Which of the following demonstrates the Distributive Property?
3(4a + 2) = 12a + 6
3(4a + 2) = 12a + 2
3(4a + 2) = 4a + 6
3(4a + 2) = 7a + 5

Answer:

the answer will be similar, because if you divide 2, 3, 2.25 by 3 you will get the answer.

Answer:

1400 units

$26,600

1700 units

$32,300

Step-by-step explanation:

yw :))

Answer:

Day 5 ................

Answer:

easy = 391 items

hard = there are 8 more girls than there are boys

Answer:

sorry I don't know the answer

Answer:

Linear Equations In One Variable =

[6(x-1)6] / 7 = 12

[(6x - 6)6] = 84

[36x - 36] = 84

36x = 84 + 36

36x = 120

x = 120/36

x = 10/3

equation solved (Answer : 10/3)

Answer:

side ≈ 4.24= [tex]\frac{6}{\sqrt{2} }[/tex]

Step-by-step explanation:

sides a = b

[tex]6^{2} =a^{2} +a^{2}[/tex]

[tex]6^{2} =2a^{2}[/tex]

[tex]a^{2}= \frac{6^{2} }{2} =\frac{36}{2} =18[/tex]

[tex]a=\sqrt{18} =4.24=\sqrt{(9)(2)} =\frac{6}{\sqrt{2} }[/tex]

Hope this helps

Answer:

L1 and L2 are parallel

BAC = 58° as it is given in the question that AB bisects DAC

similarly BCE = y°

angle B = 90° ( from figure)

Y + 58° + 90° = 180 ( angle Sum property of a triangle).

Y + 148° = 180°

Y = 180 - 148

Y = 32°

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry.

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angles of a triangle is always equal to 180°.

The triangles ΔEST and ΔEFD are similar triangles, therefore, we can write,

[tex]\dfrac{ES}{EF} = \dfrac{ET}{ED} = \dfrac{ST}{FD}[/tex]

Since S and T are midpoints of EF and ED, the lines will be divided into two equal parts. Therefore,

[tex]\dfrac{ES}{EF} = \dfrac{ET}{ED} = \dfrac{ST}{FD}= \dfrac12[/tex]

Therefore, we can write it as,

[tex]FD = 2 (ST)[/tex]

In ΔEST and ΔTDR

∠T ≅ ∠T {Vertical angles}

ET ≅ TD {T is the midpoint of ED}

∠SET ≅ ∠TDR {Alternate interior angles}

Therefore,  ΔEST ≅ ΔTDR.

Since the two triangles are equal we can write,

ST ≅ TR

Further, it can be written as,

FD = 2(ST)

FD = ST + ST

FD = ST + TR

FD = SR

Hence, FD≅SR.

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The base case of [tex]n=1[/tex] is trivially true, since

[tex]\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)[/tex]

but I think the case of [tex]n=2[/tex] may be a bit more convincing in this role. We have by the inclusion/exclusion principle

[tex]\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)[/tex]

with equality if [tex]E_1\cap E_2=\emptyset[/tex].

Now assume the case of [tex]n=k[/tex] is true, that

[tex]\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)[/tex]

We want to use this to prove the claim for [tex]n=k+1[/tex], that

[tex]\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)[/tex]

The I/EP tells us

[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)[/tex]

and by the same argument as in the [tex]n=2[/tex] case, this leads to

[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})[/tex]

By the induction hypothesis, we have an upper bound for the probability of the union of the [tex]E_1[/tex] through [tex]E_k[/tex]. The result follows.

[tex]\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)[/tex]

Answer:

Another rational number.

Step-by-step explanation:

For example, [tex]\pi[/tex] is incorrect because It is a irrational number

you must add another rational number to 1/2 to make the sum rational.

Rational number are numbers that can be converted into a fraction.

all natural numbers, whole numbers, integers, are rational.

and all rational and irrational numbers are real numbers

Answer:

21, 8

Step-by-step explanation:

The second endpoint is of the form (X, 8) since it's parallel to the x axis.

