Find the median of the following frequency distribution

Answer:

98 degrees

Step-by-step explanation:

The angles of a quadrilateral add up to 360 degrees, therefore:

(25x+1)+(25x-2)+(20x-1)+82=360

70x +80=360

70x=280

x=4

the angle at C:

25*4-2 = 98 degrees

Answer:

9

Step-by-step explanation:

6x=2*x*(a number), (a number)^2=the number we need. A number=3, the number is 3^2=9

There are at most 4 distinct absolute values of elements taken from this sequence. (If there were at least 5 distinct absolute values, then you could pick [tex]a_i,a_j,a_k,a_\ell,a_m[/tex] each with different absolute values, but that would contradict the given statement "for every five indices ... at least two of ... have the same absolute value".)

The pigeonhole principle then says that 2 of any 5 numbers taken from this sequence have the same absolute value. Both |x| = x and |-x| = x, so there can be at most 8 distinct numbers in the sequence.

Answer:

18z + 24

Step-by-step explanation:

The distributive property is the multiplication of the number outside the parenthesis with the numbers inside.

All you need to do to solve this is to multiply 6 with 3z, and 6 with 4.

6 × 3 = 18 =

18z

6 × 4 = 24

Put them together:

18z + 24

Another way to do this:

Answer:

15

Step-by-step explanation:

Sum of 'n' terms formula is given by:-    

s=n/2(2a+(n-1)d)  

s=13/2[2xa+(13-1)5]    

s=13/2(2xa+12x5)    

s=13/2(2a+60)    

585=13/2(2a+60)    

585 x (2/13) = 2a + 60

90 = 2a + 60  

90-60 = 2a

30 = 2a

a = 15

Answer:

x = 24.4°

Hope this helps... Have a good day!!

Answer:

4

plug in -1 for all the x's

3 + 2 + 7 = 12 (remeber that -1 *-1 = + 1 and that -2*-1 = + 2

-2+ 5 = + 3

12/3 = 4

Step-by-step explanation:

Answer:

79°

Step-by-step explanation:

PQO is straight angle with measure of 180°

the given angles' sum makes 101° and we need 79 to complete it to 180° therefore the angle STQ = 79°

Answer:

[tex]1296x^3y^4[/tex]

Step-by-step explanation:

Given the terms:

[tex]144x^3y^2[/tex]

and [tex]81xy^4[/tex]

To find:

Greatest Common Divisor of the two terms or Least Common Multiple (LCM) of two numbers = ?

Solution:

First of all, let us find the HCF (Highest Common Factor) for both the terms.

i.e. the terms which are common to both.

Let us factorize them.

[tex]144x^3y^2 = \underline{3 \times 3} \times 16\times \underline x \times x^{2}\times \underline{y^{2} }[/tex]

[tex]81xy^4= \underline {3\times 3}\times 9 \times \underline{x} \times \underline{y^2}\times y^2[/tex]

Common terms are underlined.

So, HCF of the terms = [tex]9xy^2[/tex]

Now, we know the property that product of two numbers is equal to the product of the numbers themselves.

HCF [tex]\times[/tex] LCM = [tex]144x^3y^2[/tex] [tex]\times[/tex] [tex]81xy^4[/tex]

[tex]LCM = \dfrac{144x^3y^2 \times 81xy^4}{9xy^2}\\\Rightarrow LCM = 144x^3y^2 \times 9x^{1-1}y^{4-2}\\\Rightarrow LCM = 144x^3y^2 \times 9x^{0}y^{2}\\\Rightarrow LCM = \bold{1296x^3y^4 }[/tex]

Complete question is;

For a data set of the pulse rates for a sample of adult​ females, the lowest pulse rate is 31 beats per​ minute, the mean of the listed pulse rates is x over bar equals 71.0 beats per​ minute, and their standard deviation is s equals 12.6 beats per minute.

a. What is the difference between the pulse rate of 31 beats per minute and the mean pulse rate of the​ females?

b. How many standard deviations is that​ [the difference found in part​ (a)]?

c. Convert the pulse rate of 31 beats per minutes to a z score.

d. If we consider pulse rates that convert to z scores between minus 2 and 2 to be neither significantly low nor significantly​ high, is the pulse rate of 31 beats per minute​ significant?

Answer:

A) 40 beats per minute

B) 3.1746

C) z = -3.17

D) the pulse rate of 31 beats per minute is significantly low.

