Answer:
Step-by-step explanation:
Sugar and Salt even thought they both dissolve in water they both dissolve in different ways. When salt dissolves in water, its individual types of ions are torn apart from each other, while Sugar molecules stay together when dissolved in water, and therefore the molecules remain the same when dissolved in water.
This being said in science using your senses can be just as valuable as using calculations. In this case both Sugar and Salt taste differently. Sugar is sweet while Salt is salty. Therefore tasting the substance can be the easiest and most accurate way of determining the substance.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Final answer:
To determine if an unknown sample contains sugar, salt, or both, examine their solubility properties and chemical reactivity. Comparing the density of the sample against reference values of pure substances can provide preliminary identification, while testing for chloride ions with silver nitrate confirms the presence of salt.
Explanation:
Identifying Sugar and Salt in an Unknown Sample
To determine whether an unknown sample contains sugar, salt, or both, we must identify the physical and chemical properties that distinguish these substances. Salt (sodium chloride) and sugar (sucrose) have distinct solubility properties and chemical reactivity, which we can use to identify them when dissolved in water.
Solubility and Density Test
Both sugar and salt are highly soluble in water, but we can compare their densities to make a preliminary identification. A known volume of each substance is weighed and their densities calculated. Salt generally has a greater density than sugar. If the unknown sample has a certain mass, comparing it with the reference densities may provide an initial indication.
Chemical Reactivity Test
To confirm the identity of the substances, a chemical reactant such as silver nitrate can be introduced to the water solution of the unknown sample. If a white precipitate forms, it indicates the presence of chloride ions, which suggests the presence of salt. Since sugar does not produce a precipitate with silver nitrate, its absence would indirectly indicate the presence of sugar.
Performing these tests will allow us to determine if the unknown sample is sugar, salt, or a mixture of both. The greater the discrepancy between the calculated density and the known densities of pure sugar or pure salt, the more likely it is that the sample is a mixture.
find a general solution of
t *(dy/dt)-(y^2)*lnt+y=0
[tex]t\dfrac{\mathrm dy}{\mathrm dt}-y^2\ln t+y=0[/tex]
Divide both sides by [tex]y(t)^2[/tex]:
[tex]ty^{-2}\dfrac{\mathrm dy}{\mathrm dt}-\ln t+y^{-1}=0[/tex]
Substitute [tex]v(t)=y(t)^{-1}[/tex], so that [tex]\dfrac{\mathrm dv}{\mathrm dt}=-y(t)^{-2}\dfrac{\mathrm dy}{\mathrm dt}[/tex].
[tex]-t\dfrac{\mathrm dv}{\mathrm dt}-\ln t+v=0[/tex]
[tex]t\dfrac{\mathrm dv}{\mathrm dt}-v=\ln t[/tex]
Divide both sides by [tex]t^2[/tex]:
[tex]\dfrac1t\dfrac{\mathrm dv}{\mathrm dt}-\dfrac1{t^2}v=\dfrac{\ln t}{t^2}[/tex]
The left side can be condensed as the derivative of a product:
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac1tv\right]=\dfrac{\ln t}{t^2}[/tex]
Integrate both sides. The integral on the right side can be done by parts.
[tex]\displaystyle\int\frac{\ln t}{t^2}\,\mathrm dt=-\frac{\ln t}t+\int\frac{\mathrm dt}{t^2}=-\frac{\ln t}t-\frac1t+C[/tex]
[tex]\dfrac1tv=-\dfrac{\ln t}t-\dfrac1t+C[/tex]
[tex]v=-\ln t-1+Ct[/tex]
Now solve for [tex]y(t)[/tex].
[tex]y^{-1}=-\ln t-1+Ct[/tex]
[tex]\boxed{y(t)=\dfrac1{Ct-\ln t-1}}[/tex]
Convert 3,A5D Base 16 to Base 10
Answer:
14941.
Step-by-step explanation:
In base 16 we have that :
A=10, B=11, C=12, D=13, E=14, F=15 and the process of change is:
3: [tex]3*16^{0}= 3[/tex]
A5D= [tex]10*16^{2}+5*16^{1}+13*16^{0}= 2560+80+13=2653.[/tex]
3A5D = [tex]3*16^{3}+10*16^{2}+5*16^{1}+13*16^{0}= 12288+2560+80+13=14941.[/tex]
Find the volumes of the solids generated by revolving the triangle with vertices (2, 2), (2, 6), and (5, 6) about a) the x-axis, b) the y-axis, c) the line x=7, and d) the line y=2.
[tex]\displaystyle\pi\int_2^5\left(6^2-\left(\frac43x-\frac23\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(20+x-x^2)\,\mathrm dx=\boxed{56\pi}[/tex]
About the [tex]y[/tex]-axis (shell method):[tex]\displaystyle2\pi\int_2^5x\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5x(5-x)\,\mathrm dx=\boxed{36\pi}[/tex]
About [tex]x=7[/tex] (shell method):[tex]\displaystyle2\pi\int_2^5(7-x)\left(6-\left(\frac43x-\frac23\right)\right)\,\mathrm dx=\frac{8\pi}3\int_2^5(35-12x+x^2)\,\mathrm dx=\boxed{48\pi}[/tex]
About [tex]y=2[/tex] (washer method):[tex]\displaystyle\pi\int_2^5\left((6-2)^2-\left(\frac43x-\frac23-2\right)^2\right)\,\mathrm dx=\frac{16\pi}9\int_2^5(5+4x-x^2)\,\mathrm dx=\boxed{32\pi}[/tex]
The y-coordinates of the two intersection points of the triangle and the line y=2 (Option d).
