Answer:
a) The percentage of adults who smoke are decreasing with time. b) the equation that best described this data is y=-0.3364x+22.809 (R^2=0.859) in which y is the percentage of adults who smoke and x the number of years. c) the percentage of adults who smoke will be 19.8% and it will not meet the expected 12%, it would take 32 years to reach that value.
Step-by-step explanation:
The data can be plotted to which years is the independent variable and percentage of adults who smoke is the dependent variable. The linear trendline that described this data has a negative slope which indicates that the percentage of adults is decreasing with time. In order to determine if the OSH target is being met, the x is replaced by 9 which is the goal period of nine years. The y is 19% which is higher than the 12% goal. In order to know the period it will take to the reach the goal of 12%, the y is replaced by 12 in the curve and the x is the answer in years = 32 years.
You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of each color. Draw a marble out of the jar. If it's red, put it back in the jar, and add three red marbles to the jar from the supply of extras. If it's blue, put it back into the jar, and add five blue marbles to the jar from the supply of extras. Do this two more times. Now, pull a marble from the jar, at random. What's the probability that this last marble is red? What's the probability that we actually drew the same marble all four times?
Answer:
There is a 57.68% probability that this last marble is red.
There is a 20.78% probability that we actually drew the same marble all four times.
Step-by-step explanation:
Initially, there are 50 marbles, of which:
30 are red
20 are blue
Any time a red marble is drawn:
The marble is placed back, and another three red marbles are added
Any time a blue marble is drawn
The marble is placed back, and another five blue marbles are added.
The first three marbles can have the following combinations:
R - R - R
R - R - B
R - B - R
R - B - B
B - R - R
B - R - B
B - B - R
B - B - B
Now, for each case, we have to find the probability that the last marble is red. So
[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}[/tex]
[tex]P_{1}[/tex] is the probability that we go R - R - R - R
There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is [tex]\frac{30}{50} = \frac{3}{5}[/tex].
Now the red marble is returned to the bag, and another 3 red marbles are added.
Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is [tex]\frac{33}{53}[/tex]
Again, the red marble is returned to the bag, and another 3 red marbles are added
Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is [tex]\frac{36}{56}[/tex]
Again, the red marble sorted is returned, and another 3 are added.
Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is [tex]\frac{39}{59}[/tex]. So
[tex]P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588[/tex]
[tex]P_{2}[/tex] is the probability that we go R - R - B - R
[tex]P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788[/tex]
[tex]P_{3}[/tex] is the probability that we go R - B - R - R
[tex]P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076[/tex]
[tex]P_{4}[/tex] is the probability that we go R - B - B - R
[tex]P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511[/tex]
[tex]P_{5}[/tex] is the probability that we go B - R - R - R
[tex]P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733[/tex]
[tex]P_{6}[/tex] is the probability that we go B - R - B - R
[tex]P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493[/tex]
[tex]P_{7}[/tex] is the probability that we go B - B - R - R
[tex]P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476[/tex]
[tex]P_{8}[/tex] is the probability that we go B - B - B - R
[tex]P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419[/tex]
So, the probability that this last marble is red is:
[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768[/tex]
There is a 57.68% probability that this last marble is red.
What's the probability that we actually drew the same marble all four times?
[tex]P = P_{1} + P_{2}[/tex]
[tex]P_{1}[/tex] is the probability that we go R-R-R-R. It is the same [tex]P_{1}[/tex] from the previous item(the last marble being red). So [tex]P_{1} = 0.1588[/tex]
[tex]P_{2}[/tex] is the probability that we go B-B-B-B. It is almost the same as [tex]P_{8}[/tex] in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.
[tex]P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490[/tex]
[tex]P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078[/tex]
There is a 20.78% probability that we actually drew the same marble all four times
Find the solution of the given initial value problem:
(a) y' + 2y = te^{-2t}, y(1) = 0
(b) t^{3}y' + 4t^{2}y = e^{-t}, y(-1) = 0
Answer:
[tex](a)\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
[tex](b)\ y(t)=\ (1-t)e^{-t}\ -\ 2e[/tex]
Step-by-step explanation:
(a) [tex]y'\ +\ 2y\ =\ te^{-2t},\ y(1)\ =\ 0[/tex]
[tex]=>\ (D+2)y\ =\ te^{-2t}[/tex]
To find the complementary function
D+2 = 0
=> D = -2
So, the complementary function can by given by
[tex]y_c(t)\ =\ C.e^{-2t}[/tex]
Now, to find particular integral
[tex](D+2)y_p(t)\ =\ te^{-2t}[/tex]
[tex]=>y_p(t)\ =\ \dfrac{ te^{-2t}}{D+2}[/tex]
[tex]=\ \dfrac{ te^{-2t}}{-2+2}[/tex]
= not defined
So,
[tex]y_p(t)\ =\ \dfrac{ t^2e^{-2t}}{D^2}[/tex]