Step-by-step explanation:

A) The mean pulse rate is given as x bar = 71

Thus difference between this and pulse rate of 31 beats per minute is;

Difference = 71 - 31 = 40 beats per minute

B) number of standard deviations of the difference found in part a is given as;

number = difference/standard deviation(s)

number = 40/12.6

number = 3.1746

C) The z-score is calculated from;

z = (x - xbar)/s

z = (31 - 71)/12.6

z = -3.17

D) from the question we are told that the z scores between minus2 and 2 to be neither significantly low nor significantly​ high. However, we have a Z-score of -3.17 which doesn't fall into that range. Thus, the pulse rate of 31 beats per minute is significantly low.

Answer:

Option C

Step-by-step explanation:

Options A, B, C, and D all satisfy the base case of V(0) = 10; however, they also all fail the next step case of V(4) = 452.

A at t = 4, results in 163047.361

B at t = 4, results in 1770

C at t = 4, results in 450 (close but not 452)

D at t = 4, results in 41740124.42

Note that at option C, we had the closet value 450 which is only 2 from 452 whereas the next closet was 1218 away.

Choose option C as the curve of best fit.

Cheers.

Answer:

c

Step-by-step explanation:

Answer:

16

Step-by-step explanation:

2 (l+b)​

Let l=5 and b=3

2 (5+3)​

Parentheses first

2(8)

Multiply

16

Answer:

16

Step-by-step explanation:

Plug in:

2 (l + b)

2 (5 + 3)

2 (8)

 16

Hope this helped.

Answer:

enlarge it

Step-by-step explanation:

I ft = 12 inches

Thus 8 in → 12 in makes the transformation larger.

Thus going from 8 in to 12 in is an enlargement

Answer:

U in the line x=-2: (-4,0)

U in the line x=-1: (-2-0)

Step-by-step explanation:

...............

Answer:

C

Step-by-step explanation:

10×-28=-280

35-8=27

35×(-8)=-280

10x²+27x-28

=10x²+(35-8) x-28

=10x²+35x-8x-28

=5x(2x+7)-4(2x+7)

=(2x+7)(5x-4)

=========================================================

Explanation:

One way we can factor is through use the of the quadratic formula.

Let [tex]10x^2+27x-28 = 0[/tex]

For now, the goal is to find the two roots of that equation.

Plug a = 10, b = 27, c = -28 into the quadratic formula

[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(27)\pm\sqrt{(27)^2-4(10)(-28)}}{2(10)}\\\\x = \frac{-27\pm\sqrt{1849}}{20}\\\\x = \frac{-27\pm43}{20}\\\\x = \frac{-27+43}{20} \ \text{ or } \ x = \frac{-27-43}{20}\\\\x = \frac{16}{20} \ \text{ or } \ x = \frac{-70}{20}\\\\x = \frac{4}{5} \ \text{ or } \ x = -\frac{7}{2}\\\\[/tex]

The two roots are x = 4/5 and x = -7/2

For each root, rearrange the equation so we have 0 on the right hand side, and it's ideal to get rid of the fractions

x = 4/5

5x = 4

5x-4 = 0 gives us one factor

and

x = -7/2

2x = -7

2x+7 = 0 gives the other factor

The two factors are 5x-4 and 2x+7

Note how (5x-4)(2x+7) = 0 leads to the two separate equations of 5x-4 = 0 and 2x+7 = 0 due to the zero product property. Solving each individual equation leads to the two roots we found earlier.

Alternative methods to solve this problem are the AC factoring method (which leads to factor by grouping), using the box method, or you could use guess and check.

Answer:

The strength of a correlation is independent of whether the correlation of a scatterplot is positive or negative. A scatterplot with positive correlation can have either a weak correlation or strong correlation.

Step-by-step explanation:

Answer:

The answer to your question would be A.) The strength of a correlation is independent of whether the correlation of a scatterplot is positive or negative. A scatterplot with positive correlation can have either weak correlation or strong correlation.

Step-by-step explanation:

I got it right on edge 2020

The solution to each of the question takes different approach, as the questions are taken from different concepts; however, a common operation among all questions, is factorization.

