Barry runs at an average rate of 8 mi/hr. He walks at an average rate of 3 mi/hr. If x represents the time spent running and y represents the time spent walking, write a linear equation that relates the time he could spend running and walking if he travels a total distance of 16 miles

Answer:

The number of ways Tony and Maria can both be selected for the committee is 8 ways.

Step-by-step explanation:

i) Tony and Maria have to be on the committee together or not at all.

ii) Let us consider Tony and Maria as combined as one person.

 Therefore now we can say that the number of eligible people for the committee = 9 - 1 = 8.

iii) therefore the number of ways that both Tony and Maria can be selected for the committee are

 = 8C1   =  [tex]\hspace{0.2cm}\binom{8}{1} = \frac{8!}{1! (8-1)!} = \frac{8!}{1!\times 7!} = \frac{8}{1} = \hspace{0.1cm}8 \hspace{0.1cm}ways[/tex]

The number of ways to select exactly one of Tony and Maria is 70.

It is given that,

Explanation:

Excluding Tony and Maria from the 9 people. The number of remaining people is 7.

Exactly one of Tony and Maria be selected for the committee. So, one person is selected from 2 and 3 people are selected from the remaining 7 people.

[tex]\text{Number of ways}=^2C_1\times ^7C_3[/tex]

[tex]\text{Number of ways}=\dfrac{2!}{1!(2-1)!}\times \dfrac{7!}{3!(7-3)!}[/tex]

[tex]\text{Number of ways}=2\times 35[/tex]

[tex]\text{Number of ways}=70[/tex]

Thus, the number of ways to select exactly one of Tony and Maria is 70.

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Answer:

The answer to your question is  [tex]\frac{3b}{a}[/tex] , check your options, maybe you forgot one.

Step-by-step explanation:

Original operation

                                [tex]\frac{6a^{2}}{5b^{2}} \frac{45b^{3}}{18a^{3}}[/tex]

Process

1.- Simplify 6 and 18 and 45 and 5

   [tex]\frac{6}{18} = \frac{3}{9} = \frac{1}{3}[/tex]

  [tex]\frac{45}{5} = \frac{9}{1} = 9[/tex]

 Result

  [tex]\frac{9}{3} = \frac{3}{1} = 3[/tex]

2.- Simplify a² and a³

    [tex]\frac{a^{2}}{a^{3}} = \frac{1}{a}[/tex]

3.- Simplify b³ and b²

     [tex]\frac{b^{3}}{b^{2}} = b[/tex]

4.- Join the results [tex]\frac{3b}{a}[/tex]

Answer:

Therefore the measure of∠ A is 60.07.

Step-by-step explanation:

Given:

In Right Angle Triangle ABC

∠ B = 90°

BC = 13   ....Side opposite to angle A

AC = 15  .... Hypotenuse

To Find:

m∠A = ?

Solution:

In Right Angle Triangle ABC ,Sine Identity,

[tex]\sin A = \dfrac{\textrm{side opposite to angle A}}{Hypotenuse}\\[/tex]

Substituting the values we get

[tex]\sin A = \dfrac{BC}{AC}=\dfrac{13}{15}=0.8666\\\\A=\sin^{-1}(0.8666)=60.065\\\\m\angle A=60.07\°[/tex]

Therefore the measure of∠ A is 60.07

Answer:

y = 0.5x - 5

Step-by-step explanation:

(0,-5) (6,-2)

m = (y2-y1)/(x2-x1)

= (-2-(-5))/(6-0)

= (-2+5)/6

= 3/6

= 1/2 or 0.5

Y-intercept is clearly -5, so c = -5

in y = mx + c

y = 0.5x - 5

Answer:

In this case the Central Tendency measure that would be appropriate to report is Mode.

Step-by-step explanation:

Central Tendency measures are listed as follows:

i.) Mean

ii) Median

iii) Mode.

In the case that the data collected of a population is qualitative and not quantitative then the best Central Tendency measure to qualify the data is Mode of the data.

In the given example the data collected is of the students' racial classification which is not quantitative and purely qualitative. Therefore in case it is proper to take the Central Tendency measure to be reported as the Mode.

