Which of the fractions below are less than 2/5? Select two.

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

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:

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.

⇛6(2 - 6x) = -24

⇛12 - 36x = -24

⇛-36x = -24 - 12

⇛-36x = -36

⇛x = -36/-36

⇛x = 1

Answer:

D

Step-by-step explanation:

More than 40 miles per gallon

Open circle at 40 and line goes to the left

Answer: the answer is C

Step-by-step explanation:

because it is a open circle going to the left

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.

Is this question complete?

[tex]\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}[/tex]

[tex]\boxed{\sf \dfrac{cosA}{sinA}=cotA}[/tex]

[tex]\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}[/tex]

[tex]\\ \sf\longmapsto 1-cot2A[/tex]

[tex]\\ \sf\longmapsto 1-\dfrac{1}{tan2A}[/tex]

[tex]\\ \sf\longmapsto \dfrac{tan2A-1}{tan2A}[/tex]

[tex]\\ \sf\longmapsto tan2A[/tex]

Answer:

multiples of 2 2,4,6,8,10,12,14,16,18,20

multiples of 3 3,6,9,12,15,,18,21,24,27,30

Step-by-step explanation:

Lcm is 6

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:

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

Answer:

yep it's D

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:  [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]

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]

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]

Answer: LCM = 1320

Step-by-step explanation:

2 | 220, 440, 660

2 | 110, 220, 330

2 | 55, 110, 165

3 | 55, 55,165

5 | 55, 55 , 55

11 | 11, 11, 11

| 1, 1, 1

= 2 × 2 × 2 × 3 × 5 × 11

= 1320

Therefore the LCM is 1320

Must click thanks and mark brainliest

Answer:

16.25

Step-by-step explanation:

first do 12 -5 = 7. then 7/4 = 1.75 then 9+1.75 = 10.75 and finally 27-10.75= 16.25

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.

[tex]\\ \sf\longmapsto Area=Length\times Breadth[/tex]

[tex]\\ \sf\longmapsto Area=48(36)[/tex]

[tex]\\ \sf\longmapsto Area=1728in^2[/tex]

[tex]\\ \sf\longmapsto Area=144ft^2[/tex]

[tex]\\ \sf\longmapsto Area=48yard^2[/tex]

Answer:

2 years

Step-by-step explanation:

population in 1 year= 24400*105%=25620

population in 2 year= 25620*105%=26901

Answer:

m=-1 I think

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

In this question, we have to identify the zeros of the polynomial, along with a point, and then we get that the formula for the polynomial is:

[tex]p(x) = -0.5(x^3 - x^2 + 6x)[/tex]

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

Equation of a polynomial, according to it's zeros:

Given a polynomial f(x), this polynomial has roots such that it can be written as: , in which a is the leading coefficient.

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

Identifying the zeros:

Given the graph, the zeros are the points where the graph crosses the x-axis. In this question, they are:

[tex]x_1 = -2, x_2 = 0, x_3 = 3[/tex]

Thus

[tex]p(x) = a(x - x_{1})(x - x_{2})(x-x_3)[/tex]

[tex]p(x) = a(x - (-2))(x - 0)(x-3)[/tex]

[tex]p(x) = ax(x+2)(x-3)[/tex]

[tex]p(x) = ax(x^2 - x + 6)[/tex]

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

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

Leading coefficient:

Passes through point (2,-8), that is, when [tex]x = 2, y = -8[/tex], which is used to find a. So

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

[tex]-8 = a(2^3 - 2^2 + 6*2)[/tex]

[tex]16a = -8[/tex]

[tex]a = -\frac{8}{16} = -0.5[/tex]

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

Considering the zeros and the leading coefficient, the formula is:

[tex]p(x) = -0.5(x^3 - x^2 + 6x)[/tex]

A similar problem is found at https://brainly.com/question/16078990

The formula that represents the polynomial in the figure is [tex]p(x) = x^{3}-x^{2}-6\cdot x[/tex].

Based on the Fundamental Theorem of Algebra, we understand that Polynomials with real Coefficient have at least one real Root and at most a number of Roots equal to its Grade. The Grade is the maximum exponent that Polynomial has and root is a point such that [tex]p(x) = 0[/tex]. By Algebra we understand that polynomial can be represented in this manner known as Factorized form:

[tex]p(x) = \Pi\limits_{i=0}^{n} (x-r_i)[/tex] (1)

Where:

[tex]n[/tex] - Grade of the polynomial.

[tex]i[/tex] - Index of the root binomial.

[tex]x[/tex] - Independent variable.

We notice that polynomials has three roots in [tex]x = -2[/tex], [tex]x = 0[/tex] and [tex]x = 3[/tex], having the following construction:

[tex]p(x) =(x+2)\cdot x \cdot (x-3)[/tex]

[tex]p(x) = (x^{2}+2\cdot x)\cdot (x-3)[/tex]

[tex]p(x) = x^{3}+2\cdot x^{2}-3\cdot x^{2}-6\cdot x[/tex]

[tex]p(x) = x^{3}-x^{2}-6\cdot x[/tex]

The formula that represents the polynomial in the figure is [tex]p(x) = x^{3}-x^{2}-6\cdot x[/tex].

Here is a question related to the determination polynomials: https://brainly.com/question/10241002

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]

3385.8

Step-by-step explanation:

Answer:

y=-1/3x+7

Step-by-step explanation:

y=mx+c

m=-1/3, c=7

y=-1/3x+7

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