Identifying properties used to solve liner equation.
Options are:
Addition property of Equality
subtraction property of Equality
multiplication property of Equality
division property of Equality
simplifying
distributive

Answer:

C=6cm

D=3cm

Step-by-step explanation:

C=6×6cm

36cm

D=3×3cm

=9cm

Answer:

1/y^3

Step-by-step explanation:

We know that a^-b = 1/a^b

y ^-3 = 1/y^3

9514 1404 393

Answer:

  B(8, 4)

Step-by-step explanation:

Reflection across the origin negates both coordinate values.

  (x, y) ⇒ (-x, -y) . . . . . reflection across the origin

  A(-8, -4) ⇒ B(8, 4)

Solution :

Given :

There are five cities in a network and the cost of [tex]\text{building}[/tex] a road directly between [tex]i[/tex] and [tex]j[/tex]  is the entry [tex]a_{i,j}[/tex]

[tex]a_{i,j}[/tex]  refers to the matrix.

Road cannot be built because there is a mountain.

The given matrix :

[tex]\begin{bmatrix}0 & 3 & 5 & 11 & 9\\ 3 & 0 & 3 & 9 & 8\\ 5 & 3 & 0 & \infty & 10\\ 11 & 9 & \infty & 0 & 7\\ 9 & 8 & 10 & 7 & 0\end{bmatrix}[/tex]

The matrix on the left above corresponds to the weighted graph on the right.

Using the [tex]\text{Kruskal's algorithm}[/tex] we can select the cheapest edge that is not creating a cycle.

Starting with 2 edges of weight 3 and the edge of weight 5 is forbidden but the edge is 7  is available.

The edge of the weight 8 completes a minimum spanning tree and total weight 21.

If the edge of weight 8 had weight 10 then either of the edges of weight 9 could be chosen the complete the tree and in this case there could be 2 spanning trees with minimum value.

Answer:

first tile: X²-55=9

second tile:2x²-32=0

third tile:4x²-100=0

fourth tile:x²-140=-19

Step-by-step explanation:

apply difference of two squares to all i.e (a+b)(a-b)=(a²-b²)=0

x²-55-9=0

x²-64=0

x-8,x+8=0

x=8,x=-8

2x²-32=0

divide through by two

x²-16=0

x=4,x=-4

4x²-100=0

divide through by 4

x²-25=0

x=5 or -5

x²-140=-19

x²-140+19=0

x²-121=0

x=11 or -11

If the scale factor is 30, then all you have to do is multiply each measurement by the scale factor. In this case, 4 · 30 = 120.

Answer:

1:-

[tex]\\ \sf\longmapsto 23123+11001+21302[/tex]

[tex]\\ \sf\longmapsto 55426[/tex]

2:-

[tex]\\ \sf\longmapsto 21031+12301+32211[/tex]

[tex]\\ \sf\longmapsto 65543[/tex]

3:-

[tex]\\ \sf\longmapsto 21003+12346+21220[/tex]

[tex]\\ \sf\longmapsto 54569[/tex]

4:-

[tex]\\ \sf\longmapsto 32101+12301+1032[/tex]

[tex]\\ \sf\longmapsto 45434[/tex]

5:-

[tex]\\ \sf\longmapsto 301242+123310+10002[/tex]

[tex]\\ \sf\longmapsto 434554[/tex]

Answer:

0.05x + 0.1(55 - x) = 3.9

Step-by-step explanation:

There are 55 coins.

Let x = number of nickels.

The number of dimes is 55 - x.

The value of a nickel is $0.05, and the value of a dime is $0.10.

The value of x nickels is 0.05x

The value of 55 - x dimes is 0.1(55 - x)

The total value of the coins is 0.05x + 0.1(55 - x)

The total value of the coins is $3.90

0.05x + 0.1(55 - x) = 3.9

Answer:

4842.24 cubic feet

Step-by-step explanation:

Use the formula for the volume of a cone, V = [tex]\pi[/tex]r²[tex]\frac{h}{3}[/tex]

The diameter of the cone is 34 ft, so the radius is 17 ft.