Answer: D

Step-by-step explanation:

The similarity statement relating the three triangles in the diagram is 'YHB ~ YDH ~ HDB'. This is because they are all similar triangles, sharing the same shape but differing in sizes due to scale.

In this case, the answer would be 'YHB ~ YDH ~ HDB'. We are looking for a similarity statement, which states that all three triangles are similar. Similar triangles are triangles that have the same shape, but can be different sizes, i.e. they are scaled versions of each other. Since HD is a perpendicular drawn from right angle H in triangle YHB, this creates two triangles (YDH, HDB) that are respectively similar to the original triangle YHB as each of the two triangles include one of the acute angles of triangle YHB and their own right angles. Therefore, YHB ~ YDH ~ HDB.

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Answer:

The value of the house after adding the garage is $135,700.

Step-by-step explanation:

Given,

value of house before adding garage = $118,200.00

we need to find the value of house after adding two car garage.

Solution,

Since after adding two car garage the value of the house increased by 15%.

So firstly we will find out the 15% of the value of the house after adding garage.

So we can say that;

15% of the value of the house after adding garage is equal to 15 divided by 100 the multiplied with the value of the house before adding garage.

15% of the value of the house after adding garage = [tex]\frac{15}{100}\times118,000=\$17,700[/tex]

Now, The value of the house after adding garage is equal to the sum of value of house before adding garage and 15% of value of house before adding garage.

We can frame it in equation form as;

The value of the house after adding garage = [tex]\$118,000+\$17,000=\$135,700[/tex]

Hence The value of the house after adding the garage is $135,700.

Answer:

Step-by-step explanation:

On pet day 18 children brought a pet to school.

2/3 of the pets were dogs. This means that the number of pets that were dogs would be

2/3 × 18 = 12 dogs

1/9 of the pets were cats. This means that the number of pets that were cats would be

1/9 × 18 = 2 dogs

Total number of dogs and cats is

12 + 2 = 14

The number of animals that were neither dogs nor cats is

18 - 14 = 4

Final answer:

There were 12 dogs, 2 cats, and 4 animals that were neither dogs nor cats among the 18 pets that children brought to school on pet day.

Explanation:

On pet day, 18 children brought a pet to school. To find out how many dogs were there, we calculate two-thirds of 18, and for the cats, we calculate one-ninth of 18.

Dogs: ⅓ × 18 = 12 dogs

Cats: ⅙ × 18 = 2 cats

To determine how many animals were neither dogs nor cats, we subtract the number of dogs and cats from the total number of pets: 18 - (12 + 2) = 4 animals neither dogs nor cats.

So, there were 12 dogs, 2 cats, and 4 animals that were neither dogs nor cats.

Answer: Charlotte finished the word search in 16 minutes.

Step-by-step explanation:

Frank finished the word search in 32 minutes.

Terrence finished a word search in 3/4 the time it took Frank. This means that the time it took Terrence to finish the word search would be

3/4 × 32 = 24 minutes.

Charlotte finished the word search in 2/3 the time it took Terrence. This means that the time it took Charlotte to finish the word search would be

2/3 × 24 = 16 minutes

Answer:

Each knight will guard 288 minutes in one day.

Step-by-step explanation:

Given:

A castle has to be guarded 24 hours a day. Five knights are ordered to split each day's guard duty equally.

Now, to find the minutes will each night spend on guard duty in one day.

As, in 1 hour there are 60 minutes.

Thus, in 24 hour there are 60 × 24 = 1440 minutes.

Total minutes for guarding = 1440 minutes.

So, there are knights ordered for guarding are = 5.

And each day's guard duty equally.

Now, to get the minutes will each night spend on guard duty in one day we divide the total minutes for guarding by number of knights that is 5:

[tex]1440\div5[/tex]

[tex]=288\ minutes.[/tex]

Therefore, each knight will guard 288 minutes in one day.

Answer: 321 adult tickets and 227 children tickets were sold.

Step-by-step explanation:

Let x represent the number of adult tickets that were sold.

Let y represent the number of children tickets that were sold.

