Answer:
(a) [tex]P_{t}=3(2)^{6t}[/tex]
(b) [tex]3(2)^{42}[/tex]
(c) 28.07 minutes
Step-by-step explanation:
A cell of some bacteria divides itself into 2 cells in every 10 minutes and initial population of the bacteria was 3.
That means sequence formed will be 3, 6, 12, 24............
We can easily say that this sequence is a geometric sequence having common ratio (r) = [tex]\frac{T_{2}}{T_{1}}=\frac{6}{3}[/tex]
r = 2
Now we know the explicit formula of a geometric sequence is given by
[tex]P_{t}=P_{0}(r)^{\frac{60t}{10}}=P_{0}(r)^{6t}[/tex]
Where a = Initial population = 3 bacteria
r = common ratio = 2
and t = time in hours
So explicit formula will be [tex]P_{t}=3(2)^{6t}[/tex]
(a) Now we have to calculate the size of population after t hours
[tex]P_{t}=3(2)^{6t}[/tex]
(b) We have to find the size of population after 7 hours or 420 minutes
[tex]P_{t}=3(2)^{6\times7}[/tex]
= [tex]3(2)^{42}[/tex]
After 7 hours bacteria population will be [tex]3(2)^{42}[/tex]
(c) Time to reach population as 21
By the explicit formula
[tex]21=3(2)^{6t}[/tex]
[tex]2^{6t}=\frac{21}{3}=7[/tex]
Now we take log on both the sides of the equation
[tex]log(2^{6t})=log(7)[/tex]
6t log2 = log 7
6t(0.301) = 0.845
t(1.806) = 0.845
t = [tex]\frac{0.845}{1.806}=0.468[/tex] hours
Or t = 0.468×60 = 28.07 minutes
Therefore, after 28.07 minutes bacterial population will be 21
The bacteria population grows following an exponential pattern, therefore the population after t hours can be calculated using the exponential growth formula with the initial population as 3 and each cell dividing every 10 minutes. To calculate the time when the population reaches a certain size, solve the exponential growth equation for t.
Explanation:The growth of bacteria population can be described as exponential growth, with each cell dividing into two every 10 minutes. Given the initial population as 3 bacteria, we would need to calculate the number of divisions that occur within the specified time frame to calculate the population after t hours.
(a) To find the population after t hours, we convert the hours to minutes (since each division occurs every 10 minutes) and then calculate the number of divisions. Each bacterial division results in a doubling of the population, so we use the formula for exponential growth: N = N0 * 2^n, where N0 is the initial population (3), and n is the number of divisions (6t, because t hours is 60t minutes and each division occurs every 10 minutes, making a total of 6t divisions per hour). So the population after t hours is N = 3 * 2^(6t).
(b) To find the size of the population after 7 hours, we substitute t = 7 into the formula, to get N = 3 * 2^(6*7) = 3 * 2^42.
(c) To find out when the population reaches 21, we equate N to 21 in the formula and solve for t. So, 21 = 3 * 2^(6t). Solving this equation gives the time t in hours when the population will reach 21.
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A company that manufactures and bottles apple juice uses a machine that automatically fills 16–ounce bottles. There is some variation in the amounts of liquid dispensed into the bottles. The amount dispensed is approximately normally distributed with mean 16 ounces and standard deviation 1 ounce. What proportion of bottles will have more than 17 ounces?
Answer: 0.1587
Step-by-step explanation:
Given : The amount dispensed is approximately normally distributed with Mean : [tex]\mu=\ 16[/tex]
Standard deviation : [tex]\sigma= 1[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 17
[tex]z=\dfrac{17-16}{1}=1[/tex]
The p-value =[tex] P(17<x)=P(1<z)[/tex]
[tex]=1-P(z<1)=1-0.8413447\\\\=0.1586553\approx0.1587[/tex]
The proportion of bottles will have more than 17 ounces = 0.1587
A particular dam contains approximately 1,100,000,000,000 cubic feet of water. For a week-long spike flood, water was released at a rate of 25,100 cubic feet per second.
The amount of water released during the week-long flood was?
Answer:
The amount of water released during the week-long flood was 15,180,400,000 cubic feet per second.
Step-by-step explanation:
How many seconds are there in a week?
Each minute has 60 seconds
Each hour has 60 minutes
Each day has 24 hours
Each week has 7 days. So
60*60*24*7 = 604,800
A week has 604,800 seconds.
Water was released at a rate of 25,100 cubic feet per second.
In a week(604,800 seconds)
604,800*25,100 = 15,180,400,000
The amount of water released during the week-long flood was 15,180,400,000 cubic feet per second.
Let R be a communtative ring and a, b elements in R. Prove that if a and b are units, then so is ab. What can we say about ab when a is a unit and b is a zero divisor? Prove your claim.
