Create a question with this scenario you could ask that could be answered only by graphing or using logarithm.


David estimated he had about 20 fish in his pond. A year later, there were about 1.5 times as many fish. The year after that, the number of fish increased by a factor of 1.5 again. The number of fish is modeled by f(x)=20(1.5)^x.

Answer:yes

Step-by-step explanation:because there are more marbles than jars

Answer:

water : 90 ounces

salt solution: 90 ounces

Step-by-step explanation:

Call w the amount of water and call the solution containing 30% salt.

We want to get 180 ounces of a mixture with 15% salt.

So:

The amount of mixture will be:

[tex]w + s = 180[/tex]

the amount of salt will be

[tex]0w + 0.3s = 180 * 0.15[/tex]

[tex]0.3s = 27[/tex]

[tex]s = 90\ ounces[/tex]

Now we substitute the value of s in the first equation and solve for w

[tex]w + 90 = 180[/tex]

[tex]w = 90\ ounces[/tex]

Answer:

i say option (a) is the answer its correct.

Step-by-step explanation:

Answer:

A) [tex]\angle C \text{ is opposite } \overline{AB}[/tex]

Step-by-step explanation:

We need to name the side of the triangle that is across from angle C. A side of a triangle, if not otherwise given, is named based on the two points that form it. In this case, the side is formed by points B and C, so the side is called [tex]\overline{BC}[/tex].

If you evaluate it, it's −9≤x≤2 and trying to find the absolute maximum/minimum of it then you'll get nothing due to it being an improper fraction of some sorts.. And there's still nothing when trying to find it all together.. Sorry that I wasn't that much help.

Answer:

A= 4 , B= 3

Step-by-step explanation:

Simply put, use systems of equations to solve for A and B. First make y equal 4, and to make A equal to four, while being multiplied by 1 it has to be 4. Then solve backwards and now you know that A equals four, what multiplied by 4 is equal to 36, 9. What is the square root of 9? 3!

Answer:

The range of this data set is 41.8

The standard deviation of the data set is 16.42

Step-by-step explanation:

* Lets read the information and use it to solve the problem

- There is a sample of size n = 10,  is drawn from a population

- The data are: 97 , 100.4 , 101.9 , 114.2 , 116.4 , 117.6 , 128.8 , 138.8 ,

  138.8 , 138.8

- The range is the difference between the largest number and

  the smallest number

∵ The largest number is 138.8

∵ The smallest number is 97

∴ The range = 138.8 - 97 = 41.8

* The range of this data set is 41.8

- Lets explain how to find the standard deviation

# Step 1: find the mean of the data set

∵ The mean = the sum of the data ÷ the number of the data

∵ The data set is 97 , 100.4 , 101.9 , 114.2 , 116.4 , 117.6 , 128.8 , 138.8 ,

  138.8 , 138.8

∵ Their sum = 97 + 100.4 + 101.9 + 114.2 + 116.4 + 117.6 + 128.8 + 138.8 +

  138.8 + 138.8 = 1192.7

∵ n = 10  

∴ The mean = 1192.7 ÷ 10 = 119.27

# Step 2: subtract the mean from each data and square the answer

∴ (97 - 119.27)² = 495.95

∴ (100.4 - 119.27)² = 356.08

∴ (101.9 - 119.27)² = 301.72

∴ (114.2 - 119.27)² = 25.70

∴ (116.4 - 119.27)² = 8.24

∴ (117.6 - 119.27)² = 2.79

∴ (128.8 - 119.27)² = 90.82

∴ (138.8 - 119.27)² = 381.42

∴ (138.8 - 119.27)² = 381.42

∴ (138.8 - 119.27)² = 381.42

# Step 3: find the mean of these squared difference

∵ A Sample: divide by n - 1 when calculating standard deviation of

  a sample

∵ The mean = the sum of the data ÷ (the number of the data - 1)

∵ The sum = 495.95 + 356.08 + 301.72 + 25.70 + 8.24 + 2.79 + 90.82 +

   381.42 + 381.42 + 381.42 = 2425.56

∴ The mean = 2425.56 ÷ (10 - 1) = 269.51

# Step 4: the standard deviation is the square root of this mean

∴ The standard deviation = √(269.51) = 16.416658 ≅ 16.42

* The standard deviation of the data set is 16.42

Answer:

The probability of getting a jack, a three, a club or a diamond is 0.58.

