21. In the figure given below, AC is parallel to DE. Find the valuesof xy and z and hence find the 2DBE.
21-70X
509​

21. In The Figure Given Below, AC Is Parallel To DE. Find The Valuesof Xy And Z And Hence Find The 2DBE.21-70X509

Answers

Answer 1

Answer:

X= 50°

Y= 70°

Z= 30°

BDE= 30°

2BDE= 60°

Step-by-step explanation:

(2x -70 )+z+(2x+20)=180...(sum of angle on a straight line)

2x -70 = BDE... alternate angles

Y + (2x-70)+(50+x-20) = 180...(sum of angles in a triangle)

X-20 = z ... alternate and opposite angles

(2x -70 )+z+(2x-+20)=180

2x-70 + x-20 +2x +20= 180

5x -70= 180

5x = 250

X= 50°

X-20 = z

50-20= z

30° = z

2x -70 = BDE

2(50) -70 = BDE

100-70 = BDE

30°= BDE

Y + (2x-70)+(50+x-20)

Y + 100-70 +50 +50 -20 = 180

Y + 200-90=180

Y= 70°

2BDE = 2*30

2BDE= 60°


Related Questions

A rectangle's length and width are in a ratio of 10:1. The perimeter is 66 feet. What are the length and width?

Answers

hii

Step-by-step explanation:

length-10x

width-x

perimeter-2(l+b)

66=2(10x+x)

66-2=10x+x

64=11x

x=11/64

lenght-11

width-64

Length-11
Width-64
Hope it helps

What is the equation of the following line? Be sure to scroll down first to see all answer options.



A.
y = 18x

B.
y = 9x

C.
y = -9x

D.
y = - x

E.
y = -18x

F.
y = x

Answers

Answer:

y=9x

Step-by-step explanation:

rise over run the rise is the y=9 and run is x=1.

9/1=9x

A sector with a central angle measure of 200 degrees has a radius of 9 cm. What is the area of the sector?

Answers

Answer:

[tex]\boxed{Area\ of\ sector = 141.4\ cm^2}[/tex]

Step-by-step explanation:

Radius = r = 9 cm

Angle = θ = 200° = 3.5 radians

Now,

[tex]Area \ of \ sector = \frac{1}{2} r^2 \theta[/tex]

Area = 1/2 (9)²(3.5)

Area = 1/2 (81)(3.5)

Area = 282.7 / 2

Area of sector = 141.4 cm²

Answer:

45 pi cm^2 or 141.3 cm^2

Step-by-step explanation:

First find the area of the circle

A = pi r^2

A = pi (9)^2

A = 81 pi

A circle has 360 degrees

The shaded part has 200

The fraction that is shaded is

200/360 =5/9

Multiply by the total area

5/9 * 81 pi

45 pi

Using 3.14 for pi

141.3

45 pi cm^2 or 141.3 cm^2

Find the measures of the angles in the figure.

Answers

Answer:

[tex]120^o,\,120^o,\,60^o,\,\,\,and\,\,\,60^o[/tex]

which agrees with the first answer in the list of possible options.

Step-by-step explanation:

We can use the fact that the addition of all four internal angles of a quadrilateral must render [tex]360^o[/tex]. Then we can create the following equation and solve for the unknown "h":

[tex]2h+2h+h+h = 360^o\\6h=360^o\\h=60^o[/tex]

Therefore the angles of this quadrilateral are:

[tex]120^o,\,120^o,\,60^o,\,\,\,and\,\,\,60^o[/tex]

Answer:60,60,120,120

Step-by-step explanation:All qualdrilaterals equal to 360, so if you add all of the different numbers you should get 360

I need help with this !!

Answers

Answer:

A

Step-by-step explanation:

When subtracting 7 on the left of the equation, he also needs to subtract 7 from the right of the equation.

Step 2 should be:

⅓X +7 -7= 15 -7

What he is trying to do here by subtracting 7 is to move all the constants, that is numbers without any variables such as x, to one side of the equation.