In this explanation, we will explore how to find the volumes of solids formed by revolving a triangle with given vertices about different axes and lines. We'll use basic calculus principles to calculate the volumes and understand the concept of rotation in three-dimensional space.
a) To find the volume of the solid generated by revolving the triangle about the x-axis, we imagine rotating the triangle in a circular motion around the x-axis. This forms a three-dimensional shape known as a "solid of revolution."* To calculate the volume, we integrate the cross-sectional area of each infinitesimally thin slice of the solid perpendicular to the x-axis, from the x-coordinate of the leftmost point to the rightmost point.
Let's use the "disk method" to integrate the cross-sectional areas. Each disk has a radius equal to the y-coordinate of the triangle at a particular x-coordinate. The formula for the volume using the disk method is:
Vx = ∫[from a to b] π * (y)² dx
Where (a, b) are the x-coordinates of the leftmost and rightmost points of the triangle, and y represents the y-coordinate of the triangle at a specific x.
b) Similarly, to find the volume of the solid formed by revolving the triangle about the y-axis, we use the "washer method". In this case, the inner radius of each washer is given by the x-coordinate of the triangle at a particular y-coordinate. The formula for the volume using the washer method is:
Vy = ∫[from c to d] π * (x)² dy
Where (c, d) are the y-coordinates of the bottommost and topmost points of the triangle, and x represents the x-coordinate of the triangle at a specific y.
c) To find the volume of the solid formed by revolving the triangle about the line x=7, we use the "shell method".
We integrate the circumference of each cylindrical shell formed between the triangle and the line x=7. The formula for the volume using the shell method is:
V7 = ∫[from e to f] 2π * (x-7) * y dx
Where (e, f) are the x-coordinates of the two intersection points of the triangle and the line x=7.
d) Lastly, to find the volume of the solid formed by revolving the triangle about the line y=2, we can use the shell method as well, considering cylindrical shells formed between the triangle and the line y=2. The formula for the volume using the shell method is:
Vy=2 = ∫[from g to h] 2π * (y-2) * x dy
Where (g, h) are the y-coordinates of the two intersection points of the triangle and the line y=2.
By calculating the integrals using these formulas, we can find the volumes of the solids generated by revolving the triangle about the specified axes and lines. Remember to always set up the integral limits correctly based on the x or y coordinates of the triangle's vertices.
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Write a differential equation to represent each situation below. Do NOT solve them. a. A new technology is introduced into a community of 5000 people. If the rate at which the technology is adopted in the community is jointly proportional to the number of people who have adopted the technology and the number of people who have not adopted it, write a differential equation to represent the number of people, x(t), who have adopted the technology by time t. b. A tank with a capacity of 1000 gal originally contains 800 gal of water with 200 lbs of salt in the solution. Water containing 3.5 lbs of salt per gal is entering at a rate of 2 gal/min, and the mixture is allowed to flow out of the tank at a rate of 5 gal/min. Write a differential equation reflecting the information above, clearly stating the requested intermediate results below. Let A(t) represent the amount of salt (in pounds) in the tank after t minutes. Do NOT solve the DE! Show units with each factor in the three setup steps below, and simplify each expression, showing the resulting units as well. *Be sure to include the initial conditions for this DE in your final equation. R_in = Concentration of salt in the tank: c(t) = R_out = Differential equation: A(0) =
Answer:
a. x'(t) = kx(5000-x)
b. R_in = 7 lbs/min
R_out = 5c(t) lbs/min
c(t) = A(t)/(800 -3t) lbs/gal
A'(t) = R_in - R_out; A(0) = 200 lbs
Step-by-step explanation:
A. When proportionality is joint between x and y, the expression describing that is kxy, where k is the constant of proportionality. Here, the rate of change is jointly proportional to number who have adopted (x) and number of the community of 5000 who have not (5000-x). The corresponding equation is ...
x'(t) = kx·(5000-x)
We know the initial number of adopters must be greater than 0, or the equation's only solution would be x=0.
__
B. The rate of influx of salt is ...
R_in = (3.5 lbs/gal)×(2 gal/min) = 7 lbs/min
The number of gallons in the tank at time t is ...
v = 800 gal + (2 gal/min - 5 gal/min)(t min) = (800 -3t) gal
For amount of salt A(t) pounds, the concentration c(t) will be ...
c(t) = A(t)/v = A(t)/(800 -3t) . . . . lbs/gal
The outflow rate will be the product of outflow volume and concentration:
R_out = (c(t) lbs/gal)(5 gal/min) = 5c(t) lbs/min
And the differential equation for A(t) is ...
A'(t) = R_in -R_out . . . with initial condition A(0) = 200 lbs.
__
When you go to solve the equation for A'(t), it will need to be cast in terms of A(t):
A'(t) = 7 -5A(t)/(800 -3t)
if ( 43.65 ) ( 8.79 ) / x = ( 0.4365 ) ( 87.9 ) then value of x is:
(a) .01 (b) 0.1 (c) 1 (d) 10 (e) 100
Answer:
(d)
Step-by-step explanation:
Multiple your numbers out first. This always makes it less hectic to look at. Multiply both sides by x to move the x to the right side. Then divide both sides by ((0.4365)(87.9)). You should get 10.
