Perimeter =68 Length (L) is 4 less than twice the width (W)

Answer:

D

Step-by-step explanation:

Option D is correct. Slope of the line shown in the graph is 3.

The slope of the line is the ratio of the rise to the run, or rise divided by the run.

It describes the steepness of line in the coordinate plane.

The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.

The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is

m=(y₂-y₁)/(x₂-x₁)

The line is passing through point (2, 2) and (4, 8).

Lets find the corresponding point values y₂= 8, y₁ = 2, x₂= 4 and x₁ =2.

Plug in the values in slope formula:

Slope = (8-2)/(4-2)  

=6/2

=3

Hence, slope of the line shown in the graph is 3. Option D is correct.

To learn more on slope of line click:

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Answer:

1/3(a+b)

hope it helps :>

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Answer:

[tex]\Large \boxed{\mathrm{C) \ D: (-7, 3] \ R: (-10, 9.2]}}[/tex]

Step-by-step explanation:

The domain is the set of all possible x values.

The range is the set of all possible y values.

For the domain, we observe the graph, the graph will contain all the x values shown on the x-axis.

[tex]\mathrm{D= (-7,3] }[/tex]

For the range, we observe the graph, the graph will contain all the y values shown on the y-axis.

[tex]\mathrm{R= (-10,9.2] }[/tex]

[tex]\displaystyle\\(a+b)^n\\T_{r+1}=\binom{n}{r}a^{n-r}b^r\\\\\\(x+2)^7\\a=2x\\b=3\\r+1=4\Rightarrow r=3\\n=5\\T_4=\binom{5}{3}\cdot (2x)^{5-3}\cdot3^3\\T_4=\dfrac{5!}{3!2!}\cdot 4x^2\cdot27\\T_4=\dfrac{4\cdot5}{2}\cdot 4x^2\cdot27\\\\T_4=1080x^2[/tex]

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Answer:

2x^2 + x - 6 = rectangular garden: length = 2x – 3 feet; width = x + 2 feet

Step-by-step explanation:

(2x - 3)(x + 2) = 2x^2 + x - 6 =

2x^2 + 4x - 3x - 6 = 2x^2 + x - 6 =

2x^2 + x - 6

You get the original equation from the two sides multiplied. :)

Hope this helps, have a good day.

The length and width of the rectangle will be (2x – 3) and (x + 2). Then the correct option is D.

Let W be the rectangle's width and L its length.

The area of the rectangle is the multiplication of the two different sides of the rectangle. Then the rectangle's area will be

Area of the rectangle = L × W square units

The area is 2x² + x – 6 square feet. Then the factor of the equation is given as,

A = 2x² + x – 6

A = 2x² + 4x – 3x – 6

A = 2x(x + 2) – 3(x + 2)

L × W = (2x – 3)(x + 2)

The length and width of the rectangle will be (2x – 3) and (x + 2). Then the correct option is D.

More about the area of the rectangle link is given below.

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Answer: ABCD is a parallelogram.

Step-by-step explanation:

First we plot these point on a graph as given in attachment.

From the attachment we can observe that AD || BC || x-axis .

also, AB ||CD, that will make ABCD a parallelogram ,  but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".

Mid point of AC= [tex](\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Mid point of BD= [tex](\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Thus, Mid point of AC=Mid point of BD

i.e. diagonals bisect each other.

That means ABCD is a parallelogram.

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First, we plot these points on a graph as given in the attachment. From the attachment, we can observe that AD || BC || x-axis. Also, AB ||CD, which will make ABCD a parallelogram, but to confirm, we check the parallelogram property "diagonals bisect each other," i.e., "Midpoint of both diagonals is equal."

The midpoint of AC=. The midpoint of BD=. Thus, the Midpoint of AC=Mid point of BD diagonals bisects each other. That means ABCD is a parallelogram.

Answer: y=(4/3)x+2/3

Step-by-step explanation:

Slope-intercept form is expressed as y=mx+b

First, find the slope (m):

m= rise/run or vertical/horizontal or y/x (found between the highlighted points)

m = 4/3

Second, find b:

Use one of the highlighted points for (x, y)

2=4/3(1)+b

6/3=4/3+b

2/3=b

b=2/3

Plug it into the equation:

You get y=(4/3)x+2/3 :)

volume of a cone

.

.

.

volume of sphere

.

.

number of spheres that can be made......

.

