Quadrilateral RSTV is dilated with respect to the origin by a scale factor of 1.5 to produce quadrilateral R'S'T'V' . Vertex R is located at (6, -9). Which ordered pair represents R' after the dilation?

Answer:

50%

Step-by-step explanation:

22 is half of 44.

So, this means 50% of 44 is 22.

Answer:

2√2

Step-by-step explanation:

hope you understand.

Answer:

[tex]a_{i} = \frac{(-3)^{i}}{4\cdot i!}[/tex] converges.

Step-by-step explanation:

The convergence analysis of this sequence is done by Ratio Test. That is to say:

[tex]r = \frac{a_{n+1}}{a_{n}}[/tex], where sequence converges if and only if [tex]|r| < 1[/tex].

Let be [tex]a_{i} = \frac{(-3)^{i}}{4\cdot i!}[/tex], the ratio for the expression is:

[tex]r =-\frac{3}{n+1}[/tex]

[tex]|r| = \frac{3}{n+1}[/tex]

Inasmuch [tex]n[/tex] becomes bigger, then [tex]r \longrightarrow 0[/tex]. Hence, [tex]a_{i} = \frac{(-3)^{i}}{4\cdot i!}[/tex] converges.

Answer:

See below.

Step-by-step explanation:

He does not have enough to loose 2,000,000 at that point, so this whole problem is nonsense.

Answer:

Step-by-step explanation:

x + 30 + 25 = 180

x + 55 = 180

x = 125

y + 125 = 180

y = 55

(a) Take the Laplace transform of both sides:

[tex]2y''(t)+ty'(t)-2y(t)=14[/tex]

[tex]\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s[/tex]

where the transform of [tex]ty'(t)[/tex] comes from

[tex]L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)[/tex]

This yields the linear ODE,

[tex]-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s[/tex]

Divides both sides by [tex]-s[/tex]:

[tex]Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}[/tex]

Find the integrating factor:

[tex]\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C[/tex]

Multiply both sides of the ODE by [tex]e^{3\ln|s|-s^2}=s^3e^{-s^2}[/tex]:

[tex]s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}[/tex]

The left side condenses into the derivative of a product:

[tex]\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}[/tex]

Integrate both sides and solve for [tex]Y(s)[/tex]:

[tex]s^3e^{-s^2}Y(s)=7e^{-s^2}+C[/tex]

[tex]Y(s)=\dfrac{7+Ce^{s^2}}{s^3}[/tex]

(b) Taking the inverse transform of both sides gives

[tex]y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right][/tex]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that [tex]\frac{7t^2}2[/tex] is one solution to the original ODE.

[tex]y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7[/tex]

Substitute these into the ODE to see everything checks out:

[tex]2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14[/tex]

[tex]|\Omega|=_{12}C_3=\dfrac{12!}{3!9!}=\dfrac{10\cdot11\cdot12}{2\cdot3}=220\\|A|=4\cdot3\cdot5=60\\\\P(A)=\dfrac{60}{220}=\dfrac{3}{11}\approx0,273[/tex]

Answer:

0.273

Step-by-step explanation:

Total number of balls is 4+3+5 = 12

There are 6 ways to draw 3 different colors (WBR, WRB, BWR, BRW, RWB, RBW) each with a chance of 4/12 · 3/11 · 5/10 = 1/22

So the total chance is 6 · 1/22 = 6/22 = 3/11 ≈ 0.273

Answer:

(a) Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 30 points

    Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 30 points

(b) Type I error is that we conclude that test-takers who had completed the training showed a mean increase smaller than 30 points but in actual, the program raises SAT scores by an average of at least 30 points.

Step-by-step explanation:

We are given that a test-preparation company advertises that its training program raises SAT scores by an average of at least 30 points.

A random sample of test-takers who had completed the training showed a mean increase smaller than 30 points.

Let [tex]\mu[/tex] = average SAT score.