The total number of tickets that the theatre sold is 548. This means that

x + y = 548

Adult tickets sell for $6.50 each, and children's tickets sell for $3.50 each. The total ticket sales was $2881. This means that

6.5x + 3.5y = 2881 - - - - - - - - - - -1

Substituting x = 548 - y into equation 1, it becomes

6.5(548 - y) + 3.5y = 2881

3562 - 6.5y + 3.5y = 2881

- 6.5y + 3.5y = 2881 - 3562

- 3y = - 681

y = - 681/ -3

y = 227

x = 548 - y = 548 - 227

x = 321

To Determine:

So, here is the complete proof of the Pythagorean theorem.

Given: Δ ABC is a right triangle, with

a right angle at ∠C

Prove: A²+B² =C²

Answer:

Note: All the answers for the questions marks are filled with in bold text.

Statement

1. ΔABC is a right triangle, with a right angle at ∠C

2. Draw an altitude from point C to line AB

3. ∠CDB and ∠CDA are right angles.

4. ∠BCA ≅ ∠BDC

5. ∠B ≅ ∠B

6. AA Similarity Postulate          

7.   [tex]\frac{a}{x}\:=\:\frac{c}{a}[/tex]

8. a² = cx

9. ∠CDA ≅ ∠BCA

10. ∠A ≅ ∠A

11. ΔCBA ~ ΔDBA      

12.  [tex]\frac{b}{y}\:=\:\frac{c}{b}[/tex]

13. b² = cy

14. a² + b² = cx + cy

15. [tex]\left(CB\right)^2+\left(CA\right)^2=\left(AB\right)\left(DB+BA\right)[/tex]

16. x + y = c

17. a² + b² = c²

Reason

1. Given

2. From a point not on a line, exactly one perpendicular can be drawn through the point to the line.

3. Definition of altitude

4. All right angles are congruent.

5. Reflexive Property

6. AA Similarity Postulate

7. Polygon Similarity Postulate

8. Cross Multiply and Simplify

9. All right angles are Congruent

10. Reflexive Property

11. AA similarity Postulate

12. Polygon Similarity Postulate

13. Cross Multiply and Simplify

14. Addition Property of Equality

15. Distributive Property

16. Segment Addition Postulate

17. Substitution Property

Keywords: statement, reason

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Answer:

Vanilla milkshake = 1/3 and Chocolate milkshake = 2/3

Step-by-step explanation:

Given data:

Total ounce of milkshake = 12 ounce

Vanilla milkshake = 4

Chocolate is the rest which can be interpreted as 12 - 4= 8 ounce

Representing as fractions

Vanilla milkshake = 4/12 (Reducing to lowest terms that is diving numerator and denominator by common factor in this case 4)

Vanilla milkshake = 1/3

Chocolate milkshake = 8/12 (Reducing to lowest terms that is diving numerator and denominator by common factor in this case 4)

Chocolate milkshake = 2/3

Answer:

a) [tex] t = 2 *ln(\frac{82}{5}) =5.595[/tex]

b) [tex] t = 2 *ln(-\frac{820}{p_0 -820}) [/tex]

c) [tex] p_0 = 820-\frac{820}{e^6}[/tex]

Step-by-step explanation:

For this case we have the following differential equation:

[tex] \frac{dp}{dt}=\frac{1}{2} (p-820)[/tex]

And if we rewrite the expression we got:

[tex] \frac{dp}{p-820}= \frac{1}{2} dt[/tex]

If we integrate both sides we have:

[tex]ln|P-820|= \frac{1}{2}t +c[/tex]

Using exponential on both sides we got:

[tex] P= 820 + P_o e^{1/2t}[/tex]

Part a

For this case we know that p(0) = 770 so we have this:

[tex] 770 = 820 + P_o e^0[/tex]

[tex] P_o = -50[/tex]

So then our model would be given by:

[tex] P(t) = -50e^{1/2t} +820[/tex]

And if we want to find at which time the population would be extinct we have:

[tex] 0=-50 e^{1/2 t} +820[/tex]

[tex] \frac{820}{50} = e^{1/2 t}[/tex]

Using natural log on both sides we got:

[tex] ln(\frac{82}{5}) = \frac{1}{2}t[/tex]