Given the congruence of the two segments, we can tell that its length is 16 units (since [tex]|4-20| = |-16|= 16[/tex])

At this point the two possible endpoints for the segment are [tex](5-16; 8)[/tex] or [tex](5+16, 8)[/tex]. The fact that the segment has to sit in the first quadrant rules out the first option (it's endpoint will be at (-11, 8) which will set most of it in the second quadrant) and we're left with (21, 8)

Answer:

A

Step-by-step explanation:

note that i = [tex]\sqrt{-1}[/tex]

given

- 1 + 2i[tex]\sqrt{3}[/tex]

= - 1 + [tex]\sqrt{2^2(-1)3}[/tex]

= - 1 + [tex]\sqrt{-12}[/tex]

Answer:

10

Step-by-step explanation:

24/30 = 0.8

24+x/30+x =0.85

10=x

If you want more clarification just ask

Total games Chase has to win in a row to achieve an 85% win percentage would be 10.

We can write the total number game won by Chase as -

w = 80% of 30

w = 80/100 x 30

w = 4/5 x 30

w = 24

Assume that he wins {x} games after winning 24 games to achieve 85% win percentage. Now, we can write that out of (30 + x) number of games played, Chase won (24 + x) games to achieve 85% win percentage. So, we can write that -

(24 + x)/(30 + x) = 85/100

(24 + x)/(30 + x) = 0.85

(24 + x) = 0.85(30 + x)

24 + x = 25.5 + 0.85x

0.15x = 1.5

x = 10

Total games Chase has to win in a row to achieve an 85% win percentage would be 10.

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Answer:

C

Step-by-step explanation:

The correct statement is option C.

A real number and an imaginary number are effectively combined to create a complex number. The complex number is written as a+ib, where a and ib are real and imaginary numbers, respectively. Additionally, i = √-1 and both a and b are real numbers.

Since we know that

Complex number is of the form a+ib

Where,

a is real number belongs to real axis

And b is also a real number belongs to imaginary axis.

And the value of i = √-1

Thus,

The set of all numbers of the form a+bi, where a and b are any

real numbers and i equals √-1 is the correct statement.

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Answer:

360

Step-by-step explanation:

using the definition

n [tex]P_{r}[/tex] = [tex]\frac{n!}{(n-r)!}[/tex]

where n! = n(n - 1)(n - 2)..... × 3 × 2 × 1

then

6[tex]P_{4}[/tex]

= [tex]\frac{6!}{(6-4)!}[/tex]

= [tex]\frac{6!}{2!}[/tex]

= [tex]\frac{6(5)(4)(3(2)(1)}{2(1)}[/tex] ← cancel 2(1) on numerator / denominator

=  6 × 5 × 4 × 3

= 360

Answer:

need more info

Step-by-step explanation:

what I can tell you is that when you change all the x's in the equation to 6s, both sides of the equation will equal eachother.

Answer:

Step-by-step explanation:

Answer:

x² + 3 and x² + 4x + 5

Step-by-step explanation:

(a)

to find f(g(x)) substitute x = g(x) into f(x)

f(g(x))

= f(x² + 1)

= x² + 1 + 2

= x² + 3

(b)

to find g(f(x)) substitute x = f(x) into g(x)

g(f(x))

= g(x + 2)

= (x + 2)² + 1 ← expand factor using FOIL

= x² + 4x + 4 + 1

= x² + 4x + 5

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[tex] \: \: [/tex]

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[tex]\Large{ \boxed{ \tt{ \blue{S}}}}\Large{ \boxed{ \tt{ \pink{O}}}}\Large{ \boxed{ \tt{ \green{L}}}}\Large{ \boxed{ \tt{ \purple{U}}}}\Large{ \boxed{ \tt{ \red{T}}}}\Large{ \boxed{ \tt{ \pink{I}}}}\Large{ \boxed{ \tt{ \blue{O}}}}\Large{ \boxed{ \tt{ \pink{N}}}}[/tex]

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keep learning

The maximum number of data points that may not actually be on the line is 16. so, the correct option is D.

We know that a line of best fit is basically a straight line drawn for a given data that may or may not pass through data points.

It is given that the line is drawn for 16 data points.

Hence, the line may or may not pass through all these 16 points.

Therefore, the maximum number of data points that may not actually be on the line is 16.

so, the correct option is D.

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Answer:

67.13°

Step-by-step explanation:

First separate triangles then you use Pythagoras theorem to get the third side and then apply the sine rule

Answer:

The minimum cost of the X-ray machines is 12,197dollars.