Answer with explanation:
Let R be a communtative ring .
a and b elements in R.Let a and b are units
1.To prove that ab is also unit in R.
Proof: a and b are units.Therefore,there exist elements u[tex]\neq0[/tex] and v [tex]\neq0[/tex] such that
au=1 and bv=1 ( by definition of unit )
Where u and v are inverse element of a and b.
(ab)(uv)=(ba)(uv)=b(au)(v)=bv=1 ( because ring is commutative)
Because bv=1 and au=1
Hence, uv is an inverse element of ab.Therefore, ab is a unit .
Hence, proved.
2. Let a is a unit and b is a zero divisor .
a is a unit then there exist an element u [tex]\neq0[/tex]
such that au=1
By definition of unit
b is a zero divisor then there exist an element [tex]v\neq0[/tex]
such that bv=0 where [tex]b\neq0[/tex]
By definition of zero divisor
(ab)(uv)=b(au)v ( because ring is commutative)
(ab)(uv)=b.1.v=bv=0
Hence, ab is a zero divisor.
If a is unit and b is a zero divisor then ab is a zero divisor.
8) The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the probability that there will be (a) at least 2 such accidents in the next month; (b) at most 1 accident in the next month? Explain your reasoning!
Answer: (a) 0.8641
(b) 0.1359
Step-by-step explanation:
Given : The monthly worldwide average number of airplane crashes of commercial airlines [tex]\lambda= 3.5[/tex]
We use the Poisson distribution for the given situation.
The Poisson distribution formula for probability is given by :-
[tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
a) The probability that there will be at least 2 such accidents in the next month is given by :-
[tex]P(X\geq2)=1-(P(X=1)+P(X=0))\\\\=1-(\dfrac{e^{-3.5}(3.5)^0}{0!}+\dfrac{e^{-3.5}(3.5)^1}{1!})\\\\=1-(0.1358882254)=0.8641117746\approx0.8641[/tex]
b) The probability that there will be at most 1 accident in the next month is given by :-
[tex]P(X\leq1)=(P(X=1)+P(X=0))\\\\=\dfrac{e^{-3.5}(3.5)^0}{0!}+\dfrac{e^{-3.5}(3.5)^1}{1!}\\\\=0.1358882254\approx0.1359[/tex]
Use the given data to find the 95% confidence interval estimate of the population mean μ. Assume that the population has a normal distribution. IQ scores of professional athletes: Sample size n=10 Mean x¯=104 Standard deviation s=10
With a 95% confidence level, the population mean is estimated to be between approximately 96.85 and 111.15 based on a sample size of 10, a mean of 104, and a standard deviation of 10.
With a sample size (n) of 10, a mean \bar{x}104, and a standard deviation (s) of 10, we can find the 95% confidence interval for the population mean (μ).
First, we calculate the standard error of the mean (SE). The standard error of the mean can be calculated by dividing the standard deviation by the square root of the sample size.
SE = s/√n.
By substituting s = 10 and n = 10 into the equation, we get SE = 3.162277660168379.
Next, we need to find the critical value (t) for a 95% confidence interval based on a t-distribution. Since we're using a confidence level of 95% and the sample size is 10, which means degree of freedom is n-1=9, the critical value (t) is 2.2621571627409915 based on the t-distribution table.
To calculate the lower bound and the upper bound of the 95% confidence interval, you should subtract and add to the mean the product of the critical value and the standard error respectively.
So,
Lower Bound = \bar{x} - t * SE
Upper Bound = \bar{x} + t * SE
Substituting from our known values, we get:
Lower Bound = 104 - 2.2621571627409915 * 3.162277660168379 = 96.84643094047428
Upper Bound = 104 + 2.2621571627409915 * 3.162277660168379 = 111.15356905952572
So, with a 95% confidence level, the confidence interval estimate of the population mean is (96.84643094047428, 111.15356905952572). This means we are 95% confident that the true population mean lies somewhere between approximately 96.85 and 111.15.
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The 95% confidence interval for the population mean IQ score of professional athletes, based on a sample size of 10 with a mean of 104 and standard deviation of 10, is estimated to be between 96.83 and 111.17.
To find the 95% confidence interval estimate of the population mean [tex](\( \mu \))[/tex] given the sample data, we'll use the formula for the confidence interval for a population mean when the population standard deviation is unknown:
[tex]\[ \text{Confidence interval} = \bar{x} \pm t \left( \frac{s}{\sqrt{n}} \right) \][/tex]
Where:
-[tex]\( \bar{x} \)[/tex] is the sample mean,
- s is the sample standard deviation,
- n is the sample size, and
- t is the critical value from the t-distribution for the desired confidence level and degrees of freedom.
