 - 130 - 87 = 43
Therefore, the total number of luxury cars are 43.
A given binomial distribution has mean-4 and variance-2. What is the probability of success? Explain your answer.
Answer: 0.5
Step-by-step explanation:
We know that the mean and variance of a binomial distribution with probability of success p is given by :-
[tex]\text{Mean}:\mu=np\\\\\text{Variance}:\sigma^2=np(1-p)[/tex], where n is the total number of trials .
Given : A given binomial distribution has
[tex]\text{Mean}:\mu=np=4.......(1)\\\\\text{Variance}:\sigma^2=np(1-p)=2............(2)[/tex]
Now we substitute , the value of np from (1) in (2), we get
[tex]4(1-p)=2\\\\\Rightarrow\ 1-p=\dfrac{2}{4}\\\\\Rightarrow\ p=1-\dfrac{1}{2}\\\\\Rightarrow\ p=\dfrac{1}{2}=0.5[/tex]
Hence, the probability of success (p) = 0.5
Assume that there are 365 days in a year, the probability that a person is born on any given day is the same for all days, and everyone’s birthdays are independent.
(a) If there are n people, what is the probability that no one is born in July?
(b) If there are 15 people who are all born in July or August, what is the probability that at least two of these people share the same birthday?
For part a, wouldn't it just be 365^n - 365^31 since there are only 31 days in July? Is there a way to make it more order specific? For b I am still confused on how to make order matter if we are just taking days from July or August. Explanations well appreciated!
Answer:
Step-by-step explanation:
If each day equal chance then p = Prob that a person is borne on a particular day = 1/365
Each person is independent of the other and there are two outcomes either borne in July or not
p = prob for one person not borne in July = (365-31)/365 = 334/365
a)Hence prob that no one from n people borne in July = [tex](\frac{334}{365} )^n[/tex]
b) p = prob of any one borne in July or Aug = [tex]\frac{(31+31) }{365} =\frac{62}{365}[/tex]=0.1698
X- no of people borne in July or Aug
n =15
P(X>=2) =[tex]15C2 (0.1698)^2*(1-0.1698)^{13} +15C3 (0.1698)^3*(1-0.1698)^{12} +...+15C15 (0.1698)^{15}[/tex]
=0.7505
3 is what percent of 40
Answer:
1.2
Step-by-step explanation:
Find the? inverse, if it? exists, for the given matrix.
[4 3]
[3 6]
Answer:
Therefore, the inverse of given matrix is
[tex]=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}[/tex]
Step-by-step explanation:
The inverse of a square matrix [tex]A[/tex] is [tex]A^{-1}[/tex] such that
[tex]A A^{-1}=I[/tex] where I is the identity matrix.
Consider, [tex]A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right][/tex]
[tex]\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}[/tex]
[tex]\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0[/tex]
[tex]\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}[/tex]
[tex]=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}[/tex]
[tex]\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc[/tex]
[tex]4\cdot \:6-3\cdot \:3=15[/tex]
[tex]=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}[/tex]
[tex]=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}[/tex]
Therefore, the inverse of given matrix is
[tex]=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}[/tex]
A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 99% confident that the true mean is within 3 ounces of the sample mean? The standard deviation of the birth weights is known to be 5 ounces.
Answer: 18
Step-by-step explanation:
Given : Margin of error = 3 ounces
Significance level : [tex]\alpha = 1-0.99=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=2.576[/tex]
Standard deviation :[tex]\sigma=5\text{ ounces}[/tex]
The formula to calculate the sample size is given by :-
[tex]n=(\dfrac{z_{\alpha/2}\ \sigma}{E})^2[/tex]
[tex]\Rightarrow\ n=(\dfrac{2.576\times5}{3})^2\\\\\Rightarrow\ n=18.4327111111\approx18[/tex]
Hence, the minimum sample size should be 18.
To estimate the birth weight of infants, the doctor must select a sample size of at least 19 infants to be 99% confident that the true mean is within 3 ounces of the sample mean. This calculation is based on the given standard deviation of 5 ounces.
Explanation:To determine the sample size needed to estimate the birth weight of infants, we will use the formula for a margin of error:
Margin of Error = (Z-value) * (Standard Deviation / Square Root of Sample Size)
Since the doctor wants to be 99% confident that the true mean is within 3 ounces of the sample mean, the Z-value will be 2.576 (corresponding to a 99% confidence level). The given standard deviation is 5 ounces. Plugging these values into the formula, we get:
3 = 2.576 * (5 / Square Root of Sample Size)
Solving for the sample size:
Square Root of Sample Size = 2.576 * 5 / 3
Square Root of Sample Size ≈ 4.293
Sample Size ≈ 18.41
Since the sample size must be a whole number, the doctor must select a sample size of at least 19 infants to be 99% confident that the true mean is within 3 ounces of the sample mean.
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A thief steals an ATM card and must randomly guess the correct four-digit pin code from a 9-key keypad. Repetition of digits is allowed. What is the probability of a correct guess on the first try?