[tex]=\ \dfrac{t^2e^{-2t}}{(-2)^2}[/tex]
[tex]=\ \dfrac{t^2e^{-2t}}{4}[/tex]
So, complete solution can be given by
[tex]y(t)\ =\ y_c(t)\ +\ y_p(t)[/tex]
[tex]=> y(t) =\ C.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
As given in question
[tex]=>\ y(1)\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]
[tex]=>\ 0\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]
[tex]=>\ C\ =\ 4e^2[/tex]
Hence, the complete solution can be give by
[tex]=>\ y(t) =\ 4e^2.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
[tex]=>\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
(b) [tex]t^{3}y'\ +\ 4t^{2}y\ =\ e^{-t},\ y(-1)\ =\ 0[/tex]
[tex]=>\ y'\ +\ 4t^{-1}y\ =\ t^{-3}e^{-t}[/tex]
Integrating factor can be given by
[tex]I.F\ =\ e^{\int (4t^{-1})dt}[/tex]
[tex]=\ e^{log\ t^4}[/tex]
[tex]=\ t^4[/tex]
Now , the solution of the given differential equation can be given by
[tex]y(t)\times t^4\ =\ \int t^{-3}e^{-t}t^4dt\ +\ C[/tex]
[tex]=>\ y(t)\ =\ \int t.e^{-t}dt\ +\ C[/tex]
[tex]=\ (1-t)e^{-t}\ +\ C[/tex]
According to question
[tex]y(-1)\ =\ (1-(-1))e^1\ +\ C[/tex]
[tex]=>\ 0\ =\ 2e\ +\ C[/tex]
[tex]=>\ C\ =\ -2e[/tex]
Now, the complete solution of the given differential equation cab be given by
[tex]y(t)\ =\ (1-t)e^{-t}\ -\ 2e[/tex]
Answer:
a. [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]
b.[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]
Step-by-step explanation:
We are given that
a.[tex]y'+2y=te^{-2t},y(1)=0[/tex]
Compare with [tex]y'+P(t)y=Q(t)[/tex]
We have P(t)=2,Q(t)=[tex]te^{-2t}[/tex]
Integration factor=[tex]\int e^{2dt}=e^{2t}[/tex]
[tex]y\cdot I.F=\int Q(t)\cdot I.F dt+C[/tex]
Substitute the values then, we get
[tex]y\cdot e^{2t}=\int te^{-2t}\cdot e^{2t} dt+C[/tex]
[tex]y\cdot e^{2t}=\int tdt+C[/tex]
[tex]ye^{2t}=\frac{t^2}{2}+C[/tex]
Substitute the values x=1 and y=0
Then, we get [tex]0\cdot e^2=\frac{1}{2}+C[/tex]
[tex]C=-\frac{1}{2}[/tex]
Substitute the value in the given function
[tex]ye^{2t}=\frac{t^2}{2}-\frac{1}{2}[/tex]
[tex]y=\frac{t^2}{2}e^{-2t}-\frac{1}{2}e^{-2t}[/tex]
Hence, [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]
b.[tex]t^3y'+4t^2y=e^{-t},y(-1)=0[/tex]
[tex]y'+\frac{4}{t}y=\frac{e^{-t}}{t^3}[/tex]
[tex]P(t)=\frac{4}{t},Q(t)=\frac{e^{-t}}{t^3}[/tex]
I.F=[tex]\int e^{\frac{4}{t}dt}=e^{4lnt}=e^{lnt^4}=t^4[/tex]
[tex]y\cdot \frac{t^4}=\int e^{-t}\frac{t^4}{t^3} dt+C[/tex]
[tex]y\cdot t^4=\int te^{-t}dt+C[/tex]
[tex]yt^4=-te^{-t}+\int e^{-t} dt+C[/tex]
[tex]u\cdot v dt=u\int vdt-\int (\frac{du}{dt}\cdot \int vdt)dt[/tex]
[tex]yt^4=-te^{-t}-e^{-t}+C[/tex]
Substitute the values x=-1,y=0 then, we get
[tex]0=-(-1)e-e+C[/tex]
[tex]C+e-e=0[/tex]
C=0
Substitute the value of C then we get
[tex]yt^4=-te^{-t}-e^{-t}[/tex]
[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]
use cramers Rule to solve the following system:
5x - 3y + z = 6
2y - 3z = 11
7x + 10y = -13
Answer:
The solution to the system is [tex]x=1[/tex],[tex]y=-2[/tex] and [tex]z=-5[/tex]
Step-by-step explanation:
Cramer's rule defines the solution of a system of equations in the following way:
[tex]x= \frac{D_x}{D}[/tex], [tex]y= \frac{D_y}{D}[/tex] and [tex]z= \frac{D_z}{D}[/tex] where [tex]D_x[/tex], [tex]D_y[/tex] and [tex]D_z[/tex] are the determinants formed by replacing the x,y and z-column values with the answer-column values respectively. [tex]D[/tex] is the determinant of the system. Let's see how this rule applies to this system.
The system can be written in matrix form like:
[tex]\left[\begin{array}{ccc}5&-3&1\\0&2&-3\\7&10&0\end{array}\right]\times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}6&11&-13\end{array}\right][/tex]
Then each of the previous determinants are given by:
[tex]D_x = \left|\begin{array}{ccc}6&-3&1\\11&2&-3\\-13&10&0\end{array}\right|=199[/tex] Notice how the x-column has been substituted with the answer-column one.
[tex]D_y = \left|\begin{array}{ccc}5&6&1\\0&11&-3\\7&-13&0\end{array}\right|=-398[/tex] Notice how the y-column has been substituted with the answer-column one.
[tex]D_z = \left|\begin{array}{ccc}5&-3&6\\0&2&11\\7&10&-13\end{array}\right|=-995[/tex]
Then, substituting the values:
[tex]x= \frac{D_x}{D}=\frac{199}{199}\\ x=1[/tex]
[tex]x= \frac{D_y}{D}=\frac{-398}{199}\\ y=-2[/tex]
[tex]x= \frac{D_z}{D}=\frac{-995}{199}\\ x=-5[/tex]
One milligram of streptomycin sulfate contains the antibiotic activity of 650 μg of streptomycin base. How many grams streptomycin sulfate would be the equivalent of 1 g of streptomycin base?
Answer:
1.538 g of streptomycin sulfate
Step-by-step explanation:
As we know, we have 650 μg of streptomycin base in 1 milligram of streptomycin sulfate.
If we convert everithing to grams:
650 μg= 0.00065 g of Streptomycin base for every 0.001 grams of Streptomycin Sulfate so we have :0.001 grs Streptomycin Sulfate/0.00065 gr Streptomycin base=1.538 gr Streptomycin Sulfate/Streptomycin base
Now if we want 1 gram of Streptomycin base we will need:
1 g of Streptomycin base*1.538 gr Streptomycin Sulfate/Streptomycin base= 1.538 gr Streptomycin Sulfate
At a restaurant a menu has 5 salads and 6 entrees. How many ways can you order a dinner that contains 1 salad and 1 entree?
Answer: 30
Step-by-step explanation:
The combination of n things taking r at a time is given by :-
[tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
Given : At a restaurant a menu has 5 salads and 6 entrees.