Given:
- Sample size n = 10
- Sample mean [tex](\( \bar{x} \))[/tex]= 104
- Sample standard deviation s = 10
First, we need to find the critical value t for a 95% confidence level with 9 degrees of freedom (since n - 1 = 10 - 1 = 9 ).
Using a t-table or statistical software, [tex]\( t \approx 2.262 \)[/tex] for a 95% confidence level and 9 degrees of freedom.
Now, let's plug in the values into the formula:
[tex]\[ \text{Confidence interval} = 104 \pm 2.262 \left( \frac{10}{\sqrt{10}} \right) \][/tex]
Now, let's calculate the margin of error:
[tex]\[ \text{Margin of error} = 2.262 \left( \frac{10}{\sqrt{10}} \right) \]\[ \text{Margin of error} \approx 7.17 \][/tex]
Finally, let's calculate the confidence interval:
[tex]\[ \text{Lower bound} = 104 - 7.17 \]\[ \text{Upper bound} = 104 + 7.17 \]\[ \text{Lower bound} \approx 96.83 \]\[ \text{Upper bound} \approx 111.17 \][/tex]
So, the 95% confidence interval estimate of the population mean IQ score of professional athletes is approximately between 96.83 and 111.17.
2. Let A be a 3 x 3 matrix such that det(A)= -4. Find det(3A)
Answer:
The value of det (3A) is -108.
Step-by-step explanation:
If M is square matrix of order n x n, then
[tex]|kA|=k^n|A|[/tex]
Let as consider a matrix A or order 3 x 3. Using the above mentioned property of determinant we get
[tex]|kA|=k^3|A|[/tex]
We need to find the value of det(3A).
[tex]|3A|=3^3|A|[/tex]
[tex]|3A|=27|A|[/tex]
It is given that the det(A)= -4. Substitute |A|=-4 in the above equation.
[tex]|3A|=27(-4)[/tex]
[tex]|3A|=-108[/tex]
Therefore the value of det (3A) is -108.
Write the equation in the slope-intercept form 7x +4y - 20 0 Find the y-intercept of the corresponding line. (x, y)
Step-by-step explanation:
write the equation in slope intercept form
7x + 4y - 20=0
find the slope of corresponding line.
????
then
find the y intercept of corresponding line
(x,y)=???.
What would the seasons be like if the axis of Earth's rotation was tilted 0 degrees to the ecliptic, instead of the 23.5 degrees we find it today? What about if it was tilted 0 degrees?
Answer:
The seasons would become constant. It would be equinox throughout the year.
Step-by-step explanation:
The earth would be in a state of constant equinox i.e., the length of day and night would be same in a particular place.
The season of a place would be what it is when it is normally titled at equinox.
The animal and plant life which depend on the seasons would be affected.
Snow would only occur at parts where it normally snows at equinoxes.
Tickets for a play cost 2 pounds for a child, and 4 pounds for an adult. one adult brought 4 children with him and the remaining adults each brought 2 children with them. The total ticket sales were 60 pounds. how many adults and children were present in that play?
Solve using augmented matrix.
Answer:
Number of adults = 7
Number of children = 16
Step-by-step explanation:
Tickets for a play cost 2 pounds for a child, and 4 pounds for an adult.
Let x number of adults and y number of children.
1 child ticket cost = 2 pound
y children ticket cost = 2y pound
1 adult ticket cost = 4 pound
x adults ticket cost = 4x pound
Total number of ticket sales were 60 pounds
Therefore, 4x + 2y = 60 ------------- (1)
One adult brought 4 children with him and the remaining adults each brought 2 children with them.
Remaining number of adult whose brought 2 children = x-1
Number children = 2(x-1)
Total number of children = 2(x-1)+4
Therefore, y=2x+2 ---------------------(2)
System of equation,
2x + y = 30
-2x + y = 2
Using augmented matrix to solve system of equation.
[tex]\begin{bmatrix}2&1&\ |30\\-2&1&|2\end{bmatrix}\\\\R_2\rightarrow R_2+R_1\\\\\begin{bmatrix}2&1& |30\\0&2&|32\end{bmatrix}\\\\R_2\rightarrow\dfrac{1}{2}R_2\\[/tex]
[tex]\begin{bmatrix}2&1&\ |30\\0&1&|16\end{bmatrix}\\\\R_1\rightarrow R_1-R_2\\\\\begin{bmatrix}2&0&\ |14\\0&1&|16\end{bmatrix}\\\\\\[/tex]
[tex]R_1\rightarrow \dfrac{1}{2}R_1\\\\\begin{bmatrix}1&0&|7\\0&1&|16\end{bmatrix}\\\\[/tex]
Now, we find the value of variable.
[tex]x=7\text{ and }y=16[/tex]
Hence, Number of adults are 7 and Number of children are 16.