[tex](1)\ (-xy)^3(xz)[/tex]

Expand

[tex](-xy)^3(xz) = (-x)^3* y^3*(xz)[/tex]

[tex](-xy)^3(xz) = -x^3* y^3*xz[/tex]

Rewrite as:

[tex](-xy)^3(xz) = -x^3*x* y^3*z[/tex]

Apply law of indices

[tex](-xy)^3(xz) = -x^4y^3z[/tex]

[tex](2)\ (\frac{1}{3}mn^{-4})^2[/tex]

Expand

[tex](\frac{1}{3}mn^{-4})^2 =(\frac{1}{3})^2m^2n^{-4*2}[/tex]

[tex](\frac{1}{3}mn^{-4})^2 =\frac{1}{9}m^2n^{-8[/tex]

[tex](3)\ (\frac{1}{5x^4})^{-2}[/tex]

Apply negative power rule of indices

[tex](\frac{1}{5x^4})^{-2}= (5x^4)^2[/tex]

Expand

[tex](\frac{1}{5x^4})^{-2}= 5^2x^{4*2}[/tex]

[tex](\frac{1}{5x^4})^{-2}= 25x^{8[/tex]

[tex](4)\ -x(2x^2 - 4x) - 6x^2[/tex]

Expand

[tex]-x(2x^2 - 4x) - 6x^2 = -2x^3 + 4x^2 - 6x^2[/tex]

Evaluate like terms

[tex]-x(2x^2 - 4x) - 6x^2 = -2x^3 -2x^2[/tex]

Factor out x^2

[tex]-x(2x^2 - 4x) - 6x^2 = (-2x-2)x^2[/tex]

Factor out -2

[tex]-x(2x^2 - 4x) - 6x^2 = -2(x+1)x^2[/tex]

[tex](5)\ \sqrt{\frac{4y}{3y^2}}[/tex]

Divide by y

[tex]\sqrt{\frac{4y}{3y^2}} = \sqrt{\frac{4}{3y}}[/tex]

Split

[tex]\sqrt{\frac{4y}{3y^2}} = \frac{\sqrt{4}}{\sqrt{3y}}[/tex]

[tex]\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}}[/tex]

Rationalize

[tex]\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}} * \frac{\sqrt{3y}}{\sqrt{3y}}[/tex]

[tex]\sqrt{\frac{4y}{3y^2}} = \frac{2\sqrt{3y}}{3y}[/tex]

[tex](6)\ \frac{8}{3 + \sqrt 3}[/tex]

Rationalize

[tex]\frac{8}{3 + \sqrt 3} = \frac{3 - \sqrt 3}{3 - \sqrt 3}[/tex]

[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{(3 + \sqrt 3)(3 - \sqrt 3)}[/tex]

Apply difference of two squares to the denominator

[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{3^2 - (\sqrt 3)^2}[/tex]

[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{9 - 3}[/tex]

[tex]\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{6}[/tex]

Simplify

[tex]\frac{8}{3 + \sqrt 3} = \frac{4(3 - \sqrt 3)}{3}[/tex]

[tex](7)\ \sqrt{40} - \sqrt{10} + \sqrt{90}[/tex]

Expand

[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4*10} - \sqrt{10} + \sqrt{9*10}[/tex]

Split

[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4}*\sqrt{10} - \sqrt{10} + \sqrt{9}*\sqrt{10}[/tex]

Evaluate all roots

[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =2*\sqrt{10} - \sqrt{10} + 3*\sqrt{10}[/tex]

[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =2\sqrt{10} - \sqrt{10} + 3\sqrt{10}[/tex]

[tex]\sqrt{40} - \sqrt{10} + \sqrt{90} =4\sqrt{10}[/tex]

[tex](8)\ \frac{r^2 + r - 6}{r^2 + 4r -12}[/tex]

Expand

[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r^2 + 3r-2r - 6}{r^2 + 6r-2r -12}[/tex]

Factorize each

[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r(r + 3)-2(r + 3)}{r(r + 6)-2(r +6)}[/tex]

Factor out (r+3) in the numerator and (r + 6) in the denominator

[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{(r -2)(r + 3)}{(r - 2)(r +6)}[/tex]

Cancel out r - 2

[tex]\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r + 3}{r +6}[/tex]

[tex](9)\ \frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14}[/tex]

Cancel out x

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 5x - 14}[/tex]

Expand the numerator of the 2nd fraction

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 7x+2x - 14}[/tex]

Factorize

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x(x - 7)+2(x - 7)}[/tex]

Factor out x - 7

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{(x + 2)(x - 7)}[/tex]

Factor out 4 from 4x + 8

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4(x + 2)}{x} \cdot \frac{1}{(x + 2)(x - 7)}[/tex]

