I need help ASAP
Geometry​

Answer:

\(300^{\circ}\)

Step-by-step explanation:

\(\text{In quadrant IV, both ratios of tangent and sine are negative.}\\ \\300^{\circ}~ \text{lies in the fourth quadrant, so}~ \tan 300^{\circ} ~ \text{and}~ \sin 300^{\circ}~ \text{are negative.}\)

The possible number of ways of selection by using combination formula is 3060.

What is combination?

A grouping of items where the order in which they were chosen is irrelevant is known as combination.

Given that the number of people in the group is 18. Only 4 people will be member of the council.

The formula of combination is \(^{n}C_{r}=\frac{n!}{r!(n-r)!}\) , where r number of objects are selected from n objects.

In the given question the value of n is 18 and the value of r is 2.

Substitute the value of n and r in \(^{n}C_{r}=\frac{n!}{r!(n-r)!}\).

\(^{18}C_{4}=\frac{18!}{4!(18-4)!}\)

Solve the above expression:

18^C_4 = 18x17x16x15x14/4x14

18^C_4 = 18x17x16x15/4x3x2x1

3060 18^C_4 = 3060

The number of ways such selection is 3060.

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Answer:

x = 40

smaller angle is 80

larger angle is 100

Step-by-step explanation:

Supplementary angles add up to equal 180 degrees

Hence, x + 40 + 3x - 20 = 180

( note that we just created an equation that we can use to solve for x )

We now solve for x using the equation created

x + 40 + 3x - 20 = 180

step 1 combine like terms

x + 3x = 4x

40 - 20 = 20

we now have 4x + 20 = 180

step 2 subtract 20 from each side

180 - 20 = 160

we now have 160 = 4x

step 3 divide each side by 4

160 / 4 = 40

4x / 4 = x

we're left with x = 40

we now plug in 40 for x to find the value of each angle

x + 40

* substitute 40 for x *

40 + 40 = 80

3x - 20

* substitute 40 for x *

3 ( 40 ) - 20

3 * 40 = 120

120 - 20 = 20

so the larger angle is 100 degrees and the smaller angle is 80 degrees

Answer:

11

Step-by-step explanation:

-14-3x=-5x+8

-14+2x=8

2x=22

x=11

Hope this helps!

Answer:

x=11

Step-by-step explanation:

-14-3x=-5x+8

-14+2x=8

2x=22

x=11

Option D. The median cost of course materials for nursing majors is over $300 more than the median cost of course materials for non-nursing majors.

The given boxplots show the circulation of the expense obviously materials for nursing majors and non-nursing majors. From the plots, we can reason that the scope of the circulation of the expense obviously materials for nursing majors is like that of non-nursing majors, as the most extreme and least qualities are around at a similar level. We can likewise infer that the most extreme expense for non-nursing majors is more prominent than the middle expense for nursing majors.

Also, the inconstancy of the expense obviously materials for the center half of nursing majors is more noteworthy than the fluctuation of the center half for non-nursing majors, as the cases for nursing majors are more extensive. In any case, we can't reason that the middle expense obviously materials for nursing majors is more than $300 more than the middle expense obviously materials for non-nursing majors, as the medians are not straightforwardly named and their division isn't plainly shown. At last, we can see that the study included 17 nursing majors and 17 non-nursing majors.

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The complete question is:

A college professor conducted a survey in order to assess how much money nursing majors spend on course material compared to all other majors. To do so, she selected a random sample of 34 students. Each student was classified as a nursing major or as a non-nursing major. They were then asked how much they spent on books and other materials required for their courses this semester. Shown above are parallel boxplots summarizing the responses. Based upon the boxplots, which of the following statements cannot be concluded?

A. The range of the distribution of the cost of course materials for nursing majors is about the same as that of non-nursing majors.

B. The maximum cost for non-nursing majors is greater than the median cost for nursing majors.

C. The variability of the cost of course materials for the middle 50% of nursing majors is greater than the variability of the middle 50% for non-nursing majors.

D. The median cost of course materials for nursing majors is over $300 more than the median cost of course materials for non-nursing majors.