Step-by-step explanation:

In a standard deck of 52 cards have 13 club, 13 spade, 13 diamond, 13 heart cards. Each suit has one jack and 3.

Number of club cards = 13

Number of diamond cards = 13

Number of jack = 4

Number of 3 = 4

Jack of club  and diamond = 2

3 of club  and diamond = 2

Total number of cards that are either a jack, a three, a club or a diamond is

[tex]13+13+4+4-2-2=30[/tex]

The probability of getting a jack, a three, a club or a diamond is

[tex]Probability=\frac{\text{A jack, a three, a club or a diamond}}{\text{Total number of cards}}[/tex]

[tex]Probability=\frac{30}{52}[/tex]

[tex]Probability=0.576923076923[/tex]

[tex]Probability\approx 0.58[/tex]

Therefore the probability of getting a jack, a three, a club or a diamond is 0.58.

The correct answer is 0.50.

To determine the probability of getting a jack, a three, a club, or a diamond from a standard 52-card deck, we can calculate the probability of each individual event and then combine them, taking care to avoid double-counting any cards.

First, let's calculate the probability of drawing a jack. There are 4 jacks in the deck (one for each suit). Since there are 52 cards in total, the probability of drawing a jack is:

[tex]\[ P(\text{jack}) = \frac{4}{52} \][/tex]

Next, we calculate the probability of drawing a three. There are also 4 threes in the deck, one for each suit. So, the probability of drawing a three is:

[tex]\[ P(\text{three}) = \frac{4}{52} \][/tex]

Now, let's calculate the probability of drawing a club. There are 13 clubs in the deck (since there are 13 cards in each suit). Thus, the probability of drawing a club is:

[tex]\[ P(\text{club}) = \frac{13}{52} \][/tex]

Similarly, there are 13 diamonds in the deck, so the probability of drawing a diamond is:

[tex]\[ P(\text{diamond}) = \frac{13}{52} \][/tex]

However, we must be careful not to double-count the cards that are both a jack or a three and a club or a diamond. There are 2 jacks and 2 threes that are also clubs or diamonds (one jack and one three of clubs, and one jack and one three of diamonds).

 To find the total probability, we add the probabilities of each event and subtract the probabilities of the events that have been counted twice (the jack and three of clubs and diamonds):

[tex]\[ P(\text{total}) = P(\text{jack}) + P(\text{three}) + P(\text{club}) + P(\text{diamond}) - 2 \times P(\text{jack or three of clubs or diamonds}) \] \[ P(\text{total}) = \frac{4}{52} + \frac{4}{52} + \frac{13}{52} + \frac{13}{52} - 2 \times \frac{2}{52} \] \[ P(\text{total}) = \frac{4 + 4 + 13 + 13 - 4}{52} \] \[ P(\text{total}) = \frac{30}{52} \] \[ P(\text{total}) = \frac{15}{26} \] \[ P(\text{total}) = \frac{5}{8} \] \[ P(\text{total}) = 0.625 \][/tex]

Since we need to round to the nearest hundredth, the final answer is:

[tex]\[ P(\text{total}) \approx 0.63 \][/tex]

[tex]\[ P(\text{total}) = P(\text{jack}) + P(\text{three}) + P(\text{club}) + P(\text{diamond}) \] \[ P(\text{total}) = \frac{4}{52} + \frac{4}{52} + \frac{13}{52} + \frac{13}{52} \] \[ P(\text{total}) = \frac{4 + 4 + 13 + 13}{52} \] \[ P(\text{total}) = \frac{34}{52} \] \[ P(\text{total}) = \frac{17}{26} \] \[ P(\text{total}) = \frac{1}{2} \] \[ P(\text{total}) = 0.50 \][/tex]

Answer:

  2nd choice: Counterclockwise rotation about the origin by 180 degrees followed by a reflection about the y-axis

Step-by-step explanation:

A simple reflection across the x-axis will do.

A rotation of 180 degrees about the origin is equivalent to a reflection across both axes. Then a reflection back across the y-axis leaves the net effect being the desired reflection across the x-axis.

ANSWER

See below

EXPLANATION

We want to verify that,

[tex] { \sin ^{4} x} - { \sin^{2} x} = { \cos ^{4} x} - { \cos^{2} x}[/tex]

To verify this identity, we can take the left hand side simplify it to get the right hand side or vice versa.

