The​ x-value(s) for which ​f(x)​g(x) ​is/are ___

Answer:

Pvalue > 0.10

Step-by-step explanation:

Given the hypothesis :

H0 : β1 = β4 = 0

H1 : Atleast one of βj is not 0

F statistic = 1.482 ;

Degree of freedom = 2 and 21 ;

DFnumerator = 2

DFdenominator = 21

Using the Pvalue calculator from Fstatistic ;

Pvalue(1.482, 2, 21) = 0.24999 = 0.25

Hence, Pvalue for the test is 0.25

Pvalue > 0.10

Answer:

Step-by-step explanation:

Answer: $1,223.40.

Step-by-step explanation:

Since she earns $22.5 per hour, for 34 hours, she would earn:

$22.5 × 34 = $765

She earn 3% of her sales, therefore find 3% of $15,280:

$15280(3%) = $15280(0.03) = $458.4

Add them together:

$765 + $458.4 = $1223.4

9514 1404 393

Answer:

  $1,223.40

Step-by-step explanation:

Zoe's total pay is the sum of the products of hours and hourly rate, and sales and commission rate.

  Pay = (34 h)($22.50/h) +($15,280)(.03) = $765.00 +458.40

  Pay = $1,223.40

Zoe's pay for the week is $1,223.40.

Answer:

$3792116

Step-by-step explanation:

that's the answer above

Answer:

5 big sleds and 1 small sled

⇛L = W + 5

⇛L × W = 150 sq. units

⇛W (W + 5) = 150

⇛W² + 5W = 150

⇛W² + 5W - 150 = 0

⇛[(W - 10) (W + 15)] = 0

⇛W = -10 , 15 (reject negative width)

☞ Width of rectangle is 10 units

☞ Length of rectangle is 15 units

Sector area

Area of whole = 51.313

Area of unshaded = 9.424

Area of shaded = 41.8886

Answer:

Step-by-step explanation:

Find the area of the bigger circle:

Find the area of 120° sector AOC:

Find the area of smaller circle:

Find the area of 120° sector of DOB:

Now find the shaded area, the difference of areas of sectors:

Answer:

Step-by-step explanation:

Answering a comes from simplification, and answering b and c are done all in one step: completing the square on the quadratic that results from a.

(a)  If Martina has 240 m of fencing and is only utilizing one side for the length and 2 sides for the width, the perimeter formula is

240 = x + 2w where x is a length and w is the width. Solving this for w in terms of x:

240 - x = 2w so

[tex]w=120-.5x[/tex] The area for a rectangle is L * W, so our area using the lengths we have is

A(x) = x(120 - .5x) and we simplify:

A(x) = 120x - .5x²  That's the answer to a.

Now for b and c, we will complete the square on this to get the vertex.

Begin by factoring out the -.5:

[tex]A(x)=-.5(x^2-240x)[/tex] Now we take half the linear term, square it and add it both inside the parenthesis and outside the parenthesis. Our linear term is 240. Half of 240 is 120, and 120 squared is 14400:

[tex]A(x)=-.5(x^2-240x+14400)+7200[/tex] (The 7200 comes from multiplying the 14400 times the -.5; -.5 times 14400 is -7200 so to balance things out, we have to add 7200).

The perfect square binomial that results from this is

A(x) = -.5(x - 120)² + 7200. From this we determine that our vertex is

(120, 7200). The 120 is the value of x, the length we are asked to find in b; the 7200 is the max area we are asked to find in c.

The required solutions are,,
(a) area = 240x - 2x²
(b) the side adjacent to the rivers gives the maximum length of the field.
(c) the maximum area could be 6400-meter square.

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
length of the field is x,
The perimeter of the field, = 240
x + x + width = 240
width = 240 - 2x
now,
(a)
area of the field,
= length * width,

= x(240-2x)
= 240x - 2x²

Similarly,
(b) the side adjacent to the rivers gives the maximum area of the field.
(c) the maximum area could be 6400-meter square.

Thus, the required solutions are mentioned above.