Then, the number of ways you can order a dinner that contains 1 salad and 1 entree will be :_
[tex]^5C_1\times ^6C_1\\\\=5\times6=30[/tex]
Hence, the number of ways you can order a dinner that contains 1 salad and 1 entree = 30
The slope f′(x) at each point (x,y) on a curve y=f(x) is given, along with a point (a,b) on the curve. Use this information to find f(x). f′(x) = 4x/(1 + 7x^2) (0,10) NOTE: OF absolute value symbols, | | , are needed for the answer, then use abs(expression). For example, ln|x| must be entered as ln(abs(x))
[tex]f'(x)=\dfrac{4x}{1+7x^2}[/tex]
Integrating gives
[tex]f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx[/tex]
To compute the integral, substitute [tex]u=1+7x^2[/tex], so that [tex]\frac27\,\mathrm du=4x\,\mathrm dx[/tex]. Then
[tex]f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C[/tex]
Since [tex]u=1+7x^2>0[/tex] for all [tex]x[/tex], we can drop the absolute value, so we end up with
[tex]f(x)=\dfrac27\ln(1+7x^2)+C[/tex]
Given that [tex]f(0)=10[/tex], we have
[tex]10=\dfrac27\ln1+C\implies C=10[/tex]
so that
[tex]\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}[/tex]
To get to school you can travel by car, bus or bicycle. If you travel by car, there is a 50% chance you will be late because the roads are very busy. If you travel by bus, which uses special reserved lanes and the busway, the probability of being late is only 20%. If you travel by bicycle you are only late 1% of the time.
(a) Suppose that you are late one day to class. Since your teacher does not know which mode of transportation you usually use, he assumes each of the three possibilities are equally likely. If you are late, find the probability that you travelled to school that day by car?
(b) Suppose that a friend tells your teacher that you almost always ride your bicycle to school, never take the bus, but 10% of the time travel by car. If you are late, what is the new probability that you travelled to school that day by car?
Answer:
(a) 0.704
(b) 0.8475
Step-by-step explanation:
(a) Let 'A' be the event that you travel by car and late
Let 'B' be the event that you travel by bus and late
Let 'C' be the event that you travel by Bicycle and late
Then, P (A) = 50% = [tex]\frac{50}{100}[/tex] = [tex]\frac{1}{2}[/tex]
P (B) = 20% = [tex]\frac{20}{100}[/tex] = [tex]\frac{1}{5}[/tex]
P (C) = 1% = [tex]\frac{1}{100}[/tex] = [tex]\frac{1}{100}[/tex]
A₁ = Student travels by car
B₁ = Student travels by bus
C₁ = Student travels by bicycle
Then according to teacher P(A₁) = [tex]\frac{1}{3}[/tex], P(B₁) = [tex]\frac{1}{3}[/tex], P(C₁) = [tex]\frac{1}{3}[/tex]
Now we have to find "Student is already late and traveled to school that day by car." which will be given as [tex]P(\frac{A}{L})[/tex]
where L : student is late
By using Bay's Theorem :
[tex]P(\frac{A}{L})[/tex] = [tex]\frac{P(A)\times P(A_1)}{P(A)\times P(A_1)+P(B)\times P(B_1)+P(C)\times P(C_1)}[/tex]
= [tex]\frac{\frac{1}{2}\times \frac{1}{3}}{\frac{1}{2}\times \frac{1}{3}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{100}\times \frac{1}{3}}[/tex]
= [tex]\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{15}+\frac{1}{300}}[/tex]
= [tex]\frac{\frac{1}{6}}{\frac{50+20+1}{300}}[/tex]
= [tex]\frac{1}{6}\times \frac{300}{71}[/tex]
= [tex](\frac{50}{71})[/tex]
= 0.704
(b) Here P(A₁) = [tex](\frac{10}{100})[/tex]
P(C₁) = [tex](\frac{90}{100})[/tex]
[tex]P(\frac{A}{L})[/tex] = We have to find and known student is late and traveled by car.
[tex]P(\frac{A}{L})[/tex] = [tex]\frac{P(A)\times P(A_1)}{P(A)\times P(A_1)+(P(C)\times P(C_1)}[/tex]
= [tex]\frac{\frac{1}{2}\times \frac{1}{10}}{\frac{1}{2}\times \frac{1}{10}+\frac{1}{100}\times \frac{9}{10}}[/tex]
= [tex]\frac{\frac{1}{20} }{\frac{1}{20}+\frac{9}{1000}}[/tex]
= [tex]\frac{\frac{1}{20}}{\frac{50+9}{1000}}[/tex]
= [tex]\frac{1}{20}\times \frac{1000}{59}[/tex]
= [tex](\frac{50}{59})[/tex]
= 0.8475
Suppose you pick 6 different numbers in [10]. Prove that 2 of the numbers are next to each other. (Hint: use the pigeonhole principle. What are the pigeons and what are the holes?)
Step-by-step explanation:
We are picking 6 numbers from the numbers 1,2,3,4,5,6,7,8,9,10. Since we care about numbers being next to each other, we might think of the 10 numbers as being distributed in 5 boxes (which you can think of as the holes):
| 1 2 | 3 4 | 5 6 | 7 8 | 9 10 |
So on the first box we have the numbers 1 and 2, on the second box we have the numbers 3 and 4, and so on. Since we are picking 6 numbers from those 10 numbers, that means we'll have to pick 6 boxes (and inside each box we pick a number), but we only have 5 available boxes, so by the pigeonhole principle, we'll have to pick 1 same box at least two times. Since on each picked box we'll need to pick a number, on this box which was picked two times, we will have to pick both of its numbers. And so those 2 numbers inside that box will be next to each other (meaning they're consecutive numbers).
A pot is being used to boil off 1 kg of water. The specific energy required to cause the phase change is 2297 kJ/kg. Assuming the stovetop supplies 20 kJ/s to the water and the liquid is at boiling temperature, how long will it take to vaporize half of the water? Report your answer in seconds to the nearest whole number don't knou 2 attemots
Answer:
58 seconds
Step-by-step explanation:
Given:
Initial mass of water = 1 kg
Specific energy = 2297 kJ/kg
Heat supplied by the stove = 20 kJ/s
Now,
Half water is to be vaporized i.e 0.5 kg
Thus, heat required for vaporizing 0.5 kg water = mass × specific heat
or
heat required for vaporizing 0.5 kg water = 0.5 × 2297 = 1148.5 kJ
Therefore,
time taken to provide the required heat = [tex]\frac{\textup{Heat required}}{\textup{Heat supplied per second}}[/tex]
or
time taken to provide the required heat = [tex]\frac{\textup{1148.5 kJ}}{\textup{20 kJ/s}}[/tex]
or
time taken to provide the required heat = 57.425 ≈ 58 seconds
It will take approximately 58 seconds to vaporize half a kilogram of water with a heat supply of 20 kJ/s.