A pair of dice is rolled, and the sum of the numbers is either 7 or 11. What is the probability of this event?
Answer: [tex]\dfrac{2}{9}[/tex]
Step-by-step explanation:
Let A be the event that the sum is 7 and and B be the event that the sum is 11 .
When two pair of dices rolled the total number of outcomes = [tex]n(S)=6\times6=36[/tex]
The sample space of event A ={(1,6), (6,1), (5,2), (2,5), (4,3), (3,4)}
Thus n(A)= 6
The sample space of event B = {(5,6), (6,5)}
n(B)=2
Since , both the events are independent , then the required probability is given by :-
[tex]P(A\cup B)=P(A)+P(B)\\\\=\dfrac{n(A)}{n(S)}+\dfrac{n(B)}{n(S)}=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}[/tex]
Hence, the required probability = [tex]\dfrac{2}{9}[/tex]
Answer:
Probability that sum of numbers is either 7 or 11 is:
0.22
Step-by-step explanation:
A pair of dice is rolled.
Sample Space:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Total outcomes= 36
Outcomes with sum of numbers either 7 or 11 are in bold letters=8
i.e. number of favorable outcomes=8
So, P(sum of numbers is either 7 or 11 )=8/36
=0.22
a customer is owed $30.00. how many different combinations of bills,using only five, ten, and twenty dollars bills are possible to give his or her change?
Answer:
1. 6 fives.
2. 1 ten and 4 fives.
3. 2 tens and 2 fives.
4. 3 tens.
5. 1 twenty and 2 fives.
6. 1 twenty and 1 ten.
Step-by-step explanation:
Given : A customer is owed $30.00.
To find : How many different combinations of bills,using only five, ten, and twenty dollars bills are possible to give his or her change?
Solution :
We have to split $30 in terms of only five, ten, and twenty dollars.
1) In terms of only five we required 6 fives as
[tex]6\times 5=30[/tex]
So, 6 fives.
2) In terms of only ten and five,
a) We required 1 ten and 4 fives as
[tex]1\times 10+4\times 5=10+20=30[/tex]
So, 1 ten and 4 fives.
b) We required 2 tens and 2 fives as
[tex]2\times 10+2\times 5=20+10=30[/tex]
So, 2 tens and 2 fives
3) In terms of only tens we require 3 tens as
[tex]3\times 10=30[/tex]
So, 3 tens.
4) In terms of only twenty and five, we required 1 twenty and 2 fives as
[tex]1\times 20+2\times 5=20+10=30[/tex]
So, 1 twenty and 2 fives.
5) In terms of only twenty and ten, we required 1 twenty and 1 ten as
[tex]1\times 20+1\times 10=20+10=30[/tex]
So, 1 twenty and 1 ten.
Therefore, There are 6 different combinations.
Determine whether the given procedure results in a binomial distribution. If not, state the reason why. Choosing 3 marbles from a box of 40 marbles (20 purple, 12 red, and 8 green) one at a time with replacement, keeping track of the colors of the marbles chosen.
Hey there!:
Here , we choose the 10 marbles from the box of 40 marbles without replacement
Therefore , probability is changes for every time
Also , the trials are dependent
Therefore ,the assumptions of binomial distributions are not satisfied
Therefore , Not binomial : the trials are not independent
Hope this helps!
The given procedure does not follow the characteristics of a binomial distribution.
The procedure of choosing marbles with replacement from a box with different colored marbles does not meet the criteria for a binomial distribution.
The given procedure does not result in a binomial distribution because in a binomial distribution, the trials must be independent, there must be a fixed number of trials, and there can only be two outcomes (success and failure).
In this case, choosing marbles from a box with replacement and tracking their colors does not meet the criteria for a binomial experiment, as the trials are not independent, the number of trials is not fixed, and there are more than two possible outcomes (purple, red, green).
Therefore, the given procedure does not follow the characteristics of a binomial distribution.
How many dfferent strings can be made from the letters in MISSISSIPPI, using all letters? 15 013837
Answer: 34650
Step-by-step explanation:
The number of permutations of n objects, where one object is repeated [tex]n_1[/tex] times , another is repeated [tex]n_2[/tex] times and so on is :
[tex]\dfrac{n!}{n_1!n_2!....n_k!}[/tex]
Given : The number of letters in string MISSISSIPPI = 11
Here I is repeated 4 times, S is repeated 4 times and P is repeated 2 times.
Then , the number of different strings can be made from the letters in MISSISSIPPI, using all letters is given by :-
[tex]\dfrac{11!}{4!4!2!}=34650[/tex]
Therefore , there are 34650 different strings can be made from the letters in MISSISSIPPI.