⅓X= 8

X= 8 ×3

X= 24

The circumference of the base of a cylinder is 24π mm. A similar cylinder has a base with circumference of 60π mm. The lateral area of the larger cylinder is 210π mm2. What is the lateral area of the smaller cylinder? 17.1π mm2 33.6π mm2 60π mm2 84π mm2

Answers

Answer:

84π mm^2

Step-by-step explanation:

formula for circumference is 2πr where r is the radius of circle

Given,The circumference of the base of a cylinder is 24π mm

Thus,

2πr= 24π mm

=> r = 24π mm/2π = 12 mm

________________________________________

A similar cylinder has a base with circumference of 60π mm.

radius for this cylinder will be

2πr= 60π mm

r =   60π mm/2π = 30mm

______________________________________________

Given

The lateral area of the larger cylinder is 210π mm2

lateral area of cylinder is given by  2πrl

where l is the length of cylinder

thus,

r for larger cylinder = 30mm

2π*30*l  = 210π mm^2

=> l = 210π mm^2/2π*30 = 3.5 mm

___________________________________________

the lateral area of the smaller cylinder

r = 12 mm

l = 3.5 mm as both larger and smaller cylinder are same

2πrl  =  2π*12*3.5 mm^2 = 84π mm^2 answer

Answer:

33.6pi mm2 is the correct answer

edge 2021

Step-by-step explanation:

The circumference of the base of a cylinder is 24π mm. A similar cylinder has a base with circumference of 60π mm. The lateral area of the larger cylinder is 210π mm2.

What is the lateral area of the smaller cylinder?

17.1π mm2

33.6π mm2

60π mm2

84π mm2

Need Answers ASAP!!!!

Answers

Answer:

15.9degrees

Step-by-step explanation:

in photo above

Answer:

[tex]\boxed{15.95\°}[/tex]

Step-by-step explanation:

The angle can be found by using trigonometric functions.

tan (θ) = [tex]\frac{opposite}{adjacent}[/tex]

tan (θ) = [tex]\frac{4}{14}[/tex]

θ = [tex]tan^{-1} \frac{4}{14}[/tex]

θ = 15.9453959

θ ≈ 15.95

The solutions to the inequality ys-x+1 are shaded on
the graph. Which point is a solution?
(2, 3)
(3,-2)
(2.1)
(-1,3)

Answers

Answer:

the solutions to the inequality ys-x+1 are shaded on the graph. which point is B. (3 ,-2)

All sides of the building shown above meet at right angles. If three of the sides measure 2 meters, 7 meters, and 11 meters as shown, then what is the perimeter of the building in meters?

Answers

Answer:

Perimeter= 40 units

Step-by-step explanation:

Ok

We are asked to look for the perimeter.

We have some clue given.

All at right angle and some sides are given it's full length.

We have the bae to be 11 unit

The height to be 7 unit.

What this mean is that taking either the base or the height should sum up to either 11 or 7 respectively.

Let's go for the other side of the height.

Let's take all the vertical height and sum it up to 7 because the right side is equal to 7.

So we have 7+7+11

But it's not complete yet.

We are given a dimension 2.

And the 2 is in two places so it's total 2*2= 4

The two is for a small base .

The base is actually an extra to the 11 of the other base.

So summing up

We have 2*11 + 2*7 + 2*2

Perimeter= 22+14+4

Perimeter= 40 units

Please help
ASAP

ANSWERS

A-48.21
B-66.35
C-53.68
D-28.34

Answers

Answer:

B

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos54° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{AC}{AB}[/tex] = [tex]\frac{39}{AB}[/tex] ( multiply both sides by AB )

AB × cos54° = 39 ( divide both sides by cos54° )

AB = [tex]\frac{39}{cos54}[/tex] ≈ 66.35 → B

Find the difference of functions at x= - 3, (g - f)(-3), given f(x) and g(x): g(x) = x^2−15, and f(x) =2x

Answers

Answer:

0

Step-by-step explanation:

Solution:-

We are given two functions as follows:

                      [tex]f ( x ) = x^2 - 15\\\\g ( x ) = 2x[/tex]

We need to determine the composite function defined as ( g - f ) ( x ). To determine this function we need to make sure that both function exist for all real positive value of x.

The function f ( x ) is a quadratic function which has real values for all values of x. Similarly, function g ( x ) is a linear line that starts from the origin. Hence, both functions are defined over the domain ( -∞, ∞ )

We will perform arithmetic operation of subtracting function f ( x ) from g ( x ) as follows:

                       [tex][ g - f ] ( x ) = g ( x ) - f ( x )\\\\\\( g - f ) ( x ) = x^2 - 15 - 2x\\\\[/tex]

Now evaluate the above determined function at x = -3 as follows:

                       [tex]( g - f ) ( -3 ) = ( -3 )^2 - 2 ( -3 ) - 15\\\\( g - f ) ( -3 ) = 9 + 6 - 15\\\\( g - f ) ( -3 ) = 0[/tex]

need help thanksssssssss

Answers

Answer:

Volume: 112 m³.