Cancel out x + 2

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x} \cdot \frac{1}{(x - 7)}[/tex]

[tex]\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x(x - 7)}[/tex]

[tex](10)\ (3x^3 + 15x^2 -21x) \div 3x[/tex]

Factorize

[tex](3x^3 + 15x^2 -21x) \div 3x = 3x(x^2 + 5x -7) \div 3x[/tex]

Cancel out 3x

[tex](3x^3 + 15x^2 -21x) \div 3x = x^2 + 5x -7[/tex]

[tex](11)\ \frac{m}{6m + 6} - \frac{1}{m+1}[/tex]

Take LCM

[tex]\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m(m + 1) - 1(6m + 6)}{(6m + 6)(m + 1)}[/tex]

Expand

[tex]\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 + m- 6m - 6}{(6m + 6)(m + 1)}[/tex]

[tex]\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 - 5m - 6}{(6m + 6)(m + 1)}[/tex]

[tex](12)\ \frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}}[/tex]

Rewrite as:

[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} \div \frac{2}{y^2 - 9}[/tex]

Express as multiplication

[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 9}{2}[/tex]

Express y^2 - 9 as y^2 - 3^2

[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 3^2}{2}[/tex]

Express as difference of two squares

[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{(y - 3)(y+3)}{2}[/tex]

[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{1} * \frac{(y+3)}{2}[/tex]

[tex]\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{y+3}{2}[/tex]

Read more at:

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Answer:

B

Step-by-step explanation:

if it is shaded under than it is <

the slope is 1/2x so that is the answer

Answer:

[tex]\displaystyle [0, 6][/tex]

Step-by-step explanation:

[tex]\displaystyle 1 = -\frac{1}{2}[10] + b \\ \\ 1 = -5 + b; 6 = b \\ \\ [0, 6][/tex]

I am joyous to assist you at any time.

Answer:

The answer is B.

I think I answered the wrong answer.

Answer:

0.25%

Step-by-step explanation:

Rate of commission

= (commission*100)/cost of land

=( 20000*100)/8000000

= 2000000/8000000

=2/8

= 0.25%

Answer:

The correct option is;

y < x - 2, and y > x + 1

Step-by-step explanation:

The given graph of inequalities is made up of parallel lines. Therefore, the slope of the inequalities are equal

By examination of the graph, the common slope = (Increase in y-value)/(Corresponding increase in x-value) = (0 - 1)/(-1 - 0) = 1

Therefore, the slope = 1

We note that the there are three different colored regions, therefore, the different colored regions opposite to each inequalities should be the areas of interest

The y-intercept for the upper bounding linear inequality, (y >) is 1

The y-intercept for the lower bounding linear inequality, (y <) is -2

The two inequalities are y > x + 1 and y < x - 2

The correct option is y < x - 2, and y > x + 1.

The system of linear inequalities is represented by  the graph is y> x-2 and y < x + 1

Inequalities is an expression that shows the non equal comparison of two or more variables and numbers.

Given that:

The system of linear inequalities is represented by  the graph is y> x-2 and y < x + 1

Find out more on linear inequalities at: https://brainly.com/question/21103162

Answer:

19

Step-by-step explanation:

Answer:

Hey there!

We use the distance formula to find the distances between each of these points.

From -2, 1 to 6, 1 is a total of 8 units.

From 6, 1 to 6, -3 is a total of 4 units.

From 6, -3 to 9, -3 is a total of 3 units.

8+4+3=15 units.

Let me know if this helps :)

Answer:

see below

Step-by-step explanation:

3, 12, 48, 192

We are multiplying by 4 each time  (12/3 =4)

The 5th term

192 *4 =768

The 6th term

768 *4 =3072

The th term

3072 *4 =12288

To find the common ratio, take the second term and divide by the first

2/4 = 1/2

The common ratio is 1/2

Problem 1

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

Explanation:

The first term is a = 3 and the common ratio or multiplier is r = 4. We start with 3 and multiply each term by 4 to get the next one. The nth term of this geometric sequence is

a(n) = a*(r)^(n-1)

a(n) = 3*(4)^(n-1)

Plug in n = 7 to get the seventh term

a(7) = 3*(4)^(7-1)

a(7) = 3*(4)^6

a(7) = 3*4096

a(7) = 12288

====================================================

Problem 2

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

Explanation:

To find the common ratio, we pick any term but the first one and divide it over its previous term

r = common ratio

r = (second term)/(first term) = 2/4 = 1/2 = 0.5

r = (third term)/(second term) = 1/2 = 0.5

r = (fourth term)/(third term) = 0.5/1 = 0.5

Each term is cut in half to get the next one.

