Answer:
The equation of tangent line is [tex]y=5x-33 [/tex]
Step-by-step explanation:
We need to find out the equation of tangent line.
Given :- g(6)=−3 and g'(6)= 5
If g(6)=−3
then the point on the line for the required tangent is (6,−3)
If g'(6)= 5
then the slope of the tangent at that point is 45
The tangent line can be specified by the slope-point form of the equation:
[tex](y-y_1)=m(x-x_1)[/tex]
which in this case is
[tex](y-(-3))=5(x-6)[/tex]
[tex](y+3)=(5x-30)[/tex]
subtract both the sides by 3,
[tex]y+3-3=5x-30-3[/tex]
[tex]y=5x-33[/tex]
Therefore, the equation of tangent line is [tex]y=5x-33[/tex]
Final answer:
The equation of the tangent line to the graph of y = g(x) at the point where x = 6 is y = 5x - 33, using the point-slope form and the given point (6, -3) with the slope of 5.
Explanation:
To find the equation of the tangent line to the graph at a particular point, we use the point-slope form of a line, given by y - y1 = m(x - x1), where (x1, y1) is the point on the graph and m is the slope at that point. Given that g(6) = -3 and g'(6) = 5, we can substitute these values into the point-slope form to get the equation of the tangent line. The equation is then y + 3 = 5(x - 6), which simplifies to y = 5x - 33.
In a classroom of 32 Students , 14 are male . What % of the Class is female ????
Answer:
56.25%
Step-by-step explanation:
If 14 are male, that means that 32 - 14 = 18 are female. If 18/32 are female, do the division and then multiply by 100 to get the percentage.
18/32 = .5625
.5625 × 100% = 56.25%
x^3-y^3
Factor completely, then place the answer in the proper location on the grid.
For this case we must factor the following expression:
[tex]x ^ 3-y ^ 3[/tex]
We have that since both terms are perfect cubes, we factor using the cube difference formula.
[tex]a ^ 3-b ^ 3 = (a-b) (a ^ 2 + ab + b ^ 2)[/tex]
Where:
[tex]a = x\\b = y[/tex]
So, we have to:
[tex]x ^ 3-y ^ 3 = (x-y) (x ^ 2 + xy + y ^ 2)[/tex]
Answer:
[tex]x ^ 3-y ^ 3 = (x-y) (x ^ 2 + xy + y ^ 2)[/tex]
What is the solution to the system of equations below when graphed y= -2x +3, y= -4x + 15
Answer:
(6, -9) → x = 6, y = -9Step-by-step explanation:
These are linear functions. The graph is a straight line. We only need two points to draw a graph.
We choose any two values of x, substitute and calculate the value of y.
For y = -2x + 3
x = 0 → y =-2(0) + 3 = 0 + 3 = 3 → (0, 3)
x = 1 → y = -2(1) + 3 = - 2 + 3 = 1 → (1, 1)
For y = -4x + 15
x = 3 → y = -4(3) + 15 = -12 + 15 = 3 → (3, 3)
x = 4 → y = -4(4) + 15 = -16 + 15 = -1 → (4, -1)
Look at the picture.
The solution is the coordinates of the point of intersection of lines.
The solution to the system of equations below when graphed is (6, -9).
In order to graphically determine the solution for this system of linear equations on a coordinate plane, we would make use of an online graphing calculator to plot the given system of linear equations while taking note of the point of intersection.
When x is 0, 3, and 6, the y-value is given by;
y= -2(0) +3 = 3
y= -2(3) +3 = -3
y= -2(6) +3 = -9
y= -4(0) + 15 = 11
y= -4(3) + 15 = 3
y= -4(6) + 15 = -9
Since the y-intercept is at (0, 3) and the slope is 3, we would start with the point (0, 3), move 1 unit to the right and 3 units upward for the remaining points. For the second equation, the y-intercept is at (0, 15) and the slope is -4, we would start with the point (0, 15), move 1 unit to the right and 4 units downward for the remaining points.
Based on the graph shown, the solution set to the system is represented by the purple shaded region. Hence, a possible solution is (6, -9).
If two samples A and B had the same mean and standard deviation, but sample A had a larger sample size, which sample would have the wider 95% confidence interval? Sample A as it comes first Sample B as its sample is more dispersed Sample A as it has the larger sample Sample B as it has the smaller sample
===========================================================
Explanation:
Recall that the margin of error (MOE) is defined as
MOE = z*s/sqrt(n)
The sample size n is located in the denominator, meaning that as n gets bigger, the MOE gets smaller. The same happens in reverse: as n gets smaller, the MOE gets bigger.
Put another way, a small sample size means we have more error because small samples mean they are less representative of the population at large. The bigger a sample is, the better estimate we will have of the parameter.
We are told that "sample A had a larger sample size" indicating that sample A has a more narrow confidence interval.
Therefore, sample B would have a wider confidence interval.
This is true regardless of what the confidence level is set at.
Sample B as it has the smaller sample size.Therefore, sample B, with the smaller sample size, will have a wider 95% confidence interval compared to sample A.
The width of a confidence interval is inversely proportional to the sample size. A larger sample size results in a narrower confidence interval, and a smaller sample size results in a wider confidence interval. Since both samples have the same mean and standard deviation, the only factor affecting the width of the confidence interval is the sample size.