Surface area: 172 m².

Step-by-step explanation:

The volume is the base times height times length. So, the volume will be 2 * 8 * 7 = 16 * 7 = 112 m³.

The surface area is 2lw + 2lh + 2wh. l = 8; w = 7; h = 2.

2(8)(7) + 2(8)(2) + 2(7)(2) = 2 * 56 + 2 * 16 + 2 * 14 = 112 + 32 + 28 = 112 + 60 = 172 m².

Hope this helps!

The cost of plastering the 4 walls of a room which is 4m high and breadth one third of its length is Rs. 640 at the rate of Rs. 5/m². What will be the cost of carpeting its floor at the rate of Rs. 250/m².​

Answers

Answer:

Rs. 32,000

Step-by-step explanation:

height = 4m

let length = x m

breadth = x/3 m

Area of the 4 walls = 2(length × height) + 2(breadth × height)

Area = 2(4×x) + 2(4 × x/3) = 8x + (8x)/3

Area = (32x)/3 m²

1 m² = Rs. 5

The cost for an area that is (32x)/3 m²=  (32x)/3  × 5 Rs.

The cost of plastering 4 walls at Rs.5 per m² = 640

(32x)/3  × 5  = 640

(160x)/3 = 640

x = length = 12

Area =  (32x)/3 m² = (32×12)/3 = 128m²

The cost of carpeting its floor at the rate of Rs. 250/m²:

= 128m² × Rs. 250/m² = 32,000

The cost of carpeting its floor at the rate of Rs. 250/m² = Rs. 32,000

In a soccer league, the ratio of boys to girls is 4 to 6. There are a total of 50 players in the soccer league. Determine how many girls play in the soccer league.

Answers

Answer:

30

Step-by-step explanation:

We can call the number of boys 4x and girls 6x so we can write:

4x + 6x = 50

10x = 50

x = 5, therefore the number of girls is 6x = 6 * 5 = 30.

Answer:

30

Step-by-step explanation:

In the ratio 4:6, we can think of this like 4 boys and 6 girls out of 10 team members.

We can find how many girls play by multiplying 6 by 5, since 50 divided by 10 is 5.

6(5) = 30, so 30 girls play in the soccer league.

The exact heights of different elephants Choose the correct answer below. A. The data are continuous because the data can only take on specific values. B. The data are discrete because the data can take on any value in an interval. C. The data are discrete because the data can only take on specific values. D. The data are continuous because the data can take on any value in an interval.

Answers

Answer:

Option d: The data are continuous because the data can take on any value in an interval.

Step-by-step explanation:

The data are continuous if they can take on any value within a range. In this case study, there are different elephants including small/young ones and big ones/old ones.

Thus, their heights will vary and can take on any value within a particular range.

Hypothesis Testing
Problem 1. Adults saving for retirement
In a recent survey conducted by Pew Research, it was found that 156 of 295 adult Americans without a high school diploma were worried about having enough saved for retirement. Does
the sample evidence suggest that a majority of adult Americans without a high school diploma are worried about having enough saved for retirement? Use a 0.05 level of significance
1. State the null and alternative hypothesis.
2. What type of hypothesis test is to be used?
3. What distribution should be used and why?
4. Is this a right, left, or two-tailed test?
5. Compute the test statistic.
6. Compute the p-value.
7. Do you reject or not reject the null hypothesis? Explain why.
8. What do you conclude?
Problem 2: Google Stock
Google became a publicly traded company in August 2004. Initially, the stock traded over 10 million shares each day! Since the initial offering, the volume of stock traded daily has
decreased substantially. In 2010, the mean daily volume in Google stock was 5.44 million shares, according to Yahoo!Enance. A random sample of 35 trading days in 2014 resulted in a
sample mean of 3.28 million shares with a standard deviation of 1.68 million shares. Does the evidence suggest that the volume of Google stock has changed since 2007? Use a 0.05 level of
significance
1. State the null and alternative hypothesis.
2. What type of hypothesis test is to be used?
3. What distribution should be used and why?
4. Is this a right, left, or two-tailed test?
5. Compute the test statistic.
6. Compute the p-value.
7. Do you reject or not reject the null hypothesis? Explain why
8. What do you conclude?