[tex]{ \sin ^{4} x} - { \sin^{2} x} =( { \sin ^{2} x} )^{2} - { \sin^{2} x}[/tex]

[tex]{ \sin ^{4} x} - { \sin^{2} x} ={ \sin ^{2} x}({ \sin ^{2} x} - 1)[/tex]

[tex]{ \sin ^{4} x} - { \sin^{2} x} ={ \sin ^{2} x} \times - (1 - { \sin ^{2} x})[/tex]

[tex]{ \sin ^{4} x} - { \sin^{2} x} =({1 - \cos^{2} x} )\times - ({ \cos^{2} x})[/tex]

[tex]{ \sin ^{4} x} - { \sin^{2} x} =({ \cos^{2} x} - 1 )\times ({ \cos^{2} x})[/tex]

We now expand the right hand side to get:

[tex] { \sin ^{4} x} - { \sin^{2} x} = { \cos ^{4} x} - { \cos^{2} x}[/tex]

Answer:

[tex]\large\boxed{8m^2n^3-24m^2n^2+4m^3n=4m^2n(2n^2-6n+m)}[/tex]

Step-by-step explanation:

[tex]8m^2n^3-24m^2n^2+4m^3n\\\\8m^2n^3=\boxed{(2)}\boxed{(2)}(2)\boxed{(m)}\boxed{(m)}\boxed{(n)}(n)(n)\\\\24m^2n^2=\boxed{(2)}\boxed{(2)}(2)(3)\boxed{(m)}\boxed{(m)}\boxed{(n)}(n)\\\\4m^3n=\boxed{(2)}\boxed{(2)}\boxed{(m)}\boxed{(m)}(m)\boxed{(n)}\\\\8m^2n^3-24m^2n^2+4m^3n\\\\=\boxed{(2)}\boxed{(2)}\boxed{(m)}\boxed{(m)}\boxed{(n)}\bigg((2)(n)(n)-(2)(3)(n)+(m)\bigg)\\\\=4m^2n(2n^2-6n+m)[/tex]

Answer is 'a'.(4;8;8)

All the details are provided in the attachment; the answer is marked with green colour.

Answer:

The number of key rings sold on that day is 4000 key rings

Step-by-step explanation:

* Lets explain the information in the problem

- The profit earned (in thousands of dollars) per day by selling n number

  of key rings is given by the function P(n) = n² - 2n - 3, where n is the

  number of key rings in thousands and P is the profit in thousands

  for one day

- On a particular day the total profit is $5,000

∵ 5000 = 5 in thousands

∵ The function P(n) is the profit of n key ring in thousands

∴ P(n) = 5

- Lets solve the function to find the number of key rings

∵ P(n) = n² - 2n - 3

∴ 5 = n² - 2n - 3 ⇒ subtract 5 from both sides

∴ 0 = n² - 2n - 8 ⇒ factorize it

∵ n² = n × n ⇒ 1st terms in the 2 brackets

∵ -8 = -4 × 2 ⇒ 2nd terms in the 2 brackets

∵ n × -4 = -4n ⇒ nears

∵ n × 2 = 2n ⇒ extremes

∵ -4n + 2n = -2n ⇒ the middle term

∴ (n - 4)(n + 2) = 0 ⇒ equate each bracket by 0 to find n

∴ n - 4 = 0 ⇒ add 4 to both sides

∴ n = 4 key ring in thousands = 4000 key rings

- OR

∴ n + 2 = 0 ⇒ subtract 2 from both sides

∴ n = -2 ⇒ we will refused this value because number of key rings

   must be positive

∴ The number of key rings sold on that day is 4000 key rings

To find the number of key rings sold on a particular day when the total profit is $5,000, we need to solve the given equation for n.

The owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by the equation P(n) = n^2-2n-3. To find the number of key rings sold on a particular day when the total profit is $5,000, we need to solve the equation P(n) = 5000 for n.

Step 1: Set the equation equal to 5000: n^2-2n-3 = 5000.

Step 2: Rearrange the equation and set it equal to zero: n^2-2n-5003 = 0.

Step 3: Solve the quadratic equation using factoring, completing the square, or the quadratic formula to find the values of n.

