What is the area of this composite figure?

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.

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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]

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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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) t sampling distribution because B the population is normal, and standard deviation is unknown

b)  H0: mu <= 21

HA mu > 21

alpha = 0.05

t critical value at 4 df and alpha 0.05 is 2.132

The rejection region is t > 2.132

t = (xbar - µ)/(s/√n)

t = (19 - 21 )/(4/√5)

t = -2 / (4/2.2361)

t = -1.118

t does not fall into the rejection region, so we have insufficient evidence to reject the null hypothesis. The claim cannot be verified.

We tested the manufacturer's claim that the mean mpg is greater than 21, at alpha = 0.05. We used a one-tailed one-sample t-test (4 df). We placed the rejection region in the right tail of the t-distribution because we were only interested in the claim that the mileage was more than 21. The test result showed that the claim could not be validated. The sample mean was 19, which was less than the claim, so no calculations were needed to reject the null hypothesis. We were not able to find that the mean was statistically greater than 21.

Based on hypothesis testing in statistics, there isn't enough evidence to support the car company’s claim that the average gas mileage for its luxury sedan is at least 24 miles per gallon.

This question involves the use of hypothesis testing in statistics. The null hypothesis for this test is that the mean gas mileage is at least 24 miles per gallon (μ >= 24), and the alternative hypothesis is that the mean gas mileage is less than 24 miles per gallon (μ < 24).  

With a calculated sample mean of 23 miles per gallon and a sample standard deviation of 1.1 miles per gallon for 7 cars, we use the standard error formula SE = σ/√n = 1.1/√7 = 0.415 to calculate the standard error. The t value is then calculated as (X - μ) / SE = (23 - 24) / 0.415 = -2.41.

Using a t-distribution table, we find that the critical value for a one-tailed test with degrees of freedom = n - 1 = 6 and α = 0.05 is -1.943. Since our calculated t value (-2.41) is less than the critical value (-1.943), we reject the null hypothesis. Therefore, we cannot support the company’s claim that the mean gas mileage for its luxury sedan is at least 24 miles per gallon.

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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]

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]

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:

(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.

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.

Answer:yes

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

answer 40

Step-by-step explanation:

because you added all together to make one

Step-by-step answer:

Answer to problems of this kind is the reciprocal of the harmonic mean of the time required.

We need to find the average of the speeds, not the average of the time.

The respective speeds are 1/3 and 1/4.

The average of the speeds is therefore (1/3+1/4)/2 = 7/24  (harmonic mean of the time taken).

The time required is therefore the reciprocal of the unit speed,

T = 1/(7/24) = 24/7 = 3 3/7 minutes, or approximately 3.43 minutes.

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

Based on the study results, the percentage of girls born in China could range from 32.98% to 47.02%.

To determine if the percentage of girls born in China is 46.7%, we can calculate the confidence interval for the proportion of girls in the population using a binomial distribution. Based on the study, out of 150 births, 60 were girls and 90 were boys.

The confidence interval suggests that the true proportion of girls born in China could range from 32.98% to 47.02%. Since the reported figure of 46.7% falls within this interval, it is plausible based on the study results.

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:

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:

[tex]\boxed{x<2}[/tex]

Step-by-step explanation:

You subtract by 1 from both sides of equation.

[tex]2x+1-1<5-1[/tex]

Simplify.

[tex]5-1=4[/tex]

[tex]2x<4[/tex]

Divide by 2 from both sides of equation.

[tex]\frac{2x}{2}<\frac{4}{2}[/tex]

Simplify, to find the answer.

[tex]4\div2=2[/tex]

X<2 is the correct answer.

Answer: [tex]x<2[/tex]

Step-by-step explanation:

Given the inequality [tex]2x + 1 < 5[/tex] you can follow this procedure to solve it:

The first step is to subtract 1 from both sides on the inequaltity.

[tex]2x + 1-(1) < 5-(1)\\\\2x < 4[/tex]

Now, the second  and final step is to divide both sides of the inequality by 2. Therefore, you get this result:

[tex]\frac{2x}{2}<\frac{4}{2} \\\\(1)x<2\\\\x<2[/tex]

Hello!

Answer:

[tex]\boxed{m=\frac{8}{15}}[/tex]

Step-by-step explanation:

First, you switch sides.

[tex]m+\frac{1}{3}=\frac{13}{15}[/tex]

Then, you subtract by 1/3 from both sides.

[tex]m+\frac{1}{3}-\frac{1}{3}=\frac{13}{15}-\frac{1}{3}[/tex]

Simplify and solve.

[tex]\frac{13}{15}=\frac{8}{15}[/tex]

Therefore, [tex]\boxed{\frac{8}{15}}[/tex], which is our final answer.

I hope this helps you!