The question is asking how long it will take to vaporize half a kilogram of water with a heat supply of 20 kJ/s, assuming the water is at its boiling point and the specific energy required for the phase change is 2297 kJ/kg. To calculate the time required, we can use the formula:
Time (s) = Amount of energy required (kJ) / Energy supply rate (kJ/s).
Since it takes 2297 kJ to vaporize 1 kg, half of this amount is required to vaporize 0.5 kg, which is 1148.5 kJ. Hence, the time taken can be calculated as follows:
Time (s) = 1148.5 kJ / 20 kJ/s = 57.425 s.
So, it would take approximately 58 seconds to vaporize half of the water.
use grouping symbols to make each equation true
1. 9 + 3 - 2 + 4 = 6
2. 4^2 - 5 x 2 + 1 = 1
Answer:
1. 9 + 3 - ( 2 + 4) = 6
2. 4^2 - (5 x (2 + 1)) = 1
Step-by-step explanation:
Here we must follow order of operations - that is commonly expressed as PEDMAS - First do parenthesis, then exponents, then divisions and multiplications from left to right and finally addition and subtraction from left to right.
If we follow this rule on 1)
9+3-2+4= 12-2+4= 10+4 = 14
Sow lets do it by parts
9+3-2+4= 12-2+4
if we can subtract 6 from 12 we would arrive to 6. This can be done id 2 and 4 are added first by 12-(2+4). So the result would be at:
9 + 3 - ( 2 + 4) = 6
In 2)
4^2 - 5 x 2 + 1 = 16-5x2+1 = 16-10 + 1 = 6+1 = 7
4^2 is always the first operation
16-5x2+1
Now if from 16 we subtract 15 we would obtain 1 so 5x2+1 must be equal 15 that can be done if we express it as:
16- 5x2+1
16- (5*(2+1)) = 5x3 = 15
So we have at the end:
4^2 - (5 x (2 + 1)) = 16 - 15 = 1
The probability that a customer's order is not shipped on timeis 0.05. A particular customer places three orders, and the ordersare placed far enough apart in time that they can be considered tobe independent events. Round the answers to 3 significantdigits.a) What is the probability that all are shipped on time?b) What is the probability that exactly one is not shipped ontime?c) What is the probability that two or more orders are notshipped on time?
Answer:
Step-by-step explanation:
Given that the three orders are independent of each other.
X - no of ships not shipped on time
X is binomial with p = 0.05 and q = 0.95, n = 3
a) the probability that all are shipped on time=[tex]0.95^3 =0.854[/tex]
b) the probability that exactly one is not shipped ontime
=P(X=1) =[tex]3C1(0.05)(0.95)^2 = 0.135[/tex]
c) the probability that two or more orders are notshipped on time
[tex]= P(X=2)+P(x=3)\\= 1-P(x=0)-P(x=1)\\=1-0.854-0.135\\=0.011[/tex]
Final answer:
The probabilities for the shipping orders, assuming independence, are calculated: all on time is 0.857, exactly one not on time is 0.135, and two or more not on time is 0.008.
Explanation:
The probability that a customer's order is not shipped on time is 0.05. Since the orders are independent events, we can calculate the following:
Probability that all are shipped on time: Since the orders are independent, the probability of all orders shipped on time is the product of their individual probabilities: (1-0.05)3 = (0.95)3 = 0.857.
Probability that exactly one is not shipped on time: There are three scenarios where exactly one order can be not shipped on time (NSO for not shipped, SO for shipped): NSO-SO-SO, SO-NSO-SO, SO-SO-NSO. The probability for each scenario is 0.05 * 0.95 * 0.95. Since there are three such scenarios, multiply by 3: 3 * (0.05 * 0.95 * 0.95) = 0.135.
Probability that two or more orders are not shipped on time: The probability of at least one order being shipped on time is 1 minus the probability of none being shipped on time, which is 1 - (0.05)3. From this, we subtract the probability of all being shipped on time and the probability of exactly one not being shipped on time to get our answer: 1 - (0.95)3 - (3 * 0.05 * 0.95 * 0.95) = 0.008.
What is 7810 divided by 215 is?
Answer:
36.3255814 or 36 (when rounded)
Step-by-step explanation:
Calculator
Consider a particle moving around a circle with a radius of 38cm. It rotates from 10 degrees to 100 degrees in 11 seconds. Calculate the instantaneous velocity of the particle.
Step-by-step explanation:
Given that,
Radius of circle, r = 38 cm = 0.38 m
It rotates form 10 degrees to 100 degrees in 11 seconds i.e.
[tex]\theta_i=10^{\circ}=0.174\ rad[/tex]
[tex]\theta_f=100^{\circ}=1.74\ rad[/tex]
Let [tex]\omega[/tex] is the angular velocity of the particle such that, [tex]\omega=\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]\omega=\dfrac{1.74-0.174}{11}[/tex]
[tex]\omega=0.142\ rad/s[/tex]
We need to find the instantaneous velocity of the particle. The relation between the angular velocity and the linear velocity is given by :
[tex]v=r\times \omega[/tex]
[tex]v=0.38\times 0.142[/tex]
v = 0.053 m/s
So, the instantaneous velocity of the particle is 0.053 m/s. Hence, this is the required solution.
Suppose you are in a game show and there are 10 doors in front of you. You know that there is a prize behind one of them, and nothing behind the other 9. You have to choose a door containing the prize in order to win the prize. However, before you choose, the game show host promises that rather than immediately opening the door of your choice to reveal its contents, he will open one of the other 9 doors to reveal that it is an empty door. He will then give you the option to change your choice. You may assume that the host is completely impartial – not malicious in any way. For instance, if you choose door 3, he will open one door, say door 5, to reveal that it is empty. Should you continue with door 3 or choose another door? Please compute the probability of finding the prize behind your chosen door before the game show host reveals that one door is empty, and the probability of you finding the prize by changing to a different door after seeing the revealed empty door.
Answer:
The probabilities are [tex]\frac{1}{10}[/tex] and [tex]\frac{9}{80}[/tex]
Step-by-step explanation:
There are 10 doors. 9 of wich have no prizes and 1 with the prize. So the probability to choose the winner one is 1 out of 10. So:
The probability of finding the prize behind your chosen door before the game show host reveals that one door is empty is [tex]\frac{1}{10}[/tex].