Answers

Answer:

Problem 1: We conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

Problem 2: We conclude that the volume of Google stock has changed.

Step-by-step explanation:

Problem 1:

We are given that in a recent survey conducted by Pew Research, it was found that 156 of 295 adult Americans without a high school diploma were worried about having enough saved for retirement.

Let p = proportion of adult Americans without a high school diploma who are worried about having enough saved for retirement

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 50%    {means that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement}

Alternate Hypothesis, [tex]H_A[/tex] : p > 50%     {means that a majority of adult Americans without a high school diploma are worried about having enough saved for retirement}

This is a right-tailed test.

The test statistics that would be used here is One-sample z-test for proportions;

                       T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of adult Americans who were worried about having enough saved for retirement = [tex]\frac{156}{295}[/tex] = 0.53

           n = sample of adult Americans = 295

So, the test statistics =  [tex]\frac{0.53-0.50}{\sqrt{\frac{0.50(1-0.50)}{295} } }[/tex]

                                    =  1.03

The value of z-test statistics is 1.03.

Also, the P-value of the test statistics is given by;

              P-value = P(Z > 1.03) = 1 - P(Z [tex]\leq[/tex] 1.03)

                           = 1 - 0.8485 = 0.1515

Now, at a 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.03 < 1.645, so we insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that less than or equal to 50% of adult Americans without a high school diploma are worried about having enough saved for retirement.

Problem 2:

We are given that a random sample of 35 trading days in 2014 resulted in a  sample mean of 3.28 million shares with a standard deviation of 1.68 million shares.

Let [tex]\mu[/tex] = mean daily volume in Google stock

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 5.44 million shares    {means that the volume of Google stock has not changed}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 5.44 million shares     {means that the volume of Google stock has changed}

This is a two-tailed test.

The test statistics that would be used here is One-sample t-test statistics because we don't know about the population standard deviation;

                       T.S.  =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean volume in Google stock = 3.28 million shares

            s = sample standard deviation = 1.68 million shares

           n = sample of trading days = 35

So, the test statistics =  [tex]\frac{3.28-5.44}{\frac{1.68}{\sqrt{35} } }[/tex]  ~ [tex]t_3_4[/tex]

                                    =  -7.606

The value of t-test statistics is -7.606.

Also, the P-value of the test statistics is given by;

              P-value = P([tex]t_3_4[/tex] < -7.606) = Less than 0.05%

Now, at a 0.05 level of significance, the t table gives a critical value of -2.032 and 2.032 at 34 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the volume of Google stock has changed.

2.CommerceThe weight distribution of parcels sent in a certain manner is normal with meanvalue 12 pounds and standard deviation 3.5 pounds. The parcel service wishes to establish aweight valuecbeyond which there will be a surcharge. What value ofcis such that 99% ofall parcels are under the surcharge weight

Answers

Answer:

the value of c is  20.155 such that 99% of all parcels are under the surcharge weight.

Step-by-step explanation:

Given that :

The mean value [tex]\mu[/tex] = 12

The standard deviation [tex]\sigma[/tex] = 3.5

Let Consider Q to be the weight of the parcel that is normally distributed .

Then;

Q [tex]\sim[/tex] Norm(12,3.5)

The objective is to determine thewight  value of c under which there is a surcharge

Also, let's not that 99% of all the parcels are below the surcharge

However ;

From the Percentiles table of Standard Normal Distribution;

At 99th percentile; the value for Z = 2.33

The formula for the Z-score is:

[tex]Z = \dfrac{X- \mu}{\sigma}[/tex]

[tex]2.33 = \dfrac{X - 12}{3.5}[/tex]

2.33 × 3.5 = X - 12

8.155 = X - 12

- X = - 12 - 8.155

- X = -20.155

 X = 20.155

the weight  value of c under which there is a surcharge = X + 1 (0) since all the  pounds are below the surcharge

c = 20.155 + 1(0)

c = 20.155

Thus ; the value of c is  20.155 such that 99% of all parcels are under the surcharge weight.