Step 4: The solutions will give us the possible values of n, representing the number of key rings sold on the particular day when the total profit is $5,000.

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Answer:

The price of the home = 232,000

20% is down payment.

Part A:

[tex]0.20\times232000=46400[/tex]

So, the loan amount will be =[tex]232000-46400=185600[/tex]

Loan amount or p = $185,600

Part B:

p = 185600

r = [tex]3.6/12/100=0.003[/tex]

n = [tex]10\times12=120[/tex]

The EMI formula is :

[tex]\frac{p\times r\times(1+r)^{n} }{(1+r)^{n}-1 }[/tex]

Now putting the values in formula we get

[tex]\frac{185600\times 0.003\times(1+0.003)^{120} }{(1+0.003)^{120}-1 }[/tex]

=> [tex]\frac{185600\times 0.003\times(1.003)^{120} }{(1.003)^{120}-1 }[/tex]

Monthly payments = $1844.02

Part C:

p = 185600

r = [tex]3.6/12/100=0.003[/tex]

n = [tex]20\times12=240[/tex]

Now putting the values in formula we get

[tex]\frac{185600\times 0.003\times(1+0.003)^{240} }{(1+0.003)^{240}-1 }[/tex]

=> [tex]\frac{185600\times 0.003\times(1.003)^{240} }{(1.003)^{240}-1 }[/tex]

Monthly payments = $1085.96

Part D:

For 10 year loan you have to pay = [tex]120\times1844.02=221282.40[/tex]

For 20 years loan you have to pay =[tex]240\times1085.96=260630.40[/tex]

So, you ended up paying [tex]260630.40-221282.40=39348[/tex] dollars more in longer loan.

The difference is $39,348.

Answer:

a) $42,580

b) $43,900

Step-by-step explanation:

Recall, "mode" refers to the value which occurs most frequently.

In this case, the question says that there are 2 modes,

this means $34,000 and $50,500 both share the spot for the most frequently appearing salary.

because there are only 5 employees (and hence 5 salaries), the only possible way that there are two modes is if there are two of each mode, leaving only the last salary unknown.

i.e  if we list the 5 salaries (in no particular order)

$34,000   $34,000  $50,500   $50,500   $ x

where x is the 5th unknown salary.

Given that the mean is $42,100

Then (34,000 + 34,000 + 50,500 + 50,500 + x) / 5 = 42,100

solving for x gives x = $41,500

Now we know all the values, we can rearrange them in increasing value:

$34,000   $34,000  $41,500   $50,500   $50,500  

from this, we can see that the median salary is $41,500

Given that the median salary gets a $2400 raise,

the new median salary = $41,500 + $2400 = $43,900 (Ans for part b)

new mean salary,

= ($34,000  + $34,000 + $43,900 +  $50,500  + $50,500  ) / 5

= $42,580 (answer for part a)

For this case we can raise a rule of three:

600 employees -------------> 100%

270 employees -------------> x

Where the variable "x" represents the percentage of employees who are satisfied with their salary. So, we have:

[tex]x = \frac {270 * 100} {600}\\x = 45[/tex]

Thus, 45% of employees are satisfied with their salary.

Answer:

45%

ANSWER

[tex]45\%[/tex]

EXPLANATION

The total number of employees is 600.

The number of employees who are happy with their pay is 270.

The percentage of employees who are happy with their pay is the number who are happy with their pay divided by total number of employees times 100%

[tex] \frac{270}{600} \times 100\%[/tex]

This simplifies to

[tex]45\%[/tex]

Final answer:

The directional derivative of the function f(x, y, z) = [tex]xe^y + ye^z + ze^x[/tex] at the point (0, 0, 0) in the direction of the vector v = 6, 3, −3 is 0.

Explanation:

To find the directional derivative of the function f(x, y, z) = [tex]xe^y + ye^z + ze^x[/tex] at the point (0, 0, 0) in the direction of the vector v = 6, 3, −3, we first need to find the gradient of f. The gradient of f, denoted as ∇f, is a vector of partial derivatives with respect to each variable. We calculate the partial derivatives as follows:

At the point (0, 0, 0), the gradient ∇f is (0 + 0, 0 + 1, 0 + 1) = (0, 1, 1).