And solving for t we got:

[tex] t = 2 *ln(\frac{82}{5}) =5.595[/tex]

Part b

For this case we know that p(0) = p0 so we have this:

[tex] p_0 = 820 + P_o e^0[/tex]

[tex] P_o = p_0 -820[/tex]

So then our model would be given by:

[tex] P(t) = (p_o -820)e^{1/2t} +820[/tex]

And if we want to find at which time the population would be extinct we have:

[tex] 0=(p_o -820)e^{1/2 t} +820[/tex]

[tex] -\frac{820}{p_0 -820} = e^{1/2 t}[/tex]

Using natural log on both sides we got:

[tex] ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t[/tex]

And solving for t we got:

[tex] t = 2 *ln(-\frac{820}{p_0 -820}) [/tex]

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

[tex] 12 = 2 *ln(\frac{820}{820-p_0}) [/tex]

[tex] 6 = ln (\frac{820}{820-p_0}) [/tex]

Using exponentials we got:

[tex] e^6 = \frac{820}{820-p_0}[/tex]

[tex] (820-p_0) e^6 = 820[/tex]

[tex] 820-p_0 = \frac{820}{e^6}[/tex]

[tex] p_0 = 820-\frac{820}{e^6}[/tex]

For the given case of increment of mice' population, we get following figures:

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations. The derivatives might be of any order, some terms might contain product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

For the considered case, the population of mice with respect to time passed in months is given by the differential equation:

[tex]\dfrac{dp}{dt} = 0.5p - 410[/tex]

Taking same variable terms on same side, and then integrating, we get:

[tex]\dfrac{dp}{0.5p - 410} = dt\\\\\int \dfrac{dp}{0.5p - 410} = \int dt\\\\\dfrac{\ln(|0.5p - 410|)}{0.5} = t + C_1\\\ln(|0.5p - 410|) = 0.5t + 0.5C_1 = 0.5t + C[/tex]

where C₁ is integration constant.

Since it is specified that at time t = 0, the population p = 770, therefore, putting these values in the equation obtained above, we get:

[tex]\ln(|0.5p - 410|) = 0.5t + 0.5C_1 = 0.5t + C\\\\\ln(|0.5 \times 770 - 410|) = 0.5 \times 0 + C\\\\\ln(|-25|) = C\\C = \ln(25) \approx 3.22[/tex]

Therefore, we get the relation between p and t as:

[tex]\ln(|0.5p - 410|) = 0.5t + 0.5C_1 = 0.5t + C\\\ln(|0.5p - 410|) \approx 0.5t + 3.22\\\\|0.5p - 410| \approx e^{0.5t + 3.22}\\\text{Squaring both the sides}\\\\(0.5p - 410)^2 \approx e^{t+6.44}\\(p-820)^2 \approx 4e^{t+6.44}\\\\p^2 -1640p + 672400 \approx 4e^{t+6.44}[/tex]

Calculating the needed figures for each sub-parts of the problem:

a): The time at which the population becomes extinct.

Let it be t at which p becomes 0, then, from the equation obtained, we get:

[tex]p^2 -1640p + 672400 \approx 4e^{t+6.44}\\\text{At p = 0}\\672400 \approx 4e^{t+6.44}\\\\t \approx \ln{(\dfrac{672400}{4}) - 6.44 = \ln(168100) - 6.44 \approx 5.59 \text{\: (In months)}[/tex]

Thus, after 5.59 months approx, the population of mice will extinct.

b) Find the time of extinction if p(0) = p0, where 0 < p0 < 820

From the equation [tex]\ln(|0.5p - 410|) = 0.5t + C[/tex]

putting [tex]p = p_0[/tex] when t = 0, we get the value of C as:

[tex]\ln(|0.5p_0 - 410|) = C[/tex]

Thus, the equation becomes

[tex]\ln(|0.5p - 410|) = 0.5t + \ln(|0.5p_0 - 410|)[/tex]