Have a nice day! :)

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:

Step-by-step explanation:

X/6 +4 = 20

X/6 = 20 - 4

X/6 = 16

X = 16/6

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

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 with explanation:

The equation which we have to solve by Newton-Raphson Method is,

 f(x)=log (3 x) +5 x²

[tex]f'(x)=\frac{1}{3x}+10 x[/tex]

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

  [tex]x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248[/tex]

[tex]x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012[/tex]

[tex]x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057[/tex]

[tex]x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052[/tex]

So, root of the equation =0.205 (Approx)

Approximate relative error

                [tex]=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41[/tex]

 Approximate relative error in terms of Percentage

   =0.41 × 100

   = 41 %

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

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

The maximum rate of change occurs in the direction of the gradient vector at (1, 1, 1).

[tex]T(x,y,z)=e^{-x^2-2y^2-3z^2}\implies\nabla T(x,y,z)=\langle-2x,-4y,-6z\rangle e^{-x^2-2y^2-3z^2}[/tex]

At (1, 1, 1), this has a value of

[tex]\nabla T(1,1,1)=\langle-2,-4,-6\rangle e^{-6}[/tex]

so the captain should move in the direction of the vector [tex]\langle-1, -2, -3\rangle[/tex] (which is a vector pointing in the same direction but scaled down by a factor of [tex]2e^{-6}[/tex]).

The direction Captain Ralph should proceed in order to decrease the temperature most rapidly is towards the direction of the steepest temperature decrease gradient. This direction is given by the negative gradient of the temperature function.

In this case, the negative gradient of T(x, y, z) = e^(-x^2 - 2y^2 - 3z^2) at the point (1, 1, 1) would be (-2e^(-6), -4e^(-6), -6e^(-6)).

Therefore, Captain Ralph should proceed in the direction (-2e^(-6), -4e^(-6), -6e^(-6)) to decrease the temperature most rapidly at his current location.

Answer:

The set of vertices of quadrilateral EFGH with the transformation 7 units right is E(4 , 4) , F(8 , 3) , G(10 , 6) , and H(8 , 6)

The set of vertices of quadrilateral EFGH with a reflection across the y-axis is E(3 , 4) , F(-1 , 3) , G(-3 , 6) , and H(-1 , 6)

The set of vertices of quadrilateral EFGH with a reflection across the x-axis is E(-3 , -4) , F(1 , -3) , G(3 , -6) , and H(1 , -6)

Step-by-step explanation:

Lets revise some transformation

- If point (x , y) reflected across the x-axis

 then Its image is (x , -y)

- If point (x , y) reflected across the y-axis

 then Its image is (-x , y)

- If the point (x , y) translated horizontally to the right by h units

 then its image is (x + h , y)

- If the point (x , y) translated horizontally to the left by h units

 then its image is (x - h , y)

* Now lets solve the problem

- The vertices of the quadrilateral ABCD are:

  A = (-3 , 4) , B = (1 , 3) , C = (3 , 6) , D = (1 , 6)

- The quadrilateral ABCD translated 7 units right to form

 quadrilateral EFGH

- We add each x-coordinates in ABCD by 7

∵ A = (-3 , 4)

∴ E = (-3 + 7 , 4) = (4 , 4)

∵ B = (1 , 3)

∴ F = (1 + 7 , 3) = (8 , 3)

∵ C = (3 , 6)

∴ G = (3 + 7 , 6) = (10 , 6)

∵ D = (1 , 6)

∴ H = (1 + 7 , 6) = (8 , 6)

* The set of vertices of quadrilateral EFGH with the transformation

  7 units right is E(4 , 4) , F(8 , 3) , G(10 , 6) , and H(8 , 6)

- The quadrilateral ABCD reflected across the y-axis to form

 quadrilateral EFGH

- We change the sign of the x-coordinate

∵ A = (-3 , 4)

∴ E = (3 , 4)

∵ B = (1 , 3)

∴ F = (-1 , 3)

∵ C = (3 , 6)

∴ G = (-3 , 6)

∵ D = (1 , 6)

∴ H = (-1 , 6)

* The set of vertices of quadrilateral EFGH with a reflection across the

  y-axis is E(3 , 4) , F(-1 , 3) , G(-3 , 6) , and H(-1 , 6)

- The quadrilateral ABCD reflected across the x-axis to form

 quadrilateral EFGH

- We change the sign of the y-coordinate

∵ A = (-3 , 4)

∴ E = (-3 , -4)

∵ B = (1 , 3)

∴ F = (1 , -3)

∵ C = (3 , 6)

∴ G = (3 , -6)

∵ D = (1 , 6)

∴ H = (1 , -6)

* The set of vertices of quadrilateral EFGH with a reflection across the

  x-axis is E(-3 , -4) , F(1 , -3) , G(3 , -6) , and H(1 , -6)

Answer:

here is the answer

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].

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