To find the number of different strings that can be formed from the letters in MISSISSIPPI, we use the formula for permutations of a multi set. The total is 34,650 unique permutations.
The question asks how many different strings can be made from the letters in MISSISSIPPI, using all letters.
This is a problem of calculating permutations of a multi set.
The word MISSISSIPPI has 11 letters with the following counts of each letter: M occurs 1 time, I occurs 4 times, S occurs 4 times, and P occurs 2 times.
To find the total number of unique permutations, we use the formula for the permutations of a multiset:
The formula is:
Permutations = n! / (n1! * n2! * ... * nk!),
where n is the total number of items to arrange, and n1, n2, ..., nk are the counts of each distinct item.
For MISSISSIPPI:
Total permutations = 11! / (1! * 4! * 4! * 2!) = 34650
Therefore, there are 34,650 different strings that can be made from the letters in MISSISSIPPI using all the letters.
If a population is recorded at 1,200 in the year 2000 and the rate of increase is a steady 50 people each year, what will be the population in 2018?
Answer:
2100
Step-by-step explanation:
50*18=900
900+1,200=2100
To find the population in 2018, we calculate the total increase from 2000 to 2018 by multiplying the yearly increase of 50 people by 18 years, resulting in an additional 900 people. Adding this to the initial population of 1,200 people gives us a total population of 2,100 people in 2018.
If a population is recorded at 1,200 in the year 2000 and increases at a steady rate of 50 people each year, we can calculate the population in 2018 using a linear growth model. First, we need to determine the number of years between 2000 and 2018, which is 18 years. Next, we multiply the annual increase (50 people) by the number of years (18) to find the total increase over this period.
The calculation would be as follows:
Total Increase = Annual Increase times Number of YearsTotal Increase = 50 people/year times 18 yearsTotal Increase = 900 peopleWe then add this total increase to the initial population to get the population in 2018:
Population in 2018 = Initial Population + Total IncreasePopulation in 2018 = 1,200 people + 900 peoplePopulation in 2018 = 2,100 peopleThe population in 2018 would be 2,100 people.
The distribution of cholesterol levels in teenage boys is approximately normal withnbsp mu equals 170 and sigma equals 30 (Source: U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that a teenage boy has a cholesterol level greater than 225.
Answer: 0.0336
Step-by-step explanation:
Given : The distribution of cholesterol levels in teenage boys is approximately normal with mean :[tex]\mu= 170[/tex]
Standard deviation : [tex]\sigma= 30[/tex]
The formula for z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x=225
[tex]z=\dfrac{225-170}{30}=1.83[/tex]
The p-value =[tex]P(z>1.83)=1-P(z<1.83)[/tex]
[tex]=1-0.966375=0.033625\approx0.0336[/tex]
The probability that a teenage boy has a cholesterol level greater than 225 =0.0336
Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/(partial p), (partial u)/(partial r), (partial u)/(partial theta) when p = 2, r = 2, θ = 0
[tex]u(x,y,z)=x^2+yz[/tex]
[tex]\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}[/tex]
At the point [tex](p,r,\theta)=(2,2,0)[/tex], we have
[tex]\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}[/tex]
Denote by [tex]f_x:=\dfrac{\partial f}{\partial x}[/tex] the partial derivative of a function [tex]f[/tex] with respect to the variable [tex]x[/tex]. We have
[tex]\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}[/tex]
The Jacobian is
[tex]\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}[/tex]
By the chain rule,
[tex]u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y[/tex]
[tex]u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}[/tex]
[tex]u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y[/tex]
[tex]u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}[/tex]
[tex]u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta[/tex]
[tex]u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}[/tex]
This problem is about using the Chain Rule to compute the partial derivatives of a function with respect to different variables, followed by substitution of specific values into the obtained derivatives.
Explanation:The problem involves finding partial derivatives using the Chain Rule on the given equations with given parameters: p = 2, r = 2, θ = 0. By substituting the equations for x, y, z into u which gives us u = (prcosθ)² + prsinθ(p+r). The next step is to compute (partial u)/(partial p), (partial u)/(partial r), (partial u)/(partial theta) by using the Chain Rule to find each partial derivative. After computing, you just substitute the given values of p, r, θ into the obtained derivates to get the final answers.