E. The boxplots reveal that 17 students are nursing majors and 17 students are non-nursing majors

Answer:

El tercer ángulo en un triángulo será 16 ° si los otros dos ángulos son 112º y 52º.

El cuarto ángulo en un cuadrilátero será 213 ° si los otros tres ángulos son 105º, 32º y 10º.

Step-by-step explanation:

Spanish

La suma de los ángulos de un triángulo siempre es igual a 180 grados. Si conocemos dos de ellos, podemos encontrar el tercero. Dejemos que el tercero se denote por x entonces

112 ° + 52 ° + x = 180 °

x = 180 ° -112 ° -52 °

x = 16 °

El tercer ángulo en un triángulo será 16 ° si los otros dos ángulos son 112º y 52º.

La suma de los ángulos de un cuadrilátero siempre es igual a 360 grados. Si conocemos tres de ellos, podemos encontrar el cuarto. Dejemos que el cuarto se denote por y luego

105º + 32º + 10º + y = 360 °

o

y = 360 ° -105º- 32º- 10º

y = 213 °

El cuarto ángulo en un cuadrilátero será 213 ° si los otros tres ángulos son 105º, 32º y 10º.

English

The sum of angles in a triangle is always equal to 180 degrees. If we know any two of them we can find the third one. Let the third one be denoted by x then

112°+52°+x= 180°

x=  180°-112°-52°

x= 16°

The third angle in a triangle will be 16° if the other two angles are 112º and 52º.

The sum of angles in a quadrilateral is always equal to 360 degrees. If we know any three of them we can find the fourth one. Let the fourth one be denoted by y then

105º+ 32º+  10º+y= 360°

or

y = 360°-105º- 32º- 10º

y= 213°

The fourth angle in a quadrilateral  will be 213° if the other three angles are 105º, 32º and 10º.

The conjecture that describes the pattern in the sequence is current term is 2 added to the previous term and the next term is 10

The sequence is given as:

0, 2, 4, 6, 8

In the above sequence, we can see that each current term is 2 added to the previous term

i.e.

Tn = 2 + Tn-1

Using the above conjecture, we have

Next term = 8 + 2

Evaluate

Next term = 10

Hence, the conjecture that describes the pattern in the sequence is current term is 2 added to the previous term and the next term is 10

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c = pi/2, and the value of c > 1 such that the average value of f(x) on the interval [2, 20] is equal to c is c = pi/2.

The average value of a function f(x) on the interval [a, b] is given by:

Avg = 1/(b-a) * ∫[a, b] f(x) dx

We want to find a value of c > 1 such that the average value of the function \(f(x) = (9pi/x^2)cos(pi/x)\) on the interval [2, 20] is equal to c.

First, we find the integral of f(x) on the interval [2, 20]:

\(∫[2, 20] (9pi/x^2)cos(pi/x) dx\)

We can use u-substitution with u = pi/x, which gives us:

-9pi * ∫[pi/20, pi/2] cos(u) du

Evaluating this integral gives us:

\(-9pi * sin(u) |_pi/20^pi/2 = 9pi\)

Therefore, the average value of f(x) on the interval [2, 20] is:

\(Avg = 1/(20-2) * ∫[2, 20] (9pi/x^2)cos(pi/x) dx\)

= 1/18 * 9pi

= pi/2

Now we set c = pi/2 and solve for x:

Avg = c

\(pi/2 = 1/(20-2) * ∫[2, 20] (9pi/x^2)cos(pi/x) dx\)

pi/2 = 1/18 * 9pi

pi/2 = pi/2

Therefore, c = pi/2, and the value of c > 1 such that the average value of f(x) on the interval [2, 20] is equal to c is c = pi/2.

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Answer:

76.275

Step-by-step explanation:

First - the entire circles area is 78.54. \(\pi r^{2}\) with r being 5

6 triangles of that size could fit in that circle.

A=3\(\sqrt{3}\)/(2) * a2

The area of that would equal 64.95

78.54-64.95 = 13.59

then divide that number by 6

which is 2.265.