Determine the t critical value for a lower or an upper confidence bound in each of the following situations. (Round your answers to three decimal places.)

a. Confidence level = 95%, df = 10
b. Confidence level = 95%, df = 15
c. Confidence level = 99%, df = 15
d. Confidence level = 99%, n = 5
e. Confidence level = 98%, df = 23
f. Confidence level = 99%, n = 32

Answers

Answer:

A. 1.812

B. 1.753

C. 2.602

D. 3.747

E. 2.069

F. 2.453

Step-by-step explanation:

A. 95% confidence level, the level of significance = 5% or 0.05

Using t-table, the critical value for a lower or an upper confidence bound at 0.05 significance level with 10 degrees of freedom = 1.182

B. 95% confidence interval = 0.05 level of significance

Using t-table, the critical value for a lower or an upper confidence bound at 0.05 significance level with 15 degrees of freedom = 1.753

C. 99% confidence interval = 0.01 level of significance

Using t-table, the critical value for a lower or an upper confidence bound at 0.01 significance level with 15 degrees of freedom = 2.602

D. 99% confidence interval = 0.01 level of significance; DF (n - 1) = 5- 1 = 4

Using t-table, the critical value for a lower or an upper confidence bound at 0.01 significance level with 4 degrees of freedom = 3.747

E. 98% confidence interval = 0.02 level of significance

Using t-table, the critical value for a lower or an upper confidence bound at 0.02 significance level with 23 degrees of freedom = 2.069

F. 99% confidence interval = 0.01 level of significance; df (n - 1) = 32 - 1 = 31

Using t-table, the critical value for a lower or an upper confidence bound at 0.01 significance level with 31 degrees of freedom = 2.453

PLEASE HELP QUICK! Determine x value of: sqrt x + 8 - sqrt x - 4 = 2

Answers

Answer:

x=8

Step-by-step explanation:

[tex]\sqrt{x+8}-\sqrt{x-4}=2\\\sqrt{x+8}=2+\sqrt{x-4}\\\left(\sqrt{x+8}\right)^2=\left(2+\sqrt{x-4}\right)^2\\x+8=x+4\sqrt{x-4}\\8=4\sqrt{x-4}\\8^2=\left(4\sqrt{x-4}\right)^2\\64=16x-64\\x=8[/tex]

Copy the problem, mark the givens in the diagram. Given: CS ≅ HR, ∠CHS ≅ ∠HCR, ∠CSH ≅ ∠HRC, Prove: CR ≅ HS

Help urgently needed

Answers

Explanation:

1. CS ≅ HR, ∠CHS ≅ ∠HCR, ∠CSH ≅ ∠HRC — given

2. ∆CRH ~ ∆HSC — AA similarity theorem

3. ∠SCH ≅ ∠RHC — corresponding angles of similar triangles are congruent

4. CH ≅ HC — reflexive property of congruence

5. ∆CRH ≅ ∆HSC — SAS congruence theorem

6. CR ≅ HS — CPCTC

2) A basketball player scores 70% of his shots on average. What is the probability that he scores at least 18 successful shots tonight if he gets 20 shots?

Answers

Answer:

3.54%

Step-by-step explanation:

This question represents a binomial distribution. A binomial distribution is given by:

[tex]P(x)=\frac{n!}{(n-x)!x!} p^xq^{n-x}[/tex]

Where n is the total number of trials, p is the probability of success, q is the probability of failure and x is the number of success.

Given that:

A basketball player scores 70% of his shots on average, therefore p = 70% = 0.7. Also q = 1 - p = 1 - 0.7 = 0.3.

The total number of trials (n) = 20 shots

The probability that he scores at least 18 successful shots tonight if he gets 20 shots = P(x = 18) + P(x = 19) + P(x = 20)

P(x = 18) = [tex]\frac{20!}{(20-18)!18!}*0.7^{18}*0.3^{20-18}=0.0278[/tex]

P(x = 19) = [tex]\frac{20!}{(20-19)!19!}*0.7^{19}*0.3^{20-19}=0.0068[/tex]

P(x = 20) = [tex]\frac{20!}{(20-20)!20!}*0.7^{20}*0.3^{20-20}=0.0008[/tex]

The probability that he scores at least 18 successful shots tonight if he gets 20 shots = P(x = 18) + P(x = 19) + P(x = 20) = 0.0278 + 0.0068 + 0.0008 = 0.0354 = 3.54%

Sam invest $4000 in an account that compounds interest continuously and earns 5.5% how long will it take for his money to reach $80,000 round to the nearest 10th of a year

Answers

Answer:

54.5 years.