At time of extension t months, p becomes 0, thus,

[tex]\ln(|0.5p - 410|) = 0.5t + \ln(|0.5p_0 - 410|)\\\text{At p = 0, we get}\\\\\ln(410)=0.5t + \ln(|0.5p_0 - 410|)\\\\t = 2\ln(\dfrac{410}{0.5p_0 - 410}) = 2\ln(\dfrac{820}{|p_0-820|})\\\\\text{Since 0 } < p_0 < 820, \text{ thus, we get }\\\\t = 2\ln(\dfrac{820}{820-p_0})[/tex]

Thus, the extinction time (in months) of population of mice when its given that [tex]p(0) = p_0[/tex] is given by:

[tex]t = 2\ln(\dfrac{820}{820-p_0})[/tex]

c) Find the initial population [tex]p_0[/tex] if the population is to become extinct in 1 year.

Putting t = 12 (since t is measured in months, and that 1 year = 12 months) in the equation obtained in the second part, we get the value of initial population as:

[tex]t = 2\ln(\dfrac{820}{820-p_0})\\\\12 = 2\ln(\dfrac{820}{820-p_0})\\e^{6} = \dfrac{820}{820-p_0}\\1 - \dfrac{p_0}{820} = \dfrac{1}{e^6}\\p_0 \approx 820(1 - \dfrac{1}{e^6}}) \approx 818[/tex]

Thus, the initial population of mice for given conditions would be approx 818

Therefore, for the given case of increment of mice' population, we get following figures:

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Answer:

the gcf is 21r^2t^3

Step-by-step explanation:

21r^2t^3(3+2rt^2)

The greatest common factor of the expression 63r²t³+42r³t⁵ is 21r²t³, determined by finding the highest common power of each factor.

To find the greatest common factor (GCF) of the expression 63r²t³+42r³t⁵, we need to identify the highest powers of each factor that divide both terms.

Thus, the GCF of the expression 63r²t³+42r³t⁵ is 21r²t³.

Answer: the number of multiple choice questions in the test is 12.

the number of short response questions in the test is 8.

Step-by-step explanation:

Let x represent the number of multiple choice questions in the test.

Let y represent the number of short response questions in the test.

The total number of questions in the test is 20. It means that

x + y = 20

The multiple choice questions are worth 3 points each and the short response question are worth 8 points each. The total number of points is 100. It means that

3x + 8y = 100 - - - - - - - - - - 1

Substituting x = 20 - y into equation 1, it becomes

3(20 - y) + 8y = 100

60 - 3y + 8y = 100

- 3y + 8y = 100 - 60

5y = 40

y = 40/5 = 8

x = 20 - y = 20 - 8

x = 12

Answer:

[tex]FJ=18\ units[/tex]

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

If two triangles are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

In this problem

△FEG is similar with △FHJ -----> by AA Similarity Theorem

so

[tex]\frac{FE}{FH}=\frac{FG}{FJ}[/tex]

we have

[tex]FE=HF+EH=12+8=20\ units\\FH=HF=12\ units\\FG=30\ units[/tex]

substitute the given values

[tex]\frac{20}{12}=\frac{30}{FJ}\\\\FJ=12(30)/20\\FJ=18\ units[/tex]

Answer: Fj = 18

Step-by-step explanation:

Hello! So i just took the test and I got this wrong but the answer is 18 down below is a screenshot so you know 18 is the correct answer :) Hope this helps ^.^

Answer:

Answer in explanation

Step-by-step explanation:

The ratio of their ownership is 4:3 I.e J to T

Now we know that Terrence has 36 model cars. To find the number of model cars Jim own, we need to find the unit ownership. This is the same as 36/3 which is 12 cars.

This means Jim has 12 * 4 = 48 model cars.

Now we are looking at them selling 10 of their cars each. This would bring the number of model cars owned to be 38 and 26 respectively.

The ratio here would now be 38:26 which is same as 19:13. Of course this is different from 4:3, hence we can conclude that the ratio will indeed change

Final answer:

Jim owns 48 model cars.

If both Jim and Terrence sell ten cars each, Jim will have 38 cars, Terrence will have 26, and the ratio of Jim's cars to Terrence's cars will change to 19:13.