Take that number from the original 78.54

78.54-2.265 = 76.275

Answer:

x=8.1 /hour during the week

y=10.4 /hour in  the weekend

Step-by-step explanation:

x=earnings during the week

y= earnings in the wekend

13x+14y=250.90  |*-4

15x+8y=204.70    |*7

-52x-56y=-1003.6

105x+56y=1432.9

---------------------------

53x=429.3

x=8.1 /hour during the week

15*8.1+8y=204.7

121.5+8y=204.7

-121.5        -121.5

8y=83.2

:8    :8

y=10.4 /hour in  the weekend

Question:

Solve for x.

2

——

X +6. =4

—

6

Answer:  

1

,

5

)

Step-by-step explanation:

{

2

x

−

6

<

4

x

≥

3

−

2

x

+

6

<

4

x

<

3

Solve  

2

x

−

6

<

4

when  

x

≥

3

.

3

≤

x

<

5

Solve  

−

2

x

+

6

<

4

when  

x

<

3

.

1

<

x

<

3

Find the union of the solutions.

1

<

x

<

5

The result can be shown in multiple forms.

Inequality Form:

1

<

x

<

5

Interval Notation:

(

1

,

5

)

Answer:

32 seconds

Step-by-step explanation:

Answer:

2.5 meters in a second.

Step-by-step explanation:

I'm assuming it's how many meters per second.

meters : seconds

  40     :      16

  10      :       4

  2.5    :     1

The probability of choosing a 5 and then a 6 is 1/49

From the question, we have the following parameters that can be used in our computation:

The tiles

Where we have

Total = 7

The probability of choosing a 5 and then a 6 is

P = P(5) * P(6)

So, we have

P = 1/7 * 1/7

Evaluate

P = 1/49

Hence, the probability of choosing a 5 and then a 6 is 1/49

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Question

You randomly choose one of the tiles. Without replacing the first tile, you choose a second tile. Find the probability of the compound event. Write your answer as a fraction or percent rounded to the nearest tenth. The probability of choosing a 5 and then a 6

The reasoning provided, "If you build it, they will come. You build it. Therefore, they will come," is an example of deductive reasoning.

Deductive reasoning is a logical process that involves drawing specific conclusions based on general principles, premises, or known facts. In this example, the premise is "If you build it, they will come," which establishes a cause-and-effect relationship. The subsequent statement, "You build it," serves as a confirmation of the premise. From these premises, the conclusion is drawn: "Therefore, they will come." The conclusion logically follows from the premises, making it an example of deductive reasoning.

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The scatterplot with the least squares line provides insights into the relationship between average annual global surface temperature and the years from 2000 to 2015, allowing us to assess trends, strength of correlation, and make predictions within certain limitations.

The scatterplot represents the relationship between the average annual global surface temperature, in degrees Celsius, and the corresponding years from 2000 to 2015. The line drawn on the plot is the least squares line, which is the best fit line that minimizes the overall distance between the observed data points and the line.

The least squares line is determined using a statistical method called linear regression. It calculates the equation of a straight line that represents the trend in the data. This line serves as a mathematical model to estimate the average temperature based on the year.

By analyzing the scatterplot and the least squares line, we can make several observations. Firstly, we can see whether the temperature has been increasing, decreasing, or remaining relatively stable over the given years. If the slope of the line is positive, it indicates a positive correlation, implying that the temperature has been increasing. Conversely, a negative slope suggests a decreasing trend.

Additionally, we can evaluate the strength of the relationship between temperature and time by examining how closely the data points cluster around the line. If the points are closely grouped around the line, it suggests a strong correlation, indicating that the line is a good representation of the data. On the other hand, if the points are more scattered, the correlation may be weaker.

Furthermore, the line can be used to predict the average annual global surface temperature for future years beyond the data range of 2000 to 2015. However, it's important to note that such predictions should be made with caution and considering other factors that may affect global temperatures, such as climate change and natural variability.

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Question

The given scatterplot shows the average annual global surface temperature, in degrees celsius, for each year from 2000 to 2015. the line drawn is the least squares line for the data set.