Step-by-step explanation:

First, we check whether it’s a Quadratic Equation or not

The general form of the Quadratic equation [tex]f(x) =[/tex] [tex]ax^{2} + bx + c[/tex]

Let’s compare it with the given equation, we get

a = 1

b = -520

c = 79,79

Hence, it’s a quadratic equation.

Now, we will use the VERTEX FORMULA as we are asked to find the ‘minimum’ unit cost.

X = [tex]\frac{-b}\[2a[/tex]  =  [tex]\frac{-(-520)}\[2(1)[/tex] = [tex]260[/tex]

So , number of X-ray machines are = 260 (value of x)

To find the minimum unit cost, plug the value of x into the given equation, and we get

[tex]f(260) = (260)^{2} -520(260) + 79,797[/tex]

[tex]=12,197[/tex]

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Answer:

P(E1 ∪ E2) = P(E1) + P(E2) - P(E1 ∩ E2)

Step-by-step explanation:

As we know that, if A and B are two events then

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(E1 ∪ E2) = P(E1) + P(E2) - P(E1 ∩ E2)

Answer:

[tex]-6.5x+11[/tex]

Step-by-step explanation:

Expand The Brackets:

[tex]x-1+12-7.5x[/tex]

[tex]=x+(-1)+12+(-7.5x)[/tex]

Combine Like Terms:

[tex]=x+(-1)+12+(-7.5x)[/tex]

[tex]=(x+-7.5x)+(-1+12)[/tex]

Answer:

[tex]-6.5x+11[/tex]

I Hope This Helps  

Answer:

[tex]\boxed{\sf x=12}[/tex]

Step-by-step explanation:

By applying Similar Triangles Theorem:-

[tex]\sf \cfrac{x}{6} =\cfrac{6}{3}[/tex]

We'll multiply both sides of the equation by 6, the LCM of 6,3.

[tex]\sf x=2 \times 6[/tex]

[tex]\sf x=12[/tex]

If m VB = 60°  and m BS = 30°, then m ∠3 =15°.

A tangent is a line passing by touching the perimeter of the circle, perpendicular to the line joining the center and touch point.

Given, PB tangents PV, PU secants

If m VB = 60°  and m BS = 30°, then m ∠3 =

The measurement of the external angles is the semi difference of the arcs it consists.

∠3 = 1/2 (60 - 30)

∠3 = 15°

Thus, If m VB = 60°  and m BS = 30°, then m ∠3 =15°.

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Using the normal approximation to the binomial, it is found that the probabilities are given as follows:

a) 0.0055 = 0.55%.

b) 0.2296 = 22.96%.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The parameters of the binomial distribution are given as follows:

n = 150, p = 0.889.

Hence the mean and the standard deviation of the approximation are:

Item a:

Using continuity correction, the probability is P(123.5 < X < 124.5), which is the p-value of Z when X = 124.5 subtracted by the p-value of Z when X = 123.5, hence:

X = 124.5:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{124.5 - 133.35}{3.8473}[/tex]

Z = -2.3

Z = -2.3 has a p-value of 0.0107.

X = 123.5:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{123.5 - 133.35}{3.8473}[/tex]

Z = -2.56

Z = -2.56 has a p-value of 0.0052.

Hence the probability is 0.0107 - 0.0052 = 0.0055 = 0.55%.

Item b:

The probability is P(112.5 < X < 130.5), which is the p-value of Z when X = 130.5 subtracted by the p-value of Z when X = 112.5, hence:

X = 130.5:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{130.5 - 133.35}{3.8473}[/tex]

Z = -0.74

Z = -0.74 has a p-value of 0.2296.

X = 112.5:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{112.5 - 133.35}{3.8473}[/tex]

Z = -5.42

Z = -5.42 has a p-value of 0.

Hence the probability is 0.2296 - 0 = 0.2296 = 22.96%.

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Applying the negative exponent, the equivalent expression  [tex]3^{-3}[/tex]  is 1/27.

The exponents of a number are defined as the representation of a number that shows how many times a number is multiplied by itself.

When we have a negative exponent, we use a fraction, with the term with the exponent going to the denominator.

Hence, the equivalent expression is:

[tex]3^{-3} = \dfrac{1}{3^3} = \dfrac{1}{27}[/tex]

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