Please help!!! Find the domain of the function y = 2 cot(5∕8x).
A) All real numbers except odd integer multiples of 8π∕5
B) All real numbers except 0 and integer multiples of 8π∕5
C) All real numbers except 0 and integer multiples of 4π∕5
D) All real numbers except odd integer multiples of 4π∕5

Answer:

1+i

Step-by-step explanation:

I do believe i to be the imaginary unit.

Let's write out some partial sums from power=0 to power=7 or whatever we need to see a pattern.

i^0=1

i^0+i^1=1+i

i^0+i^1+i^2=1+i+-1=i

i^0+i^1+i^2+i^3=i+i^3=i+-i=0

i^0+i^1+i^2+i^3+i^4=0+i^4=0+1=1

Hmmm.... we might see 1+i, then i, then 0 again.... let's see.

i^0+i^1+i^2+i^3+i^4+i^5=1+i

Coolness so we should see a pattern

Sum from power=0 to power=multiples of 4 will give us 1.

Sum from power=0 to power=remainder of 1 when final power is divided by 4 gives us 1+i.

Sum from power=0 to power=remainder of 2 when final power is divided by 4 gives us i.

Sum from power=0 to power=remainder of 3 when final power is divided by 4 gives us 1

0.

So 2021 divided by 4....

Since 2020 is a multiple of 4, then 2021 has a remainder of 1 when divided by 4.

So the answer is 1+i.

Answer:

Z =  -1.60

it is low ... it appears that for this problem 2 standard deviations below must be reached to be considered "unusual"

Step-by-step explanation:

Answer:

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Step-by-step explanation:

We have the mean, which means that the Poisson distribution is used to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

A mean of 3.5 customers arrive hourly at the drive-through window.

This means that [tex]\mu = 3.5[/tex]

What is the probability that, in any hour, more than 5 customers will arrive?

This is:

[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]

In which

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

Then

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3.5}*3.5^{0}}{(0)!} = 0.0302[/tex]

[tex]P(X = 1) = \frac{e^{-3.5}*3.5^{1}}{(1)!} = 0.1057[/tex]

[tex]P(X = 2) = \frac{e^{-3.5}*3.5^{2}}{(2)!} = 0.1850[/tex]

[tex]P(X = 3) = \frac{e^{-3.5}*3.5^{3}}{(3)!} = 0.2158[/tex]

[tex]P(X = 4) = \frac{e^{-3.5}*3.5^{4}}{(4)!} = 0.1888[/tex]

[tex]P(X = 5) = \frac{e^{-3.5}*3.5^{5}}{(5)!} = 0.1322[/tex]

Finally

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0302 + 0.1057 + 0.1850 + 0.2158 + 0.1888 + 0.1322 = 0.8577[/tex]

[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8577 = 0.1423[/tex]

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

9514 1404 393

Answer:

  (d)  -1/32

Step-by-step explanation:

It may be easier to rearrange the expression so it has positive exponents.

  [tex]\dfrac{1}{2^{-2}x^{-3}y^5}=\dfrac{2^2x^3}{y^5}=\dfrac{4(2)^3}{(-4)^5}=-\dfrac{4\cdot8}{1024}=\boxed{-\dfrac{1}{32}}[/tex]

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

For each subset it can either contain or not contain an element. For each element, there are 2 possibilities. Multiplying these together we get 27 or 128 subsets. For generalisation the total number of subsets of a set containing n elements is 2 to the power n.

total subsets

Answer:

(1238.845 ;1285.376)

Step-by-step explanation:

Conditions for constructing a confidence interval :

Data must be random

Distribution should be normal and independent ;

Based on the conditions above ; data meets initial conditions ;

C. I = sample mean ± margin of error

Given the data :

1241 1210 1267 1314 1211 1299 1246 1280 1291

Mean, xbar = Σx / n = 11359 / 9 = 1262.11

The standard deviation, s = [√Σ(x - xbar)²/n - 1]

Using a calculator ; s = 37.525

The confidence interval :

C.I = xbar ± [Tcritical * s/√n]

Tcritical(0.10 ; df = n - 1 = 9 - 1 = 8)

Tcritical at 90% = 1.860

C. I = 1262.11 ± [1.860 * 37.525/√9]

C.I = 1262.11 ± 23.266

(1238.845 ;1285.376)

± 23.266

The margin of error :

[Tcritical * s/√n]

[1.860 * 37.525/√9]

C.I = ± 23.266

The length of wire used for the equilateral triangle is approximately 5.61 meters.