Let A be a 5 × 4 matrix with a pivot in each column. Is A invertible? Why or why not?
Answer:
Matrix A is not invertible.
Step-by-step explanation:
Invertible matrix are those matrix which are square matrix or non singular matrix.
For any matrix to be invertible, matrix should be non singular i.e. det(x)[tex]\neq[/tex] 0.
But for the question given above we cannot find determinant of matrix A as it is not square matrix. so inverse of given matrix does not exist. so it is not possible to have non trival solutions.
The diameters of red delicious apples of an orchard have a normal distribution with a mean of 3 inches and a standard deviation of 0.5 inch. One apple will be randomly chosen. What is the probability of picking an apple with diameter greater than 3.76 inches?
To find the probability of picking an apple with a diameter greater than 3.76 inches from a normally distributed population with a mean of 3 inches and a standard deviation of 0.5 inches, we calculate a z-score and then find the corresponding probability using a z-table or calculator.
Explanation:The question is about finding the probability in a situation that follows a normal distribution, specifically, the likelihood of picking an apple with a diameter greater than 3.76 inches given a mean of 3 inches and a standard deviation of 0.5 inches.
To solve this, we must first calculate the z-score, which is the number of standard deviations a data point (in this case, the apple size of 3.76 inches) is from the mean. The z-score is calculated as follows: Z = (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation. Substituting the given values, we get Z = (3.76 - 3) / 0.5, which results in a z-score of 1.52.
Next, we will refer to the z-table to find the probability corresponding to this z-score. However, most z-tables give the probability from the mean to our z-score (the left-hand side). But we want the probability that the apple's diameter is greater than 3.76 inches, i.e., the right-hand side of the distribution curve. As a result, we have to subtract the value we get from the z-table from 1 (because the cumulative probability under the entire normal curve is 1).
If a z-table is not readily available, there are many online calculators available to find this probability.
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swimsuit buyer reduced a group of designer swimwear from $75.00 to $50.00 for a special sale. If 40 swimsuits sold at the reduced price and the remaining 25 swimsuits were returned to the original price after the sale, calculate the total markdowns, markdown cancellations, and net markdown achieved.
Answer:
The original price is = $75
The reduced price = $50
So, price reduced is = [tex]75-50=25[/tex] dollars
Total swimsuits are = [tex]40+25=65[/tex]
Total markdown = [tex]65\times25=1625[/tex] dollars
Now, 25 swimsuits were returned to the original price. Means 25 swimsuits were returned to $75, increasing $25 again.
So, markdown cancellation = [tex]25\times25=625[/tex] dollars
Net markdown = total markdown - markdown cancellation
= [tex]1625-625=1000[/tex] dollars
Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct): 79, 64, 84, 82, 92, and 77. Instead of averaging the six scores, the instructor indicated he would randomly select two grades and compute the final percent correct based on the two percents. How many different samples, without replacement, of two test grades are possible
Answer:
15 samples
Step-by-step explanation:
The total sample space consists of 6 items
{79,64,84,82,92,77}
So,
n=6
The instructor has to randomly select 2 test scores out of 6.
So, r=6
The arrangement of scores selection doesn't matter so combinations will be used.
[tex]C(n,r)=\frac{n!}{r!(n-r)!} \\C(6,2)=\frac{6!}{2!(6-2)!}\\=\frac{6!}{2!*4!}\\=\frac{6*5*4!}{2!*4!} \\=\frac{30}{2}\\=15\ ways[/tex]
Therefore, there are 15 different samples are possible without replacement ..
At Southern States University (SSU) there are 399 students taking Finite Mathematics or Statistics. 238 are taking Finite Mathematics, 184 are taking Statistics, and 23 are taking both Finite Mathematics and Statistics. How many are taking Finite Mathematics but not Statistics?
Answer:
215
Step-by-step explanation:
The 238 taking Finite Math includes those taking Finite Math and Statistics. Subtracting out the 23 who are taking both leaves 215 taking Finite Math only.
215 students are taking Finite Mathematics but not Statistics.
To find out how many students are taking Finite Mathematics but not Statistics at Southern States University (SSU), let's break down the information given and use set theory concepts.
Total number of students taking either Finite Mathematics or Statistics: 399
Number of students taking Finite Mathematics: 238
Number of students taking Statistics: 184
Number of students taking both Finite Mathematics and Statistics: 23
First, we need to figure out how many students are taking only Finite Mathematics. We can do this by subtracting the number of students taking both Finite Mathematics and Statistics from the total number of students taking Finite Mathematics.
Number of students taking only Finite Mathematics = Total taking Finite Mathematics - Total taking both Finite Mathematics and Statistics
So,
Number of students taking only Finite Mathematics = 238 - 23
Number of students taking only Finite Mathematics = 215
A television network commissioned a telephone poll of randomly sampled men. Of the 708 respondents who had children, 19% said "yes" to the question "Are you a stay-at-home dad?" To help market commercial time, the network wants an accurate estimate of the true percentage of stay-at-home dads. Construct a 90% confidence interval. left parenthesis nothing % comma nothing % right parenthesis
Answer: (16.6%, 21.4%)
Step-by-step explanation:
The confidence interval for proportion is given by :-
[tex]p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]
Given : Sample size : n= 708
The proportion of respondents who had children = [tex]p=0.19[/tex]
Significance level : [tex]\alpha=1-0.90=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=z_{0.05}=\pm1.645[/tex]
Now, the 90% confidence interval for proportion will be :-
[tex]0.19\pm (1.645)\sqrt{\dfrac{0.19(1-0.19)}{708}}\approx0.19\pm 0.024\\\\=(0.19-0.024,0.19+0.024)=(0.166,\ 0.214)=(16.6\%,\ 21.4\%)[/tex]
Hence, the 90% confidence interval for the proportion = (16.6%, 21.4%)
A poll showed that 50.3% of Americans say they believe that some people see the future in their dreams. What is the probability of randomly selecting someone who does not believe that some people see the future in their dreams.