(a) So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 30 points

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 30 points

Here, the null hypothesis states that the training program raises SAT scores by an average of at least 30 points.

On the other hand, the alternate hypothesis states that test-takers who had completed the training showed a mean increase smaller than 30 points.

(b) Type I error states the probability of rejecting the null hypothesis given the fact that null hypothesis is true.

According to the question, the Type I error is that we conclude that test-takers who had completed the training showed a mean increase smaller than 30 points but in actual, the program raises SAT scores by an average of at least 30 points.

The consequence of a Type I error is that we conclude the test-takers have low SAT scores but in actual they have an SAT score of at least 30 points.

Answer:

95 times digit 9 appears in all integers from 1 to 500.

Step-by-step explanation:

No. of 9 from

1-9: 1 time

10-19:  1 time

20-29: 1 time

30-39: 1 time

40-49: 1 time

50-59: 1 time

60-69: 1 time

70-79: 1 time

80-89 : 1 time

from 90 to 99

there will be one in 91 to 98

then two 9 in 99

thus, no of 9 from 90 to 99 is 10

Thus, total 9's from 1 to 99 is 9+10 = 19

Thus there 19 9's in 1 to 99

similarly

there will be

19 9's in 100 to 199

19 9's in 200 to 299

19 9's in 300 to 399

19 9's in 400 to 499

Thus, total 9's will be

19 + 19 + 19+ 19 + 19 + 19 = 95

Thus, 95 times digit 9 appears in all integers from 1 to 500.

Answer:

35 or 25

Step-by-step explanation:

Answer:

30% is the correct answer.

Step-by-step explanation:

Total number of boys = 2

Total number of girls = 3

Total number of students = 5

To find:

Probability that the pianist will be a boy and the alternate will be a girl?

Solution:

Here we have to make 2 choices.

1st choice has to be boy (pianist) and 2nd choice has to be girl (alternate).

[tex]\bold{\text{Required probability }= P(\text{boy as pianist first}) \times P(\text{girl as alternate})}[/tex]

Formula for probability of an event E is given as:

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]

For [tex]P(\text{boy as pianist})[/tex], number of favorable cases are 2 (total number of boys).

Total number of cases = Total number of students i.e. 5

So, [tex]P(\text{boy as pianist})[/tex] is:

[tex]P(\text{boy as pianist}) = \dfrac{2}{5}[/tex]

For [tex]P(\text{girl as alternate})[/tex], number of favorable cases are 3 (total number of girls).

Now, one boy is already chosen as pianist so Total number of cases = Total number of students left i.e. (5 - 1) = 4

[tex]P(\text{girl as alternate}) = \dfrac{3}{4}[/tex]

So, the required probability is:

[tex]\text{Required probability } = \dfrac{2}{5}\times \dfrac{3}{4} = \dfrac{3}{10} = \bold{30\%}[/tex]

Answer:

30% A

Step-by-step explanation:

Answer:

B. 259

Step-by-step explanation:

6^(i - 1) for i = 1 to 4

sum = 6^(1 - 1) + 6^(2 - 1) + 6^(3 - 1) + 6^(4 - 1) =

= 6^0 + 6^1 + 6^2 + 6^3

= 1 + 6 + 36 + 216

= 259

Answer: B. 259

Answer:

Find a common denominator on the right side of the equation

Step-by-step explanation:

The equation before the problem is

X² + b/a(x) + (b/2a)²= -c/a + b²/4a²

The next step in solving the above equation is to fibd tge common denominator on the right side of the equation.

X² + b/a(x) + (b/2a)²= -c/a + b²/4a²

X² + b/a(x) + (b/2a)²= -4ac/4a² + b²/4a²

X² + b/a(x) + (b/2a)²=( b²-4ac)/4a²

The right side of the equation now has a common denominator

The next step is to factorize the left side of the equation.