The car that cost $\($35750\) in \(2014\) will be valued $\(14,919.48\) in \(2022\) if the annual depreciation rate is \(10.5%\)%

The depreciation rate is the percentage at which an object depreciates over its anticipated useful life.

It can also be described as the portion of an organization's long-term investment in an asset that the organization claims as a tax-deductible expenditure over the course of the asset's useful life.

Given.

Value for the \(1\)st Year (\(2014-2015\)) is $ \($35,750\) -$(\(350,750*10.5\)%) =

$\(31,981.25\)

Value for \(2 years\) \((2015-2016)\) =$\(31,981.25-(\)$\(31,981.25*10.5\)%) =

$\(28,604.41\)

Value for \(8 years\) ($\(16550.09\) -$\(16550.09*10.5\)%) = $\(14919.48\)

Therefore the car that cost $\($35750\) in \(2014\) will be valued $\(14,919.48\) in \(2022\) if the annual depreciation rate is \(10.5%\)%

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Answer:

false

Step-by-step explanation

Step 1: Evaluate the left side of the equal sign:

3q < 18ifq = 3

Step 2: Evaluate the right side of the equal sign:

6 = 6

Step 3: since 3 does not equal 6 then 3q<18ifq=6

two polynomials are relatively prime if the greatest common divisor of the two polynomial is 1

How do you prove two polynomials are relatively prime?

Two polynomials from Z[x] are called evaluationally relatively prime if the greatest common divisor of the two polynomials in Z[x] is 1 and gcd(f(t),g(t)) = 1 for all t ∈ Z.

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Answer:

small car

Step-by-step explanation:

if they rent the small car for 5 days

(39+21)×5÷4=75

This is an example of interviewer bias where the gender of the interviewer may have impacted the replies of the unmarried males in the poll.

This is an example of interviewer bias, where the gender of the interviewer may have influenced the responses of the unmarried men in the survey. The results may not accurately reflect the true opinions of the participants as their answers could have been affected by their desire to impress or please the interviewer.

This situation is an example of response bias, specifically interviewer bias, which occurs when the respondent's answer is influenced by the gender or characteristics of the interviewer, rather than their true preferences or opinions.

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Part A) Estimated Populations of Cities A and B are given.

Part B)  City A's estimated population is increasing at a faster rate than City B's estimated population for each decade.

Part C) the estimated population of City A first exceeds the estimated population of City B at the start of the 4th decade (t = 3)

Part D) The models may not realistically predict population growth over very long time scales, as they do not take into account other factors  

Part A:

Estimated Populations of Cities A and B

Decades Since 1980  City A (in thousands)          City B (in thousands)

0                                       15                                           16

1                                  19.5                                   16.4

2                                  25.2                                   17.8

3                                  32.7                                   19.2

4                                  42.4                                    20.6

Part B:

Population Increases from the previous Decade for Cities A and B

Decades Since 1980 City A (in hundreds)           City B (in hundreds)

0                                         -                                                   -

1                                       450                                              40

2                                       570                                               140

3                                       750                                               140

4                                       970                                               140

Interpretation:

From the table, we can see that City A's estimated population is increasing at a faster rate than City B's estimated population for each decade. This indicates that City A is growing more rapidly than City B, and may continue to do so in the future.

Part C:

To find the start of the decade when the estimated population of City A first exceeds the estimated population of City B, we need to solve the equation:

15\((1.03)^t\) > 0.4t + 16

where t is the number of decades since 1980.

One way to solve this is to use a graphing calculator to graph both functions and find their intersection point. Alternatively, we can use trial and error to test different values of t until we find the smallest one that satisfies the inequality.

Using a graphing calculator, we find that the estimated population of City A first exceeds the estimated population of City B at the start of the 4th decade (t = 3), when the estimated population of City A is approximately 32.7 thousand people, and the estimated population of City B is approximately 19.2 thousand people.

Part D:

The models may not realistically predict population growth over very long time scales, as they do not take into account factors such as changes in birth and death rates, migration patterns, and economic or environmental changes that may affect population growth. Additionally, the models assume that population growth will continue to follow a linear or exponential pattern, which may not be accurate in the long term. Therefore, while the models can provide useful estimates for short- to medium-term population trends, they should be used with caution when making long-term projections.