Answer:
The probability of randomly selecting someone who does not believe that some people see the future in their dreams =0.497.
Step-by-step explanation:
Given
Percent of Americans who Say they believe that some people see the future in their dreams=50.3%
Total percentages=100%
Therefore, Number of americans who say they believe that some people see the future in their dreams=50.3
The probability of randomly selecting someone who say they believe that some people see the future in their dreams =[tex]\frac{50.3}{100}[/tex]
Hence, the probability of randomly selecting someone who believe that some people see the future in their dreams, P(E)=0.503
Now, the probability of randomly selecting someone who does not believe that some people see the future in their dreams ,P(E')= 1-P(E)
The probability of randomly selecting someone who does not believe that some people see the future in their dreams =1-0.503
Hence,the probability of randomly selecting someone who does not believe that some people see the future in their dreams=0.497.
Answer: 0.497
Step-by-step explanation:
Let A be the event that Americans believe that some people see the future in their dreams.
Then , the probability that Americans believe that some people see the future in their dreams is given by :-
[tex]P(A)=50.3\%=0.503[/tex]
We know that the complement of a event X is given by :-
[tex]P(X')=1-P(X)[/tex]
Hence, the probability of randomly selecting someone who does not believe that some people see the future in their dreams is
[tex]P(A')=1-P(A)\\\\=1-0.503=0.497[/tex]
Dave is buying pizza and soda. Suppose that a slice of pizza andd a can of soda cost $3. Let E be the amount in dollars that Dave spends on pizza and soda. If Dave buys P slices of pizza and S cans of soda, which equation correctly describes the amount of money he spends? E=P+3S E=3P+S E=P+S E=3P+3S 3E=P+S 2. Now rearrange the equation you wrote above so that S is written in terms of E and P. Which of the following is correct? S=P+E S=(1/3)E-P S=E-3P S=3E-3P S=E-P
Answer:
E = 3P + 3S,
S=(1/3)E-P
Step-by-step explanation:
Given,
A slice of pizza and a can of soda cost $3,
That is, total cost of P slices of pizza and S cans of soda = ( 3P + 3S ) dollars,
According to the question,
Total cost of P slices of pizza and S cans of soda = E dollars,
⇒ 3P + 3S = E
Which is the required equation that correctly describes the amount of money spent,
For rearranging the equation in terms of E and P,
We need to isolate S,
Subtract 3P on both sides,
3S = E - 3P
Divide both sides by 3,
[tex]S=\frac{E-3P}{3}[/tex]
[tex]\implies S=(\frac{1}{3})E-P[/tex]
Which is the required equation.
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes.
Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-
[tex]f(x)=\dfrac{1}{5-0}=\dfrac{1}{5}=0.2\ \ ,\ 0<x<5[/tex]
The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
[tex]P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75[/tex]
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
The probability that a randomly selected passenger has a waiting time greater than 1.25 minutes is 1.
Explanation:To find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes, we need to find the area under the probability density function (PDF) curve for values greater than 1.25. Since the waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 minutes, the PDF is a rectangle with height 1/5 and base 5. The area of the rectangle represents the probability.
The probability of a waiting time greater than 1.25 minutes is the ratio of the area of the rectangle representing waiting times greater than 1.25 minutes to the total area of the rectangle representing all waiting times.
To calculate this probability, we first need to find the area of the rectangle representing waiting times greater than 1.25 minutes. Since the base of the rectangle is 5 minutes and the height is 1/5, the area is given by:
Area = base * height = 5 * (1/5) = 1
The total area of the rectangle representing all waiting times is the area of the entire rectangle, which is also equal to:
Area = base * height = 5 * (1/5) = 1
Therefore, the probability of a randomly selected passenger having a waiting time greater than 1.25 minutes is:
Probability = Area of waiting times greater than 1.25 minutes / Total area = 1/1 = 1
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The distance between my house and Memphis is 150 km. If I drive in my car, it takes me 72 min to make the first 100 km of the drive. If I keep traveling at the same rate, how many more minutes will it take to drive the remaining distance? Round your answer to the nearest tenth.
Set up a ratio:
You drove 72 minutes and 100 km = 72/100
You want the number of minutes (x) to drive 150 km = x/150
Set the ratios to equal each other and solve for x:
72/100 = x/150
Cross multiply:
(72 * 150) = 100 * x)
Simplify:
10,800/100x
Divide both sides by 100:
x = 10800/100 = 108
This means it would take 108 minutes to drive 150 km.
Now subtract the time you have already driven to fin how much more you need:
180 - 72 = 36 more minutes.
Answer:
36 min
Step-by-step explanation:
It takes 72 min to drive 100 km. 50 km are left to drive.