Next, we need to normalize the given vector v. The normalization process involves dividing v by its magnitude to obtain a unit vector u in the direction of v. The magnitude of v is √(6² + 3² + (-3)²) = √(36 + 9 + 9) = √54. Therefore, the unit vector u is (6/√54, 3/√54, -3/√54).

Finally, the directional derivative of f at (0, 0, 0) in the direction of v is the dot product of ∇f and u, which is (0, 1, 1) ⋅ (6/√54, 3/√54, -3/√54) = 0*6/√54 + 1*3/√54 + 1*(-3)/√54 = 0.

Final answer:

The probability that a coffee maker will have a defective cord is calculated by using the principle of inclusion-exclusion. Subtracting the percentage of coffee makers with faulty switches from the total percentage with any defect and adding the percentage with both defects, we find that the probability is 1.6%.

Explanation:

To calculate the probability that a coffee maker will have a defective cord, we can use the principle of inclusion-exclusion. According to the question, 4.0% of coffee makers have either a faulty switch or a defective cord. Of these, 2.5% have a faulty switch, and 0.1% have both defects. The probability that a coffee maker will have a defective cord is the total probability of any defect minus the probability of a faulty switch, plus the probability of having both defects, since those with both defects were counted in both the faulty switch and defective cord categories.

The formula we will use is: Probability(defective cord) = Probability(faulty switch or defective cord) - Probability(faulty switch) + Probability(both defects).

Plugging in the values we have: Probability(defective cord) = 4.0% - 2.5% + 0.1% = 1.6%

Therefore, the probability that a coffee maker will have a defective cord is 1.6%.

ANSWER

{x|x < -2 or x > 8}

EXPLANATION

The given absolute inequality is

[tex] |2x - 6| \: > \: 10[/tex]

By the definition of absolute value,

[tex] (2x - 6)\: > \: 10 \: or \: \: - (2x - 6)\: > \: 10[/tex]

Multiply through the second inequality by -1 and reverse the inequality sign

[tex]2x - 6\: > \: 10 \: or \: \: 2x - 6\: < \: - 10[/tex]

[tex]2x \: > \: 10 + 6\: or \: \: 2x \: < \: - 10 + 6[/tex]

Simplify

[tex]2x \: > \: 16\: or \: \: 2x \: < \: -4[/tex]

Divide through by 2

[tex]x \: > \: 8\: or \: \: x \: < \: -2[/tex]

Answer:

{x|x < -2 or x > 8}

Step-by-step explanation:

|2x - 6| > 10

We split the inequality into two functions, one positive and one negative.  The negative one flips the inequality.  since this is greater than, this is an or problem

2x-6 >10                 or  2x-6 < -10

Add 6 to each side

2x-6+6 > 10+6         2x-6+6 < -10+6

2x   > 16                    2x < -4

Divide by 2

2x/2 > 16/2                2x/2 < -4/2

x >8                 or        x < -2

answer 40

Step-by-step explanation:

because you added all together to make one

Answer:

No, the point [tex](2,3)[/tex] is not a solution to the system of equation [tex]x2+y2=13[/tex] and [tex]2x-y=4[/tex].

Step-by-step explanation:

To make sure it is, it has to work on both equation. Plugging this into the first and second equations respectivly gives us:

[tex]13=x2+y2=(2)2+(3)2=4+6=10[/tex]

[tex]4=2x-y=2(2)-(3)=4-3=1[/tex]

Since [tex]13\neq10[/tex] and [tex]4\neq 1[/tex], the point [tex](2,3)[/tex] is not a solution to the system of equation [tex]x2+y2=13[/tex] and [tex]2x-y=4[/tex].

Answer:

That is False.

Step-by-step explanation:

The equation of the least-squares regression line can be used to predict the beak heat loss at a specific temperature. The coefficient of determination indicates the proportion of variation explained by the relationship between beak heat loss and temperature. The correlation coefficient quantifies the strength and direction of the linear relationship.

The given data represents the relationship between the beak heat loss of the toco toucan and the outdoor temperature. To find the equation of the least-squares regression line, we need to calculate the slope, which represents the change in beak heat loss for every degree Celsius increase in temperature, and the y-intercept, which represents the predicted beak heat loss at 0 degrees Celsius. Using these values, we can then predict the beak heat loss at a temperature of 25 degrees Celsius. To determine the percent of variation explained by the straight-line relationship, we can calculate the coefficient of determination (r^2). Finally, the correlation coefficient (r) represents the strength and direction of the linear relationship between beak heat loss and temperature.