The confidence interval is a range of values that is likely to contain the true population parameter with a certain level of confidence. In other words, it is an estimate of the range of values within which the true population parameter lies.
The width of the confidence interval is influenced by several factors, including the level of confidence, sample size, and variability of the data. When the sample size is larger, the estimate of the population parameter becomes more precise, resulting in a narrower confidence interval. Conversely, when the sample size is smaller, the estimate becomes less precise, resulting in a wider confidence interval.
In this scenario, both samples have the same mean and standard deviation, indicating that they have similar variability. However, sample A has a larger sample size than sample B. Therefore, sample A's estimate of the population parameter is likely to be more precise than sample B's estimate. This increased precision results in a narrower confidence interval for sample A compared to sample B. Therefore, sample B will have a wider 95% confidence interval compared to sample A.
Learn more about confidence here: brainly.com/question/29048041
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Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye^(x+y)
Answer:
The mass of the lamina is 1
Step-by-step explanation:
Let [tex]\rho(x,y)[/tex] be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:
[tex]m=\int\limits \int\limits_D \rho(x,y) \, dA[/tex].
From the question, the given density function is [tex]\rho (x,y)=xye^{x+y}[/tex].
Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
The mass of the lamina can be found by evaluating the double integral:
[tex]I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx[/tex].
Since D is a rectangular region, we can apply Fubini's Theorem to get:
[tex]I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx[/tex].
Let the inner integral be: [tex]I_0=\int\limits^1_0xye^{x+y}dy[/tex], then
[tex]I=\int\limits^1_0(I_0)dx[/tex].
The inner integral is evaluated using integration by parts.
Let [tex]u=xy[/tex], the partial derivative of u wrt y is
[tex]\implies du=xdy[/tex]
and
[tex]dv=\int\limits e^{x+y} dy[/tex], integrating wrt y, we obtain
[tex]v=\int\limits e^{x+y}[/tex]
Recall the integration by parts formula:[tex]\int\limits udv=uv- \int\limits vdu[/tex]
This implies that:
[tex]\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy[/tex]
[tex]\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}[/tex]
[tex]I_0=\int\limits^1_0 xye^{x+y}dy[/tex]
We substitute the limits of integration and evaluate to get:
[tex]I_0=xe^x[/tex]
This implies that:
[tex]I=\int\limits^1_0(xe^x)dx[/tex].
Or
[tex]I=\int\limits^1_0xe^xdx[/tex].
We again apply integration by parts formula to get:
[tex]\int\limits xe^xdx=e^x(x-1)[/tex].
[tex]I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)[/tex].
[tex]I=\int\limits^1_0xe^xdx=0-1(0-1)[/tex].
[tex]I=\int\limits^1_0xe^xdx=0-1(-1)=1[/tex].
No unit is given, therefore the mass of the lamina is 1.
The mass of the lamina is (1 − e)/2 units.
To find the mass of the lamina occupying the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xy[tex]e^{(x+y)[/tex] , we need to calculate the double integral of the density function over the given region.
Set up the double integral:M = ∬D ρ(x,y) dA
Since D is a rectangular region, we can write:
M = ∫01 ∫01 xy [tex]e^{(x+y)[/tex] dy dx
Evaluate the inner integral with respect to y first:∫01 xy [tex]e^{(x+y)[/tex] dy = x ∫01 y [tex]e^{(x+y)[/tex] dy
Using integration by parts where u = y, dv = [tex]e^{(x+y)[/tex] dy:
du = dyv = [tex]e^{(x+y)[/tex] / eˣ∫ y [tex]e^{(x+y)[/tex] dy = [y [tex]e^{(x+y)[/tex] / ex] 01 - ∫ [tex]e^{(x+y)[/tex] / ex dy
= [y [tex]e^{(x+y)[/tex]] 01 - [[tex]e^{(x+y)[/tex]] 01
= [[tex]e^{(x+1)[/tex]) - eˣ - [tex]e^{(x+1)[/tex] + e⁰]/eˣ
= x (1 − e¹)
Next, evaluate the outer integral:∫01 x (1 − e¹) dx = (1 − e¹) ∫01 x dx
= (1 − e¹) [x2 / 2] 01
= (1 − e¹)(1/2)
= (1 − e¹)/2
The mass of the lamina is (1 − e)/2 units.
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=6−x2. What are the dimensions of such a rectangle with the greatest possible area?
Answer:
length: 2sqrt(2)
width: 4
Step-by-step explanation:
Let the x-coordinate of the right lower vertex be x. Then the x-coordinate of the left lower vertex is -x. The distance between x and -x is 2x, so the lower side of the rectangle, that rests on the x-axis, measures 2x. The length of the rectangle is 2x.
We now use the equation of the parabola to find the y-coordinate of the upper vertices. y = 6 - 2^x. If you plug in x for x, naturally you get y = 6 - x^2, so the y-coordinates of the two upper vertices are 6 - x^2. Since the y-coordinates of the lower vertices is 0, the vertical sides of the rectangle have length 6 - x^2. The width of the rectangle is 6 - x^2.
We have a rectangle with length 2x and width 6 - x^2.
Now we can find the area of the rectangle.
area = length * width
A = 2x(6 - x^2)
A = 12x - 2x^3
To find a maximum value of x, we differentiate the expression for the area with respect to x, set the derivative equal to zero, and solve for x.
dA/dx = 12 - 6x^2
We set the derivative equal to zero and solve for x.