Now. If the game show host opens one of the other 9 doors to reveal that it is an empty door, there are 2 posibilities:
1) Do not change your chosen door: In this case the probability reamins the same, [tex]\frac{1}{10}[/tex].
2) Change your chosen door. Lets compute the probability to loose: There are two posibilities.
2a) If your initial door is the one with the prize. In this case you are going to loose (because you will change your door). The probability for this to happen is [tex]\frac{1}{10}[/tex].
2b) If your initial door is not the one with the prize (the probability of this is [tex]\frac{9}{10}[/tex]). In this case we will loose if, after the game show host opens an empty door, we choose an empty door. The probability of choosing an empty door in this case is [tex]\frac{7}{8}[/tex].
So the probability to loose is:
[tex]\frac{1}{10}+\frac{7}{8}\frac{9}{10}=\frac{1}{10}+\frac{63}{80}=\frac{71}{80}[/tex]
Then, the probability to win is [tex]1-\frac{71}{80}=\frac{9}{80}>\frac{1}{10}[/tex]
In conclusion: Changing the door improves the probability to win.
Tomas Briggs and Sons reviewed their first year operations. Gross sales were $204,000 Customer returns and allowances were $18,000 The cost of the merchandise they sold was $90,000 First-year expenses were $84,000 The overall profit before taxes was $12,000 Represent the following in percentages: Total net sales in dollars were The cost of the merchandise sold was First-year expenses were Overall profit was Don't forget how to calculate net sales!
Answer:
Percentage of total net sales = 91.17%
The percentage of cost of merchandise sold = 44.11 %
Percentage of First year expenses = 41.17 %
Percentage of Overall profit = 5.88 %
Step-by-step explanation:
Given:
Gross sales= $204,000
Customer returns and allowances = $18,000
Cost of the merchandise they sold = $90,000
overall profit before taxes = $12,000
Now,
The Net sales = Gross sales - sales returns
or
The net sales = $204,000 - $18,000 = $186,000
Thus,
Percentage of total net sales = [tex]\frac{\textup{Net sales}}{\textup{Gross sales}}\times100[/tex]
or
Percentage of total net sales = [tex]\frac{186,000}{204000}\times100[/tex]
or
Percentage of total net sales = 91.17%
Now,
The percentage of cost of merchandise sold = [tex]\frac{\textup{cost of the merchandise sold }}{\textup{Gross sales}}\times100[/tex]
or
The percentage of cost of merchandise sold = [tex]\frac{\textup{90,000}}{\textup{204,000}}\times100[/tex]
or
The percentage of cost of merchandise sold = 44.11 %
And,
Percentage of First year expenses = [tex]\frac{\textup{Expenses}}{\textup{Gross sales}}\times100[/tex]
or
Percentage of First year expenses = [tex]\frac{\textup{84000}}{\textup{204000}}\times100[/tex]
or
Percentage of First year expenses = 41.17 %
also,
Percentage of Overall profit = [tex]\frac{\textup{Overall profit}}{\textup{Gross sales}}\times100[/tex]
or
Percentage of Overall profit = [tex]\frac{\textup{12,000}}{\textup{204,000}}\times100[/tex]
or
Percentage of Overall profit = 5.88 %
Write the following as a ratio: $72 for 488 photos
Answer:
The required ratio is 9 : 61
Step-by-step explanation:
Given,
$72 for 488 photos,
That is, the price of 488 photos = 72 dollars,
So, the ratio of price of photos and number of photos = [tex]\frac{72}{488}[/tex]
∵ HCF(72, 488) = 8,
Thus, the ratio of price of photos and number of photos = [tex]\frac{72\div 8}{488\div 8}[/tex]
= [tex]\frac{9}{61}[/tex]
Simplify negative 5 minus the square root of negative 44
A negative 5 minus 4 times the square root of 11 i
B negative 5 minus 4 i times the square root of 11
C negative 5 minus 2 i times the square root of 11
D negative 5 minus 2 times the square root of 11 i
Answer:
Option C.
Step-by-step explanation:
Negative 5 represents (-5)
Square root of negative 44 is [tex]\sqrt{(-44)}[/tex]
By the statement "negative 5 minus the square root of negative 44." will be
(-5) - [tex]\sqrt{(-44)}[/tex]
= -5 -[tex]\sqrt{(-1)(44)}[/tex]
= [tex]-5-\sqrt{(-1)(4\times11)}[/tex]
= [tex]-5-2\sqrt{11(-1)}[/tex]
= [tex]-5-2(\sqrt{11} )[\sqrt{(-1)} ][/tex]
= [tex]-5-2i\sqrt{11}[since\sqrt{(-1)=i}][/tex]
Option c will be the answer.
Ima Neworker requires 30 minutes to produce her first unit of output. If her learning curve rte is 65%, how many units will be produced before the output rate exceeds 12 units per hour?
To find when Ima Neworker's rate will exceed 12 units per hour, given a learning curve rate of 65%, we analyze the improvement in production rate from the initial 2 units per hour up to the target, using the learning curve concept.
Explanation:The question relates to the concept of a learning curve, which represents how new workers or processes improve in efficiency as experience is gained. Ima Neworker can produce her first unit in 30 minutes (which is half an hour), so when she starts, her production rate is 2 units per hour. The question asks how many units will be produced before her production rate exceeds 12 units per hour, given a learning curve rate of 65%. This means that each time the cumulative production doubles, the time taken to produce each unit falls to 65% of the previous time.
Since the initial production rate is 2 units per hour, we want to know how many units she has to produce before her production rate exceeds 12 units per hour. 12 units per hour is 6 times faster than her initial rate, and we can reference a learning curve table or use the formula to calculate the necessary doubling periods required to achieve this.
To determine when Ima Neworker's production rate exceeds 12 units per hour, we use a 65% learning curve. By calculations, production time per unit drops below 5 minutes per unit between producing 8 and 16 units, indicating she exceeds the rate at around 12 units. Thus, she will need to produce approximately 12 units before reaching this threshold.
Calculating Production Using a Learning Curve
Ima Neworker requires 30 minutes to produce her first unit, which translates to 2 units per hour initially. The learning curve rate of 65% indicates that with each doubling of previously produced units, the time required to produce another unit will be 65% of the time it took for the previous set.