Explanation:

The ratio of the number of model cars that Jim owns to the number of cars Terrence owns is 4:3. If Terrence owns 36 model cars, we can set up a proportion to find out how many model cars Jim owns.

Because Terrence's part of the ratio corresponds to 36 cars, we have:

Jim's cars / Terrence's cars = 4/3
Jim's cars / 36 = 4/3

Cross-multiplying to solve for Jim's cars:

(Jim's cars) * 3 = 4 * 36
Jim's cars = (4 * 36) / 3
Jim's cars = 144 / 3
Jim's cars = 48

Therefore, Jim owns 48 model cars.

If Jim and Terrence each sell ten of their model cars, Jim will have 38 model cars and Terrence will have 26. The new ratio will be:

38 / 26, which simplifies to 19 / 13, different from the original ratio of 4/3.

So, yes, the ratio will change if they both sell ten of their cars.

Question was Incomplete;Complete question is given below;

Trey drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 12 hours. When Trey drove home, there was no traffic and the trip only took 8 hours. If his average rate was 20miles per hour faster on the trip home, how far away does Trey live from the mountains?

Do not do any rounding.

Answer:

Trey lives 480 miles from the mountain.

Step-by-step explanation:

Given:

Time taken to drove the mountain =12 hours

Time taken to return back from mountain = 8 hours.

Let the speed at which he drove to mountain be denoted by 's'.

Speed on the trip to home = [tex]s+20[/tex]

We need to find the distance Trey live from the mountains.

Solution:

Let the distance be denoted by 'd'.

Now we know that;

Distance is equal to speed times Time.

framing in equation form we get;

distance from home to mountain [tex]d=12s[/tex]

Also distance from mountain to home [tex]d = (s+20)8=8s+160[/tex]

Now distance is same for both the trips;

so we can say that;

[tex]12s=8s+160[/tex]

Combining the like terms we get;

[tex]12s-8s=160\\\\4s=160[/tex]

Dividing both side by 4 we get;

[tex]\frac{4s}{4}=\frac{160}{4}\\\\s=40\ mph[/tex]

Speed while trip to mountain = 40 mph

Speed while trip to home = [tex]s+20=420+20=60\ mph[/tex]

So Distance [tex]d=12s=12\times40 = 480\ miles[/tex]

Hence Trey lives 480 miles from the mountain.

Answer: the ratio that represents Cos C = a/b

Step-by-step explanation:

Triangle ABC is a right angle triangle.

From the given right angle triangle,

AC represents the hypotenuse of the right angle triangle.

With m∠C as the reference angle,

BC represents the adjacent side of the right angle triangle.

AB represents the opposite side of the right angle triangle.

To determine m∠C, we would apply

the cosine trigonometric ratio.

Cos θ = adjacent side/hypotenuse. Therefore,

Cos C = a/b

Answer:

Interval data

Step-by-step explanation:

Brian's score is an interval data because it appears within the GMAT range of score, which is 200-800

Answer:

pictured and shown and solved

Answer:

5x - 4

Step-by-step explanation:

2( x + 4 ) + 3( x - 4)

Distribute the 2 into ( x + 4 ) so it comes out to be 2x + 8

now distribute 3 into ( x - 4 ) = 3x - 12

now combine like terms 2x + 8 + 3x - 12

2x and 3x can combine into 5x

-12 and 8 can combine into -4

so it comes out to be 5x - 4

Answer:

72 = 2h

Step-by-step explanation:

6 dozens = 6×12 = 72

72 = 2h

Each hen lays eggs, so h hens will lay 2×h eggs

The equation you've made implies every hen lays 36 eggs

72 = 2h is the equation,

Which simplifies to

h = 36

Answer:

[tex]m\angle CED= 64\°[/tex]  

[tex]m\angle ACD=124\°[/tex]  

Step-by-step explanation:

In the figure given:

∠ABC = 93°

∠BAC = 31°

∠CDE = 60°

To find ∠CED and ∠ACD.

Solution:

In triangle ABC, we are given two vertex angles. We can find the third angle as angle sum of triangle = 180°.