12 - 6x^2 = 0
2 - x^2 = 0
x^2 = 2
x = +/- sqrt(2)
length = 2x = 2 * sqrt(2) = 2sqrt(2)
width = 6 - x^2 = 6 - (sqrt(2))^2 = 6 - 2 = 4
Final answer:
The greatest possible area of a rectangle inscribed under the parabola y = 6 - x² is achieved with dimensions 2√3 (width) by 3 (height), obtained by optimizing the area function using calculus.
Explanation:
To find the dimensions of the rectangle with the greatest possible area that is inscribed with its base on the x-axis and its upper corners on the parabola y = 6 - x², we apply the optimization technique in calculus.
Let the x-coordinate of the rectangle's upper corners be x, which means that the width of the rectangle is 2x (since the parabola is symmetric with respect to the y-axis). The height of the rectangle is given by the y-coordinate on the parabola, which is 6 - x².
Therefore, the area A of the rectangle can be expressed as A = 2x(6 - x²). To find the maximum area, we take the derivative of A with respect to x and set it to zero, solving for x.
A'(x) = 2(6 - x²) - 4x²(x). Setting A'(x) = 0 gives us x = √3. The maximum area occurs when x = √3, so the width of the rectangle will be 2√3 and the height will be 6 - 3, which simplifies to 3.
Hence, the rectangle with the greatest possible area has dimensions of 2√3 by 3.
Solve the following system. (Use (x,y) format in a single answer space.)
x2 + y2 = 25
y2 - x2 = 7
The solution set is
Answer:
(3;4); (-3;-4); (3;-4); (-3;4)
Step-by-step explanation:
for more information see the attached picture.
Answer: The required solution set is
(x, y) = (3, 4), (-3, 4), (3, -4) and (-3, -4).
Step-by-step explanation: We are given to solve the following system :
[tex]x^2+y^2=25~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y^2-x^2=7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
We will be using the method of Elimination to solve the problem.
Adding equations (i) and (ii), we have
[tex](x^2+y^2)+(y^2-x^2)=25+7\\\\\Rightarrow 2y^2=32\\\\\Rightarrow y^2=16\\\\\Rightarrow y=\pm\sqrt{16}~~~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow y=\pm4.[/tex]
From equation (ii), we get
[tex](\pm4)^2-x^2=7\\\\\Rightarrow 16-x^2=7\\\\\Rightarrow x^2=16-7\\\\\Rightarrow x^2=9\\\\\Rightarrow x=\pm\sqrt9~~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow x=\pm3.[/tex]
Thus, the required solution set is
(x, y) = (3, 4), (-3, 4), (3, -4) and (-3, -4).
Use a Venn diagram to answer the question. A survey of 180 families showed that 67 had a dog; 52 had a cat; 22 had a dog and a cat; 70 had neither a cat nor a dog, and in addition did not have a parakeet; 4 had a cat, a dog, and a parakeet. How many had a parakeet only?
Answer:
There are 13 families had a parakeet only
Step-by-step explanation:
* Lets explain the problem
- There are 180 families
- 67 families had a dog
- 52 families had a cat
- 22 families had a dog and a cat
- 70 had neither a cat nor a dog, and in addition did not have a
parakeet
- 4 had a cat, a dog, and a parakeet (4 is a part of 22 and 22 is a part
of 67 and 520
* We will explain the Venn-diagram
- A rectangle represent the total of the families
- Three intersected circles:
C represented the cat
D represented the dog
P represented the parakeet
- The common part of the three circle had 4 families
- The common part between the circle of the cat and the circle of the
dog only had 22 - 4 = 18 families
- The common part between the circle of the dog and the circle of the
parakeet only had a families
- The common part between the circle of the cat and the circle of the
parakeet only had b families
- The non-intersected part of the circle of the dog had 67 - 22 - a =
45 - a families
had dogs only
- The non-intersected part of the circle of the cat had 52 - 22 - b =
30 - b families
had cats only
- The non-intersected part of the circle of the parakeet had c families
had parakeets only
- The part out side the circles and inside the triangle has 70 families
- Look to the attached graph for more under stand
∵ The total of the families is 180
∴ The sum of all steps above is 180
∴ 45 - a + 18 + 4 + 30 - b + b + c + a + 70 = 180 ⇒ simplify
- (-a) will cancel (a) and (-b) will cancel (b)
∴ (45 + 18 + 4 + 30 + 70) + (-a + a) + (-b + b) + c = 180
∴ 167 + c = 180 ⇒ subtract 167 from both sides
∴ c = 180 - 167 = 13 families
* There are 13 families had a parakeet only
Data were collected on the amount spent by 64 customers for lunch at a major Houston restaurant. These data are contained in the file named Houston. Based upon past studies the population standard deviation is known with = $9. Click on the datafile logo to reference the data. a. At 99% confidence, what is the margin of error? b. Develop a 99% confidence interval estimate of the mean amount spent for lunch.
Amount
20.50
14.63
23.77
29.96
29.49
32.70
9.20
20.89
28.87
15.78
18.16
12.16
11.22
16.43
17.66
9.59
18.89
19.88
23.11
20.11
20.34
20.08
30.36
21.79
21.18
19.22
34.13
27.49
36.55
18.37
32.27
12.63
25.53
27.71
33.81
21.79
19.16
26.35
20.01
26.85
13.63
17.22
13.17
20.12
22.11
22.47
20.36
35.47
11.85
17.88
6.83
30.99
14.62
18.38
26.85
25.10
27.55
25.87
14.37
15.61
26.46
24.24
16.66
20.85
Final answer:
At a 99% confidence level, the estimated mean amount spent for lunch is given by the sample mean with a margin of error of $2.900, resulting in a confidence interval.