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Answer:

The words from parsley are PAR, YES, RAP, PAY,and SLAP

The words from pepper are PEEP and that is it

Step-by-step explanation:

Answer:

  -18, -26

Step-by-step explanation:

The differences are decreasing by 1 each time, so the next differences will be -7 and -8.

  -11 -7 = -18

  -18 -8 = -26

Answer:

The solution is [tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]

Step-by-step explanation:

We need to find the solution of IVP for differential equation [tex]\frac{dy}{dx}=(x+2)^{2}e^{y}[/tex] when [tex]y(1)=0[/tex]

[tex]\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}[/tex]

[tex]\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)[/tex]

[tex]\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}[/tex]

[tex]\frac{1}{e^y}y'\:=\left(x+2\right)^2[/tex]

[tex]N\left(y\right)\cdot y'\:=M\left(x\right)[/tex]

[tex]N\left(y\right)=\frac{1}{e^y},\:\quad M\left(x\right)=\left(x+2\right)^2[/tex]

[tex]\mathrm{Solve\:}\:\frac{1}{e^y}y'\:=\left(x+2\right)^2[/tex]

[tex]\frac{1}{e^y}y'\:=x^{2}+4+4x[/tex]

Integrate both the sides with respect to dx

[tex]\int\frac{1}{e^y}y'dx\:=\intx^{2}dx+4\int dx+4\int x dx[/tex]

[tex]\int e^{-y}dy\:=\intx^{2}dx+4\int dx+4\int x dx[/tex]

[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+4\frac{x^{2}}{2}+c_1[/tex]

[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+2x^{2}+c_1[/tex]

Since, IVP is y(1)=0

put x=1 and y=0 in above equation

[tex]-\frac{1}{e^{0}}\:=\frac{1^{3}}{3}+4(1)+2(1)^{2}+c_1[/tex]

[tex]-1\:=\frac{1}{3}+4+2+c_1[/tex]

[tex]-1\:=\frac{19}{3}+c_1[/tex]

add both the sides by [tex]-\frac{19}{3}[/tex]

[tex]-1-\frac{19}{3}\:=\frac{19}{3}-\frac{19}{3}+c_1[/tex]

[tex]-1-\frac{19}{3}\:=c_1[/tex]

[tex]-\frac{22}{3}\:=c_1[/tex]

so,

[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+2x^{2}-\frac{22}{3}[/tex]

Multiply both the sides by '-1'

[tex]\frac{1}{e^{y}}\:=-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3}[/tex]

[tex]e^{-y}\:=-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3}[/tex]

Take natural logarithm both the sides,

[tex]-y\:=\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]

Multiply both the sides by '-1'

[tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]

Therefore, the solution is [tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]

Simplify 3 + 3 to 6

6^2/10 - 4 × 3

Simplify 4 × 3 to 12

6^2/10 - 12

Simplify 10 - 12 to -2

6^2/-2

Simplify 6^2 to 36

36/-2

Move the negative sign to the left

-36/2

Simplify 36/2 to 18

= -18

Answer: Hence, Probability of more than 1 death in a corps in a year is 0.126.

Step-by-step explanation:

Since we have given that

Mean for a poisson distribution (λ) = 0.61

Number of years = 20 years

We need to find the probability of more than 1 death  in a corps in a year.

P(X>1)=1-P(X=0)-P(X=1)

Here,

[tex]P(X=0)=\dfrac{e^{-0.61}(0.61)^0}{0!}=0.543\\\\and\\\\P(X=1)=\dfrac{e^{-0.61}(0.61)}{1}=0.331[/tex]

So,

P(X>1)=1-0.543-0.331=0.126

Hence, Probability of more than 1 death in a corps in a year is 0.126.

Using the Poisson distribution with a mean of 0.61, we calculate the probability of 0 or 1 death and subtract that from 1 to get the probability of more than 1 death in a Prussian cavalry corps in a year.