Step-by-Step Calculation
Initial production time for the first unit: 30 minutes (0.5 hours)After producing the first unit, producing 2 units will take 0.65 × 0.5 = 0.325 hoursProducing 4 units: 0.65 × 0.325 = 0.21125 hoursProducing 8 units: 0.65 × 0.21125 = 0.1373125 hoursProducing 16 units: 0.65 × 0.1373125 = 0.089253125 hoursWe need to produce units such that Ima's production rate exceeds 12 units per hour, meaning she should produce a unit in less than 1/12 hours (5 minutes).
At 8 units, the time per unit is about 8.24 minutes (0.1373125 hours), still above the target.At 16 units, the time per unit is about 5.36 minutes (0.089253125 hours), which is below the target.Thus, Ima will need to produce more than 8 but fewer than 16 units. By interpolation, it will be close to 12 units when her rate exceeds 12 units per hour.
Verify that the function(s) solve the following differential equations (DES): a) y' = -5y; y = 3e-5x b) y' = cos(3x); y = į sin(3x) + 7 c) y' = 2y; y = ce2x , where c is any real number. d) y" + y' – 6y = 0 ; yı = (2x, y2 = (–3x e) y" + 16y = 0; yı = cos(4x), y2 = sin(4x)
Answer:
In the step-by-step explanation, the verifications are made.
Step-by-step explanation:
a) [tex]y' = -5y[/tex]
This one can be solved by the variable separation method
[tex]y' = -5y[/tex]
[tex]\frac{dy}{dx} = -5y[/tex]
[tex]\frac{dy}{y} = -5dx[/tex]
[tex]\int \frac{dy}{y} = \int {-5} \, dx[/tex]
[tex]ln y = -5x + C[/tex]
[tex]e^{ln y} = e^{-5x + C}[/tex]
[tex]y = Ce^{-5x}[/tex]
The value of C is the value of y when x = 0. If [tex]y(0) = 3[/tex], then we have the following solution:
[tex]y = 3e^{-5x}[/tex]
b) [tex]y' = cos(3x)[/tex]
This one can also be solved by the variable separation method
[tex]y' = cos(3x)[/tex]
[tex]\int y' \,dy = \int {cos(3x)} \, dx[/tex]
[tex]y = \frac{sin(3x)}{3} + K[/tex]
K is also the value of y, when x = 0. So, if [tex]y(0) = 7[/tex], we have the following solution.
[tex]y = \frac{sin(3x)}{3} + 7[/tex]
c) [tex]y' = 2y[/tex]
Another one that can be solved by the variable separation method
[tex]y' = 2y[/tex]
[tex]\frac{dy}{dx} = 2y[/tex]
[tex]\frac{dy}{y} = 2dx[/tex]
[tex]\int \frac{dy}{y} = \int {2} \, dx[/tex]
[tex]ln y = 2x + C[/tex]
[tex]e^{ln y} = e^{2x + C}[/tex]
[tex]y = Ce^{2x}[/tex]
C is any real number depending on the initial conditions.
d) [tex]y'' + y' - 6y = 0[/tex]
Here, the solution depends on the roots of the following equation:
[tex]r^{2} + r - 6 = 0[/tex]
[tex]r = \frac{-1 \pm 5}{2}[/tex]
[tex]r = -3[/tex] or [tex]r = 2[/tex].
So the solution is
[tex]y(t) = c_{1}e^{-3t} + c2e^{2t}[/tex]
The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.
e) [tex]y'' + 16y = 0[/tex]
Again, we find the roots of the following equation:
[tex]r^{2} + 16 = 0[/tex]
[tex]r^{2} = -16[/tex]
[tex]r = \pm 4i[/tex]
So we have the following solution
[tex]y(t) = c_{1}cos(4t) + c_{2}sin(4t)[/tex]
The values of [tex]c_{1}, c_{2}[/tex] depends on the initial conditions.
A random variable x is uniformly distributed over the interval (-4, 6). Find the standard deviation of x. (Note: Uniform distribution is a distribution where the PDF value is the same across all x values)
Answer:
The standard deviation of x is 2.8867
Step-by-step explanation:
The standard deviation of variable x that follows a uniform distribution is calculated as:
[tex]s = \sqrt{\frac{(b-a)^{2} }{12} }[/tex]
Where (a,b) is the interval where x is defined.
So, replacing a by -4 and b by 6, the standard deviation is:
[tex]s = \sqrt{\frac{(6-(-4))^{2} }{12} }[/tex]
[tex]s = \sqrt{\frac{(10)^{2} }{12} }[/tex]
[tex]s=\sqrt{\frac{100}{12} }[/tex]
[tex]s=\sqrt{8.3333}[/tex]
[tex]s=2.8867[/tex]
Evaluate 4P3
24
7
10
12
Answer:
24
Step-by-step explanation:
4P3 = 4!/(4-3)! = 4·3·2 = 24
A random sample of 50 consumers taste tested a new snack food. Their responses were coded (0: do not like; 1: like, 2: indifferent) and recorded below: a. Test H0: p = 0.5 against Ha: p > 0.5, where p is the proportion of customers who do not like the snack food (n=17). Use α = 0.10. b. Find the observed significance level of your test.
Answer:
The level of significance observed is 0.99154
Step-by-step explanation:
Assuming that in a sample of size 50 people stated that they do not like the snack (p = 17/50), you have:
Proportion in the null hypothesis [tex]\pi_0=0.5[/tex]
Sample size [tex]n=50[/tex]
Sample proportion [tex]p=17/50=0.34[/tex]
The expression for the calculated statistic is:
[tex] = \frac{(p - \pi_0)\sqrt{n}}{\sqrt{\pi_0(1-\pi_0)}}[/tex]
[tex]= \frac{(0.34 - 0.5)\sqrt{50}}{\sqrt{0.34(0.66)}} = -2,38833[/tex]
The level of significance observed is obtained from the value of the statistic calculated:
[tex]P(Z>Z_{calculated}) = 0.99154[/tex]
A scientist is looking at 1000 germs under a microscope
andfinds that the germs double in number every 4 hours.
Using,
how many germs will there be in 7 hours.
Answer:
3500
Step-by-step explanation:
Number of germs that a scientist can see under a microscope = 1000 germs
We need to find how many germs will there be in 7 hours if the germs double in number every 4 hours .
It's given that the germs double in number every 4 hours .