Half of the driving above is: It takes 36 min to drive 50 km.
Compute the entry (the number in the second row and second column) of the product matrix resulting from the following multiplication:
[1 2] [9 6]
[3 4] [5 7]
Answer:
The entry on the second row and second column of the product matrix is [tex]c_{22} = 40[/tex].
Step-by-step explanation:
Let's define as A and B the given matrixes:
[tex]A = \left[\begin{array}{cc}1&2\\3&4\end{array}\right][/tex]
[tex]B = \left[\begin{array}{cc}9&6\\5&7\end{array}\right][/tex]
The product matrix C entry in the first row and first column [tex]c_{1,1}[/tex] or [tex]c_{11}[/tex] can be computer multiplying first row of A by first column of B (see example attached).
The product matrix C entry in the first row and second column [tex]c_{1,2}[/tex] or [tex]c_{12}[/tex] can be computer multiplying first row of A by second column of B.
The product matrix C entry in the second row and first column [tex]c_{2,1}[/tex] or [tex]c_{21}[/tex] can be computer multiplying second row of A by first column of B.
The product matrix C entry in the second row and second column [tex]c_{2,2}[/tex] or [tex]c_{22}[/tex] can be computer multiplying second row of A by second column of B.
Then, let's compute [tex]c_{22}[/tex] by doing the dot product between [3 4] and [6 7]...
[tex]c_{22} = [3 4] . [6 7] = 3*4 + 4*7 = 12 + 28 = 40[/tex]
Find the interest rate on a loan charging $855 simple interest on a principal of $3750 after 6 years.
Answer:
3.8%
Step-by-step explanation:
Simple interest formula is I=P*R*T
where:
I=interest
P=principal
R=rate
T=time.
So let's plug in our information we are given:
I=855
P=3750
T=6
R=?.
The equation becomes 855=3750*R*6.
Multiplication is commutative so we could write this as 3750*6*R=855.
After the multiplication of 3750 and 6 we obtain 22500*R=855.
Now we just divide both sides by 22500. This will give us:
R=855/22500 which when entered into the calculator as 855 division sign 22500 gives us 0.038 or 3.8%.
Let X be a continuous random variable with a uniform distribution on the interval open square brackets 0 comma 10 close square brackets. Find straight P open parentheses straight X less than 1 space or space straight X greater than 8 close parentheses.
[tex]X[/tex] has PDF
[tex]f_X(x)=\begin{cases}\frac1{10}&\text{for }0\le x\le10\\0&\text{otherwise}\end{cases}[/tex]
and thus CDF
[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\\frac x{10}&\text{for }0\le x\le10\\1&\text{for }x>10\end{cases}[/tex]
Because [tex]X[/tex] is continuous, we have
[tex]P(X<1\text{ or }X>8)=1-P(1\le X\le8)=1-(P(X\le8)-P(X\le1))[/tex]
[tex]P(X<1\text{ or }X>8)=1-P(X\le 8)+P(X\le1)[/tex]
[tex]P(X<1\text{ or }X>8)=1-F_X(8)+F_X(1)[/tex]
[tex]P(X<1\text{ or }X>8)=1-\dfrac8{10}+\dfrac1{10}=\boxed{\dfrac3{10}}[/tex]
John has won the mega-bucks lottery, which pays $1, 000, 000. Suppose he deposits the money in a savings account that pays an annual interest of 8% compounded continuously. How long will this money last if he makes annual withdrawals of $100, 000?
Answer:20.91
Step-by-step explanation:
Given
Principal amount invested=[tex]\$ 1,000,000[/tex]
Rate of interest=8%
Annual Withdrawl=[tex]\$ 100,000[/tex]
compound interest is given by
A=[tex]\left (1+ \frac{r}{100}\right )^t[/tex]
Therefore reamining Amount after certain years
Net money will become zero after t year
[tex]1,000,000\left (1+ \frac{8}{100} \right )^t - 100,000\left ( \frac{\left ( 1.08\right )^{t}-1}{0.08}\right )[/tex]=0
[tex]0.8\left ( 1.08\right )^t=\left ( 1.08\right )^{t}-1[/tex]
t=20.91 years
Find a second independent solution y1=x xy"-xy'+y=0
We can use reduction of order. Given that [tex]y_1(x)=x[/tex] is a known solution, we look for a solution of the form [tex]y_2(x)=v(x)y_1(x)[/tex]. It has derivatives [tex]{y_2}'=v'y_1+v{y_1}'[/tex] and [tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]. Substituting these into the ODE gives
[tex]x(xv''+2v')-x(xv'+v)+xv=0[/tex]
[tex]x^2v''+(2x-x^2)v'=0[/tex]
Let [tex]w(x)=v'(x)[/tex] so that [tex]w'(x)=v''(x)[/tex] and we get an ODE linear in [tex]w[/tex]:
[tex]x^2w'+(2x-x^2)w=0[/tex]
Divide both sides by [tex]e^x[/tex]:
[tex]x^2e^{-x}w'+(2x-x^2)e^{-x}w=0/tex]
Since [tex](x^2e^{-x})=(2x-x^2)e^{-x}[/tex], we can condense the left side as the derivative of a product:
[tex](x^2e^{-x}w)'=0[/tex]
Integrate both sides and solve for [tex]w(x)[/tex]:
[tex]x^2e^{-x}w=C\implies w=\dfrac{Ce^x}{x^2}[/tex]
Integrate both sides again to solve for [tex]v(x)[/tex]. Unfortunately, there is no closed form for the integral of the right side, but we can leave the result in the form of a definite integral:
[tex]v=\displaystyle C_2+C_1\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]
where [tex]x_0[/tex] is any point on an interval over which a solution to the ODE exists.