Explanation:
To calculate the margin of error at a 99% confidence level, we need to use the formula: Margin of Error = Z * (Population Standard Deviation / √Sample Size). Here, the Z-value for a 99% confidence level is 2.576. The sample size is 64 and the population standard deviation is $9. Plugging in these values, we get: Margin of Error = 2.576 * (9 / √64) = 2.576 * 1.125 = 2.900. So, the margin of error is $2.900.
To develop a 99% confidence interval estimate of the mean amount spent for lunch, we need to use the formula: Confidence Interval = Sample Mean ± Margin of Error. The sample mean is the average of the lunch amounts provided in the data file. Plugging in the values, we get: Confidence Interval = Sample Mean ± 2.900. Here, the Sample Mean is the average of the lunch amounts and the Margin of Error is $2.900.
r 2 • r • r 5
Multiply or divide as indicated.
Answer:
r^8
Step-by-step explanation:
r 2 • r • r 5
Assuming that r 2 means r^2
r ^2 • r • r^ 5
When the bases are the same and we are multiplying, we add the exponents
r ^2 • r^1 • r^ 5
r^(2+1+5)
r^(8)
Answer: [tex]r^8[/tex]
Step-by-step explanation:
You need to remember a property called "Product of powers property". This property states the following:
[tex](a^m)(a^n)=a^{(m+n)}[/tex]
You can observe that the bases of the expression [tex]r^2*r*r^5[/tex] are equal, then you can apply the property mentioned before.
Knowing this, you can make the multiplication indicated. Therefore, you get that the product is:
[tex]r^2*r*r^5=r^{(2+1+5)}=r^8[/tex]
A student's course grade is based on one midterm that counts as 10% of his final grade, one class project that counts as 25% of his final grade, a set of homework assignments that counts as 50% of his final grade, and a final exam that counts as 15% of his final grade. His midterm score is 74, his project score is 93, his homework score is 83, and his final exam score is 77. What is his overall final score? What letter grade did he earn (A, B, C, D, or F)? Assume that a mean of 90 or above is an A, a mean of at least 80 but less than 90 is a B, and so on.
Answer:
Overall Final Score = 83.7
Letter Grade = B
Step-by-step explanation:
To calculate weighted marks (grade), the rule is to multiply each weight (in decimal) with respective marks and add them together. Let's do this:
Overall final score = (0.10*74) + (0.25*93) + (0.50*83) + (0.15*77) = 83.7
From the information given, we know that score between 80 and 90 is a "B".
Find an equation of the tangent line to the curve at the given point. y=secx, (pi/3, 2)
Answer:
Equation of tangent is [tex]y-2\sqrt{3}x=2-2\frac{\pi}{\sqrt{3}}[/tex]
Step-by-step explanation:
Given:
Equation of curve, y = sec x
Passing through point = [tex](\frac{\pi}{3},2)[/tex]
We need to find Equation of tangent to the given curve and at the given point.
First we find the slope of the tangent by differentiating the given curve.
As we know that slope of the tangent, m = [tex]\frac{\mathrm{d}y}{\mathrm{d} x}[/tex]
So, consider
y = sec x
[tex]\frac{\mathrm{d}y}{\mathrm{d} x}=\frac{\mathrm{d}(sec\,x)}{\mathrm{d} x}[/tex]
[tex]\frac{\mathrm{d}y}{\mathrm{d} x}=sec\,x\:tan\,x[/tex]
Now we find value of slope at given point,
put x = [tex]\frac{\pi}{3}[/tex] in above derivate
we get
[tex]\frac{\mathrm{d}y}{\mathrm{d} x}=sec\,\frac{\pi}{3}\:tan\,{\pi}{3}=2\sqrt{3}[/tex]
Now using Slope point form, we have
Equation of tangent
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-2=2\sqrt{3}(x-\frac{\pi}{3})[/tex]
[tex]y=2\sqrt{3}x-2\sqrt{3}\frac{\pi}{3}+2[/tex]
[tex]y-2\sqrt{3}x=2-2\frac{\pi}{\sqrt{3}}[/tex]
Therefore, Equation of tangent is [tex]y-2\sqrt{3}x=2-2\frac{\pi}{\sqrt{3}}[/tex]
Thirty-five math majors, 23 music majors and 65 history majors are randomly selected from 586 math majors, 224 music majors and 254 history majors at the state university. What sampling technique is used?
Answer: Simple random sampling
Step-by-step explanation:
Simple random sampling is a technique of sampling in which an experimenter selects the group of samples or subjects randomly from a large group of population. Each sample is chosen randomly by chance and each entity of a population has equal possibility of being selected as sample.
According to the given situation, simple random sampling is the technique that should be used for sampling.