So, increase in number of germs in one hour = [tex]\frac{2}{4}=\frac{1}{2}[/tex]
Increase in number of germs in seven hours = [tex]\frac{7}{2}[/tex]
Therefore , number of germs in 7 hours = Increase in number of germs in seven hours × Number of germs initially
= [tex]\frac{7}{2}\times 1000=7\times 500=3500[/tex]
So, number of germs in 7 hours if the germs double in number every 4 hours = 3500
After 7 hours, there will be 2000 germs.
To calculate how many germs there will be in 7 hours, we need to understand the concept of exponential growth. In this scenario, the germs double every 4 hours.
Initial Number of Germs: You start with 1000 germs.
Doubling Time: The germs double every 4 hours. This means that after each 4-hour period, the population multiplies by 2.
Calculating How Many Doubling Periods in 7 Hours:
In 7 hours, there are 1 full 4-hour doubling period and 3 additional hours left.After the first 4 hours, the population doubles:Final Count:
So after 7 hours, the total number of germs will be 2000 germs. The additional 3 hours will not lead to another doubling of the population.One side of a triangular lot is 150 ft and the angel oppiste this side is 55 degrees. Another angel is 63 degrees. Determine how much fencing is needed to enclose it.
Answer:
474.84 ft of fencing is needed
Step-by-step explanation:
We know that the angles of a triangle sum up 180º. We already know 2 of the triangle's angles (55º and 63º). Therefore the third angle measures:
180 - 55 - 63 = 62.
To know how much fencing is needed, we need the perimeter of the triangle, so we need to find out how much the other sides of the lot measure.
We will use law of sins to solve this problem.
First we solve for y:
[tex]\frac{150}{sin55}= \frac{y}{sin63} \\y=150 (sin63)/(sin55)\\y=133.65/.8191\\y=163.16[/tex]
Now we solve for the other side of the lot, x:
[tex]\frac{150}{sin55}=\frac{x}{sin62}\\ x=150(sin62)/(sin55)\\x=150(.8829)/.8191\\x=132.435/.8191\\x=161.68[/tex]
Now that we have the measures of all the sides we sum them up
total fencing needed= 150 + 163.16 + 161.68 = 474.84
A vial contains 80 mg of drug in 2 mL of injection. How many milliliters of the injection should be administered to obtain 0.02 g of the drug?
Answer:
0.5mL of the injection should be administered to obtain 0.02 g of the drug.
Step-by-step explanation:
First step: The first step of this problem is the conversion of 0.02g to mg.
Each gram has 1000 miligrams. So:
1g - 1000mg
0.02g - xmg
x = 1000*0.02
x = 20mg
Final step:
A vial contains 80 mg of drug in 2 mL of injection. How many mL should be administered to obtain 0.02 g = 20mg of the drug.
This can be solved as a rule of three problem.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
In this step, as the dose of the injection increases, so the quantity of the drug. So the relationship between the measures is direct. So:
80 mg - 2mL
20 mg - xmL
80x = 40
[tex]x = \frac{40}{80}[/tex]
x = 0.5mL
0.5mL of the injection should be administered to obtain 0.02 g of the drug.
To obtain 0.02 grams of the drug, the required volume to administer from the vial containing 80 mg in 2 mL is 0.5 mL.
Explanation:To solve this problem, we need to determine how many milliliters contain 0.02 grams of the drug. First, convert the amount we want from grams to milligrams since our given concentration is in milligrams: 0.02 grams is equal to 20 milligrams.
Given that the vial contains 80 mg of drug in 2 mL, we can calculate the volume required for 20 mg. The formula we will use is:
(desired dose / concentration of vial) × volume of vial = required volume
(20 mg / 80 mg) × 2 mL = 0.5 mL
Therefore, to obtain 0.02 grams (20 mg) of the drug, the required volume to administer would be 0.5 mL.
JY is a 58 year old male who was hospitalized for a total knee replacement. He was given unfractionated heparin and developed heparin-induced thrombocytopenia (HIT). Argatroban was ordered at a dose of 2 mcg/kg/min. The pharmacy mixes a concentration of 100 mg argatroban in 250 mL of DSW. JY weighs 85 kg. How many mL/hour should the nurse infuse to provide the dose? Round to the nearest whole number. .
Answer:
The nurse infuse [tex]26ml/hr[/tex] to provide the dose.
Step-by-step explanation:
Argatroban was ordered at a dose of 2 mcg/kg/min.
JY weighs 85 kg.
So, Argatroban was ordered= [tex]2 \times 85[/tex]
= [tex]170mcg/min.[/tex]
Convert the dose in mg/hr
1 hr = 60 minutes and 1 mg = 1000 mcg
So, Dose in ml/hr = [tex]170 \times \frac{60}{1000}[/tex]
= [tex]10.2 mg/hr[/tex]
Now to find in 250 mL of DSW. JY weighs 85 kg. How many mL/hour should the nurse infuse to provide the dose?
The nurse infuse to provide the dose = [tex]\text{Dose ordered} \times \frac{\text{volume available}}{\text{Dose available}}[/tex]
The nurse infuse to provide the dose = [tex]10.2 mg/hr \times \frac{250 ml}{100 mg}[/tex]
The nurse infuse to provide the dose = [tex]26ml/hr[/tex]
Hence The nurse infuse [tex]26ml/hr[/tex] to provide the dose.
A low-strength children’s/adult chewable aspirin tablet contains 81 mg of aspirin per tablet. How many tablets may be prepared from 1 kg of aspirin?
Answer:
12,345 tablets may be prepared from 1 kg of aspirin.
Step-by-step explanation:
The problem states that low-strength children’s/adult chewable aspirin tablets contains 81 mg of aspirin per tablet. And asks how many tablets may be prepared from 1 kg of aspirin.
Since the problem measures the weight of a tablet in kg, the first step is the conversion of 81mg to kg.
Each kg has 1,000,000mg. So
1kg - 1,000,000mg
xkg - 81mg.
1,000,000x = 81
[tex]x = \frac{81}{1,000,000}[/tex]
x = 0.000081kg
Each tablet generally contains 0.000081kg of aspirin. How many such tablets may be prepared from 1 kg of aspirin?
1 tablet - 0.000081kg
x tablets - 1kg
0.000081x = 1
[tex]x = \frac{1}{0.000081}[/tex]
x = 12,345 tablets
12,345 tablets may be prepared from 1 kg of aspirin.