(X+b/2a)²= ( b²-4ac)/4a²

Squaring both sides

X+b/2a= √(b²-4ac)/√4a²

Final equation

X=( -b+√(b²-4ac))/2a

Or

X=( -b-√(b²-4ac))/2a

Answer with  explanation:

A x% confidence interval interprets that a person can be x% confident thatthe true mean lies in it.

Here, Credit card companies is using the collection agency to justify the cost of , the agency must collect an average of at least $200 per customer.

i.e. [tex]H_0:\mu \geq200,\ \ \ H_a:\mu<200[/tex]

The 90% confidence interval on the mean collected amount was reported as ($190.25, $250.75) .

I recommend that we can be 90% sure that the true mean collected amount  lies in ($190.25, $250.75).

Also, $200 lies in it such that it is more far from $250.75 than $190.25, that means there are large chances of having an average is at least $200 per customer.

Answer:

Step-by-step explanation:

vector OA=a

vector OB=b

vector OX= λ vector OA=λa

vector OY=μ vector OB=μb

a.

1.vector BX=(vector OX-vector OB)=λa-b

ii. vector AY=(vector OY-vector OA)=μb-a

b.

5 vector BP=2 vector BX

5(vector OP-vector OB)=2 (vector OX-vector OB)

5(vector OP-b)=2(λa-b)

5 vector OP-5b=2λa-2b

5 vector OP=2λa-2b+5b

vector OP=1/5(2λa+3b)

ii

complete it.

Answer:

(2) If 300 lunches were sold, then 120 chose tacos.

Step-by-step explanation:

We can evaluate each option and see if it makes it true.

For 1: If 200 lunches were served, 10 more students chose pizza over hotdogs.

We can find how many pizzas/hotdogs were given if 200 lunches were served by relating it to 100.

20% chose hotdog, which is [tex]\frac{20}{100}[/tex]. Multiply both the numerator and denominator by two: [tex]\frac{40}{200}[/tex] - so 40 students chose hotdogs.

Same logic for pizza: 30% chose pizza - [tex]\frac{30}{100} = \frac{60}{200}[/tex] so 60.

60 - 40 = 20, not 10, so 1 doesn't work.

2: If 300 lunches were sold, then 120 chose tacos.

Let's set up a proportion again. 40% of 100 is 40.

[tex]\frac{40}{100} = \frac{40\cdot3}{300} = \frac{120}{300}[/tex]

So 120 tacos were chosen - yes this works!

Hope this helped!

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = [tex]\rho[/tex]

case (1)  

when 8 loop form with 2 adjacent node is [tex]\frac{\lambda }{2}[/tex]

so here

[tex]l = \frac{8 \lambda _1}{2}[/tex]      ..............1

[tex]l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}[/tex]

and we know velocity is express as

velocity = frequency × wavelength   .....................2

[tex]\sqrt{\frac{Tension}{mass\ per\ unit \length }}[/tex]   =   f × [tex]\lambda_1[/tex]

here tension = mg

so

[tex]\sqrt{\frac{mg}{\mu}}[/tex]   =   f × [tex]\lambda_1[/tex]     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is [tex]\frac{\lambda }{2}[/tex]

[tex]l = \frac{10 \lambda _1}{2}[/tex]      ..............4

[tex]l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}[/tex]

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × [tex]\rho[/tex] × g = mg

and

T = v × [tex]\rho[/tex] - v × [tex]\rho[/tex] × g    

from equation 2

f × [tex]\lambda_2[/tex] = f  × [tex]\frac{1}{5}[/tex]  

[tex]\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}[/tex]     .......................5

now we divide eq 5 by the eq 3

[tex]\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}[/tex]

solve irt we get

[tex]1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}[/tex]

so

relative density [tex]\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}[/tex]

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

Answer:

no

Step-by-step explanation:

2(4+10)+20

2(14)+20

28+20

48

Answer:

[tex]V(m) = (2 + 5m)^3[/tex]

Step-by-step explanation:

Given

Solid Shape = Cube

Edge = 2 feet

Increment = 5 feet per minute

Required

Determine volume as a function of minute

From the question, we have that the edge of the cube increases in a minute by 5 feet

This implies that,the edge will increase by 5m feet in m minutes;

Hence,

[tex]New\ Edge = 2 + 5m[/tex]

Volume of a cube is calculated as thus;

[tex]Volume = Edge^3[/tex]

Substitute 2 + 5m for Edge

[tex]Volume = (2 + 5m)^3[/tex]

Represent Volume as a function of m

[tex]V(m) = (2 + 5m)^3[/tex]

Answer:

B

Step-by-step explanation:

we take the only point we know

(0,20)

in A when x =0

[tex]f(x)=e^{20x} =e^{20*0}=1[/tex]

in B when x=0

[tex]f(x)=20e^x=20e^0=20*1=20[/tex]

fits

in C

[tex]f(x)=20^x=20^0=1[/tex]

in D

[tex]f(x)=20^{20x}=20^{20*0}=20^0=1[/tex]

so the only choice is B

Answer:

82

Step-by-step explanation:

Plug in the variables into the equation and solve for B but remember your order of operations when solving multiplication/division before addition/subtraction.

B = 4*25-9*2

B = 100-18

B = 82

Answer:

Median: 14.6, Q1: 6.1, Q3: 27.1, IR: 21, outliers:  none

Step-by-step explanation:

Step 1: order the data from the least to the largest.

2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, 14.6, 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5

Step 2: find the median.

The median is the middle value, which is the 8th value in the data set.

2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, [14.6,] 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5

Median = 14.6

Step 2: Find Q1,

Q1 is the middle value of the lower part of the data set that is divided by the median to your left.

2.8, 3.9, 5.3, (6.1), 6.5, 7.1, 12.5, [14.6], 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5

Q1 = 6.1

Step 3: find Q3.

Q3 is the middle value of the upper part of the given data set.

2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, [14.6], 16.4, 16.4, 20.8, (27.1), 28.1, 30.9, 53.5

Q3 = 27.1

Step 4: find interquartile range (IR)

IR = Q3 - Q1 = [tex] 27.1 - 6.1 = 21 [/tex]

Step 5: check if there is any outlier.

Formula for checking for outlier = [tex] Q1 - 1.5*IR [/tex]

Then compare the result you get with the given values in the data set. Any value in the data set that is less than the result we get is considered an outlier.

Thus,

[tex] Q1 - 1.5*IR [/tex]

[tex]6.1 - 1.5*21 = -25.4[/tex]

There are no value in the given data set that is less than -25.4. Therefore, there is no outlier.

Answer:

[tex]78,\!624,\!000[/tex].

Step-by-step explanation:

Note the requirements:

Because of that, permutation would be the most suitable way to count the number of possibilities.

There are [tex]\displaystyle P(26,\, 3) = \frac{26!}{(26 - 3)!} = 26 \times 25 \times 24 = 15,\!600[/tex] ways to arrange three out of [tex]26[/tex] distinct letters (without replacement.)

Similarly, there are [tex]\displaystyle P(10,\, 4) = \frac{10!}{(10 - 4)!} = 10 \times 9 \times 8 \times 7= 5,\!040[/tex] ways to arrange four out of [tex]10[/tex] distinct numbers (also without replacement.)

Therefore, there are [tex]15,\!600[/tex] possibilities for the three-letter section of this password, and [tex]5,\!040[/tex] possibilities for the four-digit section. What if these two parts are combined?

Consider: if the first three letters of the password were fixed, then there would be [tex]5,\!040[/tex] possibilities. However, if any of the first three letters was changed, the result would be another [tex]5,\!040\![/tex] possibilities, all of which are different from the previous [tex]5,\!040\!\![/tex] possibilities. These two three-letter sections along will give [tex]2 \times 5,\!400[/tex] possibilities. Since there are [tex]15,\!600[/tex] three-letter sections like that, there would be [tex]15,\!600 \times 5,\!400 = 78,\!624,\!000[/tex] possible passwords in total. That gives the number of possible passwords that satisfy these requirements.