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#SPJ2

Answer:

Step-by-step explanation:

[tex]we \ add \ \ only \ \ units \ we \ do \ not \ need \ the \ rest \\\\ \bf (3\underline 1 + 13\underline2 + 14\underline3 + 41\underline4 + 51\underline5 +15\underline6 + 6\underline1)= \\\\ 1+2+3+4+5+6+1=2\underline 2 \\\\ Answer: C) \ 2[/tex]

Answer:

(0,-4)

Step-by-step explanation:

The plot for x intercept is at -4,0 and the y intercept is at 0,5

type it in as (0,-4), some require the (#,#) format

Answer: I think the answer is A

Step-by-step explanation:

Answer:

Step-by-step explanation:

a₁ = 15

a₂/a₁ = 45/15 = 3

a₃/a₂ = 135/45 = 3

...

It is a geometric sequence with a common ratio r=3.

Sum of first 12 terms = a₁·(1-r¹²)/(1-r)

= 15·(1-3¹²)/(1-3)

= 15·(-531,441)/(-2)

= 3,985,800

Answer:

12

Step-by-step explanation:

If f(x) = 5 - x

Then f(-7) = 5 - (-7)

f(-7) = 5 + 7

f(-7) = 12

Answer:

"[tex]0.6450 < p < 0.723[/tex]" is the right solution.

Step-by-step explanation:

Given:

n = 380

x = 260

Point estimate,

[tex]\hat p = \frac{x}{n}[/tex]

  [tex]=\frac{260}{380}[/tex]

  [tex]=0.6842[/tex]

Critical value,

[tex]Zc = 1.645[/tex]

Standard error will be:

[tex]S.E = \sqrt{\frac{0.6842(1-0.6842)}{380} }[/tex]

      [tex]=0.0238[/tex]

Margin of error will be:

[tex]E = Zc\times S.E[/tex]

   [tex]=1.645\times 0.0238[/tex]

   [tex]=0.0392[/tex]

hence,

Confidence level will be:

= [tex]\hat p \pm E[/tex]

= [tex]0.6842 \pm 0.0392[/tex]

= [tex]0.6450 < p < 0.723[/tex]

Answer: surface area

Step-by-step explanation:

Answer:

A, B, and E.

Step-by-step explanation:

A. 5^x * 5^x

= 5^x+x

=5^(2)(x)

=25^x

B. 5^2x

=5^(2)(x)

=25^x

C. 5*5^2x

=5^1+2x

D. 5*5^x

=5^1+x

E. (5*5)^x

=5^x*5^x

=5^(2)(x)

=25^x

F. 5^2*5^x

=5^2+x

Answer:

0.2725

0.0431

Step-by-step explanation:

The distribution here is a poisson distribution :

λ = 1.3

The poisson distribution :

p(x) = [(e^-λ * λ^x)] ÷ x!

Expected probability of finding male with 0 accident ; x = 0

p(0) = [(e^-1.3 * 1.3^0)] ÷ 0!

p(0) = [0.2725317 * 1] ÷ 1

p(0) = 0.2725317

= 0.2725

2.)

P(x ≥ 4) = 1 - P(x < 4)

P(x < 4) = p(x = 0) + p(x. = 1) + p(x = 2) + p(x = 3)

p(x = 0) =  p(0) = [(e^-1.3 * 1.3^0)] ÷ 0! = 0.2725

p(x = 1) = p(1) = [(e^-1.3 * 1.3^1)] ÷ 1! = 0.35429

p(x = 2) = p(2) = [(e^-1.3 * 1.3^2)] ÷ 2! = 0.23029 p(x = 3) = p(3) = [(e^-1.3 * 1.3^3)] ÷ 0! = 0.09979

P(x < 4) = 0.2725 + 0.35429 + 0.23029 + 0.09979 = 0.95687

P(x ≥ 4) = 1 - 0.95687 = 0.0431

Answer:

Step-by-step explanation:

Use the point-slope equation:

The line is:

Answer: 1/6

Step-by-step explanation:

Given:

4/9 and 11/18

Solve:

STEP ONE: Make the denominators equal by determining the LCM

LCM = Least Common Multiple

First Five multiples of 9 = 9, 18, 27, 36, 45

First FIve multiples of 18 = 18, 36, 54, 72, 90

As we can see from the list above, both 18 and 36 overlap, however, 18 is less than 36. Therefore, 18 is the LCM.

STEP TWO: Compare the size and determine the greater one.