24 is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. The first part is
a) 11 b) 13 c) 16 d ) 17
Answer:
13
Step-by-step explanation:
Let the two parts are x and y
as per the question
x+y=24 -----------------(A)
Also 7 times first part (x) that is 7x ,
when added to
5 times second part (y ) that is 5y
it gives 146
Hence our second equation becomes
7x+5y=146 ------------ (B)
Now we have to solve these two equations (A) and (B) to find the values of x and y
[tex]x+y=24 \\7x+5y=146\\[/tex]
now multiplying equation A with 5 and subtracting it from B
[tex]7x+5y=146[/tex]
[tex]-5x-5y=-120[/tex]
we get
2x=26
dividing both sides by 2 we get
x = 13
And hence our first part is 13
32°
45°
48°
20°
FIND WZC
Answer:
48°
Step-by-step explanation:
In triangle CYZ, angles Y and C are complementary, so ...
(5x +22) + (12x) = 90
17x +22 = 90
17x = 68
x = 4
The angle of interest is 12x° , so has measure
12·4° = 48°
Answer:
m∠WZC=48°
Step-by-step explanation:
we know that
If ZC is an altitude
then
Triangle YZC is a right triangle
therefore
m∠CYW+m∠WZC=90° ----> by complementary angles
substitute
(5x+22)°+(12x)°=90°
17x=(90-22)°
x=4°
Find the measure of angle WZC
m∠WZC=(12x)°=12*4=48°
VWX and NOP are similar. If mV = 44° and mP = 66°, what is
mW?
Answer: 70°
Step-by-step explanation:
ΔVWX ≅ ΔNOP ⇒ ∠V=∠N, ∠W=∠O, and ∠X=∠P
Since ∠V=44° and ∠P=66°=∠X
then we can use the Triangle Sum Theorem:
∠V + ∠W + ∠X = 180°
44° + ∠W + 66° = 180°
∠W + 110° = 180°
∠W = 70°
Find S for the given geometric series. Round answers to the nearest hundredth, if necessary.
a1=0.2, a5=259.2, r=6
a) 311
b)51
c)222.2
d)624.96
Answer:
A
Step-by-step explanation:
Sum of the first n terms of a geometric series is:
S = a₁ (1 - r^n) / (1 - r)
Here, a₁ = 0.2, r = 6, and n = 5.
S = 0.2 (1 - 6^5) / (1 - 6)
S = 311
Answer:
Option A
Step-by-step explanation:
For the given geometric series
a₁ = 0.2
a₅ = 259.2
r = 6
Then we have to find the sum of initial 5 terms of this series
[tex]S_{n} =\frac{a_1(r^4-1)}{(r-1)}=\frac{0.2(6^3-1)}{(6-1)}[/tex]
[tex]=\frac{0.2(7776-1)}{5}[/tex]
[tex]\frac{0.2\times 7775}{5}[/tex]
= 311
Option A is the answer.
A clothing store has just advertised at a bargain price, that a customer tries on a t-shirt is 0.40. a new range of t-shirts. The probability If a customer tries a t-shirt, the probability that they will buy it is 0.70 while if a customer does not try on a t-shirt, the probability that they will buy it is 0.15. A customer has just bought a t-shirt. Find the probability that the customer did not try on the t- shirt.
The probability that the customer did not try on the t- shirt is:
0.2432
Step-by-step explanation:It is given that the probability that a customer tries a t-shirt is: 0.40
After he will try
The probability that he will buy the t-shirt is: 0.70
The probability that he will not try t-shirt is: 1-0.40=0.60
If he will not try
then the probability of buying a t-shirt is: 0.15
This means that the total probability of Buying a t-shirt is:
0.40×0.70+0.60×0.15
= 0.37
It is given that the customer bought a t-shirt, we need to find the probability that he did not try the t-shirt:
The probability is given by:
[tex]\dfrac{0.60\times 0.15}{0.37}\\\\\\=\dfrac{0.09}{0.37}\\\\\\=0.2432[/tex]
The question asks for the conditional probability that a customer did not try on a t-shirt given that they bought it. This probability value can be calculated using the total probability theorem and Bayes' theorem. It is found to be approximately 0.159.
Explanation:This question falls under the topic of probability, specifically conditional probability and Bayes' theorem. The probability values are given in the problem statement but it seeks the probability that a customer did not try on a t-shirt given that they bought it.
Let A represent the event that a customer buys a t-shirt and B1 represent the event that a customer tries on a t-shirt whereas B2 represent the event that the customer did not try on a t-shirt.
We are given that P(B1) = 0.40, P(A|B1) = 0.70 and P(A|B2) = 0.15. We are asked to find P(B2|A).
To find P(B2|A), we use Bayes' theorem:
P(B2|A) = (P(B2) * P(A|B2)) / P(A)
We know that P(B1) + P(B2) = 1, so P(B2) = 1 - P(B1) = 1 - 0.40 = 0.60.
The probability that a customer tried a t-shirt is 0.40 and bought it is 0.70, that yields the joint probability P(A, B1) = P(B1) * P(A|B1) = 0.40 * 0.70 = 0.28. Similarly the joint probability P(A, B2) = P(B2) * P(A|B2) = 0.60 * 0.15 = 0.09.
P(A) is the probability that a customer buys a t-shirt and is calculated using the total probability theorem as P(A) = P(A, B1) + P(A, B2) = 0.28 + 0.09 = 0.37.
Substituting the values we calculated into Bayes' theorem, we find:
P(B2|A) = (0.60 * 0.15) / 0.37 ≈ 0.24.
So, the probability that a customer did not try the t-shirt given that they bought it is approximately 0.24.