.

hence a hemisphere can be formed

Answer:

Yes it can be concluded that state employees earn on average less than federal employees

  The critical value is  [tex]Z_{\alpha } = - 2.33[/tex]

Step-by-step explanation:

From the question we are told that

   The  population mean is  [tex]\mu = \$ 59593[/tex]

   The sample size is  n =  30

    The  sample mean is [tex]\= x = \$ 58800[/tex]

     The  standard deviation is  [tex]\sigma = \$ 1500[/tex]

     The significance level is  [tex]\alpha = 0.01[/tex]

The null hypothesis is  [tex]H_o : \mu = \$ 59593[/tex]

 The  alternative hypothesis is  [tex]H_a : \mu < \$ 59593[/tex]

The critical value of [tex]\alpha[/tex] from the normal distribution table is  [tex]Z_{\alpha } = - 2.33[/tex]

 Generally the test statistics is  mathematically evaluated as

            [tex]t = \frac{\= x - \mu}{ \frac{ \sigma }{ \sqrt{n} } }[/tex]

=>         [tex]t = \frac{ 58800 - 59593 }{ \frac{ 1500 }{ \sqrt{30} } }[/tex]  

=>          [tex]t = -2.896[/tex]

The  p-value is obtained from the z-table

   [tex]p-value = P(t < -2.896) = 0.0018898[/tex]

Since [tex]p-value < \alpha[/tex] , we reject the null hypothesis, hence it can be concluded that state employees earn on average less than federal employees  

Answer:

The answer is D)90

Hope I helped

Answer:

The upper bound of the confidence interval is 0.34

Step-by-step explanation:

Here in this question, we want to calculate the upper bound of the confidence interval.

We start by calculating the margin of error.

Mathematically, the margin of error = 0.29 -0.24 = 0.05

So to get the upper bound of the confidence interval, we simply add this margin of error to the mean

That would be 0.05 + 0.29 = 0.34

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option C is the correct option

Step-by-step explanation:

From the question we are told that

   The equation is  [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]

    The constraint is  [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]

Now using Lagrange multipliers  we have that  

   [tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]

=>       [tex]x = \frac{ \lambda }{2}[/tex]

          [tex]y = \frac{ \lambda }{2}[/tex]

         [tex]z = \frac{ \lambda }{2}[/tex]

From the constraint  we have

      [tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]

=>   [tex]\frac{3 \lambda }{2} = 24[/tex]

=>   [tex]\lambda = 16[/tex]

substituting for x, y, z

=>   x =  8

=>  y =  8

=>   z =  8        

Hence

    [tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]

    [tex]f (8, 8 , 8 ) = 192[/tex]

Answer:

A) C1 = 0.00187 m = 0.187 cm,  C2 = 0.0062 m = 0.62 cm

B)  A sample of how the graph looks like is attached below ( periodic sine wave )

C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum

Step-by-step explanation:

Given data :

mass = 5kg

length of spring = 10 cm = 0.1 m

f(t) = 10sin(t) N

viscous force = 2 N

speed of mass = 4 cm/s = 0.04 m/s

initial velocity = 3 cm/s = 0.03 m/s

Formulating initial value problem

y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m

spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m

f(t) = 10sin(t/2) N

using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion

the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)

A) finding the solution of the initial value

attached below is the solution and

B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like

C attached below

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

Answer:

Perimeter = 42 cm

Step-by-step explanation:

A square has all equal sides so you would just add 10.5 + 10.5 + 10.5 + 10.5 to get 42 cm.

Answer:

42 cm

Step-by-step explanation:

Side of square = 10.5 cm (given)

Perimeter of square = Side X 4

                                  = 10.5 X 4

                                  = 42 cm

HOPE THIS HELPED YOU !

:)

Answer:

this? hope it helps ........

Answer:

The answer is area=32pi-64 and the perimeter is 8pi

Step-by-step explanation:

Answer:

Step-by-step explanation:

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.

(a) How many different samples of size 4 pens are possible?

12C4=12!/(4!*8!)=495

(b) How many samples have 3 red pens and 1 black pen?

5C3*7C1

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

=>5C3*7C1=10*7=70

(c) How many samples of size 4 contain at least two red pens?

(5C2*7C2)+(5C3*7C1)+(5C4*7C0)

5C2=5!/(2!*3!)=10

7C2=7!/(2!*5!)=21

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

5C4=5!/(4!*1!)=5

7C0=7!/(0!*7!)=1

=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285

(d) How many samples of size 4 contain at most one black pen?