Step-by-step explanation:

From the above question, we are asked to find the time

The formula for Time(t) =

t = log(A/P) / n[log(1 + r/n)]

A = Amount accumulated after a particular interest and period of time = $80,000

P = Principal (Money invested) = $4,000

r = rate = 5.5% = 0.055

n = compounding frequency = compounding continuously

n = number of days in a year × number of hours in a day

= 365 days × 24 hours = 8760

t = log(A/P) / n[log(1 + r/n)]

t = log(80,000/4,000) /8760[log(1 + 0.055/8760)]

t = log(80000 ÷ 4000) ÷ (8760 × [log(1 + 0.0000062785)]

t = 54.468367222 years

Approximately to the nearest tenth of a year, therefore, the length of time it will it take for his money to reach $80,000 is 54.5 years

Answer:

54.5

Step-by-step explanation:

How to calculate a circumference of a circle?

Answers

Answer: Pi multiplied by the diameter of the circle

Step-by-step explanation:

Answer:

The formula for finding the circumference of a circle is [tex]C = 2\pi r[/tex]. You substitute the radius of the circle for [tex]r[/tex] and multiply it by [tex]2\pi[/tex].

A lottery game has balls numbered 1 through 21. What is the probability of selecting an even numbered ball or an 8? Round to nearest thousandth

Answers

Answer: 0.476

Step-by-step explanation:

Let A = Event of choosing an even number ball.

B = Event of choosing an 8 .

Given, A lottery game has balls numbered 1 through 21.

Sample space: S= {1,2,3,4,5,6,7,8,...., 21}

n(S) = 21

Then, A= {2,4,6,8, 10,...(20)}

i.e. n(A)= 10

B= {8}

n(B) = 1

A∪B = {2,4,6,8, 10,...(20)} = A

n(A∪B)=10

Now, the probability of selecting an even numbered ball or an 8 is

[tex]P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}[/tex]

[tex]=\dfrac{10}{21}\approx0.476[/tex]

Hence, the required probability =0.476

Which expressions are equivalent to: 3(−2a - 4)+3a? A: -6a - 12 +3a B: 3a+12 C: none of the above smh

Answers

Answer:

A

Step-by-step explanation:

3(−2a - 4)+3a

=-6a - 12 +3a

A: -6a - 12 +3a

[tex]hope \: this \: helps[/tex]

Answer:

the answer is A

Step-by-step explanation:

you have to distribute the number 3 throughout the parentheses so (3*-2a-3*4)+3a = -6a-12+3a

An article in Fire Technology, 2014 (50.3) studied the effectiveness of sprinklers in fire control by the number of sprinklers that activate correctly. The researchers estimate the probability of a sprinkler to activate correctly to be 0.7. Suppose that you are an inspector hired to write a safety report for a large ballroom with 10 sprinklers. Assume the sprinklers activate correctly or not independently. (a) What is the probability that all of the sprinklers will operate correctly in a fire

Answers

Answer:

probability that all of the sprinklers will operate correctly in a fire: 0.0282

Step-by-step explanation:

In order to solve this question we will use Binomial  probability distribution because:

In the question it is given that the sprinklers activate correctly or not independently. The number of outcomes are two i.e. sprinklers activate correctly or not.

A binomial distribution is a probability of a success or failures outcomes in an  repeated multiple or n times.

Number of outcomes of this distributions are two.

The formula is:

b(x; n, P) = [tex]C_{n,x}*p^{x} * (1 - p)^{n-x}[/tex]

b = binomial probability  also represented as P(X=x)

x =no of successes

P = probability of a success on a single trial

n = no of trials

[tex]C_{n,x}[/tex] is calculated as:

[tex]C_{n,x}[/tex] = n! / x!(n – x)!

        = 10!  / 10!(10-10)!

        = 1

According to given question:

probability of success i.e. p = 0.7 i.e. probability of a sprinkler to activate correctly.

number of trials i.e. n = 10 as number of sprinklers are 10

To find: probability that all of the sprinklers will operate correctly in a fire

X = 10 because we have to find the probability that "all" of the sprinklers will operate correctly and there are 10 sprinklers so all 10 of them

So putting these into the formula:

P(X=x) = [tex]C_{n,x}*p^{x} * (1 - p)^{n-x}[/tex]

           = C₁₀,₁₀ * 0.7¹⁰ * (1-0.7)¹⁰⁻¹⁰

           =  1 * 0.0282 * (0.3) ⁰

           = 1 * 0.0282 * 1

  P(X=x)        = 0.0282

A rectangular parking lot has an area of 7/10 km 2.The width is 1/3 km 2 .What is the length of the parking lot written as a improper fraction ,in kilometers

Answers

Answer:

[tex]\dfrac{21}{10}\text{ km}[/tex].