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A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50. In a random sample of 36 bins, the sample mean amount was 48.47 pounds and the sample standard deviation was 3.1 pounds. Conduct the appropriate hypothesis test using a 0.05 level of significance.\
Answer with explanation:
Given : Sample mean =[tex]\overline{x}=\text{48.47 pounds}[/tex]
Standard deviation : [tex]\sigma=\text{ 3.1 pounds.}[/tex]
Sample size : n = 36
Claim : [tex]\mu\neq50[/tex]
∴ [tex]H_0:\mu=50[/tex]
[tex]H_1:\mu\neq50[/tex]
Since the alternative hypothesis is two tail , then the test is two tail test.
By using a z statistic and a 0.05 level of significance. Reject [tex]H_0[/tex] if z < -1.960 or is z> 1.960.
Then , the test static for population mean is given by :-
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]z=\dfrac{48.47-50}{\dfrac{3.1}{\sqrt{36}}}\approx-2.96[/tex]
We reject [tex]H_0[/tex] since [tex]-2.96\leq -1.96[/tex]. We have statistically significant evidence at [tex]\alpha=0.05[/tex] to show that the mean amount of garbage per bin is not different from 50.
To test if the mean amount of garbage per bin is different from 50, we set up null and alternative hypotheses and use a z-test to calculate a z-score. The z-score of -1.49 falls within the acceptance region; hence, we fail to reject the null hypothesis, implying that the mean amount of garbage per bin is not significantly different from 50 pounds.
Explanation:First, we need to set up our null and alternative hypotheses. The null hypothesis (H0) states that the mean amount of garbage is 50, or μ = 50. The alternative hypothesis (H1) is that the mean is different from 50, or μ ≠ 50.
Second, calculate the z-score which is given by the formula Z = (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the standard deviation and n is the sample size. Substituting the given values, we get Z = (48.47 - 50) / (3.1 / √36) = -1.49
For a two-tailed test, we are checking if the z-score falls within the rejection region. Since our level of significance is 0.05, the critical values for a two-tailed test are -1.96 and 1.96. Our computed z-score of -1.49 falls within the acceptance region. Therefore, we fail to reject the null hypothesis. This suggests that the mean amount of garbage per bin is not statistically different from 50 pounds at the 0.05 level of significance.
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Consider the equation of the line, c(c2 + 2)y = c − x where c > 0 is a constant. (a) Find the coordinates of the x-intercept and the y-intercept. x-intercept ( , ) y-intercept ( , ) (b) Find a formula for the area of the triangle enclosed between the line, the x-axis, and the y-axis. A(c) =
The equation of the line indicates that the x and y-intercept and the formula for the area of the triangle are;
x-intercept (c, 0)
y-intercept (0, 1/(c² + 7))
A(c) = c/(2·(c² + 7))
The steps used to find the x-intercept and the y-intercept are presented as follows;
The equation of the line c·(c² + 7)·y = c - x, can be expressed in the slope intercept form to find the coordinates of the x-intercept and the coordinates of the y-intercept as follows;
c·(c² + 7)·y = c - x
y = (c - x)/(c·(c² + 7))
y = c/(c·(c² + 7)) - x/(c·(c² + 7))
y = 1/((c² + 7)) - x/(c·(c² + 7))
The above equation is in the slope-intercept form, y = m·x + c
Where c is the y-coordinate of the y-intercept, and (0, c) ids the coordinate of the y-intercept; Therefore, the coordinates of the y-intercept is; (0, 1/((c² + 7)))
The coordinate of the x-intercept can be found by plugging in y = 0, in the above equation to get;
0 = 1/((c² + 7)) - x/(c·(c² + 7))
x/(c·(c² + 7)) = 1/((c² + 7))
x = (c·(c² + 7))/((c² + 7))
(c·(c² + 7))/((c² + 7)) = c
x = c
Therefore coordinates of the x-intercept is; (c, 0)
The triangle enclosed by the line and the x-axis and y-axis is a right triangle, therefore;
The positive x and y-values of the x-intercept and y-intercept indicates that the area of the triangle is the product half the distance from the origin to the y-intercept and the distance from the origin to the x-intercept
Area = (1/2) × (1/((c² + 7)) - 0) × (c - 0)
(1/2) × (1/((c² + 7))) × (c) = c/(2·(c² + 7)
Area of the triangle, A(c) = c/(2·(c² + 7)
The complete question found through search can be presented as follows;
Consider the equation of the line c·(c² + 7)·y = c - x where c > 0 is a constant
(a) Find the coordinates of the x-intercept and the y-intercept
x-intercept ( , )
y-intercept ( , )
(b) Find a formula for the area of the triangle enclosed between the line, the x-axis and the y-axis
Solve |y + 2| > 6
{y|y < -8 or y > 4}
{y|y < -6 or y > 6}
{y|y < -4 or y > 4}
ANSWER
{y|y < -8 or y > 4}
EXPLANATION
The given absolute value equation is
[tex] |y + 2| \: > \: 6[/tex]
This implies that:
[tex](y + 2) \: > \: 6 \: or \: - (y + 2) \: > \: 6[/tex]
Multiply through by -1 in the second inequality and reverse the sign.
[tex]y + 2 \: > \: 6 \: or \: y + 2\: < \: - 6[/tex]
[tex]y \: > \: 6 - 2\: or \: y \: < \: - 6 - 2[/tex]
We simplify to get:
[tex]y \: > \: 4\: or \: y \: < \: - 8[/tex]
The correct answer is A.