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Find 10^(5^101) (mod 21).
note: 10^(5^101) is not 10^(501)
We have [tex]\lambda(21)=6[/tex], where [tex]\lambda[/tex] is the Carmichael function. So we have
[tex]10^{5^{101}}\equiv10^{5^{101}\pmod6}\pmod{21}[/tex]
The powers of 5 modulo 6 follow a periodic pattern
[tex]5^1\equiv5\pmod6[/tex]
[tex]5^2\equiv25\equiv1\pmod6[/tex]
[tex]5^3\equiv1\cdot5\equiv5\pmod6[/tex]
[tex]5^4\equiv5^2\equiv1\pmod6[/tex]
and so on, with odd powers of 5 equivalent to 5 modulo 6. So
[tex]10^{5^{101}}\equiv10^{5^{101}\pmod6}\equiv10^5\pmod{21}[/tex]
The rest is easy to deal with. We have
[tex]10^2\equiv16\pmod{21}[/tex]
[tex]10^3\equiv160\equiv13\pmod{21}[/tex]
[tex]10^4\equiv130\equiv4\pmod{21}[/tex]
[tex]10^5\equiv40\equiv19\pmod{21}[/tex]
and so the answer is 19.
The equation below specifies a function. Determine whether the function is linear, constant, or neither.
3x + 4y = 1
Choose the correct answer below.
A constant function is specified by the equation.
B. linear function is specified by the equation.
C. Neither a constant function nor a linear function is specified by the equation.
Answer:
Linear function is specified by the equation ⇒ answer B
Step-by-step explanation:
* Look to the attached file
Answer:
B . Linear function.
Step-by-step explanation:
3x + 4y = 1
The degree of x and y is 1 and
if we drew a graph of this function we get a straight line.
Find the area of the region enclosed by the graphs of x=10-y^2 and x=7 (Use symbolic notation and fractions where needed.)
Answer:[tex]18\sqrt{3}[/tex]
Step-by-step explanation:
Given data
we haven given a parabola and a straight line
Parabola is [tex]{y^2}={-\left ( x-10\right )[/tex]
line is [tex]x=7[/tex]
Find the point of intersection of parabola and line
[tex]y=\pm \sqrt{3}[/tex] when[tex]x=7[/tex]
Area enclosed is the shaded area which is given by
[tex]Area=\int_{0}^{\sqrt{3}}\left ( 10-y^2 \right )dy[/tex]
[tex]Area=_{0}^{\sqrt{3}}10y-_{0}^{\sqrt{3}}\frac{y^3}{3}[/tex]
[tex]Area=10\sqrt{3}-\sqrt{3}[/tex]
[tex]Area=9\sqrt{3}units[/tex]
Required area will be double of calculated because it is symmetrical about x axis=[tex]18\sqrt{3}units[/tex]
To find the area of the region enclosed by the graphs of[tex]x=10-y^2[/tex]and x=7, we need to find the points of intersection between the two equations and then integrate the curve between those points.
Explanation:To find the area of the region enclosed by the graphs of [tex]x=10-y^2[/tex] and x=7, we need to find the points of intersection between the two equations. Setting x equal to each other, we have [tex]10-y^2=7.[/tex]Solving for y, we get y=±√3.
Now we can integrate the curve between the two values of y, as y goes from -√3 to √3. So the area is given by [tex]\int (10 - y^2 - 7) \, dy[/tex] from -√3 to √3.
Evaluating the integral, we get A=√3*10-2√3/3 ≈ 30.78.
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Write the following equation in slope-intercept form and identify the slope and y-intercept 5x+3y 15 The equation in slope-intercept form is y (Simplify your answ er. Type your answer in slope-intercept form. Use integers or fractions for any numbers in the expression.)
To write the equation 5x + 3y = 15 in slope-intercept form, solve for y to get y = (-5/3)x + 5. The slope is -5/3 and the y-intercept is 5.
To convert the equation 5x + 3y = 15 into slope-intercept form, which is y = mx + b, we need to solve for y. Here are the steps:
Subtract 5x from both sides: 3y = -5x + 15.Divide every term by 3 to isolate y: y = (-5/3)x + 5.In this slope-intercept form, the coefficient of x represents the slope and the constant term represents the y-intercept. Therefore, the slope is -5/3 and the y-intercept is 5. This tells us that for every increase of 1 on the horizontal axis (x), there is a decrease of 5/3 on the vertical axis (y), and the line crosses the y-axis at the point (0, 5).A bag of 100 tulip bulbs purchased from a nursery contains 20 red tulip bulbs, 20 yellow tulip bulbs, and 60 purple tulip bulbs. (a) What is the probability that a randomly selected tulip bulb is red? (b) What is the probability that a randomly selected tulip bulb is purple? (c) Interpret these two probabilities.
Answer: a) 0.2 b) 0.6
c) The event of selecting red tulip is not likely to occur.
The event of selecting purple tulip is likely to occur.
Step-by-step explanation:
Given : Total number of tulips = 100
The number of red tulips = 20
The number of purple tulips =60
The probability that a randomly selected tulip bulb is red :-
[tex]\dfrac{\text{Number of red tulips}}{\text{Total tulips}}\\\\=\dfrac{20}{100}=0.2[/tex]
Since 0.2 is less than 0.5.