Step-by-step explanation:

It is given that,

Area of rectangular plot [tex]=\dfrac{7}{10}\text{ km}^2[/tex]

Width of rectangular plot [tex]=\dfrac{1}{3}\text{ km}[/tex]

We need to find the length of the parking lot.

We know that,

[tex]\text{Area of rectangle}=length\times width[/tex]

[tex]\dfrac{7}{10}=length\times \dfrac{1}{3}[/tex]

[tex]\dfrac{7\times 3}{10}=length[/tex]

[tex]length=\dfrac{21}{10}[/tex]

Therefore, length of the parking lot is [tex]\dfrac{21}{10}\text{ km}[/tex].

WILL MARK AS BRAINLIEST 4. Suppose there is a card game where you are dealt a hand of three cards. You have already learned that the total number of three-card hands that can be dealt from a deck of 52 cards is: 52C3=52!/49!3! 52C3=22100 Calculate the probability of getting a hand that has exactly two aces in it (A A X). Do this by finding out the number of possible hands that have exactly two aces, and then dividing by the total possible number of three-card hands that is stated above. Part A: Use the multiplication principle to tell the total number of three-card hands (permutations) that can be made with two aces. (2 points) Part B: In the answer from Part I, each two-ace hand got counted twice. For example, A A X got counted as a separate hand from A A X. Since order should not matter in a card hand, these are really the same hand. What is the actual number of two-ace hands (combinations) you can get from a deck of 52 cards?(2 points) Part C: Find the probability of drawing a three-card hand that includes two aces from a deck of 52 cards. Write your answer as a fraction. (2 points)

Answers

Answer:

Part A- 6

Part B- 3

Part C- 3/22100

Step-by-step explanation:

Part A-

Use the permutation formula and plug in 3 for n and 2 for k.

nPr=n!/(n-k)!

3P2=3!/(3-2)!

Simplify.

3P2=3!/1!

3P2=6

Part B-

Use the combination formula and plug in 3 for n and 2 for k.

nCk=n!/k!(n-k)!

3C2=3!/2!(3-2)!

Simplify.

3C2=3!/2!(1!)

3C2=3

Part C-

It is given that the total number of three-card hands that can be dealt from a deck of 52 cards is 22100. Use the fact that the probability of something equals the total successful outcomes over the sample space. In this case the total successful outcomes is 3 and the sample space is 22100.

I believe the answer is 3/22100

I honestly suck at probability but I tried my best.

What is the angle between the given vector and the positive direction of the x-axis? (Round your answer to the nearest degree.) i + 3 j

Answers

Answer:

72°

Step-by-step explanation:

Given two vectors a and b, The vector i+3j will form a right angled triangle with the x-axis (i.e the horizontal axis).

The opposite side of the triangle on the Cartesian plane will be 3units along the y axis while the adjacent will be 1 unit along the x axis.

The angle between thus two vectors is expressed as tan (theta) = opp/adj

tantheta = 3/1

theta = tan^-1(3)

Theta = 71.57° ≈ 72° to nearest degree

Find the largest integer which belongs to the following interval: [−∞, 31]

Answers

Answer:

Largest integer in the interval [−∞, 31] is 31.

Step-by-step explanation:

Given the interval: [−∞, 31]

To find: The largest integer in this interval.

Solution:

First of all, let us learn about the representation of intervals.

Two kind of brackets can be used to represent the intervals. i.e. () and [].

Round bracket means not included in the interval and square bracket means included in the interval.

Also, any combination can also be used.

Let us discuss one by one.

1. [p, q] It means the interval contains the values between p and q. Furthermore, p and q are also included in the interval.

Smallest p

Largest q

2. (p, q) It means the interval contains the values between p and q. Furthermore, p and q are not included in the interval.

Smallest value just greater than p.

Largest value just smaller than q.

3. [p, q) It means the interval contains the values between p and q. Furthermore, p is included in the interval but q is not included in the interval.

Smallest value p.

Largest value just smaller than q.

4. (p, q] It means the interval contains the values between p and q. Furthermore, p is not included in the interval but q is included in the interval.

Smallest value just greater than p.

Largest value q.

As per above explanation, we can clearly observe that:

The largest integer which belongs to the following interval: [−∞, 31] is 31.

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