The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.

Here,

To minimize the total area of both figures, we need to find the optimal cut point for the wire.

Let's assume the length of the wire used for the equilateral triangle is x meters, and the remaining length of the wire used for the circle is (9 - x) meters.

For the equilateral triangle:

An equilateral triangle has all three sides equal in length.

Let's call each side of the triangle s meters. Since the total length of the wire is x meters, each side will be x/3 meters.

The formula to find the area of an equilateral triangle with side length s is:

Area = (√(3)/4) * s²

Substitute s = x/3 into the area formula:

Area = (√(3)/4) * (x/3)²

Area = (√(3)/4) * (x²/9)

Now, for the circle:

The circumference (perimeter) of a circle is given by the formula:

Circumference = 2 * π * r

Since the remaining length of wire is (9 - x) meters, the circumference of the circle will be 2π(9 - x) meters.

The formula to find the area of a circle with radius r is:

Area = π * r²

To find the area of the circle, we need to find the radius.

Since the circumference is equal to 2πr, we can set up the equation:

2πr = 2π(9 - x)

Now, solve for r:

r = (9 - x)

Now, substitute r = (9 - x) into the area formula for the circle:

Area = π * (9 - x)²

Now, we want to minimize the total area, which is the sum of the areas of the triangle and the circle:

Total Area = (√(3)/4) * (x²/9) + π * (9 - x)²

To find the optimal value of x that minimizes the total area, we can take the derivative of the total area with respect to x, set it to zero, and solve for x.

d(Total Area)/dx = 0

Now, find the critical points and determine which one yields the minimum area.

Taking the derivative and setting it to zero:

d(Total Area)/dx = (√(3)/4) * (2x/9) - 2π * (9 - x)

Setting it to zero:

(√(3)/4) * (2x/9) - 2π * (9 - x) = 0

Now, solve for x:

(√(3)/4) * (2x/9) = 2π * (9 - x)

x/9 = (8π - 2πx) / (√(3))

Now, isolate x:

x = 9 * (8π - 2πx) / (√(3))

x(√(3)) = 9 * (8π - 2πx)

x(√(3) + 2π) = 9 * 8π

x = (9 * 8π) / (√(3) + 2π)

Now, we can calculate the value of x:

x ≈ 5.61 meters

So, the length of wire used for the equilateral triangle is approximately 5.61 meters.

The remaining length of wire used for the circle will be 9 - 5.61 ≈ 3.39 meters.

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Let x, y, and z be the amounts (in liters, L) of the 25%, 35%, and 55% solutions that the chemist used.

She ended up with 120 L of solution, so

x + y + z = 120 … … … [1]

x L of 25% acid solution contains 0.25x L of acid. Similarly, y L of 35% solution contains 0.35y L of acid, and z L of 55% solution contains 0.55z L of acid. The concentration of the new solution is 40%, so that it contains 0.40 (120 L) = 48 L of acid, which means

0.25x + 0.35y + 0.55z = 48 … … … [2]

Lastly,

z = 3y … … … [3]

since the chemist used 3 times as much of the 55% solution as she did the 35% solution.