Finally, multiply by [tex]y_1(x)[/tex] to solve for [tex]y_2(x)[/tex]:
[tex]y_2=\displaystyle C_2x+C_1x\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]
[tex]y_1(x)[/tex] already accounts for the [tex]C_2x[/tex] term above, so the second independent solution is
[tex]y_2=x\displaystyle\int_{x_0}^x\frac{e^t}{t^2}\,\mathrm dt[/tex]
A survey of 400 randomly selected high school students determined that 68 play organized sports. (a) What is the probability that a randomly selected high school student plays organized sports? (b) Interpret this probability.
Answer: a) The probability that a randomly selected high school student plays organized sports = 0.17
b) The randomly selected high school student is unlikely to play organized sports.
Step-by-step explanation:
Given : A survey of 400 randomly selected high school students determined that 68 play organized sports.
Then , the probability that a randomly selected high school student plays organized sports is given by :-
[tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{68}{400}=0.17[/tex]
In percent , the probability that a randomly selected high school student plays organized sports is 17%.
since 17% lies in the interval of unlikely events (0%, 25%).
It means that the randomly selected high school student is unlikely to play organized sports.
The probability that a randomly selected high school student plays organized sports is 0.17, which means there is a 17% chance, or a 1 in 6 chance, that a randomly picked student plays organized sports.
Explanation:This question relates to the field of Statistics, particularly to the concept of Probability. In this problem, we are provided with a total number of High School students (400), and a number of students who play organized sports (68).
To find the probability, you would divide the number of students who play sports by the total number of students. That is, Probability = Number with desired characteristic (play sports) / Total number In this case, it would be 68 / 400 = 0.17
Interpreting the probability means describing what it represents in practical terms. Here, a probability of 0.17 means that if you were to randomly select a high school student, there is a 17% chance, or around 1 in 6 chance, that this student plays organized sports.
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Find the slope and the y -intercept of the line.
Write your answers in simplest form.
-7x + y = 1
Answer:
The slope is: 7
The y-intercept is: 1
Step-by-step explanation:
The equation of the line in Slope-Intercept form is:
[tex]y=mx+b[/tex]
Where "m" is the slope of the line and "b" is the y-intercept.
To find the slope and the y-intercept of the given line, we can write it in Slope-Intercept form. We can do this by solving for "y".
Then, this is:
[tex]-7x + y = 1\\\\y=7x+1[/tex]
Therefore, you can identify that the slope of this line is:
[tex]m=7[/tex]
And the y-intercept is:
[tex]b=1[/tex]
A student takes an exam containing 18 multiple choice questions. The probability of choosing a correct answer by knowledgeable guessing is 0.3. If the student makes knowledgeable guesses, what is the probability that he will get between 8 and 12 (both inclusive) questions right? Round your answer to four decimal places.
When you have to repeatedly take the same test, with constant probability of succeeding/failing, you have to use Bernoulli's distribution. It states that, if you take [tex]n[/tex] tests with "succeeding" probability [tex]p[/tex], and you want to "succeed" k of those n times, the probability is
[tex]\displaystyle P(n,k,p) = \binom{n}{k}p^k(1-p)^{n-k}[/tex]
In your case, you have n=18 (the number of tests), and p=0.3 (the probability of succeeding). We want to succeed between 8 and 12 times, which means choosing k=8,9,10,11, or 12. For example, the probability of succeeding 8 times is
[tex]\displaystyle P(18,8,0.3) = \binom{18}{8}(0.3)^8(0.7)^{10}[/tex]
you can plug the different values of k to get the probabilities of succeeding 9, 10, 11 and 12 times, and your final answer will be
[tex]P = P(18,8,0.3) + P(18,9,0.3) + P(18,10,0.3) + P(18,11,0.3) + P(18,12,0.3)[/tex]
Final answer:
In mathematics, calculate the probability of a student guessing between 8 and 12 questions right out of 18 when the probability of guessing correctly is 0.3.
Explanation:
To calculate the probability of getting between 8 and 12 questions right through knowledgeable guessing, we can use the binomial probability formula:
P(k successes in n trials) = nCk × [tex]p^k[/tex] × (1-[tex]p^{(n-k)[/tex])
where:
k = number of successes (correct answers) = 8 to 12
n = total number of questions = 18
p = probability of success (correct answer by guessing) = 0.3
(1-p) = probability of failure (incorrect answer) = 0.7
We need to calculate the probability for each possible number of correct answers between 8 and 12 and then sum them up.
Here's the calculation:
Probability of getting 8 correct:
P(8 successes in 18 trials) = 18C8 ×0.3⁸ × 0.7¹⁰ ≈ 0.076
Probability of getting 9 correct:
P(9 successes in 18 trials) = 18C9 × 0.3⁹ ×0.7⁹ ≈ 0.124
Probability of getting 10 correct:
P(10 successes in 18 trials) = 18C10 ×0.3¹⁰ ×0.7⁸ ≈ 0.122
Probability of getting 11 correct:
P(11 successes in 18 trials) = 18C11 × 0.3¹¹ × 0.7⁷ ≈ 0.072
Probability of getting 12 correct:
P(12 successes in 18 trials) = 18C12 × 0.3¹² × 0.7⁶ ≈ 0.021
Total probability:
P(8 to 12 correct) = 0.076 + 0.124 + 0.122 + 0.072 + 0.021 ≈ 0.415
Therefore, the probability that the student gets between 8 and 12 questions right by knowledgeable guessing is approximately 41.5%.