It means that the event of selecting red tulip is not likely to occur.
The probability that a randomly selected tulip bulb is purple :-
[tex]\dfrac{\text{Number of purple tulips}}{\text{Total tulips}}\\\\=\dfrac{60}{100}=0.6[/tex]
Since 0.6 is more than 0.5.
It means that the event of selecting purple tulip is likely to occur.
Final answer:
The probability of selecting a red tulip bulb is 20%, and the probability of selecting a purple tulip bulb is 60%. These probabilities reflect the likelihood of picking a bulb of a particular color at random from the bag.
Explanation:
The question involves calculating the probability of selecting a red or purple tulip bulb from a bag.
Probability of Selecting a Red Tulip Bulb
The probability, P(Red), is calculated by dividing the number of red bulbs by the total number of bulbs:
P(Red) = Number of Red Bulbs / Total Number of Bulbs = 20 / 100 = 0.2
Probability of Selecting a Purple Tulip Bulb
Similarly, the probability, P(Purple), is:
P(Purple) = Number of Purple Bulbs / Total Number of Bulbs = 60 / 100 = 0.6
Interpretation of Probabilities
These probabilities indicate that there is a 20% chance of selecting a red bulb and a 60% chance of selecting a purple bulb from the bag. The higher the probability, the more likely it is to select a bulb of that color at random.
Two surveys were done regarding credit card debt. Survey #1: Five years ago the average credit card debt was $6618. ​Survey #2:The average credit card debt for a recent year was $9205. Assume sample sizes of 35 were used and the standard deviations of both samples were $1928. Is there enough evidence to believe that the average credit card debt has changed in the past 5 years? Assume a 5% Level of Significance
Answer:
There is enough evidence to believe that the average credit card debt has changed in the past 5 years
Step-by-step explanation:
We are to compare the means of two samples. Since only sample std deviations are used, we have to use t test for this hypothesis
H0: Means are equal
Ha: Means are not equal
(Two tailed test at 5% )
Difference between means [tex]M1-M2 = -2587[/tex]
Std deviation combined = 3856
Std error for difference = 460.88
t statistic[tex]= -2587/460.88=-5.613[/tex]
p value =0
Since p <0.05 reject null hypothesis.
There is enough evidence to believe that the average credit card debt has changed in the past 5 years
In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. The expected frequency for the Business College is
Answer:
Step-by-step explanation:
Given that in the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts College, and 30% of the students were in the Education College.
Table is prepared as follows
Bus coll. Lib Arts coll Educ. coll Total
Observed 90 120 90 300
Expected p.c. 35 35 30 100
Expected number 105 105 90 300
Percent*300/100
Hence expected frequency for business college = 105
Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 hours. IThe population is normally distributed. If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?. List your answer as a percentage to 2 decimal places without the % sign (X.XX)
Answer:
The percent of the parts are expected to fail before the 2100 hours is 0.15.
Step-by-step explanation:
Given :Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 hours.
To Find : If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?.
Solution:
We will use z score formula
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Mean value = [tex]\mu = 2500[/tex]
Standard deviation = [tex]\sigma = 135[/tex]
We are supposed to find If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?
So we are supposed to find P(z<2100)
so, x = 2100
Substitute the values in the formula
[tex]z=\frac{2100-2500}{135}[/tex]
[tex]z=−2.96[/tex]
Now to find P(z<2100) we will use z table
At z = −2.96 the value is 0.0015
So, In percent = [tex].0015 \times 100=0.15\%[/tex]
Hence The percent of the parts are expected to fail before the 2100 hours is 0.15.
Given P(A) 0.169, P(B) 0.041, and P(C) 0.172, and that events A, B, and C are mutually exclusive, what is the P(A or B or C)? Answer in decimal form. Round to 3 decimal places as needed Your Answer: Answer
Answer:
The value of P(A or B or C) is 0.382.
Step-by-step explanation:
Given,
P(A) = 0.169,
P(B) = 0.041,
P(C) = 0.172
Since, if events A, B and C are mutually events ( in which no element is common ),
Then, P(A∪B∪C) = P(A) + P(B) + P(C)
Or P(A or B or C) = P(A) + P(B) + P(C),
By substituting the values,
P(A or B or C) = 0.169 + 0.041 + 0.172 = 0.382
an irregular object with a mass of 1220g displaces 200 cubic cm of water when placed in a large overflow container. calculate the density of the object. what is the density in g/cm cubic
Answer:
[tex]6.1\frac{\text{ g}}{\text{ cm}^3}[/tex]
Step-by-step explanation:
We have been given that mass of an irregular object is 1220 g and it displaces 200 cubic cm of water when placed in a large overflow container. We are asked to find density of the object.
We will use density formula to solve our given problem.