Plug in the radius and height into the formula, and solve for the volume:

V = [tex]\pi[/tex]r²[tex]\frac{h}{3}[/tex]

V = [tex]\pi[/tex](17)²[tex]\frac{16}{3}[/tex]

V = [tex]\pi[/tex](289)[tex]\frac{16}{3}[/tex]

V = 4842.24

So, the volume of the cone is 4842.24 cubic feet

Answer:

4,841.32 ft³.

Step-by-step explanation:

Let’s assume that this is a right circular cone and that the radius of the cone is r.

For our problem, r = (1/2)d = (1/2)34 = 17.

The volume of the cone is:

V = (1/3)pi r^2 h, where r is the radius and h is the height.

So, V = (1/3)pi(17^2)16 = 4,841.32 ft³.

Answer: i) [tex]\frac{1}{3}[/tex]

              ii) [tex]\frac{5}{9}[/tex]

Step-by-step explanation:

Total length of film = 45 mins

Fraction of time left after 15 mins = [tex]\frac{15}{45}[/tex]

                                                       = [tex]\frac{1}{3}[/tex]

Fraction of time left after 25 mins = [tex]\frac{25}{45}[/tex]

                                                        = [tex]\frac{5}{9}[/tex]

9514 1404 393

Answer:

Step-by-step explanation:

The given information tells us there is one congruent side in the two right triangles. That is not sufficient to claim congruence of the triangles.

  CNBD

__

The figure shows one congruent angle in addition to one congruent side, so the figures can be shown to be congruent using the ASA theorem.

  ΔLAW ≅ ΔWKL

_____

Additional comment

We don't know which answer is expected. You should discuss this question with your teacher, since it appears to be missing the statement that

  ∠ALW ≅ ∠KWL

A = πr^2 and d = 2r.

So r = 8/2 = 4 cm.

Now use the first formula

A = π(4 cm)^2 = 50.265 cm^2

Answer:

4

Step-by-step explanation:

Pick two points on the line

(0,-5)  and (1,-1)

We can find the slope using

m = (y2-y1)/(x2-x1)

   = ( -1 - -5)/(1 - 0)

  (-1+5)/(1-0)

    4/1

  = 4

Answer:

the probability that a truck drives less than 159 miles in a day = 0.9374

Step-by-step explanation:

Given;

mean of the truck's speed, (m) = 120 miles per day

standard deviation, d = 23 miles per day

If the mileage per day is normally distributed, we use the following conceptual method to determine the probability of less than 159 miles per day;

1 standard deviation above the mean = m + d, = 120 + 23 = 143

2 standard deviation above the mean = m + 2d, = 120 + 46 = 166

159 is below 2 standard deviation above the mean but greater than 1 standard deviation above the mean.

For normal districution, 1 standard deviation above the mean = 84 percentile

Also, 2 standard deviation above the mean = 98 percentile

143 --------> 84%

159 ---------> x

166 --------- 98%

[tex]\frac{159-143}{166-143} = \frac{x-84}{98-84} \\\\\frac{16}{23} = \frac{x-84}{14} \\\\23(x-84) = 224\\\\x-84 = 9.7391\\\\x = 93.7391\ \%[/tex]

Therefore, the probability that a truck drives less than 159 miles in a day = 0.9374

Answer:

4. 53

5. 66

6. 89

7. 31

Step-by-step explanation:

4.  14 + 18 + 21

     ^      ^

        33 + 21

            53

5.  86 - 20

        66

6.  34 + 55

        89

7.  14 + 11 + 6

     ^     ^

      25 + 6

          31

A stationary point at x = 3 means the derivative dy/dx = 0 at that point. Differentiating, we have

dy/dx = 6x ² + 2ax + b

and so when x = 3,

0 = 54 + 6a + b

or

6a + b = -54 … … … [eq1]

The curve passes through the point (4, 2), which is to say y = 2 when x = 4. So we also have

2 = 128 + 16a + 4b - 30

or

16a + 4b = -96

4a + b = -24 … … … [eq2]