Substitute equation [3] into equations [1] and [2] to eliminate z :

x + y + 3y = 120

x + 4y = 120 … … … [4]

0.25x + 0.35y + 0.55 (3y) = 48

0.25x + 2y = 48 … … … [5]

Multiply through equation [5] by -2 and add that to [4] to eliminate y and solve for x :

(x + 4y) - 2 (0.25x + 2y) = 120 - 2 (48)

0.5x = 24

x = 48

Solve for y :

x + 4y = 120

4y = 72

y = 18

Solve for z :

z = 3y

z = 54

Answer: distributive property

Step-by-step explanation: the 4 is multiplied by everting in the parenthesis

Answer:

Hello

Step-by-step explanation:

The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1

and the disk ? (inside of a circle) of center (0,0) and radius 2

[tex]dom\ f(x,y)=\{(x,y) \in \mathbb{R} ^2 | \ -1\leq x \leq -1\ and \ ( -\sqrt{4-x^2} \leq \ y \leq \sqrt{4-x^2}\ ) \ \}\\[/tex]

Answer:

im not too sure but try using a cartesuan plane and measure it precisely using a protractor then key in the measurements. Im not entirely sure its the correct method tho

Answer:

A. p+q

B. (p)(q)

Step-by-step explanation:

A. The sum of the variables will be p+q, since we are just adding the two variables together.

B. The product of the variables will be (p)(q), or just pq, since you are simply multiplying the two variables.

Hope this Helps!!

9514 1404 393

Answer:

  4x+10

Step-by-step explanation:

For parallelogram adjacent sides a and b, the perimeter is ...

  P = 2(a +b)

For the given sides, the perimeter is ...

  P = 2((x +3) +(x +2)) = 2(2x +5)

  P = 4x +10 . . . perimeter of the parallelogram

Answer:

2 pages per minute

Step-by-step explanation:

Take the number of pages and divide by the number of minutes

46 pages / 23 minutes

2 pages per minute

2 Pages per Minute

46 ÷ 23 = 2

Anthony can read 2 pages per minute.

Answer:

Step-by-step explanation:

Number of electronic systems = 6

(a) Number of defected systems = 2

Probability of getting at least one system is defective

1 defective and 1 non defective + 2 defective

= (2 C 1 ) x (4 C 1) + (2 C 2) / (6 C 2)

= 3 / 5

(b) four defective

Probability of getting at least one system is defective

2 defective and 2 non defective + 3 defective and 1 non defective + 4 defective  

= (4 C 2 ) x (2 C 2) + (4 C 3 )(2 C 1) + (4 C 4) / (6 C 4)

= 1

Answer:

(a)P(At least one defective)[tex]=0.6[/tex]

P(Both are defective)[tex]=0.067[/tex]

(b)P(At least one defective)[tex]=14/15[/tex]

P(Both are defective)[tex]=0.4[/tex]

Step-by-step explanation:

We are given that

Total number of complex electronic system, n=6

(a)Defective items=2

Non-defective items=6-2=4

We have to find the  probability that at least one of the two systems tested will be defective.

P(At least one defective)=[tex]\frac{2C_1\times 4C_1}{6C_2}+\frac{2C_2\times 4C_0}{6C_2}[/tex]

Using the formula

[tex]P(E)=\frac{favorable\;cases}{total\;number\;of\;cases}[/tex]

P(At least one defective)[tex]=\frac{\frac{2!}{1!1!}\times \frac{4!}{1!3!} }{\frac{6!}{2!4!}}+\frac{\frac{2!}{0!2!}\times \frac{4!}{4!}}{\frac{6!}{2!4!}}[/tex]

Using the formula

[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]

P(At least one defective)[tex]=\frac{2\times \frac{4\times 3!}{3!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}+\frac{1}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]

P(At least one defective)[tex]=\frac{2\times 4}{3\times 5}+\frac{1}{3\times 5}[/tex]

P(At least one defective)[tex]=\frac{8}{15}+\frac{1}{15}=\frac{9}{15}[/tex]

P(At least one defective)[tex]=\frac{3}{5}=0.6[/tex]

Now, the probability that both are defective

P(Both are defective)=[tex]\frac{2C_2\times 4C_0}{6C_2}[/tex]

P(Both are defective)=[tex]\frac{\frac{2!}{0!2!}\times \frac{4!}{4!}}{\frac{6!}{2!4!}}[/tex]

P(Both are defective)[tex]=\frac{1}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]

P(Both are defective)[tex]=\frac{1}{3\times 5}[/tex]

P(Both are defective)[tex]=0.067[/tex]

(b)