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The correct question should be:

The estimated population, in thousands of people, of city A can be modeled by the function SO) = 15(1.03), where is the number of decades (10-year period) since 1980 The estimated population, in thousands of people, of the city can be modeled by the function (1)-0.41 +16, where is the number of decades since 1980

Part A)     Complete this table, showing both estimated population values, rounded to the nearest tenth for each city based on the number of decades since 1980 Estimated Populations of Cities A and B City Decades Since 1980 City B (thousands of people) (thousands of people) 0 . 2 3 4

Part B)      Using the table in Part A, complete this table showing population increases, in hundreds of people, cach decade Population Increases from previous Decade for Cities A and B Decades Since 1980, Increase of City A Increase of City B (hundreds of people), (hundreds of people), fin)-(-1) g) -R-1) 2 3 4 Interpret the differences in the increases of the estimated populations of the two cities based on the functions that model cach city's population. Provide evidence to support your answer

Part C)    At the start of which decade will the estimated population of city A first exceed the estimated population of city B? What will be the estimated populations of each city at the start of that decade? Provide evidence to support your answers.

Part D)   Based on the behavior of the functions, do the models realistically predict population growth over very long time scales? Provide evidence to support your answer.

Answer:

$144.20

Step-by-step explanation:

2.4(33) + 65

79.2 + 65

144.2

$144.20

Answer:

Step-by-step explanation:

The third one is

(10^3)^4 and equals 10^12

the 4th one is      (10*10*10*10)(10*10*10*10) or 10^8

the last one is (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10) (10*10*10*10*10*10*10*10)

or 10^88

The required volume of the wallet in m3 is 0.001393030684 m3.

Given, The volume of a wallet is 8.50 in. 3.

Convert this value to m3, using the definition 1 in. = 2.54 cm.

The given value of volume is 8.50 in.3.

But we need to convert it into m3.

We have, 1 in. = 2.54 cm.

Hence, 1 in.3 = (2.54 cm)3

                      = (2.54 × 2.54 × 2.54) cm3

                      = 16.387064 cm3

In order to convert the value from cm3 to m3, we need to follow the given steps:1

m = 100 cm

⇒ (1 m)3 = (100 cm)3

               = 106 cm3

Now, we have the relation, 16.387064 cm3 = 1 in.3

⇒ 1 cm3 = (1/16.387064) in.3

∴ 8.50 in.3 = 8.50 × 16.387064 cm3

                  = 139.3030684 cm3

∵ 1 m3 = (1/106) cm3

∴ 139.3030684 cm3= (139.3030684 × 10−6) m3

∴ 8.50 in.3= 1.393030684 × 10−3 m3

                 = 0.001393030684 m3

Hence, the required volume of the wallet in m3 is 0.001393030684 m3.

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Answer:

Yes (see below for explanation

Step-by-step explanation:

\(f(2)=2^3-6(2)+1=8-12+1=-4+1=-3\\f(3)=3^3-6(3)+1=27-18+1=9+1=10\)

Therefore, since \(f(2) < 0 < f(3)\), then the Intermediate Value Theorem guarantees that \(f(x)\) will have a root on the interval \((2,3)\).

Therefore, the measure of the central angle whose radii define the given arc is approximately 83.077 degrees.

The length of an arc on a circle is given by the formula:

Arc Length = (Central Angle / 360°) * Circumference

In this case, we know the arc length is 6π centimeters, and the radius of the circle is 13 centimeters. The circumference of the circle can be calculated using the formula:

Circumference = 2π * Radius

Substituting the radius value, we get:

Circumference = 2π * 13

= 26π

Now we can use the arc length formula to find the central angle:

6π = (Central Angle / 360°) * 26π

Dividing both sides of the equation by 26π:

6π / 26π = Central Angle / 360°

Simplifying:

6 / 26 = Central Angle / 360°

Cross-multiplying:

360° * 6 = 26 * Central Angle

2160° = 26 * Central Angle

Dividing both sides by 26:

2160° / 26 = Central Angle

Approximately:

Central Angle ≈ 83.077°

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