Answer:
[tex]\large\boxed{\{y\ |\ y<-8\ or\ y>4\}}[/tex]
Step-by-step explanation:
[tex]|y+2|>6\iff y+2>6\ or\ y+2<-6\qquad\text{subtract 2 from both sides}\\\\y+2-2>6-2\ or\ y+2-2<-6-2\\\\y>4\ or\ y<-8\Rightarrow\{y\ |\ y<-8\ or\ y>4\}[/tex]
In a sample of 8 high school students, they spent an average of 28.8 hours each week doing sports with a sample standard deviation of 3.2 hours. Find the 95% confidence interval, assuming the times are normally distributed.
Answer:
[tex]26.12\:<\:\mu\:<\:31.48[/tex]
Step-by-step explanation:
Since the population standard deviation [tex]\sigma[/tex] is unknown, and the sample standard deviation [tex]s[/tex], must replace it, the [tex]t[/tex] distribution must be used for the confidence interval.
The sample size is n=8.
The degree of freedom is [tex]df=n-1[/tex], [tex]\implies df=8-1=7[/tex].
With 95% confidence level, the [tex]\alpha-level[/tex](significance level) is 5%.
Hence with 7 degrees of freedom, [tex]t_{\frac{\alpha}{2} }=2.365[/tex]. (Read from the t-distribution table see attachment)
The 95% confidence interval can be found by using the formula:
[tex]\bar X-t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )\:<\:\mu\:<\:\bar X+t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )[/tex].
The sample mean is [tex]\bar X=28.8[/tex] hours.
The sample sample standard deviation is [tex]s=3.2[/tex] hours.
We now substitute all these values into the formula to obtain:
[tex]28.8-2.365(\frac{3.2}{\sqrt{8} } )\:<\:\mu\:<\:28.8+2.365(\frac{3.2}{\sqrt{8} } )[/tex].
[tex]26.12\:<\:\mu\:<\:31.48[/tex]
We are 95% confident that the population mean is between 26.12 and 31.48 hours.
Let F(x, y, z) = (5ex sin(y))i + (5ex cos(y))j + 7z2k. Evaluate the integral C F · ds, where c(t) = 8 t , t3, exp( t ) , 0 ≤ t ≤ 1. (Note that exp(u) = eu.)
I'm going to assume this reads
[tex]\vec F(x,y,z)=5e^x\sin y\,\vec\imath+5e^x\cos y\,\vec\jmath+7z^2\,\vec k[/tex]
and the path [tex]C[/tex] is parameterized by
[tex]\vec c(t)=8t\,\vec\imath+t^3\,\vec\jmath+e^t\,\vec k[/tex]
with [tex]0\le t\le1[/tex]. Under this parameterization,
[tex]\vec F(x,y,z)=\vec F(x(t),y(t),z(t))=5e^{8t}\sin(t^3)\,\vec\imath+5e^{8t}\cos(t^3)\,\vec\jmath+7e^{2t}\,\vec k[/tex]
and
[tex]\mathrm d\vec c=\dfrac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt=(8\,\vec\imath+3t^2\,\vec\jmath+e^t\,\vec k)\,\mathrm dt[/tex]
Then in the integral,
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec s=\int_0^1\vec F(x(t),y(t),z(t))\cdot\frac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(7e^{3t}+15e^{8t}t^2\cos(t^3)+40e^{8t}\sin(t^3))\,\mathrm dt\approx\boxed{12586.5}[/tex]
(It's unlikely that an exact answer can be found in terms of elementary functions)
Functions f(x) and g(x) are shown below:
f(x) = 2cos(x)
g(x) = (3)sin(x+pi), a graph of sine function which starts at 0 comma 0 and decreases to the minimum of pi over 2 then increases to the maximum 3 pi over 2 then decreases to 2pi where the cycle repeats.
Using complete sentences, explain how to find the maximum value for each function and determine which function has the largest maximum y-value.
Please please please help!!!
Try this option:
the functions y=sin(x) and y=cos(x) are the standart functions; it means, the period of each is 2π, maximum value is '1', minimum value is '-1'.
Using the properties described above, to find the max. value for the function y=2cos(x): 2*1=2, where 2- the amplitude of the given function, 1- maximum of the standart function;
the max. value for the function y=3sin(x+π): 3*1=3, where 3 - the amplitude of the given function, 1- maximum of the standart function.
For more details see the attached picture.
In other words, common equation for such functions is y=A*sin(ωx+Ф), where A- amplitude, ω - frequency and Ф - initial phase of the given function.
Max. and min. values depend on A, all the points of the function repeat every 2π, including max. and min. values.
[tex]f(x)=2\cos x[/tex]
We know that the function f(x) is a cosine function and the maximum value is obtained by the function when this cosine function takes the maximum value.
We know that:
[tex]-1\leq \cos x\leq 1[/tex]
This means that the cosine function takes the maximum values as: 1
and when [tex]\cos x=1[/tex] then
[tex]f(x)=2\times 1\\\\i.e.\\\\f(x)=2[/tex]
i.e. Maximum value of function f(x)=2
The function g(x) is given by:[tex]g(x)=3\sin (x+\pi)[/tex]
Again the function g(x) will attain the maximum value when the function sine will takes the maximum value.