4/9 = (4 × 2) / (9 × 2) = 8/18

11/18 = 11/18

Since 11 > 8, therefore, 11/18 is greater than 8/18

STEP THREE: Find the difference between the two fractions.

 11/18 - 4/9

=11/18 - 8/18

=(11 - 8) / 18

= 3 / 18

= 1/6

Hope this helps!! :)

Please let me know if you have any questions

Given:

Total fish (Sea bass and mackerel) = [tex]7\dfrac{2}{5}[/tex] pounds

Sea bass = [tex]4\dfrac{5}{8}[/tex] pounds

He gave away [tex]1\dfrac{7}{8}[/tex] pounds of mackerel.

To find:

The remaining mackerel.

Solution:

We know that,

Mackerel = Total fish - Sea bass

[tex]\text{Mackerel}=7\dfrac{2}{5}-4\dfrac{5}{8}[/tex]

[tex]\text{Mackerel}=\dfrac{37}{5}-\dfrac{37}{8}[/tex]

[tex]\text{Mackerel}=\dfrac{296-185}{40}[/tex]

[tex]\text{Mackerel}=\dfrac{111}{40}[/tex]

He gave away [tex]1\dfrac{7}{8}[/tex] pounds of mackerel. So, the remaining mackerel is:

[tex]\text{Remaining Mackerel}=\dfrac{111}{40}-1\dfrac{7}{8}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{111}{40}-\dfrac{15}{8}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{111-75}{40}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{36}{40}[/tex]

[tex]\text{Remaining Mackerel}=\dfrac{9}{10}[/tex]

Therefore, the remaining Mackerel is [tex]\dfrac{9}{10}[/tex] pounds or 0.9 pounds.

Answer:

The amount of Mackerel left is 9/10.

Step-by-step explanation:

total fish = 7 2/5 pounds

sea bass = 4 5/8

The amount of mackerel =

[tex]7\frac{2}{5}-4\frac{5}{8}\\\\=\frac{37}{5}-\frac{37}{8}\\\\=\frac{296-185}{40}\\\\=2 \frac{31}{40}[/tex]

Mackerel left =

[tex]2 \frac{31}{40}-1\frac{7}{8}\\\\= \frac{111}{40}-\frac{15}{8}\\\\=\frac{111-75}{40}\\\\=\frac{36}{40}\\\\=\frac{9}{10}[/tex]

Answer:

0 ≤ d ≤ 20

Step-by-step explanation:

You mention that Manatees can swim in water up to 20 feet deep. So, this means that the largest depth that he can swim is 20 feet, not more than this. Also, keep in mind that the depth can't be negative, so ----> 0 ≤ d ≤ 20 feet

We want to write an expression (an inequality actually) that defines the depth at which a manatee can swim. The inequality is: 0ft ≤ d ≤ 20ft.

We know that the manatees can swim in water up to 20 feet deep. This represents the maximum deep at which manatees can swim, the minimum is trivial, it would be 0ft (when the manatees are on the surface of the water).

Then we can write the inequality:

0ft ≤ d ≤ 20ft.

This gives the range of possible values of d, depth at which the manatee can swim.

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Answer:

6ft

Step-by-step explanation:

Answer:

6ft

Step-by-step explanation:

9514 1404 393

Answer:

  √5 +√6

Step-by-step explanation:

We know that the square of a binomial is ...

  (a +b)^2 = a^2 +2ab +b^2

Then the square root of it is ...

  a + b = √(a^2 +b^2 +2ab)

Using a=√x and b=√y, this is ...

  √x +√y = √(x + y + 2√(xy))

__

For the given expression, we need to find x and y such that ...

  xy = 30 and x+y = 11

Using x=5, y=6, we meet those requirements.

  [tex]\displaystyle \sqrt{11+2\sqrt{30}}=\sqrt{5+6+2\sqrt{5\cdot6}}=\boxed{\sqrt{5}+\sqrt{6}}[/tex]

Answer:

√5+√6 ≈ 4.68555772

Step-by-step explanation:

I hope it's correct

Answer:

A B C D

A×B×C×D

3×3×3×6

162

Answer:

Cos C = a/h

= 21/35

Step-by-step explanation:

since cos is equal to adjacent angle over hypotenuse angle, so from the question we conclude Cos C = 21/35

Leontief input output model (technology matrix) is an economic model that shows the quantitative relationship and sectorial interdependency in a national economy

