In the data set shown below, what is the value of the quartiles?

{4.3, 4.5, 4.7, 5, 5.5, 5.7, 5.9, 6, 6.1}
A. Q1 = 4.6; Q2 = 5.5; Q3 = 5.95
B. Q1 = 4.7; Q2 = 5.5; Q3 = 6
C. Q1 = 4.7; Q2 = 5.5; Q3 = 5.9
D. Q1 = 4.6; Q2 = 5.5; Q3 = 5.92

Answer:

12

Step-by-step explanation:

Since Emily eats four times the amount that Charlotte eats, then Emily will eat: = (4 × x) = 4x. Therefore, Charlotte eats 12 strawberries

The two functions should never intersect.

Since y = y, we can equate the other side of both equations together.

x^2 + 2x + 1 = x^2 + 2x - 2

Rearrange the equation to bring all the terms to one side.

0 = x^2 - x^2 + 2x - 2x - 2 - 1

0 = -3

You can see that does not make sense, so we can conclude that there are no points of intersection between the two functions.

I also graphed the functions. In the first pic you can see that the red is between the blue and they don't intersect. In the second pic I zoomed in so you can see the right side of the graph going up to y = 600 and the red is still between the blue.

22

Step-by-step explanation:

For simplicity, let

x = teary smiley

y = tongue smiley

z = plain smiley

So now our system of equations is

[tex]x + x + x = 12\:\:\:\:\:\:(1)[/tex]

[tex]y + z + x = 18\:\:\:\:\:\:\:(2)[/tex]

[tex]z + z + y = 22\:\:\:\:\:\:\:(3)[/tex]

[tex]z + y + 2x= ??\:\:\:\:\:\:(4)[/tex]

From Eqn(1), we plainly see that

[tex]3x = 12 \Rightarrow x = 4[/tex]

Now subtract Eqn(2) from Eqn(3) to get

[tex](2z + y) - (y + z + x) = 22 - 18[/tex]

[tex]\Rightarrow z - x = 4[/tex]

But we know that [tex]x = 4[/tex], which then gives us [tex]z = 8.[/tex]

Using the values of [tex]x[/tex] and [tex]z[/tex] in Eqn(2), we find that [tex]y = 6.[/tex] Now that we the values of all the variables, use them in Eqn(3) and we'll get

[tex](8) + (6) + 2(4) = 22[/tex]

Answer:

m=-1 I think

Step-by-step explanation:

Answer:

Answer is 5√6 ( none of the objectives )

Step-by-step explanation:

[tex] \sqrt{10} \times \sqrt{15} \\ = \sqrt{150} \\ = \sqrt{25 \times 6} \\ = \sqrt{25} \times \sqrt{6} \\ = 5 \times \sqrt{6} \\ = 5 \sqrt{6} [/tex]

Answer:

yep it's D

Step-by-step explanation:

Answer:

Hello,

Step-by-step explanation:

[tex]A)\\g(x)=\dfrac{x-5}{-3} =\dfrac{-x}{3} +\dfrac{5}{3} \\\\(gog)(x)=g(g(x))=g(\dfrac{-x}{3} +\dfrac{5}{3})\\\\=\dfrac{\dfrac{-x}{3} +\dfrac{5}{3} }{3}+\dfrac{5}{3} \\\\\\=\dfrac{-x}{9} +\dfrac{5}{9} +\dfrac{5}{3}\\\\=-\dfrac{x}{9}+\dfrac{20}{9} \\\\\\(gog)(2)=-\dfrac{2}{9}+\dfrac{20}{9} =\dfrac{18}{9}=2 \\\\[/tex]

[tex]B)\\f(x)=y=-3x-5\\exchanging\ y\ and\ x\\x=-3y-5\\3y=-x-5\\\\y=\dfrac{-x}{3} -\dfrac{5}{3} \\\\f^{-1}(x)=\dfrac{-x}{3} -\dfrac{5}{3} \\\\[/tex]

[tex]C)\\\\(fog)(x)\ must\ be\ equal\ to\ x\\\\\\(fog)(x)=g(f(x))=g(-3x-5)\\\\=\dfrac{-(-3x-5)}{3} +\dfrac{5}{3} \\\\=x+\dfrac{5}{3} +\dfrac{5}{3} \\\\\\=x+\dfrac{10}{3}\ and\ not\ x\ !!!\\[/tex]

f(x) and g(x) are not inverse functions.

Answer:

16x⁴+16x³-12x²-32x-16

Step-by-step explanation:

(8x²-4x-8)(2x²+3x+2)

= 16x⁴+24x³+16x²-8x³-12x²-8x-16x²-24x-16

= 16x⁴+16x³-12x²-32x-16

Answer:

c=80

Step-by-step explanation:

Based on my reading the critical damping occurs when the discriminant of the quadratic characteristic equation is 0.