Eliminate b by subtracting [eq2] from [eq1] and solve for a, then for b :

(6a + b) - (4a + b) = -54 - (-24)

2a = -30

a = -15   ===>   b = 96

Answer:

LHS.= Sin 2x /( 1 + cos2x )

We have , sin 2x = 2 sinx•cosx

And. cos2x = 2cos^2 x - 1

i.e . 1+ cosx 2x = 2cos^2x

Putting the above results in the LHSwe get,

Sin2x/ ( 1+ cos2x ) =2 sinx•cosx/2cos^2x

=sinx / cosx

= Tanx

.•. sin2x/(1 + cos2x)= tanx

Step-by-step explanation:

Answer:

main aapki madad karna chahti hun per Mujhe Ae Jahan question Nahin Aata sorry I don't know

sorry dear friend

Step-by-step explanation:

ok I don't know

Solution :

[tex]$H_0: p = 0.5$[/tex]

[tex]$H_a: p > 0.5$[/tex]

Alpha, α = 0.01

The sample proportion is :

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

   [tex]$=\frac{513}{806}$[/tex]

   = 0.636

Test statistics, [tex]$z=\frac{p'-p}{\sqrt{\frac{pq}{n}}}$[/tex]

[tex]$z=\frac{0.636-0.5}{\sqrt{\frac{0.5\times 0.5}{806}}}$[/tex]

[tex]$z=\frac{0.136}{0.0176}$[/tex]

z = 7.727

The p value = 0.00001

Here we observe that p value is less than α, and so we reject the hypothesis [tex]H_0[/tex].

Therefore, there is sufficient evidence,

Answer:  [tex]P = \frac{25}{E^2}-Q\\\\[/tex]

Work Shown:

[tex]E = 5\left(\sqrt{\frac{1}{P+Q}}\right)\\\\5\left(\sqrt{\frac{1}{P+Q}}\right) = E\\\\\sqrt{\frac{1}{P+Q}} = \frac{E}{5}\\\\\frac{1}{P+Q} = \left(\frac{E}{5}\right)^2\\\\\frac{1}{P+Q} = \frac{E^2}{25}\\\\P+Q = \frac{25}{E^2}\\\\P = \frac{25}{E^2}-Q\\\\[/tex]

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

Explanation:

Any equilateral triangle has all three angles of 60 degrees each. Splitting the triangle in half like this produces two identical copies of 30-60-90 triangles.

Any 30-60-90 triangle will have its hypotenuse twice as long compared to the short leg. The short leg here is 5 (it's opposite the smallest angle), so that doubles to 2*5 = 10 which is the value of x.

Note: the other side of this right triangle is 5*sqrt(3).

Answer:

x=10

Step-by-step explanation:

∵ Δ IS Equilateral.

∴ sides are equal.

perpendicular from vertex bisects it.

x=2×5=10

Answer:

Step-by-step explanation:

point est. 0.307824427

99% 2.58

Confidence Interval - "P" values  

(0.2914 , 0.3243 )  

Answer:

Farah: x

Ibtisam: x-3

Muna: 2(x-3) or 2x-6

Sum of all their ages: 4x-6

Step-by-step explanation:

Farah is x, so we don't need an expression for that.

Ibtisam is 3 years younger than Farah, which means that we need to subtract 3 from Farah, and that would be Ibtisam's age. x-3.

Muna is 2 TIMES Ibtisam's age, so we need to multiply whatever expression taht was used for Ibtisam by 2. Put brackets around the equation with 2 outside: 2(x-3). Solve and you get 4x-6

Now, you have all their ages in expression form, now you need to simplify by adding:

x+x+2x-6

We cannot simplify -6, so we put that aside. Add all the x's and you get 4x, insert the minus 6 at the end:

4x-6

Hope this helps!