∠ABC = 93° , ∠BAC = 31°

∠BCA=  [tex]180\°-(93\°+31\°)[/tex]

∠BCA = 56°

[tex]m\angle BCA+m\angle ACD=180\°[/tex]    [Supplementary angles forming a linear pair]

[tex]m\angle ACD=180\°-56\°[/tex]

[tex]m\angle ACD=124\°[/tex]   (Answer)

In triangle CDE:

[tex]m\angle CDE+m\angle CED = m\angle ACD[/tex]   [Exterior angle theorem :Exterior angle of a triangle is equal to sum of opposite interior angles ]

[tex]60\°+m\angle CED = 124\°[/tex]

[tex]m\angle CED= 124\°-60\°[/tex]

[tex]m\angle CED= 64\°[/tex]      (Answer)

Answer:

m\angle CED= 64\°  

m\angle ACD=124\°  

Step-by-step explanation:

get an A!

Answer:

Step-by-step explanation:

a) The two populations were i) the pregnant mothers who took the drug ii) the pregnant mothers who did not take the drugs

b) The data must have been obtained in a survey because experiment was not done.

c) 63 out of 3980 developed abnormalities in I case.

Hence out of 1000 abnormalities estimated = [tex]\frac{63}{3980} *1000\\=15.829\\[/tex]

i.e. approximately 16

d) Mothers who did not take drug

(information incomplete)

e) Medical hypothesis testing requires accurate results and hence sample sizes should be very large.

The mentioned study compares two populations: women exposed to DES during their mother's pregnancy and women who weren't. The data seems to be from a survey, is calculated with available information to be around 15.8 per 1000 women for the first population and half that for the second. Medical studies use large samples for higher statistical reliability.

a. The two populations in this study are women whose mothers took the drug DES during pregnancy and women whose mothers did not take the drug DES during pregnancy.

b. The data is most likely obtained through a survey, since medical research often relies on observations, health histories, and existing data rather than conducting an experiment. This would also protect subjects' safety and uphold ethical considerations.

c. To provide a descriptive statistic, we would use the rate of occurrence in the sample to estimate the rate in the overall population. In the sample of 3980 women, 63 developed tissue abnormalities. This is a rate of (63/3980) * 1000 ≈ 15.8 per 1000 women.

d. This question does not provide specific data for the second population, but based on the statement that women from the first group are twice as likely to develop abnormalities, we can estimate that the occurrence of abnormalities in the second population would be half as frequent. This would be approximately 7.9 out of 1000 women.

e. Large sample sizes are often used in medical studies to ensure the results are statistically significant and more reliable. This helps to avoid anomalies and provides a more accurate representation of the population.

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Answer:

Through the expression, "let freedom ring", Martin Luther King Jr. was emphasizing on the need for community effort throughout the nation in order to counteract segregation. Let freedom ring is a metaphor for the action of spreading equality

Step-by-step explanation:

Answer:this isn’t in the right spot it’s in math. But when mlk says this he wants equality for all people through the country and this is the metaphor he uses. He wants justice and freedom to be granted to all.

Step-by-step explanation:

Answer:

[tex]V=24,501.42\ m^3[/tex]

Step-by-step explanation:

we know that

The volume of the cylinder is equal to

[tex]V=\pi r^{2} h[/tex]

where

r is the radius of the circular base of cylinder

h is the height of the cylinder

we have

[tex]r=34/2=17\ m[/tex] ----> the radius is half the diameter

[tex]h=27\ m\\\pi=3.14[/tex]

substitute the given values in the formula

[tex]V=(3.14)(17)^{2} (27)=24,501.42\ m^3[/tex]

Answer:

Step-by-step explanation

Givens

r is the radius = 17

h is the height of the cylinder = 27

pi = 3.14

Unknowns = Volume

Anwser = 24,501.42 m^3

Answer:

Step-by-step explanation:

We would apply the formula for determining simple interest which is expressed as

I = PRT/100

Where

I = interest at the end of t years

r represents the interest rate.

P represents the principal or initial amount deposited.

t represents the number of years of investment.

From the information given,

P = 1000

R = 2%

T = 5 years

Therefore,

I = (1000 × 2 × 5)/100

I = $100

The total amount in the account after 5 years would be

1000 + 100 = $1100