Defective items=4

Non- defective item=6-4=2

P(At least one defective)=[tex]\frac{4C_1\times 2C_1}{6C_2}+\frac{4C_2\times 2C_0}{6C_2}[/tex]

P(At least one defective)[tex]=\frac{\frac{4!}{1!3!}\times \frac{2!}{1!1!} }{\frac{6!}{2!4!}}+\frac{\frac{4!}{2!2!}\times \frac{2!}{2!}}{\frac{6!}{2!4!}}[/tex]

P(At least one defective)[tex]=\frac{2\times \frac{4\times 3!}{3!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}+\frac{\frac{4\times 3\times 2!}{2!\times 2\times 1}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]

P(At least one defective)[tex]=\frac{2\times 4}{3\times 5}+\frac{2\times 3}{3\times 5}[/tex]

P(At least one defective)[tex]=\frac{8}{15}+\frac{6}{15}=\frac{8+6}{15}[/tex]

P(At least one defective)[tex]=\frac{14}{15}[/tex]

P(Both are defective)[tex]=\frac{4C_2\times 2C_0}{6C_2}[/tex]

P(Both are defective)[tex]=\frac{\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{2!}{2!}}{\frac{6\times 5\times 4!}{2\times 1\times 4!}}[/tex]

P(Both are defective)[tex]=\frac{\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}}{3\times 5}[/tex]

P(Both are defective)[tex]=\frac{6}{15}=0.4[/tex]

P(Both are defective)[tex]=0.4[/tex]

Answer:

106 people.

Step-by-step explanation:

Logistic equation:

The logistic equation is given by:

[tex]P(t) = \frac{K}{1+Ae^{-kt}}[/tex]

In which

[tex]A = \frac{K - P_0}{P_0}[/tex]

K is the carrying capacity, k is the growth/decay rate, t is the time and P_0 is the initial value.

Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor.

This means that [tex]K = 191, P_0 = 38[/tex], so:

[tex]A = \frac{191 - 38}{38} = 4.03[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

3 days later 68 people have heard it.

This means that [tex]P(3) = 68[/tex]. We use this to find k.

[tex]P(t) = \frac{191}{1+4.03e^{-kt}}[/tex]

[tex]68 = \frac{191}{1+4.03e^{-3k}}[/tex]

[tex]68 + 274.04e^{-3k} = 191[/tex]

[tex]e^{-3k} = \frac{191-68}{274.04}[/tex]

[tex]e^{-3k} = 0.4484[/tex]

[tex]\ln{e^{-3k}} = \ln{0.4484}[/tex]

[tex]-3k = \ln{0.4484}[/tex]

[tex]k = -\frac{\ln{0.4484}}{3}[/tex]

[tex]k = 0.2674[/tex]

Then

[tex]P(t) = \frac{191}{1+4.03e^{-0.2674t}}[/tex]

How many people are expected to have heard the rumor after 6 days total have passed since it was initially spread?

This is P(6). So

[tex]P(6) = \frac{191}{1+4.03e^{-0.2674*6}} = 105.52[/tex]

Rounding to the nearest whole number, 106 people.

Answer:

y = -x+3

Step-by-step explanation:

Slope intercept form =>  y = mx+b

To find 'm', the slope, pick 2 coordinates.

(0,3)

(2,1)

Use this equation to find the slope using these 2 coordinates: (y1 - y2)/(x1 - x2)

(3 - 1)/(0 - 2) = -1

m = slope = -1

'b' is the y-intersept, or the point when a line passes through the y-axis. That's (0,3).

b = y-intercept = 3

So the equation will be y = -1x + 3, or y = -x + 3

Step-by-step explanation:

-27 okay 3^-3 its same as 3^3

Answer: A)

[tex]3^{-3}[/tex]

[tex]3^{-3}=\frac{1}{3^3}[/tex]

[tex]=\frac{1}{3^3}[/tex]

[tex]3^3=27[/tex]

[tex]=\frac{1}{27}[/tex]

OAmalOHopeO

Answer:

Step-by-step explanation:

Answer:

-4/25

Step-by-step explanation:

3x=y+z

y=6-7=-1

z+x=8

y/z=?

solution

3x=-1 + z

x= -1 + z\3 ...eq(*)

then, z= 8- x

z= 8 - (-1 +z)\3

z= 8 +(1- z )\3

z= 8+1\3 -z \3

= 24+1\3 - z\3

z=25\3-z\3

z+z\3=25\3

4z\3=25\3

4z=25

z=25/4

then,

25/4 +x = 8

x=8- 25/4

x= 32 - 25/4

x=7/4

so that,

y/z=-1 /25/4

=-4/25

Answer:

8×1000 milligrams

8000 milligrams

Answer:

55

Step-by-step explanation:

55 appears 3 times, which is the most repetition in the data set

Answer:

55

Step-by-step explanation:

Mode = number that appears most often

The number 55 appears 3 times which is the most out of the other numbers

Hence mode = 55

Answer:

268.40

Step-by-step explanation:

We can write a ratio to solve

2.5 meters        22 meters

-----------------  = --------------

30.5 dollars       x dollars

Using cross products

2.5 * x = 30.5 * 22

2.5x =671

Divide each side by 2.5

2.5x / 2.5 = 671/2.5

x =268.4

Answer:

5 ≤ x ≤ 10             5 ≤ y ≥ -1

Step-by-step explanation:

Answer:

a)  The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

b)  P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

  P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 ,    P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6         P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4        P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0           P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24,   P(2,0) = 0,   P(0,1) = 0,  P(1,1) = 0.24,  P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36,    P(W = 1 ) = 0.48,  P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Part(a): The required joint probability of W and Z is ,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

Part(b): The pmf (marginal distribution) of W is,

[tex]P(w=0)=0.36,P(w=1)=0.48,P(w=2)=0.16[/tex]

Part(c): The pmf (marginal distribution) of Z is,

[tex]P(z=0)=0.6,P(z=1)=0.4[/tex]

Part(a):

The joint distribution is,

[tex]P(w=0\z=0)=0.6,P(w=1|z=0)=0.4,P(w=2|z=0)=0[/tex]

Also,

[tex]P(w=0\z=1)=0,P(w=1|z=1)=0.6,P(w=2|z=1)=0.4[/tex]

Therefore,

[tex]P(0,0)=0.36,P(1,0)=0.24,P(2,0)=0,P(0,1)=0,P(1,1)=0.24,\\\\P(2,1)=0.16[/tex]

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Answer:

The answer is "0.7794".

Step-by-step explanation:

Please find the complete question in the attached file.

Given:

[tex]\to n_{1}=n_{2}=25\\\\[/tex]

Hypotheses:

[tex]\to H_{0}:\mu_{B}-\mu_{A}\geq 0\\\\\to H_{a}:\mu_{B}-\mu_{A}< 0\\\\[/tex]

Testing statistics:

[tex]\to z=\frac{(\bar{x}_{B}-\bar{x}_{A})-(\mu_{B}-\mu_{A})}{\sqrt{\frac{\sigma^{2}_{B}}{n_{1}}+\frac{\sigma^{2}_{A}}{n_{2}}}}=\frac{3.5-(0)}{\sqrt{\frac{16^{2}}{25}+\frac{16^{2}}{25}}}=0.77[/tex]

The test is done just so the p-value of a test is

[tex]\to p-value = P(z < 0.77) = 0.7794[/tex]

Because the p-value of the management is large, type B can take it.

Answer:

30 clocks

Step-by-step explanation:

Set up an equation:

Variable x = number of clocks

1200 + 20x = 60x

Isolate variable x:

1200 = 60x - 20x

1200 = 40x

Divide both sides by 40:

30 = x

Check your work:

1200 + 20(30) = 60(30)

1200 + 600 = 1800

1800 = 1800

Correct!

Answer:

I think around $36

Hope it helps!

Answer:

It depends...

Step-by-step explanation:

It depends how much weeks are in the month if there are three weeks and no extra days then you would have an answer of about 1093 (exact: 1093.33333333). just divide the number of weeks by the number of money.