So let's see that characteristic equation:

20⁢r^2+c⁢r+80⁢=0

The discriminant can be found by calculating b^2-4aC of ar^2+br+C=0.

a=20

b=c

C=80

c^2-4(20)(80)

We want this to be 0.

c^2-4(20)(80)=0

Simplify:

c^2-6400=0

Add 6400 on both sides:

c^2=6400

Take square root of both sides:

c=80 or c=-80

Based on further reading damping equations in form

a⁢y′⁢′+b⁢y′+C⁢y=0

should have positive coefficients with b also having the possibility of being zero.

Answer:

y=-1/3x+7

Step-by-step explanation:

y=mx+c

m=-1/3, c=7

y=-1/3x+7

9514 1404 393

Answer:

  10

Step-by-step explanation:

The sum of terms of an arithmetic series is ...

  Sn = (2a +d(n -1))·n/2 = (2an +dn^2 -dn)/2

For the series with first term 2 and common difference 3, the sum is 155 for n terms, where ...

  155 = (3n^2 +n(2·2 -3))/2

Multiplying by 2, we have ...

  3n^2 +n -310 = 0 . . . . . arranged in standard form

Using the quadratic formula, the positive solution is ...

  n = (-1 +√(1 -4(3)(-310)))/(2(3)) = (-1 +√3721)/6 = (61 -1)/6 = 10

10 terms of the series will have a sum of 155.

Step-by-step explanation:

[tex]\displaystyle \ \Large \boldsymbol{} S_n=\frac{2a_1+d(n-1)}{2} \cdot n =155 \\\\ \frac{4+3(n-1)}{2} \cdot n =155 \\\\\\ 4n+3n^2-3n=310 \\\\ 3n^2+n-310=0 \\\\D=1+3720=3721=61^2\\\\n_1=\frac{61-1}{6} =\boxed{10} \\\\\\n_2=\frac{-61-1}{3} \ \ \o[/tex]

In mathematics, a limit is the value that a function (or sequence) approaches as the input (or index) approaches some value.

g(x) , one may look at how big f(x) and g(x) are. For example: If f(x) is close to some positive number and g(x) is close to 0 and positive, then the limit will be ∞. If f(x) is close to some positive number and g(x) is close to 0 and negative, then the limit will be −∞.

Answer:

The number is 6.

Step-by-step explanation:

[tex]4x-13=x+5\\3x-13=5\\3x=18\\x=6[/tex]

The correct option is A because

The statement makes sense. There is​ 95% probability that the confidence interval limits actually contain the true value of the population​ mean, so the probability it does not fall in this range is ​100%−​95% =​5%.

From the question we are told that:

Confidence interval [tex]CI=95\%[/tex]

Mean [tex]\=x =1.9-3.5hours[/tex]

Level of significance (of the alternative hypothesis)

[tex]\alpha=100-95[/tex]

[tex]\alpha=5\%[/tex]

[tex]\alpha=0.05[/tex]

Generally

There is​ 95% probability that the confidence interval limits actually contain the true value of the population​ mean.

In conclusion

The  it does not fall in this range is Level of significance (of the alternative hypothesis)

​100%−​95% =​5%.

For more information on this visit

https://brainly.com/question/24131141?referrer=searchResults

9514 1404 393

Answer:

  not

Step-by-step explanation:

The complement of set S is the set of elements in U and not in S.

_____

It's a definition.

Answer:

26.75 units²

Step-by-step explanation:

This shape can be split into 3 triangles and a square. Find the area of each shape then add them all up.

[tex]A(Square)=2(2)=4\\\\A(Triangle)=\frac{1}{2}(2)(2)=2\\\\A(Triangle)=\frac{1}{2}(5)(2)=5\\\\A(Triangle)=\frac{1}{2}(9)(3.5)=15.75\\\\A(Shape)=4+2+5+15.75=26.75[/tex]

Therefore, the area of the shape is 26.75 units².

Is this question complete?

Answer:

It is (x - 3)³ - 9x(3 - x)

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

[tex] = {x}^{3} - {3}^{3} [/tex]

From trinomial expansion:

[tex] {(x - y)}^{3} = (x - y)(x - y)(x - y) \\ [/tex]

open first two brackets to get a quadratic equation:

[tex] {(x - y)}^{3} = ( {x}^{2} - 2xy + {y}^{2} )(x - y)[/tex]

expand further:

[tex] {(x - y)}^{3} = {x}^{3} - y {x}^{2} - 2y {x}^{2} + 2x {y}^{2} + x {y}^{2} - {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3x {y}^{2} - 3y {x}^{2} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3xy(y - x) \\ \\ { \boxed{( {x}^{3} - {y}^{3} ) = {(x - y)}^{3} - 3xy(y - x)}}[/tex]

take y to be 3, then substitute:

[tex]( {x}^{3} - 3^3) = {(x - 3)}^{3} - 9x(3 - x)[/tex]

3385.8

Step-by-step explanation:

Answer:

432 inches

Step-by-step explanation:

We need to convert feet to inches

1 ft = 12 inches

36 ft * 12 inches/ 1 ft = 432 inches

A

(a) You're looking for

[tex]P(X\le 8) = \displaystyle \sum_{x=0}^8 P(X=x)[/tex]

where

[tex]P(X=x) = \begin{cases}\dfrac{\lambda^x e^{-\lambda}}{x!}&\text{if }x\in\{0,1,2,\ldots\}\\0&\text{otherwise}\end{cases}[/tex]

Customers arrive at a mean rate of 6 customers per 10 minutes, or equivalently 12 customers per 20 minutes, so

[tex]\lambda = \dfrac{12\,\rm customers}{20\,\rm min}\times(20\,\mathrm{min}) = 12\,\mathrm{customers}[/tex]

Then

[tex]\displaystyle P(X\le 8) = \sum_{x=0}^8 \frac{12^x e^{-12}}{x!} \approx \boxed{0.155}[/tex]

(b) Now you want

[tex]P(X\ge4) = 1 - P(X<4) = 1 - \displaystyle\sum_{x=0}^3 P(X=x)[/tex]

This time, we have

[tex]\lambda = \dfrac{6\,\rm customers}{10\,\rm min}\times(10\,\mathrm{min}) = 6\,\mathrm{customers}[/tex]

so that

[tex]P(X\ge4) = 1 - \displaystyle \sum_{x=0}^3 \frac{6^x e^{-6}}{x!} \approx \boxed{0.849}[/tex]

B

(a) In other words, you're asked to find the probability that more than 1 customer shows up in the same minute, or

[tex]P(X > 1) = 1 - P(X \le 1) = 1 - P(X=0) - P(X=1)[/tex]

with

[tex]\lambda = \dfrac{6\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 1\,\mathrm{customer}[/tex]

So we have

[tex]P(X > 1) = 1 - \dfrac{1^0 e^{-1}}{0!} - \dfrac{1^1 e^{-1}}{1!} \approx \boxed{0.264}[/tex]

C

(a) Similar to B, you're looking for

[tex]P(X \le 1) = P(X=0) + P(X=1)[/tex]

with

[tex]\lambda = \dfrac{12\,\rm customers}{6\,\rm min}\times(1\,\mathrm{min}) = 2\,\mathrm{customers}[/tex]

so that

[tex]P(X\le1) = \dfrac{2^0e^{-2}}{0!} + \dfrac{2^1e^{-2}}{1!} \approx \boxed{0.406}[/tex]

Answer:

b

Step-by-step explanation:

sinA = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{BC}{AB}[/tex] = [tex]\frac{s\sqrt{3} }{2s}[/tex] ( cancel s on numerator/ denominator ), then

sinA = [tex]\frac{\sqrt{3} }{2}[/tex] → b

Step-by-step explanation:

let George money will be X and Albert be Y

30$+x=y

x-50$=5y

30+x=y

x=y-30

(y-30)-50=5y

y-80=5y

y-5y=80

-4y=80

y=-20

x=-50

Answer:AlBERT=150; GEORGE=90

Albert-30=George+30....(.1)eq

A=(G+60)

#2 (G-50)5=A+50......(.2)eq

substitute result of #1 for A

5G-250=(G+60)+50

4G=360

G=90

substitute $90 into equation #1

A=90+60=150

Therefore Albert has $150, George has $90, and their total is $240

Answer:

5

Step-by-step explanation:

17 inches is equal to 39 inches and it's answer is 5

Cost of 17 inches of wire = 68 cents

Cost of 1 inch of wire

= 68 cents/17

= 4 cents

Cost of 39 inches of wire

= 4 cents × 39

= 156 cents

= $1.56

Answer:

13

5 - -8

When you subtract a negative number it changes to an addition so 5- -8 becomes 5 + 8 which equals 13.

Answer:

Variance is 256

Step-by-step explanation:

Variance:

[tex]var = \frac{ ({ \sum x})^{2} }{n} - {( \frac{ \sum x}{n} })^{2} [/tex]

x is the number or item in the data

n is the number of terms

[tex]{ \tt{ \sum x = (12 + 7 + 6 + 4 + 11)}} \\ { \tt{ \sum x = 40}}[/tex]

Therefore:

[tex]variance = \frac{ {40}^{2} }{5} - { (\frac{40}{5}) }^{2} \\ \\ = 320 - 64 \\ variance = 256[/tex]

I bet you meant "consecutive". If x is the smallest of the 5 numbers, then the other 4 are x + 1, x + 2, x + 3, and x + 4. If their mean is 11, then

(x + (x + 1) + (x + 2) + (x + 3) + (x + 4))/5

= (5x + 10)/5

= x + 2 = 11

==>   x = 9

Then the five numbers are {9, 10, 11, 12, 13}.

Alternatively, since we're talking about an odd number of consecutive integers, the mean among them will always be the number in the middle. So if 11 is the mean, and there are five numbers overall, then we just take the four closest integers to 11, two on either side.

Answer:  [tex]-\frac{\sqrt{2a}}{8a}[/tex]

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

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

[tex]\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\[/tex]

[tex]\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\[/tex]

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

[tex]\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\[/tex]

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

[tex]\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\[/tex]