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Since the object displaces 200 cubic cm of water, so the volume of irregular object will be equal to 200 cubic cm.
Upon substituting our given values in density formula, we will get:
[tex]\text{Density}=\frac{1220\text{ g}}{200\text{ cm}^3}[/tex]
[tex]\text{Density}=\frac{61\times 20\text{ g}}{10\times 20\text{ cm}^3}[/tex]
[tex]\text{Density}=\frac{61\text{ g}}{10\text{ cm}^3}[/tex]
[tex]\text{Density}=6.1\frac{\text{ g}}{\text{ cm}^3}[/tex]
Therefore, the density of the irregular object will be 6.1 grams per cubic centimeters.
A car dealership has 6 red, 9 silver, and 3 black cars on the lot. Six cars are randomly chosen to be displayed in front of the dealership. Find the probability that 3 cars are red and 3 are black. 0.001077 (Round to six decimal places as needed.)
Answer: Hence, our required probability is 0.001077.
Step-by-step explanation:
Since we have given that
Number of red cars = 6
Number of silver cars = 9
Number of black cars = 3
Total number of cars = 6+9+3=18
We need to find the probability that 3 cars are red and 3 are black.
So, the required probability is given by
[tex]P(3R\ and\ 3B)=\dfrac{^6C_3\times ^3C_3}{^{18}C_6}\\\\P(3R\ and\ 3B)=0.001077[/tex]
Hence, our required probability is 0.001077.
A gambler mixed a "cheat" die with all sixes into a box of eight normal dice. She chooses one at random, rolls it twice, and gets six both times. What is the probability that she chose the "cheat" die?
Answer:
The probability is [tex]\frac{1}{9}[/tex]
Step-by-step explanation:
There are a total of 9 die in the box after she added the "cheat" die. Since there is only 1 "cheat" die in the box and chooses a die at random then the probability of her having chosen the cheat die is [tex]\frac{1}{9}[/tex] . The fact that she rolled two sixes did not affect when she choose the die therefore the probability remains as [tex]\frac{number.of.cheat.die}{total.dice}[/tex].
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Beginning one month after birth of their son, Noah, the Nelsons deposited $100 each month in an annuity for his college fund. The annuity earned interest at an average rate of 6.8% compounded monthly until his 18th birthday. What was the amount of Noah's college fund on his 18th birthday? Referring to question 4, how much interest did Noah's college fund earn in total on his 18th birthday?
Answer:
$100 * (1 + 6.8%/12)^216 + $100*(1+6.8%/12)^215 + ... + $100*(1+6.8%/12)^1
Now note that
x + x^2 + x^3 + ... + x^N = x ( 1 + x + ... + x^(N-1) )
= x ( (x^N -1)/(x-1) )
Here, x = 1+6.8%1 = 1.00566666 and N = 216, so
$100 * ( 1.00566666 ( 1.00566666^216 -1) / 0.00566666 )
= $ 42398.33
The total interest earned is $42,398 - $21,600 = $20,798
Step-by-step explanation:
The generic formula used in this compound interest calculator is V = P(1+r/n)^(nt)
V = the future value of the investment
P = the principal investment amount
r = the annual interest rate
n = the number of times that interest is compounded per year
t = the number of years the money is invested for
Find each of the following for
f(x) = 8x + 3.
(a) f (x + h)
(b) f (x + h - f (x)
(c) (f (x+h - f(x))/h
(a)
[tex]f(x+ h)=8x+8h+3[/tex]
(b)
[tex]f(x+ h)-f(x)=8h[/tex]
(c)
[tex]\dfrac{f(x+ h)-f(x)}{h}=8[/tex]
Step-by-step explanation:We are given a function f(x) as :
[tex]f(x)=8x+3[/tex]
(a)
[tex]f(x+ h)[/tex]
We will substitute (x+h) in place of x in the function f(x) as follows:
[tex]f(x+h)=8(x+h)+3\\\\i.e.\\\\f(x+h)=8x+8h+3[/tex]
(b)
[tex]f(x+ h)-f(x)[/tex]
Now on subtracting the f(x+h) obtained in part (a) with the function f(x) we have:
[tex]f(x+h)-f(x)=8x+8h+3-(8x+3)\\\\i.e.\\\\f(x+h)-f(x)=8x+8h+3-8x-3\\\\i.e.\\\\f(x+h)-f(x)=8h[/tex]
(c)
[tex]\dfrac{f(x+ h)-f(x)}{h}[/tex]
In this part we will divide the numerator expression which is obtained in part (b) by h to get:
[tex]\dfrac{f(x+ h)-f(x)}{h}=\dfrac{8h}{h}\\\\i.e.\\\\\dfrac{f(x+h)-f(x)}{h}=8[/tex]