The responses with  regards to the question are;

(i) The technology matrix for the economy is presented as follows;

[tex]\mathbf{ A} =\left[\begin{array}{ccc}Agric&&Textile\\0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] \begin{array}{ccc}\mathbf{Per \ Unit}\\Agriculture\\\\Textile\end{array}\right][/tex]

(ii) The required gross production of each industry to meet the desired surplus are;

50 units of agriculture and 250 units of textile

The reason the above values are correct is as follows:

(i) The given parameters are;

The industries in the economy = Agricultural industry and textile industry

Units of agricultural input required per unit of agricultural output = 0.4

Units of textile input required per unit of agricultural output = 0.1

Units of agricultural input required per unit of textile output = 0.1

Units of textile input required per unit of textile output = 0.2

Let X represent agriculture, and let Y represent textile, we have;

[tex]Agric \ for \ agric = \dfrac{0.4 \ units \ of \ agriculture}{1\ unit \ of \ agric \ produced} \times X \ Agric \ produced= 0.4 \cdot X[/tex]

[tex]Agric \ for \ textile = \dfrac{0.1 \ units \ of \ agriculture}{1\ unit \ of \ textile \ produced} \times Y \ textile \ produced= 0.1 \cdot Y[/tex]

We also have;

Textile for agriculture = 0.1·X

Textile for textile = 0.2·Y

Therefore;

X = 0.4·X + 0.1·Y

Y = 0.1·X + 0.2·Y

Therefore;

The technology matrix for the economy is presented as follows;

[tex]\mathbf{Technology \ matrix, A} =\left[\begin{array}{ccc}Agric&&Textile\\0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] \begin{array}{ccc}\mathbf{Per \ Unit}\\Agriculture\\\\Textile\end{array}\right][/tex]

(ii)  Let P represent the production vector, and let d represent the demand vector, we have;

[tex]P = \left[\begin{array}{c}X \\Y\end{array}\right][/tex], [tex]d = \left[\begin{array}{c}5 \\195\end{array}\right][/tex]

P = A·P + d

∴ P - A·P = d

Therefore;

[tex]P = \mathbf{ \dfrac{d}{(I - A)}}[/tex]

Where I = The 2 by 2 identity matrix

We get;

[tex]I - A =\left[\begin{array}{ccc}1&&0\\&&\\0&&1\end{array}\right] - \left[\begin{array}{ccc}0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] = \mathbf{\left[\begin{array}{ccc}0.6&&-0.1\\&&\\-0.1&&0.8\end{array}\right]}[/tex]

With the use of a graphing calculator, we have;

[tex]P =\left[\begin{array}{c}X \\Y\end{array}\right] = \dfrac{\left[\begin{array}{c}5 \\195\end{array}\right]}{\left[\begin{array}{ccc}0.6&&-0.1\\&&\\-0.1&&0.8\end{array}\right]} = \left[\begin{array}{ccc}50\\\\\ 250\end{array}\right][/tex]

The required gross product of agriculture, X = 50 units

The required gross product of textile, Y = 250 units

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We have that he technology matrix for this economy and the  the gross production of each industry are

a)    [tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]

b)    [tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]          

From the Question we have told that

Each unit of agricultural output requires 0.4 unit of agricultural input

Each unit of agricultural output requires 0.1 unit of textiles input.

Each unit of textiles output requires 0.1 unit of agricultural input

Each unit of textiles output requires 0.2 unit of textiles input.

Generally the technology matrix for this economy is given below

With

X =Agricultural industry Gross output

Y= Textile industry Gross Output

Therefore

[tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]

b)

From the Question we are told that

Surpluses of 5 units of agricultural products and 195 units of textiles are desired.

Therefore, we have Desired surplus matrix of

[tex]D= \begin{vmatrix}5\\195\end{vmatrix}[/tex]

Generally the Technology equation is mathematically given as

[tex](I-X)\phi=D[/tex]

Where

X =Agricultural industry Gross output

I=A Unit matrix

\phi=Matrix of gross production

Therefore

[tex]\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}-(\begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}))\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}5\\195\end{vmatrix}[/tex]

[tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]

In conclusion

The technology matrix for this economy and the  the gross production of each industry are

[tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]

[tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]         Respectively

In conclusion

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