--Applepi101

Answer:

a) X -3

b) 4x - 9

Step-by-step explanation:

a) Farah's age is X so Ibtisam will be X - 3 old since he is 3years younger than Farah

b) Farah is X years old

Ibtisam is X - 3 years old

Muna is 2(X -3) since she is 2 times older than Ibtisam.

the sum of Thier ages will be

X + X -3 + 2(x-3)

= 2x - 3 + 2x - 6

= 4x - 9

Answer:

a = 3√6 in

Step-by-step explanation:

From the question given above, the following data were obtained:

Angle θ = 60°

Adjacent = 3√2 in

Opposite = a =?

The value of 'a' can be obtained by using the tan ratio as illustrated below:

Tan θ = Opposite / Adjacent

Tan 60 = a / 3√2

√3 = a / 3√2

Cross multiply

a = √3 × 3√2

Recall:

c√d × n√m = (c×n) √(d×m)

Thus,

√3 × 3√2 = (1×3)√(3×2)

√3 × 3√2 = 3√6

a = 3√6 in

The null and alternate hypotheses are

H0 : u = 0.44  vs   Ha: u > 0.44

Null hypothesis: 44% of readers own a personal computer.

Alternate Hypothesis : greater than 44% of readers own a personal computer.

This is one tailed test and the critical region for this one tailed test for the significance level 0.1 is  Z >  ±1.28

The given values are

p1= 0.54 , p2= 0.44 ; q2= 1-p2= 0.56

Using z test

Z = p1-p2/√p2(1-p2)/n

Z= 0.54-0.44/ √0.44*0.56/200

z= =0.1/ 0.03509

z=  2.849

Since the calculated value of Z=  2.849 is greater than Z= 1.28  reject the null hypothesis therefore there is sufficient evidence to support the executive's claim.

Null hypothesis is rejected

There is sufficient evidence to support the executive's claim at 0.10 significance level.

https://brainly.com/question/2642983

Answer:

D. Height

General Formulas and Concepts:

Geometry

Area of a Circle: A = πr²

Step-by-step explanation:

In order to find the area of a circle, we must follow the formula. Out of all the options given, height is not incorporated into the formula.

It wouldn't make sense to use height anyways since it would be 3-dimenional and we're talking 2-dimensional.

∴ our answer is D.

Answer:

a) the probability that the defective board was produced during the first hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000

b) the probability that the defective board was produced during the  last hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = [tex]\left \{ {{\frac{1}{b-a} }\\\ }} \right _0[/tex];   ( a ≤ y ≤ b )[tex]_{elsewhere[/tex]

= [tex]\left \{ {{\frac{1}{10-0} }\\\ }} \right _0[/tex];   ( 0 ≤ y ≤ 10 )[tex]_{elsewhere[/tex]

f(y) = [tex]\left \{ {{\frac{1}{10} }\\\ }} \right _0[/tex];   ( 0 ≤ y ≤ 10 )[tex]_{elsewhere[/tex]

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 )   =   [tex]\int\limits^1_0 {f(y)} \, dy[/tex]

we substitute

=    [tex]\int\limits^1_0 {\frac{1}{10} } \, dy[/tex]

= [tex]\frac{1}{10} [y]^1_0[/tex]

= [tex]\frac{1}{10} [ 1 - 0 ][/tex]

= [tex]\frac{1}{10}[/tex] or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) =    [tex]\int\limits^{10}_9 {f(y)} \, dy[/tex]

we substitute

=    [tex]\int\limits^{10}_9 {\frac{1}{10} } \, dy[/tex]

= [tex]\frac{1}{10} [y]^{10}_9[/tex]

= [tex]\frac{1}{10} [ 10 - 9 ][/tex]

= [tex]\frac{1}{10}[/tex] or 0.1000

Therefore, the probability that the defective board was produced during the  last hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000

c)

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

 [tex]= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)[/tex]

[tex]= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)[/tex]

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

Answer:

y = - x - 2

Step-by-step explanation:

y=mx+b

m refers to slope

b refers to y intercept

y = (-1)x + (-2)

y = - x - 2

Answer:

y=-1x-2

Step-by-step explanation:

plug in the slop and y intercept to the equation y=mx+b