Answer:

third option is the first step

Answer:

C

Step-by-step explanation:

It is c bro

Answer:

(DNE,DNE)

Step-by-step explanation:

-24x-12y = -16. Equation one

6x +3y = 4. Equation two

Multiplying equation two with +4 gives

4(6x +3y = 4)

24x +12y = 16...result of equation two

-24x -12y= -16...

A careful observation to the following equation will help us notice that the both equation are same thing.

Multiplying minus to equation one gives

-(-24x-12y=-16)

24x+12y = 16.

Since the both equation are same, there is no solution to it.

Answer: n is a positive odd number.

Step-by-step explanation:

Ok, we know that the function is something like:

f(x)=a(x+k)^1/n + c

In the graph we can see two thigns:

All the values of the graph are positive values (even for the negative values of x), but in the left side we can see that the function decreases and is different than the right side.

So this is not an even function, then n must be an odd number (n odd allows us to have negative values for y = f(x) that happen when x + k is negative).

Also, we can see that the function increases, if n was a negative number, like: n = -N

we would have:

[tex]f(x) = \frac{a}{(x+k)^{1/N}} + c[/tex]

So in this case x is in the denominator, so as x increases, we would see that the value of y decreases, but that does not happen, so we can conclude that the value of n must be positive.

Then n is a positive odd number.

Answer:

D) Positive Even Integer

Step-by-step explanation:

just did it

Answer:

2.086

Step-by-step explanation:

Log 122 is equal to 2.086

Answer:

Yes you can

Step-by-step explanation:

To eliminate the denominator

Answer:

No

Step-by-step explanation:

We cannot be certain that x + 3 > 0

If it was negative then the sign of the inequality would change.

To solve find the critical values of the numerator/ denominator, that is

x = 2 and x = - 3

The domain is the split into 3 intervals

(- ∞, - 3 ), (- 3, 2), (2, + ∞ )

Use test points from each interval to determine valid solution

Answer:

Area of ΔEDF = 2.7 in²

Step-by-step explanation:

It's given in the question,

ΔBAC ~ ΔEDF

In these similar triangles,

Scale factor of the sides = [tex]\frac{\text{Measure of one side of triangle BAC}}{\text{Measure of one side of triangle EDF}}[/tex]

                                        [tex]=\frac{\text{BC}}{\text{EF}}[/tex]

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

Area scale factor = (Scale factor of the sides)²

[tex]\frac{\text{Area of triangle BAC}}{\text{Area of triangle EDF}}=(\frac{3}{2})^2[/tex]

[tex]\frac{6}{\text{Area of triangle EDF}}=(\frac{9}{4})[/tex]

Area of ΔEDF = [tex]\frac{6\times 4}{9}[/tex]

                       = 2.67

                       ≈ 2.7 in²

Therefore, area of the ΔEDF is 2.7 in²                            

Answer:

B. ⅕

Step-by-step explanation:

Fraction of families having more than 2 pets = families with pets of 3 and above ÷ total number of families in the apartment

From the dot plot above, 3 families have 3 pets, and 1 family has 4 pets.

Number of families with more than 2 pets = 3 + 1 = 4

Fraction of families with more than 3 pets = [tex] \frac{4}{20} = \frac{1}{5} [/tex]

The fraction of families that have more than two pets is B. [tex]\frac{1}{5}[/tex]

Given that:

Fraction of families having more than 2 pets = families with pets of 3 and above/total number of families in the apartment

From the dot plot above:

Number of families with more than 2 pets

= 3 + 1

= 4

Fraction of families with more than 3 pets = [tex]\frac{4}{20} = \frac{1}{5}[/tex]

Read more about dot plots here:

https://brainly.com/question/25957672

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