S is the set of ordered pairs of integers and (x1, x2) R(y1, y2) means that x1= y1and x2≤ y2
Demonstrate whether R exhibits the reflexive property or not.
Demonstrate whether R exhibits the symmetric property or not.
Demonstrate whether R exhibits the transitive property or not.
Answer:
R is reflexive
R is not symmetric
R is transitive
Step-by-step explanation:
R is reflexive.
To show this, we have to verify that for any pair of integers [tex](x_1,x_2)[/tex]
[tex](x_1,x_2)R(x_1,x_2)[/tex].
But this is obvious because
[tex]x_1=x_1[/tex] and [tex]x_2\leq x_2[/tex].
R is not symmetric.
To show it, we need to find two pairs [tex](x_1,x_2)[/tex] and [tex](y_1,y_2)[/tex] such that
[tex](x_1,x_2)R(y_1,y_2)[/tex]
but [tex](y_1,y_2) \not \mathrel{R} (x_1,x_2)[/tex]
For example (1,1) and (1,2).
[tex](1,1)R(1,2)[/tex] for 1=1 and [tex]1\leq 2[/tex] but
[tex](1,2) \not \mathrel{R} (1,1)[/tex] because [tex]2\not \leq 1[/tex]
Finally, R is transitive.
If we take 3 pairs of integers [tex](x_1,x_2), (y_1,y_2)[/tex] and [tex](z_1,z_2)[/tex]
Such that
[tex](x_1,x_2)R(y_1,y_2)[/tex] and [tex](y_1,y_2)R(z_1,z_2)[/tex] then
[tex]x_1=y_1[/tex] and [tex]x_2\leq y_2[/tex]
[tex]y_1=z_1[/tex] and [tex]y_2\leq z_2[/tex]
But then,
[tex]x_1=z_1[/tex] and [tex]x_2\leq z_2[/tex]
So
[tex](x_1,x_2)R(z_1,z_2)[/tex].
EXAMPLE 2 The arc of the parabola y = 3x2 from (5, 75) to (10, 300) is rotated about the y-axis. Find the area of the resulting surface. SOLUTION 1 Using y = 3x2 and dy dx = we have, from this formula, S = 2πx ds = 10 2πx 1 + dy dx 2 dx 5 = 2π 10 x 1 + 36x2 dx 5 . Substituting u = 1 + 36x2, we have du = dx. Remembering to change the limits of integration, we have S = π 36 3601 u du 901 = π 36 3601 901 = . SOLUTION 2 Using x = y 3 and dx dy = we have S = 2πx ds = 300 2πx 1 + dx dy 2 dy 75 = 2π 300 y 3 1 + 1 12y dy 75 = π 3 300 12y + 1 dy 75 = π 36 3601 u du 901 (where u = 1 + 12y) = (as in Solution 1)
To find the surface area of the resulting surface when rotating the arc of a parabola about the y-axis, you can integrate using either x or y as the variable. Both methods yield the same result.
Explanation:The question asks for the area of the surface created by rotating the arc of the parabola y = 3x^2 from (5, 75) to (10, 300) about the y-axis.
There are two different solutions provided, both utilizing different methods of integration.
The first solution uses the given equation y = 3x^2 and integrates with respect to x, while the second solution uses the equation x = y^(1/3) and integrates with respect to y.
Both solutions arrive at the same answer to find the surface area.
Assume that 155 students are surveyed and every student takes at least one of the following languages. The results of the survey are as follows:
90 take French.
83 take German.
42 take French and German.
41 take German and Russian.
22 take French as their only foreign language.
22 take French, Russian, and German.
(1) How many take Russian?
(2) How many take French and Russian but not German?
Answer:
91 people take Russian
26 people take French and Russian but not German
Step-by-step explanation:
To solve this problem, we must build the Venn's Diagram of this set.
I am going to say that:
-The set A represents the students that take French.
-The set B represents the students that take German
-The set C represents the students that take Russian.
We have that:
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
In which a is the number of students that take only Franch, A \cap B is the number of students that take both French and German , A \cap C is the number of students that take both French and Russian and A \cap B \cap C is the number of students that take French, German and Russian.
By the same logic, we have:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
This diagram has the following subsets:
[tex]a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]
There are 155 people in my school. This means that:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155[/tex]
The problem states that:
90 take Franch, so:
[tex]A = 90[/tex]
83 take German, so:
[tex]B = 83[/tex]
22 take French, Russian, and German, so:
[tex]A \cap B \cap C = 22[/tex]
42 take French and German, so:
[tex]A \cap B = 42 - (A \cap B \cap C) = 42 - 22 = 20[/tex]
41 take German and Russian, so:
[tex]B \cap C = 41 - (A \cap B \cap C) = 41 - 22 = 19[/tex]
22 take French as their only foreign language, so:
[tex]a = 22[/tex]
Solution:
(1) How many take Russian?
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
[tex]C = c + (A \cap C) + 19 + 22[/tex]
[tex]C = c + (A \cap C) + 41[/tex]
First we need to find [tex]A \cap C[/tex], that is the number of students that take French and Russian but not German. For this, we have to go to the following equation:
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
[tex]90 = 22 + 20 + (A \cap C) + 22[/tex]
[tex](A \cap C) + 64 = 90[/tex].
[tex](A \cap C) = 26[/tex]
----------------------------
The number of students that take Russian is:
[tex]C = c + 26 + 41[/tex]
[tex]C = c + 67[/tex]
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Now we have to find c, that we can find in the equation that sums all the subsets:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 155[/tex]
[tex]22 + b + c + 20 + 26 + 19 + 22 = 155[/tex]
[tex]b + c + 109= 155[/tex]
[tex]b + c = 46[/tex]
For this, we have to find b, that is the number of students that take only German. Then we go to this eqaution:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]B = b + 19 + 20 + 22[/tex]
[tex]B = b + 61[/tex]
[tex]b + 61 = 83[/tex]
[tex]b = 22[/tex]
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[tex]b + c = 46[/tex]
[tex]c = 46 - b[/tex]
[tex]c = 24[/tex]
The number of people that take Russian is:
[tex]C = c + 67[/tex]
[tex]C = 24 + 67[/tex]
[tex]C = 91[/tex]
91 people take Russian
(2) How many take French and Russian but not German?
[tex](A \cap C) = 26[/tex]
26 people take French and Russian but not German