Based on L. J. Bortkiewicz's study, the number of soldiers killed by horse kicks in the Prussian cavalry follows a Poisson distribution with a mean (λ) of 0.61. To calculate the probability of more than one death in a corps in a year, we use the Poisson probability formula:

P(X > k) = 1 - P(X ≤ k)

Where P(X > k) is the probability of having more than k events (in this case, deaths), and P(X ≤ k) is the probability of k or fewer events. In this scenario, k equals 1. So, we need to calculate the probability of 0 or 1 death and subtract from 1 to get the probability of more than 1 death.

Using the Poisson probability formula:

Let's calculate:

The resulting calculation will give us the probability of more than one death due to horse kicks in a Prussian cavalry corps within one year.

https://brainly.com/question/33722848

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Answer:

(a) 0.110 ⇒ rounded to three decimal places

(b) 0.097 ⇒ rounded to three decimal places

(c) 0.234 ⇒ rounded to three decimal places

(d) 0.235 ⇒ rounded to three decimal places

Step-by-step explanation:

* Lets solve the problem using the combination

- The order is not important in this problem, so we can use the

  combination nCr to find the probability

- There are 22 compact fluorescent light bulbs

# 8 ⇒ rated 13 watt

# 9 ⇒ rated 18 watt

# 5 ⇒ rated 23 watt

- 3 bulbs are randomly selected

(a) Exactly two of the selected bulbs are rated 23 watt

∵ Exactly two of them are rated 23 watt

∵ The number of bulbs rated 23 watt is 5

∴ We will chose 2 from 5

∴ 5C2 = 10 ⇒ by using calculator or by next rule

# nCr = n!/[r! × (n-r)!]

- 5C2= 5!/[2! × (5 - 2)!] = (5×4×3×2×1)/[(2×1)×(3×2×1)] = 120/12 = 10

∴ There are 10 ways to chose 2 bulbs from 5

∵ The 3rd bulbs will chosen from the other two types

∵ The other two types = 8 + 9 = 17

∴ We will chose 1 bulbs from 17 means 17C1

∵ 17C1 = 17 ⇒ (by the same way above)

∴ There are 17 ways to chose 1 bulbs from 17

- 10 ways for two bulbs and 17 ways for one bulb

∴ There are 10 × 17 = 170 ways two chose 3 bulbs exact 2 of them

   rated 23 watt (and means multiply)

∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs

∵ 22C3 = 1540 ⇒ (by the same way above)

∴ The probability = 170/1540 = 17/154 = 0.110

* The probability that exactly two of the selected bulbs are rated

  23-watt is 0.110 ⇒ rounded to three decimal places

(b) All three of the bulbs have the same rating

- We can have either all 13 watt, all 18 watt or all 23 watt.

# 13 watt

∵ There are 8 are rated 13 watt

∵ We will chose 3 of them

∴ There are 8C3 ways to chose 3 bulbs from 8 bulbs

∵ 8C3 = 56

∴ There are 56 ways to chose 3 bulbs rated 13 watt

# 18 watt

∵ There are 9 are rated 18 watt

∵ We will chose 3 of them

∴ There are 9C3 ways to chose 3 bulbs from 9 bulbs

∵ 9C3 = 84

∴ There are 84 ways to chose 3 bulbs rated 18 watt

# 23 watt

∵ There are 5 are rated 23 watt

∵ We will chose 3 of them

∴ There are 5C3 ways to chose 3 bulbs from 5 bulbs

∵ 5C3 = 10

∴ There are 10 ways to chose 3 bulbs rated 23 watt

- 65 ways or 84 ways or 10 ways for all three of the bulbs have

 the same rating

∴ There are 56 + 84 + 10 = 150 ways two chose 3 bulbs have the

   same rating (or means add)

∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs

∵ 22C3 = 1540

∴ The probability = 150/1540 = 15/154 = 0.097

* The probability that all three of the bulbs have the same rating is

  0.097 ⇒ rounded to three decimal places

(c) One bulb of each type is selected

- We have to select one bulb of each type

# 13 watt

∵ There are 8 bulbs rated 13 watt

∵ We will chose 1 of them

∴ There are 8C1 ways to chose 1 bulbs from 8 bulbs

∵ 8C1 = 8

∴ There are 8 ways to chose 1 bulbs rated 13 watt

# 18 watt

∵ There are 9 bulbs rated 18 watt

∵ We will chose 1 of them

∴ There are 9C1 ways to chose 1 bulbs from 9 bulbs

∵ 9C1 = 9

∴ There are 9 ways to chose 1 bulbs rated 18 watt

# 23 watt

∵ There are 5 bulbs rated 23 watt

∵ We will chose 1 of them

∴ There are 5C1 ways to chose 1 bulbs from 5 bulbs

∵ 5C1 = 5

∴ There are 5 ways to chose 1 bulbs rated 23 watt

- 8 ways and 9 ways and 5 ways for one bulb of each type is selected

∴ There are 8 × 9 ×  5 = 360 ways that one bulb of each type is

  selected (and means multiply)

∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs

∵ 22C3 = 1540

∴ The probability = 360/1540 = 18/77 = 0.234

* The probability that one bulb of each type is selected is 0.234

  ⇒ rounded to three decimal places

(d) If bulbs are selected one by one until a 23-watt bulb is obtained,

    it is necessary to examine at least 6 bulbs

- We have a total of 22 bulbs and 5 of them are rate 23 watt

∴ There are 22 - 5 = 17 bulbs not rate 23 watt

∵ We examine at least 6 so the 6th one will be 23 watt

∴ There are 17C5 ways that the bulb not rate 23 watt

∵ 17C5 = 6188

∴ There are 6188 ways that the bulb is not rate 23 watt

∵ There are 22C5 ways to chose 5 bulbs from total 22 bulbs

∵ 22C5 = 26334

∴ The probability = 6188/26334 = 442/1881 = 0.235

* If bulbs are selected one by one until a 23-watt bulb is obtained,

 the probability that it is necessary to examine at least 6 bulbs is

  0.235 ⇒ rounded to three decimal places

OR

we can use the rule:

# P ( examine at least six ) = 1 − P ( examine at most five )  

# P ( examine at least six ) = 1 − P (1) − P (2) − P (3) − P (4) − P (5)

  where P is the probability  

∵ P(1) = 5/22

∵ P(2) = (5 × 17)/(22 × 21) = 85/462

∵ P(3) = (5 × 17× 16)/(22 ×21 × 20) = 34/231

∵ P(4) = (5 × 17 × 16 × 15)/(22 ×21 × 20 ×19) = 170/1463

∵ P(5) = (5 × 17 × 16 × 15 × 14)/(22 × 21 × 20 × 19 × 18) = 170/1881

∴ P = 1 - 5/22 - 85/462 - 34/231 - 170/1463 - 170/1881 = 442/1881

∴ P = 0.235  

These probabilities are calculated by considering the number of combinations of selections.

a) Probability of Exactly Two 23-Watt Bulbs= 0.110

b)Probability of All Bulbs Having the Same Rating = 0.097

c)Probability of One Bulb of Each Type = 0.234

d)Probability of At Least 6 Bulbs Needed to Get a 23-Watt Bulb =0.151

Given a box containing 22 lightbulbs with specific wattage ratings, we can calculate the probabilities of various selection events using combinatorial methods.

(a) Probability of Exactly Two 23-Watt Bulbs

We need the probability of selecting exactly two 23-watt bulbs out of three:

Thus, the probability is P = (10 * 17) / 1540 = 0.110.

(b) Probability of All Bulbs Having the Same Rating

Total probability: 0.036 + 0.055 + 0.006 = 0.097.

(c) Probability of One Bulb of Each Type

Number of ways to select 1 bulb from each type: 8 * 9 * 5 = 360.

Probability: P = 360 / 1540 = 0.234.

(d) Probability of At Least 6 Bulbs Needed to Get a 23-Watt Bulb

Probability of the first 5 bulbs not being 23-watt: (17/22) * (16/21) * (15/20) * (14/19) * (13/18).

Calculation: P = (17*16*15*14*13) / (22*21*20*19*18) = 0.151.

Answer:

a) 40,320

b) 384

Step-by-step explanation:

Given,

The total number of seats = 8,

Also, these 8 seats are occupied by 4 married couples or 8 people,

a) Thus, if there is no restrictions of seating ( that is any person can seat with any person ),

Then, the total number of arrangement = 8 ! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320,

b) if each married couple is seated together,

Then, the 4 couples can seat in 4 pair of seats,

Also, in a pair of seats a couple can choose any of the two seats,

So, the total number of arrangement

[tex]=4! \times 2^4[/tex]

[tex]=24\times 16[/tex]

[tex]=384[/tex]