9514 1404 393

Answer:

  c = 14

  no extraneous solutions

Step-by-step explanation:

You can subtract the right-side expression, combine fractions, and set the numerator to zero.

  [tex]\dfrac{c-4}{c-2}-\left(\dfrac{c-2}{c+2}-\dfrac{1}{2-c}\right)=0\\\\\dfrac{c-4}{c-2}-\dfrac{1}{c-2}-\dfrac{c-2}{c+2}=0\\\\\dfrac{(c-5)(c+2)-(c-2)^2}{(c-2)(c+2)}=0\\\\\dfrac{(c^2-3c-10)-(c^2-4c +4)}{(c-2)(c+2)}=0\\\\\dfrac{c-14}{(c-2)(c+2)}=0\\\\\boxed{c=14}[/tex]

__

Check

  (14 -4)/(14 -2) = (14 -2)/(14 +2) -1/(2 -14) . . . . substitute for c

  10/12 = 12/16 -1/-12

  5/6 = 3/4 +1/12 . . . . true

There is one solution (c=14) and it is a solution to the original equation. There are no extraneous solutions.

Answer:

y =  4  or y = 6

Step-by-step explanation:

2log4y - log4 (5y - 12) = 1/2

​2log_4(y) - log_4(5y-12) = log_4(2)           apply law of logarithms

log_4(y^2) + log_4(1/(5y-12)) = log_4(/2)    apply law of logarithms

log_4(y^2/(5y-12)) = log_4(2)                     remove logarithm

y^2/(5y-12) = 2                                            cross multiply

y^2 = 10y-24                                                  rearrange and factor

y^2 - 10y + 24 = 0

(y-4)(y-6) = 0

y= 4 or y=6

Answer:

TW correspondant to AD

Step-by-step explanation:

That's the right answer

Segments TW and AD are corresponding parts of the congruent triangles.

Two triangles are called congruent if their corresponding sides and angles are congruent.

Given that,  Δ MTW ≅ Δ CAD

The congruent parts are;

MT = CA

TW = AD

MW = CD

∠ M = ∠ C

∠ T = ∠ A

∠ W = ∠ D

Hence, Segments TW and AD are corresponding parts of the congruent triangles.

For more references on congruent triangles, click;

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Answer:

perimeter is  4 sqrt(29) + 4pi  cm

area is 40 + 8pi cm^2

Step-by-step explanation:

We have a semicircle and a triangle

First the semicircle with diameter 8

A = 1/2 pi r^2 for a semicircle

r = d/2 = 8/2 =4

A = 1/2 pi ( 4)^2

  =1/2 pi *16

  = 8pi

Now the triangle with base 8 and height 10

A = 1/2 bh

  =1/2 8*10

  = 40

Add the areas together

A = 40 + 8pi cm^2

Now the perimeter

We have 1/2 of the circumference

1/2 C =1/2 pi *d

         = 1/2 pi 8

        = 4pi

Now we need to find the length of the hypotenuse of the right triangles

using the pythagorean theorem

a^2+b^2 = c^2

The base is 4 ( 1/2 of the diameter) and the height is 10

4^2 + 10 ^2 = c^2

16 + 100 = c^2

116 = c^2

sqrt(116) = c

2 sqrt(29) = c

Each hypotenuse is the same so we have

hypotenuse + hypotenuse + 1/2 circumference

2 sqrt(29) + 2 sqrt(29) + 4 pi

4 sqrt(29) + 4pi  cm

Step-by-step explanation:

First we need to deal with the half circle. The radius of this circle is 4, because the diameter is 8. The formula for the circumference of a circle is 2piR.

2pi4 so the perimeter for the half circle would be 8pi/2.

The area of that half circle would be piR^2 so 16pi/2.

Now moving on the triangle part, we need to find the hypotenuse side of AC. We will use the pythagoram theorem. 4^2+10^2=C^2

16+100=C^2

116=C^2

C=sqrt(116)

making the perimeter of this triangle 2×sqrt(116)

The area of this triangle is 8×10=80, than divided by 2 which is equal to 40.

We than just need to add up the perimeters and areas for both the half circle and triangle.

The area would be equal to 8pi+40

The perimeter would be equal to 4pi+4(sqrt(29))