We know that the range of the sine function is: [-1,1]
This means that the maximum value of sine function is: 1
Hence, when [tex]\sin (x+\pi)=1[/tex] then,
[tex]g(x)=3\times 1\\\\i.e.\\\\g(x)=3[/tex]
i.e. Maximum value of function g(x)=3
The function g(x) has the largest maximum value.
( Since 3>2)
Round the following decimal to the nearest tenth.
543.67 ◯543.6 ◯543.68 ◯543.7 ◯543.670 ◯544
Answer:
543.7
Step-by-step explanation:
Answer:
543.7, 543.6, 543.7, 543.7, 543., 544.0
Step-by-step explanation:
The features of a Coverdell Education Savings Account include all of the following EXCEPT: (A) The contributions are deductible. (B) $2,000 is the maximum contribution in any one year. (C) Withdrawals are tax free. (D) Contributions are phased out for certain taxpayers who have adjusted gross income above a certain level.
Answer:
c
Step-by-step explanation:
A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangos. Imagine you stick your hand into the refrigerator and pull out a piece of fruit at random. What is the sample space for your action? ????={6, 5, 10, 3, 7, 11, 2}S={6, 5, 10, 3, 7, 11, 2} ????={apple, orange, banana, pear, peach, plum, mango}S={apple, orange, banana, pear, peach, plum, mango} ????={6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, 2 mangos}
Answer:
4th Option is correct.
Step-by-step explanation:
Given:
Number of apples in refrigerator = 6
Number of oranges in refrigerator = 5
Number of bananas in refrigerator = 10
Number of pears in refrigerator = 3
Number of peaches in refrigerator = 7
Number of plums in refrigerator = 11
Number of mangoes in refrigerator = 2
A piece is randomly taken out from refrigerator.
To find: Sample Space of the experiment
Sample Space : It is a set which contain all the possible outcome / results of the experiment.
So, here Sample Space = { 6 apples , 5 oranges , 10 bananas , 3 pears , 7 peaches , 11 plums , 2 mangoes }
Therefore, 4th Option is correct.
A city temperature is modeled as a normal random variable with mean and standard deviation both equal to 10 degrees Celsius. What is the probability that the temperature at a randomly chosen time will be less or equal to 15 degrees Celsius?
Answer: 0.6915
Step-by-step explanation:
Given : [tex]\text{Mean}=\mu=10^{\circ}C[/tex]
[tex]\text{Standard deviation}=\sigma=10^{\circ}C[/tex]
Since , the distribution follows a Normal distribution.
The formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x=[tex]15^{\circ}C[/tex]
[tex]z=\dfrac{15-10}{10}=0.5[/tex]
The p-value = [tex]P(z\leq0.5)=0.6914625\approx0.6915[/tex]
Hence, the required probability : 0.6915
The probability that the temperature will be less than or equal to 15 degrees Celsius in a city with a mean temperature of 10 degrees Celsius and a standard deviation of 10 degrees Celsius is approximately 69.15%.
To find the probability that the temperature will be less than or equal to 15 degrees Celsius when the mean and standard deviation are both 10 degrees Celsius, we can calculate the z-score.
The z-score formula is:
z = (X - μ) / Σ
For X = 15, μ = 10, and Σ = 10, the z-score is:
z = (15 - 10) / 10 = 0.5
We then look up the z-score of 0.5 in a standard normal distribution table or use a calculator to find the cumulative probability corresponding to the z-score. The probability that the temperature is less than or equal to 15 degrees Celsius is approximately 0.6915 (or 69.15%).
You find the following financial information about a company: net working capital = $735; fixed assets = $5,385; total assets = $8,190; and long-term debt = $4,281. What are the company's total liabilities?
Answer:
Company's total liabilities are $6,351.
Step-by-step explanation:
A company's financial information is given, we have to calculate the company's total liabilities.
Total assets = Fixed assets + current assets
Current assets = Total assets - Fixed assets
= $8,190 - $5,385
= $2,805
Networking capital = Current assets - Current liabilities
$735 = $2,805 - current liabilities
Current liabilities = $2,805 - $735
Current liabilities = $2,070
Total liabilities = Long term debt + current liabilities
= $4,281 + $2,070
= $6,351
Company's total liabilities are $6,351.
An old picture frame has dimensions 33 inches by 24 inches. What one length must be cut from each dimension so that the ratio of the shorter side to the longer side is 2/3?
Final answer:
To achieve a 2:3 ratio for the picture frame's dimensions, 6 inches must be cut from both the length and width.
Explanation:
To resize the picture frame while maintaining a specific ratio between its sides, we need to establish a proportion based on the desired ratio of 2:3 (shorter side to the longer side). Let's denote the amount that needs to be cut from the shorter side as x inches, and the amount to be cut from the longer side as y inches. After cutting, the dimensions of the frame will be (33 - y) inches by (24 - x) inches.
We want to set up a proportion to reflect the desired ratio: (24 - x)/(33 - y) = 2/3. To solve for x and y, we can use the property that cross-multiplying the terms in a proportion gives us an equality: 3(24 - x) = 2(33 - y). This simplifies to 72 - 3x = 66 - 2y. We also know that the amount cut off from both dimensions should be the same, thus x = y. Substituting y for x gives us 72 - 3x = 66 - 2x. Solving this equation, we get x = 6 inches. So, 6 inches must be cut from both the length and the width to achieve the 2:3 ratio.