(show the supposition, proof and conclusion)
Use proof by contradiction to show that If a and b are rational numbers with b ≠ 0 and x is an irrational number, then a + bx is irrational.
Answer:
Step-by-step explanation:
We are given that a and b are rational numbers where [tex]b\neq0[/tex] and x is irrational number .
We have to prove a+bx is irrational number by contradiction.
Supposition:let a+bx is a rational number then it can be written in [tex]\frac{p}{q}[/tex] form
[tex]a+bx=\frac{p}{q}[/tex] where [tex]q\neq0[/tex] where p and q are integers.
Proof:[tex]a+bx=\frac{p}{q}[/tex]
After dividing p and q by common factor except 1 then we get
[tex]a+bx=\frac{r}{s}[/tex]
r and s are coprime therefore, there is no common factor of r and s except 1.
[tex]a+bx=\frac{r}{s}[/tex] where r and s are integers.
[tex]bx=\frac{r}{s}-a[/tex]
[tex]x=\frac{\frac{r}{s}-a}{b}[/tex]
When we subtract one rational from other rational number then we get again a rational number and we divide one rational by other rational number then we get quotient number which is also rational.
Therefore, the number on the right hand of equal to is rational number but x is a irrational number .A rational number is not equal to an irrational number .Therefore, it is contradict by taking a+bx is a rational number .Hence, a+bx is an irrational number.
Conclusion: a+bx is an irrational number.
10 companies sponsor a tournament and decided to give M rupees collectively. Two companies dropped and remaining agreed on paying their share equally. What was the increase in share of each company ?
a) M-20/2 b) M/50 c) M/40 d) M/2 e) 2M
Answer:
c) M/40
Step-by-step explanation:
The share of each company was originally M/10. Now it is M/8. The increase is ...
M/8 -M/10 = M(1/8 -1/10) = M(10-8)/(10·8) = M/40
The increase in the share of each company after two companies dropped out is M/40.
Explanation:The total amount of money to be paid by the companies is M rupees. Initially, these M rupees were to be divided among 10 companies, but when two dropped out, they had to be divided among 8 companies instead. So, the share of each company initially was M/10 and after the drop out, the share of each company became M/8.
The increase in the amount that each company had to give is given by subtraction of the initial share from the final share, that is, (M/8) - (M/10). To solve this equation, we first must find a common denominator for the fractions, which is 40 in this case. Therefore, the equation becomes (5M/40) - (4M/40) = M/40.
So, the increase in share of each company was M/40. Therefore, the correct answer is (c) M/40.
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Let S={1,2,3,4,5,6}.
How many subsets of cardinality 4 contain at least one odd number?
Answer:
15 subsets of cardinality 4 contain at least one odd number.
Step-by-step explanation:
Here the given set,
S={1,2,3,4,5,6},
Since, a set having cardinality 4 having 4 elements,
The number of odd digits = 3 ( 1, 3, 5 )
And, the number of even digits = 3 ( 2, 4, 6 )
Thus, the total possible arrangement of a set having 4 elements out of which atleast one odd number = [tex]^3C_1\times ^3C_3+^3C_2\times ^3C_2+^3C_3\times ^3C_1[/tex]
By using [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex],
[tex]=3\times 1+3\times 3+1\times 3[/tex]
[tex]=3+9+3[/tex]
[tex]=15[/tex]
Hence, 15 subsets of cardinality 4 contain at least one odd number.
Continuing the previous problem, use the data points (1950, 0.75) and (1997, 5.15) to find the slope. Show all work necessary for your calculations. If necessary, round your answer to the hundredths place. What does the slope represent in the context of the problem?
Answer:
Slope is 0.094,
It represents the average rate of change.
Step-by-step explanation:
Since, the slope is the ratio of difference in y-coordinates and the difference in x-coordinates,
Also, in a order pair, first element shows the x-coordinate and second element shows the y-coordinate.
Here, the data points are (1950, 0.75) and (1997, 5.15),
Thus, the slope is,
[tex]m=\frac{5.15-0.75}{1997-1950}[/tex]
[tex]=\frac{4.4}{47}[/tex]
[tex]=0.0936170212766[/tex]
[tex]\approx 0.094[/tex]
Also, Slope represents the average rate of change.