Answer: [tex]\dfrac{1}{59049}[/tex]
Step-by-step explanation:
The number of keys on the keypad = 9
The number of digits in the required pin code = 4
If repetition is allowed then , the number of possible codes will be
[tex](9)^5=59049[/tex]
Then , the probability that the choosen pin code is correct is given by :-
[tex]\dfrac{1}{59049}[/tex]
The probability of Geometric distribution formula :
[tex]P(X=x)=(1-p)^{x-1}p[/tex], where p is the probability of success and x is the number of attempt.
Using geometric probability , the probability of a correct guess on the first try is given by :-
[tex]P(X=1)=(1-\dfrac{1}{59049})^{1-1}(\dfrac{1}{59049})=\dfrac{1}{59049}[/tex]
Randy deposits $9,500 in an IRA. What will be the value of his investment in 3 years if the investment is earning 1.75% per year and is compounded continuously? Round to the nearest cent.
Answer:
The value of his investment in 3 years = $10007.53
Step-by-step explanation:
Compound interest
A = P[1 +R/n]^nt
Where A - amount
P - principle amount
R = rate of interest
t - number of years
n - number of times compounded yearly
To find value if investment
Here P = $9500, R = 1.75% = 0.0175 n = 1 and t = 3
A = P[1 +R/n]^nt
= 9500[1 + 0.0175/1]^(1 * 3)
= 10007.53
Answer:
$10,012.07
Step-by-step explanation:
Identify the values of each variable in the formula. Remember to express the percent as a decimal.
APrt=?=$9,500=0.0175=3 years
For compounding continuously, use the formula A=Pert.
Substitute the values into the formula and compute the amount to find
AA=9,500e0.0175⋅3=$10,012.07.
Given the Arithmetic sequence A1,A2,A3,A4 53, 62, 71, 80 What is the value of A38?
Answer:
[tex]A_{38} = 350[/tex]
Step-by-step explanation:
The 5th term of the arithmetic sequence is 53. We can write the equation:
[tex]a + 4d = 53...(1)[/tex]
The 6th term of the arithmetic sequence is 62. We can write the equation:
[tex]a + 5d = 62...(2)[/tex]
Subtract the first equation from the second one to get:
[tex]5d - 4d = 62 - 53[/tex]
[tex]d = 9[/tex]
The first term is
[tex]a + 4(9) = 53[/tex]
[tex]a + 36 = 53[/tex]
[tex]a = 53 - 36[/tex]
[tex]a = 17[/tex]
The 38th term of the sequence is given by:
[tex] A_{38} = a + 37d[/tex]
[tex]A_{38} = 17+ 37(9)[/tex]
[tex]A_{38} = 350[/tex]
Answer:
[tex]A_{38}=386[/tex]
Step-by-step explanation:
We have been given an arithmetic sequence gas [tex]A_1,A_2,A_3,A_4[/tex] as :53,62,71,80. We are asked to find [tex]A _{38}[/tex].
We know that an arithmetic sequence is in format [tex]a_n=a_1+(n-1)d[/tex], where,
[tex]a_n[/tex] = nth term,
[tex]a_1[/tex] = 1st term of sequence,
n = Number of terms,
d = Common difference.
We have been given that 1st term of our given sequence is 53.
Now, we will find d by subtracting 71 from 80 as:
[tex]d=80-71=9[/tex]
[tex]A_{38}=53+(38-1)9[/tex]
[tex]A_{38}=53+(37)9[/tex]
[tex]A_{38}=53+333[/tex]
[tex]A_{38}=386[/tex]
Therefore, [tex]A_{38}=386[/tex].
The cost of health insurance is $1,200 per year. The cost of an operation for a dramatic injury is $500,000. The chances of someone needing such an operation are 47.3% over a twenty year period. From a purely financial point of view, are you better off taking out the insurance or not? Choose the best answer.
Question options: No. The insurance company must make a profit.
Yes. The cost of the insurance is less than the probability cost of the operation
No. The probality of needing the operation is less than 50%.
Yes. Buying insurance is always a good idea.
Answer:
yes. The cost of the insurance is less than the probability cost of the operation
Step-by-step explanation:
yes. The cost of the insurance is less than the probability cost of the operation
The cost of health insurance = $1200
Cost of dramatic injury operation= $500,000
chances of need of operation= 47.3% over a 20 years period
the amount of pay insurance after 20 years= [tex]20\times 1200= 240,00[/tex]
probable of cost operation= 0.473*500,00= $236,500
clearly the cost of insurance is less than the probable cost of operation.
Are Sin(x) and e^x linearly independent? Justify.
Answer:
yes they are linearly independent
Step-by-step explanation:
By definition of linear dependence we have if f(x) and g(x) be 2 functions
if they are linearly dependent then we can write
f(x) = αg(x)...........(i) where α is an arbitrary constant
in our case we can see that the range of sin(x) is only from [-1,+1] while as [tex]e^{x}[/tex] has range from [0,∞]
thus we cannot find any value of α for which (i) is valid