(7C1*5C3)+(7C0*5C4)

7C1=7!/(1!*6!)=7

7C0=7!/(0!*7!)=1

5C3=5!/(3!*2!)=10

5C4=5!/(4!*1!)=5

=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75

Answer:

Question 1 = D) Acute

Question 2 = C)3 feet

Question 3 = D) Obtuse

Question 4 = C)27.73 ft.

Step-by-step explanation:

Question 1: A triangle has sides with lengths 5, 6, and 7. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths

5 = a, 6 = b and c = 7

a² + b² = c²

5² + 6² = 7²

25 + 36 = 49

61 = 49

61 ≠ 49, Hence 61 > 49

Therefore, this is an Acute Triangle

Question 2: A 15-foot statue casts a 20-foot shadow. How tall is a person who casts a 4-foot-long shadow?

This is question that deals with proportion.

The formula to solve for this:

Height of the statue/ Length of the shadow of the person = Height of the person/ Length of the shadow of the person

Height of the statue = 15 feet

Length of the shadow of the person = 20 feet

Height of the person = unknown

Length of the shadow of the person = 4

15/ 20 = Height of the person/4

Cross Multiply

15 × 4 = 20 × Height of the person

Height of the person = 15 × 4/20

= 60/20

Height of the person = 3 feet

Therefore, the person is 3 feet tall.

Question 3: A triangle has sides with lengths 17, 12, and 9. Is the triangle right, acute, or obtuse?

In order to be able to accurately classify that a triangle with 3 given sides is either a right , acute or obtuse angle, we use the Pythagoras Theorem

Where:

If a² + b² = c² = Right angle triangle

If a² +b² > c² = Acute triangle.

If a² +b² < c² = Obtuse triangle.

It is important to note that the length ‘‘c′′ is always the longest.

Therefore, for the above question, we have lengths 17, 12, 9

9 = a, 12 = b and c = 17

a² + b² = c²

9² + 12² = 17²

81 + 144 = 289

225 = 289

225 ≠ 289

225 < 289

Hence, This is an Obtuse Triangle.

Question 4: Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. How far apart are the friends?

To calculate how far apart the two friends are we use the formula

Distance = √ ( Length² + Breadth²)

We are given dimensions: 12ft by 25ft

Length = 12ft

Breadth = 25ft

Distance = √(12ft)² + (25ft)²

Distance = √144ft²+ 625ft²

Distance = √769ft²

Distance = 27.730849248ft

Approximately ≈27.73ft

Therefore, the friends are 27.73ft apart.

Answer:

the answer to the question is "C"

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

Answer:

n + 1

Step-by-step explanation:

Starting with the integer 'n,' we represent the "next integer" by n + 1.

Answer:

0.4207149;0.7271136; 0.3063987; 0.04979 ; 0.01832

Step-by-step explanation:

For an exponential distribution:

IF Mean time until failure = 23100

λ = 1/ 23100 = 0.0000432900

What is the probability that a randomly selected fan will last at least 20,000 hours

x ≥ 20000

P(X ≥ 20000) = 1 - P(X ≤ 20000)

1 - P(X ≤ 20000) = [1 - (1 - e^(-λx))]

1 - P(X ≤ 20000) = [1 - (1 - e^(-0.0000432900*20000)

1 - P(X ≤ 20000) = [1 - (1 - 0.4207148)]

1 - P(X ≤ 20000) = 1 - 0.5792851

1 - P(X ≤ 20000) = 0.4207149

11) What is the probability that a randomly selected fan will last at most 30,000 hours?

x ≤ 30000

P(X ≤ 30000) = 1 - e^(-λx)

P(X ≤ 20000) = 1 - e^(-0.0000432900*30000)

= 1 - e^(−1.2987)

= 1 - 0.2728863

= 0.7271136

111) What is the probability that a randomly selected fan will last between 20,000 hours and 30,000 hours?

0.7271136 - 0.4207149 = 0.3063987

B) What is the probability that the lifetime of a fan exceeds the mean value by more than 2 standard deviations?

More than two standard deviation

X = 23100 + 2(23100) = 23100 + 46200 = 69300

Using the online exponential probability calculator :

P(X > 69300) = 0.04979

C) What is the probability that the lifetime of a fan exceeds the mean value by more than 3 standard deviations?

X = 23100 + 3(23100) = 23100 + 69300 = 92400

P(X > 92400) = 0.01832