Use the transforms in section 4.1 to find the Laplace transform of the function. t^3/2 - e^-10t
Answer:
Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t})[/tex]=[tex]\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}[/tex]
Step-by-step explanation:
Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t} )[/tex]
[tex]L(t^\frac{3}{2})=\int_{0 }^{\infty}t^\frac{3}{2}e^{-st}dt\\substitute \ u =st\\L(t^\frac{3}{2})=\int_{0 }^{\infty}\frac{u}{s} ^\frac{3}{2}e^{-u}\frac{du}{s}=\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}[/tex]
the integral is now in gamma function form
[tex]\frac{1}{s^{\frac{5}{2}}}\int_{0 }^{\infty}{u} ^\frac{3}{2}e^{-u}{du}=\frac{1}{s^{\frac{5}{2}}}\Gamma(\frac{5}{2})=\frac{1}{s^{\frac{5}{2}}}\times\frac{3}{2}\times\frac{1}{2} }\Gamma (\frac{1}{2} )=\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }[/tex]
now laplace of [tex]L(e^{-10t})[/tex]
[tex]L(e^{-10t})=\frac{1}{s+10}[/tex]
hence
Laplace transformation of [tex]L(t^\frac{3}{2}-e^{-10t})[/tex]=[tex]\frac{3\sqrt{\pi} }{4 s^\frac{5}{2} }\ -\frac{1}{s+10}[/tex]
The Laplace transform of the function [tex]t^3/2 - e^-10t[/tex] is (3/2√π)/[tex]s^5/2 - 1/(s + 10)[/tex].
To find the Laplace transform of the function[tex]t^3/2 - e-10t[/tex], we use the Laplace transform properties and tables.
Refer to the Laplace transform table: Apply to each term: Combine the results: The Laplace transform of the function is L{[tex]t^3/2 - e-10t[/tex]} = (3/2√π)/[tex]s^5/2 - 1/(s + 10)[/tex].From a shipment of 65 transistors, 6 of which are defective, a sample of 5 transistors is selected at random.
(a) In how many different ways can the sample be selected?
________ways
(b) How many samples contain exactly 3 defective transistors?
________samples
(c) How many samples do not contain any defective transistors?
________ samples
Answer:
a) 8259888
b) 34220
c) 45057474
Step-by-step explanation:
Given,
The total number of transistor = 65,
In which, the defective transistor = 6,
So, the number of non defective transistor = 65 - 6 = 59,
Since, out of these transistor 5 are selected,
a) Thus, the number of ways = the total possible combination of 5 transistors = [tex]{65}C_ 5[/tex]
[tex]=\frac{65!}{(65-5)!5!}[/tex]
[tex]=8259888[/tex]
b) The number of samples that contains exactly 3 defective transistors = the possible combination of exactly 3 defective transistors = [tex]{6}C_3\times {59}C_2[/tex]
[tex]=\frac{6!}{(6-3)!3!}\times \frac{59!}{(59-2)!\times 2!}[/tex]
[tex]=20\times 1711[/tex]
[tex]=34220[/tex]
c) The number of sample without any defective transistor = The possible combination of 0 defective transistor = [tex]^6C_0\times ^{59}C_5[/tex]
[tex]=1\times 45057474[/tex]
[tex]=45057474[/tex]
(a) 8259888 ways, (b) 34220 samples and (c) 5006386 samples.
Let's solve each part:
(a) There are 65 total transistors, and we want to select 5 of them. So, the number of ways to select a sample is the number of combinations of 65 things taken 5 at a time. This can be calculated using the combination formula:
Number of combinations = [tex]C(n, k) = \frac{n!}{ k! \times (n-k)!)}[/tex]
where:
n is the total number of items (65 transistors)
k is the number of items to select (5 transistors)
Plugging in the values:
[tex]C(65, 5) = \frac{65!}{ 5! \times (65-5)!)} = 8259888[/tex]
(b) We want samples that contain exactly 3 defective transistors. We can achieve this by selecting 3 defective transistors from the 6 available and 2 non-defective transistors from the remaining 59 (65 total - 6 defective).
So, the number of ways to select this specific scenario is:
Number of ways to choose 3 defective transistors out of 6: C(6, 3)
Number of ways to choose 2 non-defective transistors out of 59: C(59, 2)
Using the combination formula again:
[tex]C(6, 3) \times C(59, 2) = (\frac{6!}{ (3! \times 3!)}) \times (\frac{59!} { (2! \times 57!)}) = 34220[/tex]
(c) Samples with no defective transistors simply means selecting all 5 transistors from the 59 non-defective ones.
Therefore:
[tex]C(59, 5) = \frac{59!}{ (5! \times 54!)} = 5006386[/tex]
Find an equation for a circle satisfying the given conditions. (a) Center (-1, 4), passes through (3, 7) (b) The points (7, 13) and (-3, -11) are at the ends of a diameter.
Answer:
Step-by-step explanation:
In order to find the equations we need the circle's general equation:
[tex](x-h)^{2}+(y-k)^{2}=r^2[/tex] where:
(h,k) is the center and 'r' is the radius.
A. Because the center is (-1,4) then h=-1 and k=4.
Now we can find the radius as:
[tex]distance=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}[/tex]
[tex]distance=\sqrt{(3-(-1))^{2}+(7-4)^{2}}[/tex]
[tex]distance=5[/tex] so we have r=5
Then the equation is [tex](x+1)^{2}+(y-4)^{2}=25[/tex]
B. Because we have two points defining a diameter we can find the radius as follows:
[tex]diameter=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}[/tex]
[tex]diameter=\sqrt{(7-(-3))^{2}+(13-(-11))^{2}}[/tex]
[tex]diameter=26[/tex]
[tex]radius=26/2=13[/tex]
Now let's find the center of the circle as follows:
[tex]C=(\frac{x1+x2}{2} , \frac{y1+y2}{2})[/tex]
[tex]C=(\frac{7-3}{2} , \frac{13-11}{2})[/tex]
[tex]C=(2,1)[/tex]
Then the equation is [tex](x-2)^{2}+(y-1)^{2}=169[/tex]